Solution and Answer Guide: Brown et al., Organic Chemistry [9 ed.] 9780357451861

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¢ Cengage Solution and Answer Guide: Brown

et al., Organic Chemistry ® 2023, 9780357451861;

Chapter 1: Covalent Bonding and Shapes of Molecules

Solution and Answer BROWN

ET AL., ORGANIC

CHEMISTRY © 2023,

Guide

9780357451861;

CHAPTER 1: COVALENT BONDING AND SHAPES OF MOLECULES

TABLE

OF CONTENTS

IN~Chapter Problemis .. cacemiiesimmsmssasisisissssesisestossassentssssrasiass s misstussastassssstsassestane 3 Electron: Configurations: Problem V.l ainunsanansnaaivanasipamaa The Octet Rule: ProbLemy

L2

b

aiismssssimisioiisassa it b snssniosss s disii sadbinsabasssvasenvaseninssa 0

Electronagativity: ProBlem 1.3 cmsicmnsimsiisissomsisnsiisissnsi ssissiimiiniin i

O

Types of Bonds: Problem 1.4 ... iiiiiniissnsisns e sssesssnssssssssesssssssssssssssssssssnsssssssses 0

Bond Polarity: Problem 1.5 .. iiiininicsenisssinsesssst s st snsssssnss siassshesssbessshesssnesssnssssassssnss [ Lewis Structures: Problem L6 iicinmaainnnniiaiisitinnimimmsiasmias

T

Formal Charge: ProBIBM Ll cacnumamnmromisasisnsmtismvis s arsisionsmim o D Condensed Structural Formulas I: Problem 1.8 ... s

Condensed Structural Formulas 11: Problem 1.9, Condensed

Structural Formulas

[11: Problem

Condensed Structural Formulas IV: Problem

110

sssnennns 8

seeecesnesieeees 9 s

ssnesiereseennnens @

107 e

D

Condensed Structural Formulas V: Problem 112 ... VSEPR:

Problers

U3

i

i

e

e

e

e

arr

9 s

MCAT Practice Questions: FUllerenes.......cccccuuiiisisniiinssssisssnssssassissssssssssnsssssssssssssssssnssssss 10 Molecular Dipoles: Problem 118 . iimimiiiisimitoamsssiomiimisiimsiisscsimissis T1 Hybihidization and BORamgs

ProBlany 08

s

st i sepseps

Contributing Structures I: Problem 1.16 ...

it

2

emsseneees 13

Contributing Structures Il Problem 10 aicisssianimsinnniainiisiannmensls Relative Contributions to Resonance:

Problem

118 ... seeann 14

Contributing Structures and Hybridization: Problem 119

14

MCAT Practice Questions: VSEPR and ReSONANCE ....cccvivsrsessnssssssscssnnranssssssssssnsasssnasssessa19 End-of-Chapter

ProblemB. .l i iiaiasi reiasisnmsnis stnsnsiasmsirnnsdiiasasinsna ] T

Etectronic Strictine of AoMSiccumimaauanarmsmmsisissmsnimiimeimisasnesinTl

PYODIGIN 1. 20 iucmssasinsmsaminssssiisieim i PrOBIBM D2 2L

.27 oot o

T

enisisassssssssssesesssasnasssessstsasassssnssssasenssssnsnsnssssssnssssnssesssensesssasnsssessasssase 11

e YT 0o oo

TP VLT VDT R T I LRV

PR F O E Vo0 P SHep TRyl

POt Koot B TRe T IO VPP

f

«l

¢ Cengage Solution and Answer Guide: Brown

et al., Organic Chemistry ® 2023, 9780357451861;

Chapter 1: Covalent Bonding and Shapes of Molecules

Problernr L33 cuwassisssivisssiimsiaiaissisinsivsssiississssssassnmiiass s s iissismsssnsssaisssisias 18 Lewis Structuras and FONMAL ChArge......cuimsveissmvsmuensomsissismesssonsinssissasssssssssassssmsuonnsssrsN

PrODBUBIM 1.24 ouveverercreveresersnesussssssssssiesenssssasnsnssesesssassasssssessnssssassssssssssssnssassssssssensssnsssnsassssensrars 18 By

T

s i

ProblenT 126 PUODIBITY

L 2T

e

R

B

v R

i

Rt

uiiimsis i

1

sasiadsinis 19

s cnscsminssivisssransismmsssrssusntisssaasrsanssin s ss s s s Fsa i s s s st nssn )

PrOBUBIM 1.28 ovevereerreereersnesussssssssssiesensssrasnsnsse sessssssassssssessnssssasssnssssssenssasessssssessssnssansasssssenerer 20 BROBIBIY

1. 29

PrOBIeny

1.30 G unmisimiisasicisimmsiimiiimisisniansiiammiidaissaniiamiasssisin2

B

i

e

e

e

s

2T

D TBI T F T ssscswnsconirassnmassvessinsiossus oo 5 os SR s o NS NN NS T S AP A o )

Problem 1.32 ..

Polarity of Covalent PrOBIEIM 0T a1

ississnssissss sissssssssssss s benssssssssssassss sesstabtsssssssssssssinasssssssssesas 20

BONAS ..ottt st smsssnssssesssassssbess shssssessasssasessene 28

.38 it i asai e s s S

L

O

T

b

v

28

PP RTINSO

..

PrOBUBIM

1.35 ooiieeiiiieiseiiissiessssssssn b ane b b sme e smebss s s sbssnensinsnensshssssbsssebesbebsasssssassesssssanns 28

el

1 30 i

Problem

1.3 7.

R e

S R

s

e

e

e

2S

i

30

Bond Angles and Shapes of Molecules ...

s

30

PHOBEGIT T B cinessssmssissaspsssssmisnssnsns csras st I ass SRS NS IESLSSE SASSISHSb Spsiassasbans DD PrOBUBIM 1.39 ...t

PO

L

i

eteasn et aeme b sms e emeseas e e bamensasenantstess st senebesbeseatensasessssrmasss B2

S

FURCHIONAL BroBDS

s

e

B

st

daiss

S i

s s i

iy

B

s

s

i

D

PHOBEGETY ToT iivicivcasivosssionisnaviosinessinss ko st suasiosas shss oas 618 SERbmsoy bus ss b SR G g A S R i O

PrOBUBIM T2 c.oooeeeerreeeervnesissssesnss s sesesesasnanssssessssssasssessassmssssnsnansanssssensrasessasssassansassensasssssensees 3O D

g

R

PROBEBITE T

P

i i

T

A

sns s 5

N

P

LA

i SE R S S RS R

i o AR

o

o e

P s

s

e

ey . | 4

anm s s

B

PLOBIOIN T5 s ioisarisomssvasscovassiassisssos s s iasa o 1o 54 b b o VRS P SO TN Sk ban s BT PrOBUBIMY TG

...t

e st ese e s semaesaees e assessesssansassreessnes serasare s sannbsesasnstnsasessensnsensssaneas 3O

PrOblerny

BAT it ssmses s s vsinmaioni s msassiassssvonsasy 138

BB

T

B

s

Eolar and NORPOLEr

S G

S

A B

e

RS

s vassvsvavsvvwsoniia Y

MOLBCILES . cnsmummamnanvnmirsarssisn s s s g

PrOBIBM

149

vttt

RPN

LA

b

cssss s sssasnssansss sesasssrasesas s banssssassansasassenssasnssesssansassassensasssssenssars3 G

i

B

i

e

A

r

«l

¢ Cengage Solution and Answer Guide: Brown

et al., Organic Chemistry ® 2023, 9780357451861;

Chapter 1: Covalent Bonding and Shapes of Molecules

Resonance and Contributing StrUCLUIES......ccccviinrinnsnissssssnsssesinssnsssssassssssssssnssnssesssnss 45

FIEODLBIIT 1Bl cuvioeniaiorssos ssninrmintispesossssnsissssiimsannss et e s sion s obah S eSS s s S nomam s 5y sumcs s PrOBUEM 1.52 .uiiesevrcesssssssasssssssssesssssnssssessnsnssssessssmsnsnsssssssnssssssssssssssatesssesssarsssssnssersnsssnssssenseas 46 BB BNy T8

s i

s

s

S

s

o

s

s

i

die e T

PrODIONT 1:54 ccusicisuvisaisissmisiissassinsvossonisssshiduatisist siasssaiisss i snsbaapai ssisisssissismstuiviacisinssisiins 8, PrOBIBITY T B8

s ccinssnirnspinsiusssssimnsanin i s s S e e S RS A v A K v s D)

PrOBUEM 1.56 .cureisieeiesssssssassssmssserssissnssssessnsnsssssssnsmsnsnssssssinsssssssssassssatssssesssarssassnsssrsnsnsnssssenseas 49 Problem

1.57 TRINK—=POIr—SROre .........cccccmvmmsissmmasisiiisassssisssissssssssssassssssassssssssssssssssssns 90

Valence Banll TRaniy

s waimamimiimmsismiebaiii s tre e re b s i

BHOBIBITY 158 ucuvionisuivssssssinissibassisisssmsiiesssinsinssons i sassistsss soksm samssesshob s s s iy vz b isomssisiic O Problem 1.59 . b 0o o] B Problem COMBINBE

issisamssisnss sissssssssssss s besssssssssssssss sesssastsssssssssasssssnsssssssssssass 90

0 R

e e

ey

eyt

ey

R

e

Ay Oy

o U

PO

SR

PO I Ut ALAEr

b

L6l cicninanniiamimsinsamssiiisimas i MOIVB

THEOM

. 58

cannumsmsiismimsmssssenisonss st mia s DD

PrOBIBM .62 cviveesriiisssessisassssasssssiessssssssssnsss sessssssassssssstsnsssssssssssssssssnssassesssssansassasssssssssssenssas 39

e

e

s

FroBeTn

L6

o

PRODIBIN T8

i

R s

s

R

s

S

insisansisss st

mss ses s

S R

s

R S s

R

e

S

el e S

nm i S KR SAR S

i

2 M

AR

s

e

S

b

e

D) O

ansoanit OO

Additional PODLEIMS.c...cocummismismmsmssssssmsmonssiimssmstasisasesssss s srssssssssssspmasson O] PrOBUBIM

PO

.66 ...t eteasnenesaeme b me e emsses e e ssasnensasenansssessstssneresseseetensasessesssssnans BT

ey B B T

PrOBIBNT

s

L

s

e e

S

R

s

i

e

B

BB .cciisunsiiaiiusicmiiioimivisssisieisiiiinsias st iaissiisslsionsinsaioiivninissssssa O

PROBIGITN TG

Problem 170 Ol

L

PROBEBITE

s

B T2

o cosonsssmnsvnsasmissbis o

e s

o

s S saaS6

E

A

o S

eissss vetsnsnses e bsnsmsmsm st S

S

i iiasnsiiisisnaisiaicessns s o AR

A S B

st sass s

S

R S

b e S RS

oSSR

N

s

sy

OO

sb et s s s bars e sesssebstananssssenseas OO R

AR

s S

s

i

B wian i

D)

PrODIBITE T T B tiuisusauasosssoasssusaasniusnissssssnsvsssassodsionss1sasin s inssb sasa oo o Lessh s s ibonadsonsiatssssanshsnshssibons OF PLOBUBITY 1. 74

Problens L ORI

.75

A B

oot

eeeeeteeiareaeasserseasssemsesanesssansessasasensessneessnassennsesssasanssnssssassnsasesssssssensssnesans O 1

ciusiimi st e

B

e

s

B

st A

S

sasis BT

s

s

PEODIEIN E TG viaamissasnsmssiiimessins i nsn st

s

B

PrOBIBIM .77 ererereeectsiesseasnsnsssssssssesenssasssansns sesssassesesessstsnssssasesssnsasssanssesssmsnsnsnsassnssessssassasnanes B9 B

i

R

B R

e

R

e

R

T

BB

T

T

~¢ ¢ Cengag Cengage «l

Solution and Answer Guide: Brown

et al., Organic Chemistry ® 2023, 9780357451861;

Chapter 1: Covalent Bonding and Shapes of Molecules

Problemt .79

s

B IO

s swionosnshosvscinirmpeinsins o

T BN,

iisommniioviirisnsisiisssisssis shsssnsosss s masssiibassinisiseibisissisnssasismmsssisinsiniias F

o

A A

R BRSNS SR

300

MOLECULAr OrDItALS c.vvteiieciere ettt s s srsesas s sas s snanssresssrssesesresasbasserens 13 B B

B

ProblenT 182 PEOBIOTTY T,

s i

B i

S

e

T ot

R s

R

s asbrsa

g

iisiimsissmssissssadsisis 19

v s sercmniioessininsansssammsaiamrsn st s s s S A b s R s s s )

«l

¢ [ Cengage Solution and Answer Guide: Brown

et al,, Organic Chemistry @ 2023, 9780357451861;

Chapter 1: Covalent Bonding and Shapes of Molecules

IN-CHAPTER PROBLEMS ELECTRON CONFIGURATIONS: PROBLEM 1.1 Write and compare the ground-state electron configurations for each pair of elements. (a) Carbon and silicon

C (6 electrons) 1s?2s?2p? Si (14 electrons) 1s?2s?2p°3s?3p? Both carbon and silicon have four electrons in their outermost (valence) shells.

(b) Oxygen and sulfur O (8 electrons) 1s?2s?2p* S (16 electrons) 1s?25?2p°3s?3p* Both oxygen and sulfur have six electrons in their outermost (valence) shells. (c) Nitrogen and phosphorus

N (7 electrons) 1s?2s2p* P (15 electrons) 1s°2s?2p°®3s3p? Both nitrogen and phosphorus have five electrons in their outermost (valence) shells.

THE OCTET RULE: PROBLEM 1.2 Show how each chemical change leads to a stable octet. (a) Sulfur forms S*.

S (16 electrons): 1s22s?2p©3s23p* 57" (18 electrons): 1s?25?2p°3s23p® (b) Magnesium forms Mg?". Mg (12 electrons): 1s?25?2p°®3s? Mg* (10 electrons): 1s22s22p®



\

/ cl

C=C(

/

AR RN

[#1

\

H

sp

sp?

}

H—C=C—H

+

0o,

My

—»

\

:q.

sp _

H—C=C—H

\

o=C{ \

F H

sp? H + H,0

[

O

1

—> 1-|—(|:—t:—1[

H

(c) sp

s

E

H—C=C—H

X\ sp

+

spj

H

£

TN

O

H

H

)

H

+ H—0—(C—(—H

|

H

H

H

| i/ o H—C—C—0—C—C—H I | H

Ne.(= i H

H

H

H

(]

—

|

|

A

sp’

H—(C—C—0—(C—(C—H

[

H

=

—™

H\\,H G=C H H

b

0!

*

oo

H

H—O—C—0—H

[

H

+¢Cengage «l

Solution and Answer Guide: Brown

et al., Organic Chemistry ® 2023, 9780357451861;

Chapter 1: Covalent Bonding and Shapes of Molecules

Problem 1.78 Following is a structural formula of benzene, CsHs, which you will study in Chapter 21. fl! H\LZ#(:\”C/H

P H” '“‘”:I:’ Su H

(a) Using VSEPR, predict each H—C—C

and C—C—C

bond angle in benzene.

Each carbon atom in benzene has three regions of electron density around it, so according to VSEPR, the carbon atoms are trigonal planar. You should predict each H—C—C

bond angle to be 120° and each C—C—C

bond angle to be 120°.

(b) State the hybridization of each carbon in benzene. Each carbon atom is sp? hybridized because each one makes three obonds and one r

bond. (c) Predict the shape of a benzene

molecule.

Because all of the carbon atoms in the ring are sp? hybridized and thus trigonal planar, predict carbon atoms in benzene to form a flat hexagon in shape, with the hydrogen

atoms in the same plane as the carbon atoms.

(d) Draw

H. H

important resonance contributing structures.

i _C.

~c’) N X4

(El;Edbdl H

_H H

H. H

i _C.

eF R

N“cl: H

e

_H H

4¢Cengage al

Solution and Answer Guide: Brown et al.,, Organic Chemistry @ 2023, 9780357451861; Chapter 1: Covalent Bonding and Shapes of Molecules

Problem 1.79 Following are three contributing structures for diazomethane, CH;N.. This molecule is used to make methyl esters from carboxylic acids (Section 17.7C). 1 H

H

S

/

C—N=N:e—>

H

\

/

H

T (=N=N:

«—>

H

(a) Using curved arrows, show

\

/

-

w

(=N—N:

how each contributing structure is converted to the one on

its right. The arrows are indicated on the preceding structures.

(b) Which contributing structure makes the largest contribution to the hybrid? The middle and left structures have filled valence shells, so these will make a larger

contribution to the hybrid than the structure on the right, in which the terminal

nitrogen atom has an unfilled valence shell. The structure in the middle has the negative charge on the more electronegative atom, N, compared with the structure on the left (negative charge on C), so the structure in the middle will make the largest contribution to the resonance hybrid.

Problem 1.80 (a) Draw a Lewis structure for the ozone molecule, 0Os. (The order of atom

attachment is

0—0—0, and they do not form a ring.) Chemists use ozone to cleave carbon-carbon double

bonds

(Section 6.5C).

(b) Draw four contributing resonance structures; include formal charges. (e) How does the resonance model account for the fact that the length of each 0O—0 bond in ozone (128 pm) is shorter than the 0—0 single bond in hydrogen peroxide (HOOH, 147 pm) but longer than the 0—0 double bond in the oxygen molecule (123 pm)? It is not possible to draw a single Lewis structure that adequately describes the ozone

molecule. Rather, it is better to draw ozone as a hybrid of four contributing structures, each with a separation of charges. T

hd

B O._..-L

pf

0,

.

0

o T

.a

0=

TRS 4

0

o

NS

==

slP

0"

et

S,

o

Taken together, the four contributing structures present a symmetric picture of the bonding in which each 0—0 bond is an intermediate between a single bond and a double bond. Recall that bonds become shorter as bond order increases. As a result,

the bonds in ozone are shorter than the single 0—0 bond in HOOH, but longer than the 0=0 double bond in the oxygen molecule.

4¢Cengage «l

Solution and Answer Guide: Brown

et al., Organic Chemistry ® 2023, 9780357451861;

Chapter 1: Covalent Bonding and Shapes of Molecules

MOLECULAR

ORBITALS

Problem 1.81 The following two compounds are isomers; that is, they are different compounds with the same Cl

H

\ /

molecular formula. You will learn about this type of isomerism

C=C

/ N

H

H ==

Cl

Cl

A /

(=C

/ \

in Chapter

5.

H

Cl

(a) Why are these different molecules that do not interconvert? Interconversion of the two isomers involves rotation about the carbon-carbon double

bond. This cannot occur without breaking the 7 bond. The 7 bond is strong enough so that this does not happen spontaneously at room temperature and the isomers do not interconvert. (b) Absorption of light by a double bond in a molecule excites one electron from a 1 molecular orbital to a m* molecular orbital. Explain how this absorption can lead to interconversion of the two isomers.

— =

e Light

1

4

Putting electron density into an antibonding (*) orbital of a bond weakens that bond. Excitation of the electron from the 7 bond to the 7* orbital upon absorption oflight weakens the 7 bond, allowing the moleculeto rotate about the carbon-carbon bond. This

rotation interconverts the two isomers. A similar alkene rotation reaction is responsible for the mammalian photoreceptor molecules that allow us to see visible light.

~¢Cengage «l

Solution and Answer Guide: Brown et al., Organic Chemistry ® 2023, 9780357451861; Chapter 1: Covalent Bonding and Shapes of Molecules

Problem 1.82 In future chapters, you will encounter carbanions—ions in which a carbon atom has three bonds and a lone pair of electrons and bears a negative charge. Draw another contributing structure for the allyl anion. Now

using cartoon representations, draw the three orbitals

that represent the delocalized m system (look at Figure 1.26 for a hint). Which of the three orbitals are populated with electrons? H I-

“x,;':;—‘--k(‘/li

[

H

[

H

Allyl anion Here is a cartoon representation of the allyl anion 7 molecular orbitals.

il 38+

L

The lowest two molecular orbitals are filled with a pair of electrons each. Notice that the filling of the middle orbital, with lobes on only the two terminal carbon atoms,

indicates the negative charge will be found on these two atoms consistent with the contributing structures.

+¢Cengage =»l

Solution and Answer Guide: Brown

et al., Organic Chemistry ® 2023, 9780357451861;

Chapter 1: Covalent Bonding and Shapes of Molecules

Problem 1.83 Describe the bonding in PCls without using d orbitals. As a hint, the geometry of PCl; is as shown.

a, (Fl. 90° i

Pl

190°

(&

s hybridized

aA c

Qe

y

&

Q)Q

-

S S

120°

2p orbital

Three sp? orbitals

Based on the bond angles, the bonding in PCls can be explained if the P atom is sp? hybridized. The three sp® hybridized orbitals would overlap with Cl orbitals to form the three “equatorial” o bonds spaced at 120°, while the unhybridized 2p orbital would overlap with Cl orbitals to form the two “axial” o bonds.

«l

¢ Cengage Solution and Answer Guide: Brown et al., Organic Chemistry @ 2023, 978035T451861;

Chapter 2: Alkanes and Cycloalkanes

Solution and Answer Guide BROWN ET AL., ORGANIC CHEMISTRY © 2023, 9780357451861, CHAPTER 2: ALKANES AND CYCLOALKANES

TABLE OF CONTENTS In-Chapter Problems . ciiiinimiemembammmiass i isaisasrmsrisiass 1D Constitutional lsomers: Preblem: 21w uisimisanianicamibamsnnsbimismmin19 Line-Angle Formulas:

Problem 2.2......

i

ssssssssssssssssssasssssss 19

IUPAC Nomenclature

I: Problem 2.3 ...

IUPAC

Nomenclature

[1: Problem 2.4 ...

IUPAC

Nomenclature

111: Problem 2.5 ... iiceieceeiieciciieccececensseensisesssseessssenesnesns B1

ssssessssssssessenssasss 90

cscicsscsssssssssssssssessnsssssssssssssnsssssssassesssss 91

General Formulas: Problem 2.6 . i Molecular

Newman

Formulas:

Problem

2.7 ..

e

Be

eecesceseicecssscnsessanseseaesssesanenssasesanenssassnnees B

Projections: Problem 2.8 ...

sssssssssssssisssssnes 94

Axial Versus Equatorial Groups |: Problem 2.9.........ieeiiiienniicinssesiesimssssssssnnesseenss 84 Axial Versus Equatorial Groups II: Problem 2,10 .. Equilibrium Cis Versus

Populations of Conformations: Trans Isomerism:

Eyclohexane

Problem

Rings: Problem 2093

Substituted Cyclohexane

Problem

s

89

2.1 ..vccvciiesevvccissnscveesessee, 86

2.12. .. icinncsssssssessssssnssesssscssssssssssssessenss 90

s

iiuniniiaislainnmimutm8T

Rings: Problem 2.14........ienrsnsssmsrsssssesmssssssssssnsssssssss8 1

MCAT Practice Questions: TetrodOTOXiN .......cccererressrnrsressmssssssssssssesensansassssnsssanssassssssnssnsscnss S8 Alkane Boiling Points: Problem 2.15 .cinisiisssssssssassassnsesssssssssssssassssasassasseses 98

End-of-Chapter Problems.......ssssssssssssssssissssssisssssssssssssssasssssssasssassssssnssesss 99 Probletn

218 .

s,

PPOBIBIIE D T cxsviunassmnsussnsissmsinsant sisss s

sa o s o

s A

R

R SRR

88

s i it

O

PrOBIEIM 2.78 .cvviverirsssisssisssssssassssisssssnisassssssnssssssssssansssssnssssasssssasssssssssssasssssssssssssssssssssanssnsss G ConstitUtioNal

Problem 208

|SOMEFISIM .ot

s

s

ceeereecsess et ees s aesssssse sessaessanessnasssasasesnsesennsssssensnnsssasesns D1

it

BrOBIBI -2 v iisiivisusiiiisiossass s i

e e s

o

S

s oo s

S

TR

ST

e e

O v

i

PLOBIBITI 2.27 c.eeveeeeeeresecsiaessesassnsssssssssssssassssssansassassssassssnnte sanssssssssansnssssannsssssssannsesasssnssnss G

PrOBUEITT 2.22 .oooeeeeeeveeveranrssrrsreseesssvssssserssnssnssasssssssanmassssssssass snsssssassansasarsssanssssssasssansansasssens G4 Probilommy Zi20 v

caseis s

s e

aeis s

s

i

«l

¢ Cengage Solution and Answer Guide: Brown et al., Organic Chemistry

@ 2023, 9780357451867;

Chapter 2: Alkanes and Cycloalkanes

PRODISHRE 20

. oosisivmnnsiisinivnsmmssssusmisssns s sai s e s S e A e s e e msp b nii s, AP,

PrOBLOIT 2. 25 ..ttt st ssassasnsamessies s es s st e st snaansasba s st ansasnsssnsnssssssssensasssssnnannss DT Nomenclature of Alkanes and Cycloalkanes........ccoccoicicnicinciiisccinnsissrresssesessessaee: 98

PO )

e

s

B

e

S

BB

s

s s

sizscszasisie BB

PrOBIBITE 227 icssovissanvavisvsisseimsmivssirssssssivessnsississosersisssiosisssertamssiivissaseisavevistsvasins viseossonin h PrOBIOIT 2.28 ...ttt PrOBlamy 200

i

T

sansssesasmssssnsssassssssssssnsssssssssssensnssssssssnssssssnsmsssansss 100 R

TR

Problery 2.30 ;oo

s

02

s inisnisnaisinmnnisiasinaan e 105

Probloin 2.3F civnsunnassmsssvisissiisnsisssivisisssissssiisisoissyaassissimsnsiisisisisisoisssiosisasiivsinie, N0, Conformations of Alkanes and Cycloalkanes ... Problem B2

i

PO

s

BT

o B

b s

S

e

TRl

A

LR

s

bt

113 e

T

rwas i anssvadimarzavioe THE

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s

ERODOTY 28

e

s

o s

s

P

s T

e e

S B B

i

AR

s

e s s

ity TG v

i

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Problent:

28

i

sissesresssscissssnsens 120

st

120

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T 0

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i

issaisii i

i

5 s

s

sisistasesiasniiey 121

msiaiiiaiis e

127

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Probleim 248

i

s a1

24

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T,

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D,

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s

s

nomm

s

e e

N

AT

A

A A

.

e s

A A A

R

AR A AN SRR

AT

|/ &

Problem 2.83 Thittk=Fair—SRare...ceusisissisimisissssssssisssisssssassissrsssossssssssssisasissarsios 128

«l

¢ Cengage Solution and Answer Guide: Brown et al., Organic Chemistry

@ 2023, 9780357451867;

Chapter 2: Alkanes and Cycloalkanes

PrODLEH -2 B0,

s usinsisinivasuiissnsnssissanssuiniaissasosssss s asssspas st as s aiai i smnbusa s isnsmassnsssnpssaiiasappsnnisy D

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2,58 csiiisisimis o

Physical Properties

v

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R

e

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ey

G0

i

Problery 280

R

s

s

Reactions OF AKAIIEE ...t Pl OB Eroblom P

BT

12D asinaan

sy,

s 193

10

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s

s

S

o L

T

L T

e

e

S o

e A R

s

0

e asiminanss; T O

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B8

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P

i

s

e

B

e

R B

R

N R

i

40

RPYODLOIY 2 AT iiiiiiiiniiiiistiviiniiivsisibissisisissidiain s s eisivisoiavitidviiiin T

-l

¢ & Cengage

‘

Solution and Answer

Guide: Brown et al., Organic Chemistry © 2023, 97B0357451861;

Chapter 2: Alkanes and Cycloalkanes

IN-CHAPTER

PROBLEMS

CONSTITUTIONAL |ISOMERS: PROBLEM 2.1 Do the line-angle formulas in each pair represent the same compound or constitutional isomers? ?

@

rl,v_--"‘\_

and

',/"-.__"/‘\_

|1

-

These molecules are constitutional isomers. Each has six carbons in the longest chain. The first has one-carbon branches on carbons 3 and 4 of the chain; the second has one-carbon branches on carbons 2 and 4 of the chain.

and

K

(b)

N

|

These molecules are identical. Each has five carbons in the longest chain, and one-carbon branches on carbons 2 and 3 of the chalin.

LINE-ANGLE FORMULAS: PROBLEM 2.2 Draw line-angle formulas for the three constitutional isomers with the molecular formula CLHT!-

1 2

4

1

3

3

al

¢ Cengage Solution and Answer Guide: Brown et al., Organic Chemistry @ 2023, 9780357451867; Chapter 2: Alkanes and Cycloalkanes

IUPAC

NOMENCLATURE

|: PROBLEM

2.3

Write IUPAC names for these alkanes.

(a)

" lfil;lm

e 1

1-Methylethyl group

el L

7

7

o Methylgromp

- \-Methyletiyl group

-

?H,—?I—EH.—?L

CH-CH.

e [ 2-Methyl-5-(1-methylethyl)octane

m f '] 5 1

2

] 7

3




Cholic acid

4

i

o

3

5

"

(a) What is the conformation of ring A? of ring B? of ring C? of ring D? As can be seen in the following structures, the conformations of all of the sixmembered

rings, that is, A, B, and C, are all chairs. The conformation of the five-

membered

ring, ring D, is puckered. These conformations all represent preferred

conformations and the linkages between rings prevent chair-to-chair interconversion. (a) Hs

{ea}

(aelH

' CH!

2

”

;

e 2™

o

|

)

&9

G

H(a) A

=

H (a)

H e

5

&

@

Ball-and-stick model of cholic acid with all

but one of the H atoms removed for clarity.

(b) Are the hydroxyl groups on rings A, B, and C axial or equatorial to their respective rings? The hydroxyl group on ringA is equatorial, the hydroxyl groups on rings B and C are axial (c) Is the methyl group at the junction of rings A and B axial or equatorial to ring A? Is it axial or equatorial to ring B? The methyl group at the A, B ring junction

is equatorial to ring A, but axial to ring B.

(d) Is the hydrogen at the junction of rings A and B axial or equatorial to ring A? Is it axial or equatorial to ring B?

The hydrogen atom at the junction of rings A and B is axial to ring A, but equatorial to

ring B. (e) Is the methyl group at the junction of rings C and D axial or equatorial to ring C? The methyl group at the C, D junction is axial to ring C.

«l

¢ Cengage Solution

and Answer

Guide:

Brown

et al., Organic Chemistry

& 2023,

9780357451861,

Chapter 3: Stereoisomerism and Chirality

Solution and Answer Guide BROWN ET AL., ORGANIC CHEMISTRY © 2023, 9780357451861; CHAPTER 3: STEREOISOMERISM AND CHIRALITY

TABLE OF CONTENTS IN=ChEpter ProbLemIS iiainissisnicniimimnsaismbsniniminenniinnmsiimne s 18 Sterecisomers:

Problem

3.0 immiiiiiiimminimsiimmmminiiiicosmmeansiis 14

Cahn, Ingold, Prelog Priorities: Problem 3.2......ssismsssssssseaeas 144 .S contigurationa: PrOBLEM

B.3......cominssnmrmsnminnsatissaivasissnsisimssins snsissnsssssinsissossisniassatsniss. D)

R and S: Problem 3.4 ... Stersoisomerism:

Prablem

3.5

s

ssssemssmssssmassssssssssansseses 140

. iiiimiininiiaimsaiiiiimsriromiismisei YT

Eischer Projectionsi:Problem 3.6 uunnnanannastmsmeaniansimamaiaiaicas 148 Stereoisomerism

with RIiNES |1 Problem 3.7 ...t

srmessssssresenneens 148

Stereoisomerism with RiNgs I1: Problem 3.8 .....cciivieimecsnnrinmmsinsesssnssssssssassossssssassass 149 Specific Rotation: Problem 3.9 ... s sssessssenessessssssensssnseees 149 Enantiomeric

Excess:

Problem

3.0 . ianininuinisisisiisimin bt 100

MCAT Practice Questions: Amino Acid StereoChemistry..... . ecnisissssnissssssasassssasnsenss 150 End=of~CHabtor PIODIEIMNE rmsnmsssmmmnmmsnnsssmspimmsams s R

O

i

PO

i

B T

i

i

T

D L

o

A

R

e

e

e

s

s i

S

A

R

R

s

B0

e

e

psssarsniiins 1D T

iy

viiviss T3

PrEBIang 332 cavisnnasiisissiisiismaiissiiaiss s iard sy e ansisisii J Probletit SilZnyammmunsmremsanpnsmmmesmsrm o nun s ssgspaepssiny 100 ENANTIOMETS oot

P BT

TR

ittt

sy

s

R

s sb e s s sss sme st s as s s s masssssssesbesrmssnsassnsanensasss DG

A

R

S S

e

v

i

v

T8

BroBlefny 308 umviicimimissssisia s s s iy v i sk ssansss iasasiinins TG PHOBDIEITE FTB sivsrasinsisininssainsssssinintasotsssiiin s ssssssbasasspss st S v i n s b o A sk asiasisryunsss DRl PLOBIBITT 3,77 c.neoeeeeseeeereeesnseesesasisasstessssassssssasssnsssasssasssssanssssnsssansessesssssssssesasssssnssssnssssssssanane 1O0 0

L

PrOBIONM BT

B

R

O

s

R

st

A P

OO

P R T s Pt F o AR Lt SN

RS

i

L CH,CH, Bs

Identical

(hy

W

HO

H

and

/Y

H

OH

Enantiomers

Problem 3.36 Which of the following are meso compounds?

(a)

i OH

oy (b)

i OH

=»l '

¢ Cengage |

Solution and Answer

Guide: Brown

et al., Organic Chemistry & 2023, 9780357451861,

Chapter 3: Stereocisomerism and Chirality

(;H 3 H FE =OH He E =(OH (©)

CHOH

Me

Me

H

HOL(C

H

(d)

CO.H

(':Hg H= (II =OH H» (i

+ CH—C-AkCl c

pair, showing electron

«l

¢ Cengage Solution and Answer Guide: Brown

et al., Organic Chemistry ® 2023, 978035T451861;

Chapter 4: Acid and Bases

END-OF-CHAPTER

PROBLEMS

An asterisk (*) indicates an applied problem. Problem 4.9 For each conjugate acid-base pair, identify the first species as an acid or a base and the

second species as its conjugate acid or base. In addition, draw Lewis structures for each species, showing all valence electrons and any formal charge.

(a) HCOOH

HCOO-

100

-0

N H—C—0—H

I s H—C—O: C

Acid (b) NH:*

ave:t

NH;

H

| +

o

H—hll—H

H—I'Il—H

H

H

Acid

Cntl'lg:egate

(¢) CH.CH,O0™

CH:CH.OH

CHCH,—O:

CHCH,—O—H

Riis

(d) HCOs

Cnancjil:jgnte

€O

on

100

e

v w

H—0—C—O: Acid

wom

Ho N

M

cawm

:0—C—0: Cn:::eglte

¢ Cengage «l

Solution and Answer Guide: Brown

et al., Organic Chemistry ® 2023, 978035T451861;

Chapter 4: Acid and Bases

(e) H.POs

HPO,>

‘0! R | (e H—Q—IP-—OZ

10! s M ‘o = H—O—FI'—O

:0—H

10

Acid

(0

CHgCH;

(‘.n;i::egate

CH;CHQ_

o

[

o

a

H

H

H

Acid

(g8)

CH:S™

H

Co;egnte

CHsSH

CH, —S .

CH; —:..'-::—H Conjugate

B

acid

Problem 4.10

Complete a net ionic equation for each proton-transfer reaction using curved arrows to show the flow of electron pairs in each reaction. In addition, write Lewis structures for all starting materials and products. Label the original acid and its conjugate base; label the original base and its conjugate acid. If you are uncertain about which substance in each

equation is the proton donor, refer to Table 4.1 for the relative strengths of proton acids. (a)

(CHj}zNH

+:

HC—N—H |

3

Base

H30+

-

—_—

H

|')

H—O+—H

—

|+

H(—N—H

|

.

Acid

+

-

H—O0—H

& :

CH,

Conjugate Acid

Conjugate

Base

-l

¢ Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861, Chapter 4: Acid and Bases

@

NHo

o+

HCOy

==

H +I\'

10

H—N—mn

+

I

B

ll

=

-_—

0=—C—0—H

B

Base

CH,COO"

+ i

CH,OH

+

b

..

ll

°

Base

H—O0—C—0—H Conjugate

T

Acid

(0

I

:

o

Acid

+

N

-

=—=—=

| 2 4+ W—0—Cll, N BE—C—0F

2 H,50

I

Base

CH;C00

“Na*t

pK,4.76

pkK,4.76

H,CO,

pk, 6.36

(b) NH; CH;COOH

+

+

NH;

———— CH;C00

~ +

NH,*

pK,9.24

~¢ & Cengage -l

Solution and Answer Guide: Brown et al,, Organic Chemistry © 2023, 9780357451861; Chapter 4: Acid and Bases (c) H.0

CHyCOOH

+

H,0

—= _

CH,C00 ~ +

K, 476

H0"

K, L4

(d) NaOH CH,COOH

+

HO™

Na*

—_—

“kw-h.

pE,476

+

H0

PE,157

Equilibrium favors the direction that gives the weaker acid and weaker base. Therefore, equilibrium will favor the formation of an acid with a pA; value higher than 4.76, or the

formation of acetic acid if that is the weaker acid. Based on the pk; values shown, reactions (a), (b), and (d) have equilibria that lie considerably to the right, while reaction (c) has an equilibrium that lies considerably to the left.

Problem 4.27 Benzoic acid, CeHsCOOH (pA. 4.19), is only slightly soluble in water, but its sodium salt, Ce¢HsCOO~ Na*, is quite soluble in water. In which solution(s) will benzoic acid dissolve?

(a) Aqueous NaOH (b) Aqueous NaHCO; (c) Aqueous Na,CO; The pA. of benzoic acid is 4.19. The pA, values for the conjugate acids of sodium hydroxide, sodium bicarbonate (NaHCO,), and sodium carbonate (Na,CO,) are 15.7, 6.36,

and 10.33, respectively. Thus, equilibrium will favor the reaction of benzoic acid with all three of these bases to give the soluble C¢HsCOO~ Na*. Therefore, benzoic acid will

dissolve in aqueous solutions of all three bases.

~¢ & Cengage -l

Solution and Answer Guide: Brown et al,, Organic Chemistry © 2023, 9780357451861; Chapter 4: Acid and Bases Problem 4.28 4-Methylphenol, CH:CsH4OH (pA: 10.26), is only slightly soluble in water, but its sodium salt, CH,CeH4O Na', is quite soluble in water. In which solution(s) will 4-methylphenol dissolve? (a) Aqueous

NaOH

(b) Aqueous

NaHCO;

(c) Aqueous Na;CO;

The pk. of 4-methylphenol is 10.26. The pA; values for the conjugate acids of sodium hydroxide, sodium bicarbonate (NaHCO3), and sodium carbonate (Na,CO;) are 15.7, and 10.33, respectively. Thus, equilibrium will favor the reaction of 4-methylphenol sodium hydroxide (by a large amount) and sodium carbonate (only slightly) to give soluble CH3CeHsO™ Na*. 4-Methylphenol will dissolve in aqueous solutions of these

6.36, with the two

bases. Sodium bicarbonate is not a strong enough base to deprotonate 4-methylphenol, so0 4-methylphenol will not dissolve in an aqueous solution of sodium bicarbonate.

Problem 4.29 One way to determine the predominant species at equilibrium for an acid-base reaction is to say that the reaction arrow points to the acid with the higher value of pA.. For example, NH." + H,0 € NH; + H0* pK. 9.24

pK. -1.74

NH:" + OH™ & NH; + H,0

pk. 9.24

pK. 15.7

Explain why this rule works.

In acid-base reactions, the position of equilibrium favors the reaction of the stronger acid and stronger base to give the weaker acid and weaker base. The acid with the higher pA; is the weaker acid, so the arrow will point toward it. Problem 4.30 Will acetylene react with sodium hydride according to the following equation to form a salt and hydrogen, H,? Using pk; values given in Table 4.1, calculate K., for this equilibrium.

HC=CH Acetylene

+

Na'H-

=

Sodium hydride

HC=CNa" Sodium acetylide

+

H, Hydrogen

Because the pA; for hydrogen is larger than the pK; for acetylene, equilibrium will be to

the right, favoring the formation of the salt and hydrogen.

Kea

K, (Acetylene)

1x10™

K,(Hydrogen)

1x10™*

-

hx10%)

B

~¢ L Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 4: Acid and Bases

Problem 4.31 Using pA. values given in Table 4.1, predict the position of equilibrium in this acid-base reaction and calculate its A..

H,SO, +NH, —"HSO,”

H;80,

+

NHy

+NH,’

——=

HSO;

pK,=-5.2

+

NH/ PR, = 9.24

Because the p/ for the sulfuric acid (H;SO.) is smaller than the pA; for the ammonium lon (NH.*), equilibrium will be to the right, favoring formation of hydrogen sulfate and the ammonium ion species.

P “

ALV

LS

K/(NH')

575x107"

PP

LEWIS ACIDS AND BASES Problem 4.32 For each equation, label the Lewis acid and the Lewis base. In addition, show all unshared

pairs of electrons on the reacting atoms and use curved arrows to show the flow of electrons in each reaction.

™

fi H,C

P

& Gy

A

-

BF, 3

¢

e

”

e,

s0:

-’

ne” (a)

N Sen,

Lewis Base

-“’

(¥

F—8—F: Lewis Acid

—

:6/

Be ¥

g 9

~¢ Cengage -l

Solution and Answer Guide: Brown et al,, Organic Chemistry © 2023, 9780357451861; Chapter 4: Acid and Bases

flu, ®

HCH.C——0——CH,CH;

+

AICl,

NG i

i

HyCHC——0——CH,CH;

+

==

H,CHC——0——CH)CH, s

&

—

iCl——AlI—Cl:

Cewis Hikse

|

%,

Lewis

A—A—Ci: o

|

%

H,CH,C— 0 —CHCHy

Acid

Problem 4.33 Complete the equation for the reaction between each Lewis acid-base pair. In each equation, label which

starting material is the Lewis acid and which

is the Lewis base; use

curved arrows to show the flow of electrons in each reaction. In doing this problem, it is essential that you show valence electrons for all atoms participating in each reaction.

CH

= ('.II.,—(II—(ll (a)

CHy

[¢] + ,\[|—(:|.:: a

CHs

oo |

1 .. | OHS—(IZ—CI.’ + \Ai—g. CHy

_—

Ho |

Lewis

Lewis

Base

Acid

CHy

‘CI:

CH;

:ClI:

| ..¢I_..‘ CH;—(I:—CI—?I—O!.

(lzn‘ (zHA—(lr +H—O—H== (b)

CHy

?Hs

W CH:,—(I:

CH Lewis Acid

+ H-O-H i

Lewis Base

s=—=

?H?*0 H CH,-(I:—O: CcH.

2

H

~¢ Cengage -l

Solution and Answer Guide: Brown et al,, Organic Chemistry © 2023, 9780357451861; Chapter 4: Acid and Bases

© CHy—CH—CHy + Br- =

AR+

CH3-CH-CH.

3

3

Lewis Acid

)

B Bl

—s

©'

ST

CH;—CH-CH,

Lewis Base

CHy—CH—CH, + CHy—O0—H =

N+

+

CH3-CH-CHy

H,C, Y.

CH;-O—H ===

Lewis Acid

Ngé|

H

CH;—CH—CH,

Lewis Base

Problem 4.34 Each of these reactions can be written as a Lewis acid-Lewis base reaction. Label the

Lewis acid and the Lewis base; use curved arrows to show the flow of electrons in each reaction. In doing this problem, it is essential that you show valence electrons for all atoms participating in each reaction.

Ill (a) CHy—CH=CH,

+ H—Cl —

CHy—CH—CH,

+ C1 H

/\

CH3—CH=CH,

+

“H—Cl:

Lewis

Lewis

Base

Acid

Gy =G, ®)

-

Base

|

CH;—CH—CH,

S

+

:CI:

CHy

i C|13—(|3=Ct-|2 + Lewis

—

+ Hl—lir;'l.ll.‘—(;y—(vlt_.—l!r + B

CHy

CH.

+

3

s Br--Br:

Lewis Acid

—»

+ so. CHS_(I:_Q"z—B“ + CH:’

S5 Br:

B

~¢ L Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 4: Acid and Bases ADDITIONAL PROBLEMS

Problem 4.35 The

sec-butyl cation can react as both a Bronsted-Lowry

acid (a proton donor) and a

Lewis acid (an electron pair acceptor) in the presence of a water—sulfuric acid mixture. In each case, however, the product is different. The two reactions are as follows:

MN; _H CH;—CH—CH,—CH, [V}

| CH,—CH—CH,—CH,

+ H,0 =

CHy—CH=CH—CH,

sec-Butyl cation

« H.—(.H—( @

(a)

+ H,0 ==

H,—CH,;

+ H,0

sec-Butyl cation

In which

reaction(s) does this cation react as a Lewis acid? In which

reaction(s) does it

react as a Bronsted-Lowry acid?

As shown by the curved arrows that indicate the flow of electrons, in reaction (1), the sec-butyl cation Is acting as a Lewis acid by reacting with the Lewis base (water). In reaction (2), the sec-butyl cation is acting as a Brensted-Lowry acid by donating a proton to the water molecule. (b) Write Lewis structures for reactants and products and show by the use of curved arrows how each reaction occurs.

(1)

+

e

CHy-CH-CH,-CH,

! N

.

——

"1

CHy—-CH—-CH,—CH, H

H Ale—u @

CHy—

*

fl

|

CH,

X

S=——=CH,-CH=CHCH,

*+

[

H—O—H

~¢ Cengage -l

Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 4: Acid and Bases

Problem 4.36 Write equations for the reaction of each compound with H:SOs, a strong protic acid.

(a) CH;OCH;

CHy—O0—CH,

+

Hflb:—?—b:—u

:(II):

'I‘

:fi:

A

~—=

CH,—Q™CH,

+

'::o:—fi—fi—n

Mo N

Ho N

(b) CH3,CH,SCH,CH3

R + W, “HOG—S—0—H

CHiCH,~§-CHiCHy

P (2-1-Bromo-1-chloropropene

© (8)-2,3,4-Trimethyl-3-heptene NOMENCLATURE FOR CYCLIC ALKENES: PROBLEM 5.6 Write the IUPAC name of each cycloalkene.

(@) [é 1,3-Dimethyl-cyclopentene

NG Cyclooctene

©

: 4-(1,1-Dimethylethyl)-cyclohexene (IUPAC)

4- tert-Butylcyclohexene (Common)

~¢ & Cengage -l

Solution and Answer Guide: Brown et al,, Organic Chemistry © 2023, 9780357451861; Chapter 5: Alkenes: Bonding, Nomenclature, and Properties STEREOISOMERS OF POLYENES

|: PROBLEM

5.7

Draw structural formulas for the other two stereoisomers of 2,4-heptadiene.

1

HyC 2

H 4

6

5 CH,CH,

3 C=C{

c=c_

H

7

1

HyC ;

H

H4

5H

5,C=C

JC=C{

H

H

cis,trans-2,4-Heptadiene or

7

cis,cis-2,4-Heptadiene or

(2Z,4£)-2,4-Heptadiene

STEREOISOMERS OF POLYENES

H

'CHyCH, 6

(2Z,42)-2,4-Heptadiene

|I: PROBLEM

5.8

(10£122)-10,12-hexadecadien-1-ol is a sex pheromone of the silkworm. Draw a structural formula for this compound.

H\,,

10 /CHa(CHz)7CH,OH

CHyCH,CH,;5 12 C=C(

c=¢c_

H

H

H

(10£,122)-10,12-Hexadecadien-1-ol

END-OF-CHAPTER

PROBLEMS

An asterisk (*) indicates an applied problem.

STRUCTURE OF ALKENES Problem 5.9 Predict approximate bond angles about each highlighted carbon atom. To make these predictions, use valence-shell electron-pair repulsion theory (VSEPR) (Section 1.4).

109.5°

@CHJ

H

Step 2: 1,2 Shift. The 2° carbocation rearranges to the more stable 3° carbocation.

A

—E g

"

o~ Step 3: Make a new bond between a nucleophile and an electrophile.

,— )

5o—n

H

it

.lo,

_—

H

Step 4: Take a proton away.

*

iSwr

H

O\

+ 10—H | H

—_—

o

3

*

+ H—O—H !

H

~¢ Cengage LAl

(4

Solution and Answer

Guide:

Brown et al., Organic

Chemistry

© 2023,

9780357451861;

Chapter 6: Reactions of Alkenes

ADDITION

OF X2 TO AN ALKENE:

PROBLEM

6.8

Complete these reactions.

( AH,(I.( H=CH,

@

+ Br, reTr

amid-00113

s

CH,

A new chiral center is created, so the product is a racemic mixture of two enantiomers.

CH,CI CH, 2 o,

() BROMOHYDRIN FORMATION: PROBLEM 6.9 Draw the structure of the chlorohydrin formed by treating 1-methylcyclohexene with Cl,/H,0.

CH,CH,OH

Reduction; there is a gain of two hydrogens.

(b)

Q-0 Oxidation; there is a gain of one oxygen.

HS

ASH

=

(©

S

4 .

Oxidation; there is a loss of two hydrogens.

OzoNoLysis: PROBLEM 6.13 What alkene with the molecular formula

C¢Hy;, when

treated with ozone

sulfide, gives the following product(s)? (0]

Ay @

(only product)

Z

RN or

—\_/— —,

CeHyz

o R 1o

2. (CHs),S

\)L (only product)

H

and then dimethyl

-l

~¢ Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861;

Chapter 6: Reactions of Alkenes

(o]

[0

A” ®

+ )K/

(equal moles of each)

/Y\ “

o 2.Lo(CHy),8

_>=/

>

i

H

A, (only product)

-2. (CHy),8 CeHia

)K/

(equal moles of each)

Gz

©

*

o

(

(only product)

«l

~¢ 4 Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 6: Reactions of Alkenes

END-OF-CHAPTER

PROBLEMS

An asterisk (*) indicates an applied problem. ENERGETICS

OF CHEMICAL

REACTIONS

Problem 6.14 Using the table of average bond dissociation enthalpies at 25°C, determine which of the following reactions are energetically favorable at room temperature. Assume that AS = 0.

Bond Dissociation Bond

Bond Dissociation Enthalpy

Enthalpy kJ (kcal)/mol

Bond

kJ (kecal)/mol

H—H

435 (104)

C—l

238 (57)

O0—H

439 (105)

C—Si

301(72)

C—H(—CH;)

422 (101)

c=C

727 (174)

C—H(=CH,)

464 (11)

C=0 (aldehyde)

728 (174)

C—H(=CH)

556 (133)

C=0 (CO,)

803 (192)

N—H

391(93)

c=0

1075 (257)

Si—H

318 (76)

N=N

950 (227)

c—C

378 (90)

C=C

966 (231)

C—N

355 (85)

0=0

498 (119)

60

385 (92)

The following reactions can only occur to a significant extent as written if they are exothermic, that is, if the bonds that are formed are stronger than the ones that are

broken in the reaction. Recall that a catalyst increases the rate, but does not change the overall thermodynamics of a reaction. To find out if a reaction is exothermic, the dissociation enthalpy of all the bonds in the molecules on each side of the equation are

added together. If the bond dissociation enthalpy total from the right side of the equation is higher than the total from the left side of the equation, then the reaction is

exothermic (AH° for the reaction is negative).

«l

~¢ & Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 6: Reactions of Alkenes (a)

CH=CH;

+ 2H; + N> - H;NCH,CH,NH;

The bond dissociation enthalpies from the left side of the equation:

727 + (4 x 464) + (2 x 435) + 950 = 4403 kJ/mol (C=C) (4=C—H) (H—H) (N=N) The bond dissociation enthalpies from the right side of the equation:

(4 x 391) + (2 x 355) + 376 + (4 x 422) = 4338 kJ/mol (4 N—H)

(2C—N)

(C—C)(4 —C—H)

This reaction is endothermic because 4403 kJ/mol is larger than 4338 kJ/mol.

(b) CHz=CH;

+ CH4 —» CH,CH,CH;

The bond dissociation enthalpies from the left side of the equation:

727 + (4 x 464) + (4 x 422) = 4271 kJ/mol

(C=C)

(4 =C—H)

(4 —C—H)

The bond dissociation enthalpies from the right side of the equation:

(2 x 376) + (8 x 422) = 4128 kJ/mol (2C—C) (8 —C—H) This reaction is endothermic because 4271 kJ/mol is larger than 4128 kJ/mol.

(c)

CHz=CH;

+ (CH;);SiH

— CHCH,Si(CH;);

The bond dissociation enthalpies from the left side of the equation:

727 + (4 x 464) + (9 x 422) + (3 x 301) + 318 = 7602 kJ/mol (C=C)

(4=C—H) (8 —C—H)

(3 C—sSi)

(Si—H)

The bond dissociation enthalpies from the right side of the equation:

(376) + (5 x 422) + (9 x 422) + (4 x 301) = 7488 kJ/mol (C—C)

(6 —C—H)

(9 —C—H)

(4 C—si)

This reaction is endothermic because 7602 kJ/mol is larger than 7488 kJ/mol.

(d) CH,=CH, + CHl; — CH:CH,Cls The bond dissociation enthalpies from the left side of the equation:

727 + (4 x 464) + (422) + (3 x 238) = 3719 kJ/mol (C=C) (4 =C—H) (1 —C—H) (3C~—I)

The bond dissociation enthalpies from the right side of the equation: (2 x 376) + (5 x 422) + (3 x 238) = 3576 kJ/mol

(2Cc—C)

(5C—H)

(3C—I)

This reaction is endothermic because 3719 kJ/mol is larger than 3576 kJ/mol.

~¢ Cengage «l

Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 6: Reactions of Alkenes (I)

(e) CHa=CH, + CO + H, —

CHyCH,CH

The bond dissociation enthalpies from the left side of the equation:

727 + (4 x 464) + 1075 + 435 = 4093 kJ/mol

(C=C) (4 =C—H)

(C=0) (H—H)

The bond dissociation enthalpies from the right side of the equation:

(2 x 376) + (5 x 422) + 464 + 728 = 4054 kJ/mol (2C¢—C)

(5 C—H) (=C—H) (C=0)

This reaction is endothermic because 4093 kJ/mol is larger than 4054 kJ/mol. Z

® S

+ CHy= CH, —

The bond dissociation enthalpies from the left side of the equation:

(3 x 727) + 376 + (10 x 464) = 7197 kJ/mol (3C=C) (C—C) (10 =C—H)

The bond dissociation enthalpies from the right side of the equation: 727 + (5 x 376) + (8 x 422) + (2 x 464) = 6911 kJ/mol

(€=C)

(5C—C)

(8 —C—H)

(2=C—H)

This reaction is endothermic because 7197 kJ/mol is larger than 6911 kJ/mol.

)

G

+C0o, —

e

|

Y,

(8]

The bond dissociation enthalpies from the left side of the equation: (2 x 727) + 376 + (2 x 803) + (6 x 464) = 6220 kJ/mol (2c=C) (C—C) (2C=0) (6=C—H)

The bond dissociation enthalpies from the right side of the equation:

727 + (3 x 376) + (4 x 422) + (2 x 464) + (2 x 385) + 728 = 5969 kJ/mol

(c=C)

(3C—C)

(4 —C—H)

(2=C—H)

(2C—0)

(C=0)

This reaction is endothermic because 6220 kJ/mol is larger than 5969 kJ/mol.

«l

~¢ 4 Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 6: Reactions of Alkenes

o

(h)

HC=CH =

+ O, N

~

o

Il(“‘ s~ (II‘II

The bond dissociation enthalpies from the left side of the equation: 966 + (2 x 556) + 498 = 2576 kJ/mol (C=C) (2=C—H) (0=0)

The bond dissociation enthalpies from the right side of the equation: 376 + (2 x 728) + (2 x 464) = 2760 kJ/mol (C—C) (2C=0) (2=C—H) This reaction is exothermic because 2760 kJ/mol is larger than 2576 kJ/mol.

()

2CHa4 + 02 —» 2CH;0OH

The bond dissociation enthalpies from the left side of the equation:

2(4 x 422) + 498 = 3874 kJ/mol

2(4 —C—H) (0=0)

The bond dissociation enthalpies from the right side of the equation: 2(3 x 422) + 2(385) + 2(439) = 4180 kJ/mol

2(3 —C—H) 2(C—0) 2(0—H)

This reaction is exothermic because 4180 kJ/mol is larger than 3874 kJ/mol. ELECTROPHILIC ADDITIONS

Problem 6.15 Draw structural formulas for the isomeric carbocation intermediates formed on the treatment of each alkene with HCL. Label each carbocation 1°, 2°, or 3° and state which of the isomeric carbocations forms more readily.

CHLCI; M,c=c/ \L'H, (a)

Tu" —

HC——C——CH,CH, Tertiary (Formed more readily)

T"’ +

n,?—-g—-cu,('u, Primary

«l

~¢ Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 6: Reactions of Alkenes

.

O/\

)

O/\

L

O/\

Secondary

Primary

(Formed more readily)

(b)

CHy

CHy

2

-~

+

CcH,

+

Tertiary

(c)

Secondary

(Formed more readily)

H

HyC——C——C

Hy

—CHj

+

H,C—fi:fi—(:l-l‘

—_—

Secondary

(@

(Only cation formed)

Problem 6.16 Identify the carbocation intermediates first formed in the following reaction. Will rearrangements possibly occur? What major product is formed?

5

HCI

—e

(e}

. n

+

e

4

*

h Most stable carhocation Results from a 1,2-carbon shift

Major product formed

LAl

~¢ & Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 6: Reactions of Alkenes Problem 6.17 Arrange the alkenes in each set in order of increasing rate of reaction with HI and explain the basis for your ranking. Draw the structural formula of the major product formed in each case. e

and

(@)

ShNOS '

@

A tertiary,

carbocation

O~—

Major Product

i

OA

A secondary

carbacation

Major Product

New chiral conter created o racemic mivture

The structure able to form the most stable carbocation intermediate (tertiary carbocation In this case), is the fastest to react with HI. (b) HiCH.CHC=CH;

and (H,C);C=CHCH;

HOCHCHE ==CH;




NN

el

a

1. NaNH, _—

—

a

—

N

(c)

a

:

«a

—

2. Mel

(d) Problem 7.23 Show how to convert 1-butyne to each of these compounds.

(a) CH:CH.C=C Na*

CH,CH,C=CH + NaNH,

——

CH,CH,C=C Na' + NH,

The anion can be formed using sodium amide, NaNH..

(b) CH,CH,C=CD NaN| '—H’l-

CH;CH,C=CH

=

CH,CH,C=C:™

Na*

0 L?

CH;CH,C=CD

Formation of the terminal acetylide anion followed by reaction with a deuterium donor

such as D,0 gives 1-deutero-1-butyne.

X AL=C _ N

CH ©)

H

‘

Gt

C=E=CH

D

1. (sia);BH

——»

2. CH;CO,D

CHy

Cl

Qz

Hf

Cc=C

s

H

\D

Hydroboration with this disubstituted derivative of borane followed by reaction of the organoborane with deutercacetic acid gives the desired 1-deutero-1-butene.

+¢Cengage «l

Solution and Answer Guide: Brown et al., Organic Chemistry ©® 2023, 978035T7451861; Chapter 7: Alkynes

CHCth (d)

D

u

L=C

H

= CHyCH,C=CH

—— o 2. CH,CO;H

S

D

/

ic: = c’HhY

H

Hydroboration of the terminal alkyne with deuteroborane adds deuterium to the more substituted carbon of the alkyne. The reaction of the deuterated organoborane with acetic acid gives the 2-deutero-1-butene.

*Problem 7.24 Rimantadine was among the first antiviral drugs to be licensed

in the United

States to use

against the influenza A virus and to treat established illnesses. It is synthesized from adamantane

by the following sequence (we discuss the chemistry of Step

1in Chapter 8

and the chemistry of Step 5 in Section 16.8A). |

Bry,

CH,=CHBr, AlBr,

(n

Adamantane

Br

(2) Br

I-Bromoadamantane

fig

(%)

f

=

_’@\T()

Br

--""@\TA\H-)

(5)

*

Rimantadine

Rimantadine of the virus.

is thought to exert its antiviral effect by blocking a late stage in the assembly

(a) Propose a mechanism for Step 2. Hint: As we shall see in Section 22.1C, the reaction of a bromoalkane such as 1-bromoadamantane with aluminum bromide (a Lewis acid, Section

4.7) results in the formation of a carbocation and AlBr.~. Assume that adamantyl cation is formed in Step 2 and proceed from there to describe a mechanism.

The bromoethene r electrons attack the adamantyl cation, to create a new cation that captures a bromide from AlBr,~ to yield dibromoethyl-adamantane.

=»l

¢ # Cengage Solution and Answer Guide: Brown et al., Organic Chemistry & 2023, 978035T451861; Chapter 7: Alkynes

Step 1: Make a bond between a Lewis acid and Lewis base and simultaneously break a bond to give stable molecules or ions.

Q

/"ABry

—»

+

+

Q

ABr”

Br: Step 2: Make a new bond between a nucleophile (zbond) and an electrophile.

w

CH,CHEr: L

CH, =CHEr: Step 3: Make a bond between a nucleophile and an electrophile.

( Ill-Cm, )

Q

oL

—_—

Q

(b) Account for the regioselectivity of carbon-carbon

:'a-':

I

...

+

m3

bond formation in Step 2.

The carbon atom without the halogen ends up attached to the adamantane group,

because this allows the formation of the more stable intermediate with the cation adjacent to the halogen atom. The formation of this cation has a lower activation energy because of the resonance stabilization of the cation provided by the halogen. (c) Describe experimental conditions to bring about Step 3.

This reaction occurs via double dehydrohalogenation using a strong base such as NaNH.. (d) Describe experimental conditions to bring about Step 4. This transformation occurs via an acid-catalyzed hydration of the alkyne using H.0, H2S0,, and HgS04. The initially formed enol equilibrates to the more stable keto form.

al

¢ Cengage Solution and Answer Guide: Brown et al., Organic Chemistry & 2023, 978035T451861; Chapter 7: Alkynes

Problem 7.256 Show reagents and experimental conditions to bring about the following transformations. H,C

“|"

.

;

(fl CHCCH,CHy,

®

(4]

CH CH=CH, ml

S

I

(|:l (In CHCHCH,

~25

H4C

Br

® cH,e=cH %

H

NeHy

S

HC a0 St i

[0

s

o La“

i

N

v

)

CHLCHCH, i l

\(\=(,/

.

e

Br|

U

CH,

(I:| (In CH,C=CH, 2> (:H,(l‘.(:fl:‘ l

(a) (CHa):COK' (b) Cla (c) 3 NaNH; then H,0 (d) HCl (e) HCl (f)

NaNH, then CHal

(g)

HgSO.;,

“1504,

“no

ar

BH;then

(h) Li or Na in liquid NHs

(i)

Hz, Pd/CaCO; (Lindlar catalyst)

()

Bra

HzOz.

NaOH

CH, u

¢ Cengage al

L 4

Solution and Answer Guide: Brown et al., Organic Chemistry & 2023, 978035T451861; Chapter 7: Alkynes

Problem 7.26 Show reagents to bring about each conversion.

P (=)

©

\/\_/\/\r

M

\

\/\Q

\

7

B

R

°

O

o

o

TP

N

TN

g

H M

(a) HC=C" (b) NaNH; then CHs;CH,Br

(c) Hz, Pd/CaCoO; (Lindlar catalyst) (d) Li or Na in liquid NH; (e) 1. (sia);BH; 2. H,02, NaOH

(f) HgSO0s, HyS04, H20 Problem 7.27 Propose a synthesis for (2)-9-tricosene (muscalure), the sex pheromone for the common housefly (Musca domestica), starting with acetylene and haloalkanes as sources of carbon

atoms. "

9

Mn.

W\/\\/m

Alkylation of acetylene with the two alkyl halides followed by reduction using Hz; and the

Lindlar catalyst gives muscalure.

HO=CH

1. M . 2. CHy(CHy)4CH,Br

oy (CHy),C=CH (CHa),

"

______

2. CHy(CHy) ¢4 CHyBr

CHy(CHy)y.

CHy (CH,)7C=C(CHR) 12Oy e — Linds

NNt

1. NaN

,C=C_

-

H

(CHy)12CHy H

Muscalure

-l

¢ @ Cengage

‘ Solution and Answer Guide: Brown et al, Organic Chemistry ©® 2023, 5780357451861 Chapter T: Alkynes

Problem 7.28 Propose a synthesis of each compound starting from acetylene and any necessary organic

and inorganic reagents. (a) 4-Octyne =

O

1. NaNH;

_—

S ohomomar

=

—

1. NaNH,

HO=CCHCH2CHs 575, Czarzer =

CH;CH,CH,C=CCH,CH,CH,

(b) 4-Octanone

(o]

H0, H; S0, . CH,CH,CH,CH,CCH,CH,CH, i CHyCH, CH,C=CCH,CH,CH; ————2—% (From (a)) . (c) cis-4-Octene

CH;CH,CH,C=CCH,CH,CH, (From (a))

S SE— . Pdl CaCO,

CHyCH,CH,

CH,CH,CH,

/C-C\

(Lindlar catalyst)

H

H

(d) trans-4-Octene

G‘l:u'lza"n\ Lior Na

CH,CH,CH,C=CCH,CH,CH;

—

"——-

(From (%))

" ()

/C-C\

H

/

H

CH,CH,CH,

«l

¢ [ Cengage Solution and Answer Guide: Brown et al., Organic Chemistry ©® 2023, 9780357451861; Chapter 7: Alkynes

(e) 4-Octanol c"3GMZCM2\

/CHzcflzcH3

Ic=c\

H

H

(From (c)) 1. Hg(OAc),, H,0

or CH30H20H2

H

—————»

2. NaBH,

o=

OH |

CH3CH,CH,CH,CHCH,CH,CH;

H

(racemic mixture)

CH,CH,CH,

(From (d)) (f)

meso-4,5-Octanediol

Gl-lsG"lzGHz\

,CHoCHCHy

p— ,C=C_

H

HO\ 4 ————

H

OH

CH3CH,CH,/..c—c.CH, CH,CH;,

H;0,

Hi

‘H

(From (c})) Note that only the cis isomer gives the meso product, the trans isomer gives a pair of R,Rand S,S enantiomers.

Problem 7.29 Show how to prepare each compound

from 1-heptene.

(a) 1,2-Dichloroheptane

CH3CH,CH,CHy CH,CH=CH, 1-Heptene

i ——2—=

) GH, CH,CH,CH,CH,CHCH,CI (racemic)

fil‘

.2

C

Cengage

(b) 1-Heptyne

-

1 aww::-.:.ama .

fi—1.

CHy CH, CH, CH, CH, C=CH

(¢) 1-Heptanol

CHy CH, CH; CH, CHy CH=CH, -8~

Gy CH, CH, CH, CH, CH, CH, OH

(d) 2-Octyne

A

e Gl

—

1.

CHyCH;CH;CH,CH,C=CCH,

(®) cis-2-Octene

o

o, 1 O, (rem ()

(f)

f (Lindlar catalyst)

uw.w.m.w.\c_:c(as H

W

trans-2-Octene

namananancmom, (Prom @)

Lo .

CHCHCHCHCH,

o

H’°=c\at

al

¢ L 4 Cengage Solution and Answer Guide: Brown et al., Organic Chemistry & 2023, 978035T451861; Chapter 7: Alkynes

Problem 7.30 Show how to bring about the following conversion. r\/ /

W

This transformation requires the addition of a methyl group and the creation of a ¢is

double bond. = r\/

N

H,

7

1. BH,

“I“; Lindl

g o 2 cneon

Br

in”

NaNH, (excess)

[—(.:

/ }

Mel

/

NH, Halogenation of 1-butene leads to the dihalide. Treatment of the dihalide with a strong

base leads to the formation of a 1-butyne anion that upon reaction with Mel, creates

2-pentyne. Treatment with either hydrogen and Lindlar catalyst or borane followed by acetic acid leads to c/s-2-pentene.

LOOKING AHEAD Problem 7.31 Alkyne anions react with the carbonyl groups of aldehydes and ketones to form alkynyl alcohols, as illustrated by the following sequence.

A CHC=CiNa"

O

I + H—C—H

—

ey [CHyC=C—CH,0

Na'] 7=

CHyC=C—CH,0H

Propose a mechanism for the formation of the bracketed compound, using curved arrows to show the flow of electron pairs in the course of the reaction.

. fl)

Gbcfciwc——u

Z?!

Na

——> | HCC=C—C—H H

The acetylide anion acts as a nucleophile and attacks the electrophilic carbonyl carbon

atom (make a new bond between a nucleophile and an electrophile). The carbonyl =

=»l

¢ # Cengage Solution and Answer Guide: Brown et al., Organic Chemistry & 2023, 978035T451861; Chapter 7: Alkynes

bond is broken, generating the alkoxide anion intermediate corresponding to the species in the bracket. The carbonyl carbon atom is electrophilic because of the electronegative oxygen atom that provides for a rather polar C=0. 5

Z?!

H—C—H 8+

This reaction is particularly noteworthy because a new carbon-carbon bond is formed.

*Problem 7.32 Following is the structural formula of the tranquilizer meparfynol (Oblivon).

Z Oblivon HO Propose a synthesis for this compound starting with acetylene and a ketone. (Notice the -yn- and -ol in the chemical name of this compound, indicating that it contains alkyne and hydroxyl functional groups.)

_ HC=CH

fi..

HC=C:

N Na*

—o..

Z -

Na*

Z L H,0

HO

Obliven

The synthetic strategy for the construction of meparfynol (Oblivon) is based on the reaction of an acetylide anion with a carbonyl species as described in Problem 7.26. The

anion of acetylene reacts with 2-butanone followed by protonation in aqueous HCL to give the desired alcohol. In later chapters, you will see many other examples in which the carbonyl carbon atom acts as a nucleophile that makes a new carbon-carbon bond.

«l

.2 Cengage 1w

Solution and Answer Guide: Brown et al., Organic Chemistry ©® 2023, 9780357451861; Chapter 7: Alkynes

*Problem 7.33 The standard procedure for synthesizing a compound is the stepwise progress toward a target molecule by forming individual bonds through single reactions. Typically, the product of each reaction is isolated and purified before the next reaction in the sequence is carried out. One of the ways nature avoids this tedious practice of isolation and purification is by the use of a domino sequence in which each

new product is built on a

preexisting one in a stepwise fashion. A great example of a laboratory domino

reaction is

William S. Johnson’s elegant synthesis of the female hormone progesterone. Johnson first constructed the polyunsaturated monocyclic 3° alcohol (A) and then, in an acid-induced

domino reaction, formed compound

Me Me

B, which he then converted to progesterone.

Me

Me

|

Me |

|

1. CFO00H

=

Me

|

0¥

Me OH

O

0.

Chy

0

U 2. HL0, CHLOH

i

(Chaprer 16)

Me A

B

Me. Me

Me

Me,

y

(1) B

Me

1. 5% KROH/HO

i

(Chaper

Me

o

z

19)

H

=

w0

y

H

0

Me

C

Progesterone

A remarkable feature of this synthesis is that compound A, which has only one stereocenter, gives compound B, which has five stereocenters, each with the same configuration as those in progesterone. We will return to the chemistry of Step 2 in Section 16.7 and to the chemistry of Steps 3 and 4 in Chapter 19. In this problem, we focus on Step 1.

~¢ ¢ C Cengage =»l

Solution and Answer Guide: Brown et al., Organic Chemistry & 2023, 978035T451861; Chapter 7: Alkynes

(a) Assume that the domino reaction in Step 1 is initiated by protonation of the 3° alcohol

in compound A followed by loss of H:0 to give a 3° carbocation. Show how the series of reactions initiated by the formation of this cation gives compound B. As stated in the problem, the reaction begins by protonation of the hydroxyl group (add a proton).

Mo

Ko

|

o

N Me

Me

:oH

~

|

1. CFCOOH __

N

Me

A

o' N H

/

The protonated hydroxyl group makes an excellent leaving group, so H;0 departs, leaving behind a carbocation (break a bond to give stable molecules or ions). In Chapter 10, you will see many examples of reactions in which hydroxyl groups are protonated and then depart as water. This is a characteristic reaction of alcohols in acid.

Me

Me

Me

| N

Me - H;0

N

—_—

)y

Me

3 O—H

H

|

+

Me

~¢ +C Cengage al

Solution and Answer Guide: Brown et al., Organic Chemistry & 2023, 978035T451861; Chapter 7: Alkynes

The carbocation is now set up to start a domino reaction as indicated by the flow of electrons listed next. Although all of the arrows are on one figure, the timing of the individual steps may not be simultaneous as the molecule adopts various conformations. Note that in intermediate B, the cation is resonance stabilized by the three adjacent oxygen atoms, providing a driving force for the process. Me

Me

Vo

|

’

[+)

e

— |

i

[+

LT

+

s

Mo

B

The intriguing thing about this domino reaction is that the single stereocenter, which is apparently lost upon carbocation formation, can nevertheless control the

stereochemistry of five new stereocenters. This is most likely a conformation effect. The single stereocenter already present apparently influences the conformation of the starting material, so that once the carbocation forms and the reaction begins, the

molecule is conformationally set up to produce the stereocenters shown before there is time for the molecule to adopt an alternative conformation. (b) If you have access to a large enough set of molecular models or to a computer modeling program, build a model of progesterone and describe the conformation of each ring. There are two methyl groups and three hydrogen atoms at the set of ring junctions in progesterone. Which of these five groups occupies an equatorial position?

Which occupies an axial position?

I +

i

&

L 3

-

Progesterone

s

J

)

=»l

¢ L4 Cengage Solution and Answer Guide: Brown et al., Organic Chemistry & 2023, 978035T451861; Chapter 7: Alkynes

When only ring skeleton of progesterone is drawn, it is easier to see that the two internal cyclohexane rings are each in the chair conformation. The terminal cyclohexane ring has an sp? carbon and thus cannot be in a true chair conformation, but it does adopt a similar chair-like conformation. The cyclopentane ring is puckered to relieve ring strain. Notice that the rings are fused in the equatorial positions. Chair conformation Chair-like

conformation \

Chair conformation

\

Chalr-like cnnlom{li:n

0? Because the ring junctions occupy all of the equatorial positions, the methyl groups and hydrogen atoms occupy only axial positions at the junctions.

P

vl

/

CH;

07 Axial

H

fi\ /I

|

H

x \CH3

-

=»l

.2 Cengage 1w

Solution and Answer Guide: Brown et al., Organic Chemistry & 2023, 978035T451861; Chapter 7: Alkynes

ORGANIC CHEMISTRY REACTION ROADMAP Problem 7.34 Use the reaction roadmap you made for Problem 6.60 and update it to contain the reactions in this chapter. Because of their highly specific nature, do not use reactions 1 and 6 from this chapter on your roadmap. .

C-C bond forming reactions are inlicated by a box swrrounding

>

“of

the reaction number 118

/

geminal

dinaloalkanes

%/

.

)

/

|

vidnal f

allylic halides

f

T

thicls

\

»

"

dihaloalkanes | /'

W

=2k

o=

)

N

™~

e

{If

_//

amines

14

J.".

N

N

NN \ |

\ I|

-

\

\

.|

f

|

N/ S \'\ A

, X :l\ /

[N

/]

I " NN\

(\

If);

//

\@ N

\o

\e

gof

g

\

I'l

f

|

e

{

e I

head-ins

\ |

/

/

Ira“

7

/_,

@]4—f’

[ etners |

|

/

\

s

[sityt ethers

|

S SN v /) Fay. / "If;

N\

A

S

\‘

I".

|

/

alkanes

alkyl azides

s

[~

o

I

/

4]

—

[/l e

l

X\

/

WAAZ

™

]

S/

s~

/ \\

7

(2

B

/4 \

LY k

o

S

:

\\ \

| slcohols

—

acid

\

S

=~

4

"

N

i

J"

alkynes

F

e

2

/

i

e

|

vicinal

aminoalcohols

¢ @ Cengage «al

‘

Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861, Chapter

Chapter 7 Roadmap Reaction Legend Reaction 7.2

RC==C" - 82 reaction

- Must be primary haloalkane to avoid E2 - Makes a carbon-carbon bond

Reaction 7.3

2 NaNHa In NHs (1) - Last step In converting an alkene into an alkyne through a vicinal dihaloalkane - 3 NaNH, followed by protonation for terminal alkyne

Reaction 7.4 2X

- Adding only one equivalent of Xs gives an (E)-dihaloalkene Reaction 7.5 2 HX

- Markovnikov regiochemistry (with terminal alkynes)

- Adding only 1 equivalent of HX gives a haloalkene Reactlon 7.7a 1. BHy 2. H,0,, NaOH = Little regiochemical preference - Reaction involves enol formation followed by keto-enol tautomerism Reaction 7.7b 1. (sia)sBH

2. HaOa, NaOH

- Non-Markovnikov regiochemistry for terminal aliynes - Reaction Involves enol formation followed by keto-enol tautomerism Reaction 7.8 1. Hz0, H2S0,, HgSO. - Markownikov

regfochemistry for terminal alkynes

- Reaction Involves enol formation followed by keto-enol tautomerism Reaction 7.9a 2H;,, transition metal catalyst - Reaction does not stop at the alkene but continues on to the alkane Reaction 7.9b Ha, Lindlar catalyst - Syn addition

- Great way to make (Z) alkene

T: Alkynes

~¢ +C Cengage al

Solution and Answer Guide: Brown et al., Organic Chemistry & 2023, 978035T451861; Chapter 7: Alkynes

Reaction 7.10 1. R;BH 2. Protonolysis (that is, CH;CO,H)

- Can give Markovnikov regiochemistry depending on R

- Syn addition - Alternative way to make (2) alkene

Reaction 7.11 2Na® or 2Li° in NHs ()

- Radical mechanism involving two cycles of one-electron reduction followed by protonation

- Great way to make (E) alkene

Problem 7.35 Write the products of the following sequences of reactions. Refer to your reaction roadmap to see how the combined

reactions allow you to “navigate” between

the different

functional groups. For example, in part (a) below, notice how the reaction sequence results in the conversion of an alkyne into a haloalkane in two steps. — /

(a)

L. Hy / Lindlar catalyst

2. HBr

An alkyne

~

_

1. H; / Lindlar catalyst — Reaction 7.9b

M

2.HBr 45

Reactioa

ned/e\l/ mixture Br N

Br A haloalkane

1.Na® / NH,q _ (b)

—eep

2.Cly

An alkyne

cl /+/

An alkyne

1. Na® /NH, Reactien 7.11

NN

An alkene

20, i

; /\l/\/

o

e

A vicinal dihaloalkane

=l

¢ & Cengage

‘

Solution and Answer Guide: Brown et al, Organic Chemistry © 2023, 9780357451861; Chapter T: Alkynes

L. 1 equivalent of NaNH,

"\§\

2. CHyl 3. Excess H, / Pd

©

An alkyne

L1 oquivlent of NNE, _

Loxom Reaction 72

ON" An alkyne 3. Excom Hy/Pd ,

CHy

Reaction 7.9a

An alkyne (new C-C bond)

An alkans

L. 1 equivalent of NaNHy

Fg

@

s

2. A R W

T

A Hy / Lindlar catalyst

7 alkyne

e

1. Bry

o

’ Br = .. 3-"'""“.

Ll |

An alkyne

.

st Reaction 7.9

_ —mme” An alkyne (mew C-C bond)

— — —

.

=

|

racemic

Reactlon 63

An alkene

P, Br

Br

A vicinal dikaloalkane

«l

¢ L Cengage

‘

Solution and Answer

Guide: Brown et al., Organic Chemistry © 2023, 9780357451861, Chapter

T: Alkynes

1.0,

W

3

An alkene

(e)

r_-qlln.llr s

$. Milel acid

4 () Bi1

5. HO, / NaOH

L Lag.

N,

of NaNH, in NH,

An alkene

S,

racemic

‘_{"

mixture

S

M Tami—li-

[

Reaction 7.3

Analkyne

4

fir“

ool

A vicinal dihaloaikane

~Y

[+] An aldelyde

MULTISTEP

SYNTHESIS

Problem 7.36 Using your reaction roadmap as a guide, show how to convert acetylene and bromoethane into 1-butene. All of the carbon atoms of the target molecule must be derived from the given starting materials. Show all intermediate molecules synthesized along the way. =

+

Acetylene

Br —i

AN

Bromoethane

=t

I-Butene

?

O 7

Acetylene

—

Py

Bromoethane

1-Butene

NaNH,

Hy |

L

e

Pd/ CaCO,

S

\

For this synthesls, recognize that a carbon-carbon bond must be formed between both two-carbon starting materials to make the 1-butene product. The best way to accomplish this Is to deprotonate acetylene with sodium amide, followed by a reaction with bromoethane. Reduction with hydrogen and the Lindlar catalyst completes the synthesis of 1-butene.

«l

¢ Cengage '

|

Solution and Answer Guide: Brown et al., Organic Chemistry & 2023, 978035T451861; Chapter 7: Alkynes

Problem 7.37 Using your reaction roadmap as a guide, show how to convert 1-butene into hexane using ethene as a carbon source. Show all intermediate molecules synthesized along the way.

lfir;

nr

Bf\)\/

‘ Hy. Pd/C

izt NI, (sf!)

A (_—\ +*

HBr

—_—

f

Ay

Br

The carbon chain needs to be extended by two carbons, and those carbons can be added through the alkylation of an alkyne. To accomplish this, 1-butene is first converted to the dibromide. Subsequent base-catalyzed elimination and addition of a

2-carbon electrophile prior to work up yields the six-carbon chain. Hydrogenation of the alkyne produces the target alkane.

«l

¢ L 4 Cengage Solution and Answer Guide: Brown et al., Organic Chemistry ©® 2023, 9780357451861; Chapter 7: Alkynes

Problem 7.38 Using your reaction roadmap as a guide, show how to convert propyne into (R.5)-2,3-butanediol. All of the carbon atoms of the target molecule must be derived from the given starting materials. Show all intermediate molecules synthesized along the way.

OH

i

OH

Propyne

lodomethane

——

(R.85)-2,3-Butanediol OH

:

g

FE——

Propyne

é

OH NaNH,

(R,S5)-2,3-Butanediol

1. 050, 2. NHHSO},

Hzo

—_—C

Mel

S H» S—

I

ALP

catalyst

First recognize that a carbon needs to be added to the carbon chain and that the alcohol groups need to be stereoselectively formed. Removal of the hydrogen on the terminal end of the alkyne results in a nucleophile that can be added to a methyl halide

to form 2-butyne. Selective reduction to the cis-alkene is needed using hydrogen and Lindlar catalyst or borane and acetic acid. This will generate the (&,S) diol upon syn addition of hydroxyls using osmium tetroxide.

al

¢ L 4 Cengage Solution and Answer Guide: Brown et al., Organic Chemistry & 2023, 978035T451861; Chapter 7: Alkynes

Problem 7.39 Using your reaction roadmap as a guide, show how to convert the starting trans-alkene to the cis-alkene in a high yield. Show all intermediate molecules synthesized along the way.

m

2

% trans-alkene

cis-alkene

? /

_—

Hy

By

P/ CaCO,

(Lindlar catalyst)

Br

Te

-

2 NaNH,

NHy (1)

=

Br

(meso) Recognize that the only reaction covered so far that creates a cis-alkene is hydrogenation of the corresponding alkyne using the Lindlar catalyst, so propose this as the last step. The alkyne can be synthesized using the two-reaction sequence of bromination of the starting trans-alkene, followed by a reaction with NaNH in liquid ammonia.

+¢Cengage i

Solution and Answer Guide: Brown et al., Organic Chemistry & 2023, 978035T451861; Chapter 7: Alkynes

REACTIONS IN CONTEXT Problem 7.40 Functional groups such as alkynes react the same in complex molecules as they do in simpler structures. The following examples of alkyne reactions were taken from syntheses carried out in the research group of E. J. Corey at Harvard University. You can assume that the reactions listed involve only the alkyne, not any of the functional groups present in the molecules. Draw the expected products for the following reactions. H;C H4C

H-;(.;\

i

Li

! \"/\‘/\('.[I;;

CHy e

DMSO

CH;

{racemic)

CH, CHg

KyCOy, Hs0u 1,0

H

=

®) =

=

CHy

oL

_O

reactive alkyne

i =

N

=

. CHs

x

Z 40

0

0

i L0 H3C

“CHy 0.

= CHy P

i

0O

0

Y

HyLindlar

catalyst

e—_—

o

cis alkene

in product

4

CH;

0 “CH, ©

[o]

(sia)sBH, THF, 0°C

QR 6}

’

(solvent)

(racemic)

(a)

\CH CH,

w

Hi!-‘i

CHS

\Si

0/

-

/‘n\

Ir

HSC

Xulfi

(o]

HaC”

R

0

CH 3 s

)

«l

¢ Cengage Solution and Answer Guide: Brown et al., Organic Chemistry

@ 2023, 9780357451867;

Chapter 8: Haloalkanes, Halogenation, and Radical Reactions

Solution BROWN

and Answer

ET AL., ORGANIC

CHEMISTRY © 2023,

Guide

9780357451861;

CHAPTER 8: HALOALKANES, HALOGENATION, AND RADICAL REACTIONS

TABLE

OF CONTENTS

IN=CRADIBT PrODLBING it IUPAC

Nomenclature for Haloalkanes:

Problem

SO

8.1 . cicicececeecececiceseescineeeveeee. 384

Free-Radical Halogenation: Problem 8.2 .......cccciiiiisccisniccctssccissesmsnersesssssssessenses 384

Enithalpy of Reactions E Problem B niinsannnuaimsianasamsismiiiiig 385 Enthalpy of Reactions B Problomi

Bl cuissaunnnonsmssissannpismrmmasimnses oD

Allylic Bromination: Problem 8.5 .....uecereecisnismssssesmsessssrasssnsenssessessssssssassassosnsassasees 380 Radical Autoxidation: Problem 8.6 ......c.ciiinmiisssssssssssssssss mssssssssssssssissnses 390 MCAT Practice Questions: AntOXIAANTS.........ccccviiincscisteniosssisssscssssssiissssssiansssssessssensensa 38 1 Free-Radical Addition of HX to an Alkene: Problem 8.7 .......innnnenesisvissssseseeees 389 End-of-Chapter Problems. ... N O

N

U

B B

s

i

B B s

s B

ssissssssssmsssssssssssssssssssasssnsenssss 389

R

s

R

L

e

T

A

R

s

e

SD

R

B

e

i

309

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i

eimmisiim o

i i

ProBlerty B2

isiansiss e s e

o

o asinsssis

s

AR R

TSR T s

shassin i s s

s

RS

RSB

s

s s

i AR s

B

ey isaa s wnsissinss ibss sidavasaspsravasivon I AL

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. ..vieicierirssssssesssissssssssssssssnsnssssss sanssssssssssssssnsanssssnse sanssssssssnsssssssssnsssnsssansssrassssnss BT

By B

i i

mimdi

Probloin Bl

i

o

T

st

e

e

9

ssiiisiiisisiviisiisiiiisiiisi TG

PrOBIBIG BB covuicosvammiivivssiivesimitsivssaiasm i ss s vsssisspssimsssicasniimisssiruisaiisisis O 0

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s

i

Problefn B 19w

P OB

BN

i

R

imiisitaiitiisi riitei i

ovisnavmsassnsisisns aniaasss s s R o s R A AR R

RS

T

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AR ORGSR

8 399

s

s

iscvanss S

PrOBUBITI 8.27..coeeeeeeecvseeisinessssssanesssssssssssssasssssssssasssmsssasssssssesassssrssssnnssssssssssssssssansassassssnens GOT

«l

¢ Cengage Solution and Answer Guide: Brown et al., Organic Chemistry

@ 2023, 9780357451867;

Chapter 8: Haloalkanes, Halogenation, and Radical Reactions

OO

B

s

i

o

R

e

S

ProBlon B:03 ..ouvuummiiisiis s ALIE

e TS s oo

e

i i i

s

s i

i

B2 s

BT

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PrOBIOIT B 2B

i

s i

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R

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R

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s

s s

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al

¢ L 4 Cengage Solution and Answer Guide: Brown et al., Organic Chemistry

@ 2023, 9780357451867;

Chapter 8: Haloalkanes, Halogenation, and Radical Reactions

IN-CHAPTER

PROBLEMS

IUPAC NOMENCLATURE FOR HALOALKANES: PROBLEM 8.1 Write the IUPAC name,

and where

possible, the common

name

of each

compound.

stereochemistry where relevant.

(@

)\/CI 1-Chloro-2-methylpropane (Isobutyl chloride)

Cl

.. (b)

(2)-2-Chloro-2-butene

(c)

mm

Chlorocyclohexane

Cl @

}\// 2-Chloro-1,3-butadiene

FREE-RADICAL HALOGENATION: PROBLEM 8.2 Name and draw structural formulas for all monochlorination products formed by treatment of 2-methylpropane with Cl.. Predict the major product based on the regioselectivity of the reaction of Cl; with alkanes.

CH, I

CH,CHCH,

+ C

heat

———>

or light

|

CHyCCHy

monochloroalkanes

+

|

CH,CHCH,C!

a 2-Chloro-2-methylpropane

1-Chloro-2-methylpropane

HCl

Show

2 Cengage al

L 4

Solution and Answer Guide: Brown et al., Organic Chemistry

© 2023, 9780357451867;

Chapter 8: Haloalkanes, Halogenation, and Radical Reactions

There is 1 tertiary hydrogen atom and 9 primary hydrogen atoms on the molecule. The ratio of reactivity for 3°:1° chlorination is 5:1. Therefore, the predominant product will be the 1-chloro-2-methylpropane, formed in approximately:

Gl @x7)+(1x5) x100= 00 9x1

ENTHALPY

OF REACTIONS

(84% |: PROBLEM

8.3

Using the table of bond dissociation enthalpies in Appendix 3, calculate AH® for bromination of propane to give 1-bromopropane and hydrogen

CH,CH,CHjBr| +

—>

CHyCH,CH, +

bromide.

HBY]

AH® equals the difference between the bond dissociation enthalpies of bonds made vs. bonds broken in the reaction. One C—H

bond [+ 422(101) kJ(kcal)/mol)] and one Br—Br bond

[+ 192(46) kJ(kcal)/mol)] were broken while one C—Br [-301(-72) kJ(kcal)/mol)] bond and one H—Br bond [-368(-88) kJ(kcal)/mol)] were made in this reaction. Thus, for the

complete reaction A& = 422(101) + 192(46) — 301(-72) — 368(-88) = — 55(-13) kJ(kcal)/mol.

ENTHALPY OF REACTIONS |I: PROBLEM 8.4 Write a pair of chain propagation steps for the radical bromination of propane to give 1-bromopropane. Then, calculate AH*® for each propagation step and for the overall reaction.

Following is a pair of chain propagation steps for this reaction. Of these steps, the first involving hydrogen abstraction has the higher activation energy.

CHy-CH;—CH;

+.Br —>

3

CHy-CH;—CH;

+422

CHy-CH;—CHp

H-Br -368

*101) -

+

+ Br,

—>

i CHy—CH,—CH;

(-88)

AHP kJ/mol

(keal/mol) +54

(+13)

+.Br

+192

-301

-109

(+46)

72)

(-26)

Br

CHy;-CH,—CHy + Br;,

— >

|

CHy;-CH;—CH; + H-Br

=55

((13)

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Chapter 8: Haloalkanes, Halogenation, and Radical Reactions

ALLYLIC

BROMINATION:

PROBLEM

8.5

Given the solution to Example 8.5, predict the structure of the product(s) formed when 3-hexene is treated with NBS.

! CHyCHCH—CHCH,CH;

+.Br

————>

(fir)?

S

CH;CHCH—CHCH,CH;

—

CgH;,~CH,

+

H—Br

(kcal/mol)

+376

+192

-263

-368

-63

(+90)

(+46)

(-63)

(-88)

(-15)

2 L 4 Cengage al

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Chapter 8: Haloalkanes, Halogenation, and Radical Reactions

(b) Propose a pair of chain propagation steps and show that they add up to the observed reaction.

Following is a pair of chain propagation steps for this reaction.

H

|

AH®

.

CgHs —CH,

+

« Br —»

kJ/mol

CeHg—CH,

+

+376

(kcal/mol)

=368

(+99)

CeHs—CH, + Bry, —> +192

-

H—Br

4

@

(-88)

T

CgHg—CH,

+ -Br

-263

00O

8

%)

-1

@001

sum of AH® for chain propagation steps:

53

(-15) Following is an alternative pair of chain propagation steps. Because of the considerably higher activation energy of the first of these steps, the rate of chain propagation by this mechanism is so low that it is not competitive with the chain mechanism first proposed.

H

|

GHy—CH, +376

Br

AHP

i

+ -Br — > CHy—CH, + -H -263 (-63)

+90) *‘H+B,

—>

HB

kJ/mol

st +113 “27

+:-.Br

+192

-368

-176

(H46)

(-88)

(42)

sum of AH® for chain propagation steps:

(c) Calculate AH° for each chain propagation step.

See answer to part (b). (d) Which propagation step is rate-determining?

The rate-determining step corresponds to the step with the higher potential energy barrier. Thus, the first step involving hydrogen abstraction is rate-determining because it has higher activation energy.

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Selution and Answer

Guide: Brown et al., Organic Chemistry © 2023, 97B0357451861; Chapter 8: Haloalkanes, Halogenation, and Radical Reactions

Problem 8.17 Write a balanced equation and calculate AH° for the reaction of CHs and I; to give CHal and HI. Explain why this reaction cannot be used as a method of preparation of iodomethane.

The reaction is not a useful preparation method, because the formation of lodomethane (methyl lodide) by radical lodination is endothermic by 556(13) kJ(kcal)/mol. It will not

occur spontaneously.

CH,

+

.

|,

CH,l

+

AH® kJ/mol (keal/mol)

HI

+439

+151

-242

-297

+51

(+105)

(+36)

(-58)

(-71)

(+12)

Problem 8.18 Write balanced equations for the radical iodination and bromination of ethane: CHiCHs

+ |2

to give CHiCHal + Hi, and for CHiCH3 + Br; to give CH;CH.Br + HBr. Using bond dissociation enthalpies (Appendix 3), calculate the AH® for these two reactions. You will find that this reaction will not work with iodine to give iodoethane, but it is predicted to work with bromine

to give bromoethane. Comparing your calculations above, can you pinpoint the

source of this difference? Is this mostly due to a difference in the strength of the bonds in the reactants, or of those in the two products?

The balanced reaction Is (X = halogen atom): HsC—CHj + X=X — CiHg—X + H—X Bonds broken:

formed:

lodination

Bromination

CH3sCH;—H

+422 kJ/mol

CHyCHy—H

+422 kJ/mol

=i

+151 kJ/mol

Br—Br

+182 kJ/mol

CH3CHz—I

~238 kJ/mol

CHsCH;—Br

H=I

=287 kJ/mol

H—Br

=301 kJ/mol =368 kJ/mol

Overall lodination would be endothermic and is thermodynamically disfavored (AH® = (~238-297 + 422 + 151) kJ/mol = +38 kJ/mol) whereas bromination is exothermic and thermodynamically favored (A4 = (-301 - 368 + 422 + 192) kJ/mol = -55 kJ/mol). The main energetic difference is in the strength of the bonds being formed. Bromination has a slight disadvantage due to the need to break the Br—Br bond, which Is stronger than the |—I| bond; but it more than makes up for that by forming the much stronger C—Br and H—Br bonds In the products.

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Chapter 8: Haloalkanes, Halogenation, and Radical Reactions

Problem 8.19 Following are balanced equations for fluorination of propane to produce a mixture of 1-fluoropropane and 2-fluoropropane. CH,;CH,CHy

+ Fy

—

CH3CHo,CHoF

Propane

+

HF

+

HF

1-Fluoropropane

i CH4CHyCHy

+ Fy,

—

CH4CHCH;

Propane

2-Fluoropropane

Assume that each product is formed by a radical chain mechanism. (a) Calculate AH® for each reaction.

Formation of 1-fluoropropane (propyl fluoride) and 2-fluoropropane (isopropyl fluoride)

are both exothermic.

AH® CH;—CH,—CH;

+

F,

+422 (+101)

CH;—CH,—CH;

—»

+159 (+38)

+

F,

CH;—CH;—CH,—F

] —»

+

-472 (-113)

CH;—CH-CH;

kJ/mol

HF

(keal/mol)

-568 (-136)

+

-459 (-110)

HF

+414

+159

-464

-568

-459

(+99)

(+38)

(-111)

(-136)

(-110)

(b) Propose a pair of chain propagation steps for each reaction and calculate A~° for each step. AHO

o CH;-CHZ—CH;;

Ll

—_—

kJ/mol

043—042—0112

+

+422 (+101) CH;-CH,—CH,

+

F; +159 (+38)

—

CH;-CH,—CH,—F -472 (-113)

sum of AH° for chain propagation steps:

+

HF

(kcal/mol)

-568 (-136)

-146 (-35)

‘F =313 (-75)

-459

(-110)

al p

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Chapter 8: Haloalkanes, Halogenation, and Radical Reactions

AH® . C|13—CH2—CH3

+F

—

kJ/mol

CHa—CH—CHa

+

+414

HF

(kealfmo[)

-568

(+99) CH;~CH—CH;

+ F,

——®

i CH;—CH—CH;

+159

(-136)

-37)

+ 'F

-464

(+38)

-154

-305

(-111)

(-73)

sum of AH® for chain propagation steps:

-459 (-110)

(c) Reasoning from Hammond's

postulate, predict the regioselectivity of radical

fluorination

relative to that of radical chlorination

and bromination.

Because the exothermic, intermediate stabilities of

hydrogen abstraction step in each radical fluorination sequence is highly the transition state is reached very early in hydrogen abstraction and the in this step has very little radical character. Therefore, the relative primary versus secondary radicals are of little importance in the

determination of the product. Accordingly predict very low regioselectivity for fluorination of hydrocarbons. Problem

8.20

As you demonstrated in Problem 8.19, fluorination of alkanes is highly exothermic. Per Hammond’s

postulate, assume that the transition state for radical fluorination

is almost

identical to the starting material. Assuming this fact, estimate the fraction of each monofluoro product formed in the fluorination of 2-methylbutane. If it is assumed that the transition state for radical fluorination is almost identical to starting materials, then relative radical stabilities are not important. Thus, all types of carbon atoms (3¢, 2°, and 1°) will react at an approximately equal rate. The fraction of each monofluoro product will then be determined by the number of each kind of hydrogen present as shown.

50%

(6/12 1)

e

16.7% (2/12) |

8.3% (1/12)

25% (3/12)

~¢ ¢C Cengage =»l

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Chapter 8: Haloalkanes, Halogenation, and Radical Reactions

Problem 8.21 Cyclobutane reacts with bromine to give bromocyclobutane, but bicyclobutane reacts with bromine to give 1,3-dibromocyclobutane. Account for the differences between the reactions of these two compounds.

+

Br,

e —

O—Br

Cyclobutane

e

+

HBr

Bromocyclobutane

+

Br,

T

=,

Bicyclobutane

Br~8r

1,3-Dibromocyclobutane

The first reaction follows the normal radical chain mechanism. The propagation steps

consist of hydrogen atom abstraction followed by a reaction with Br; to generate the product.

) b

+

E—

o

7~ N\~ —H+ Br—Br

H

) H

——

u

+

+

HBr

Br

E

The lower reaction can be explained by a mechanism in which the highly strained bridging bond reacts with the Br radical during the first propagation step. Note that the extreme ring strain of the bicyclic molecule provides the driving force for the first propagation step.

Y

~

)

Br—Br

b

/

w — Oe Br

—

bBr

Br

+ -Br

~¢Cengage =»l

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Chapter 8: Haloalkanes, Halogenation, and Radical Reactions

Problem 8.22 The first chain propagation step of all radical halogenation reactions we considered in Section B.5B was an abstraction of hydrogen by the halogen atom to give an alkyl radical and HX, as for example

CH,CH;

+

-Br:i—> CHCH,*

+

HBr:

Suppose, instead, that radical halogenation occurs by an alternative pair of chain propagation steps, beginning with this step.

CH,CH;

+

*Br:—> CH,CHBr:

+

H-

(a) Propose a second chain propagation step. Remember that a characteristic of chain propagation steps is that they add to the observed reaction.

e "—‘-

+

‘H

——

....

H—bB.

+

‘PBr.

(b) Caleculate the heat of reaction, AH®, for each propagation step.

AHP

CHyCH; + -Br

——>

CHyCHBr

+

-H

(keal/mol)

+422

-301

+121

(+101)

72)

(+29)

Br—Br

+ ‘H

—

+192

+

-Br

=368

(H46)

CHyCH;

HBr

+ Br,

——>

(-88) CH,CHBr

=176

+

HBr

42) 55 ¢13)

(c) Compare the energetics and relative rates of the set of chain propagation steps in Section 8.58 with the set proposed here.

Here, the first propagation step has a very high activation energy barrier, so it would not compete with the first propagation step proposed in Section 8.5B.

~¢ ¢ CelCengage al

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© 2023, 9780357451867;

Chapter 8: Haloalkanes, Halogenation, and Radical Reactions

Problem 8.23 The “reactivity-selectivity principle”is a rule of thumb often used in organic chemistry to rationalize differences in selectivity for different reactants in similar reactions. It is generally said that less stable and more reactive species are less selective in their reaction outcomes

compared to more stable and less reactive species. How

Hammond’s postulate to support this principle? Hint: Would

can you use

the transition state of the

more reactive species be reactant-like or product-like, according to Hammond's postulate? What about the one for the less reactive species?

The more reactive reactant more stable one. Assuming reactive reactant (higher in resembles more closely the

is less stable, therefore higher in energy than the less reactive, that the energy of the isomeric products is similar, the more energy) will go through a pathway whose transition state starting material. The less reactive reactant on the other hand

will go through a transition state with a more significant “product-like” character. The “starting material-like" transition state (TS) will be far less affected by structural factors that could determine regio- or stereoselectivity, leading to a relatively unselective reaction. The more advanced TS of the less reactive reactant, on the other hand, will be more affected by structural differences, leading to improved selectivity. ALLYLIC

HALOGENATION

Problem 8.24 Following is a balanced equation for the allylic bromination of propene. CHz=CHCHs (a)

B

Br; ———

CHz=CHCH:Br

+

HBr

Calculate the heat of reaction, AH®, for this conversion,

See the answer to part (b). (b) Propose a pair of chain propagation steps and show that they add up to the observed stoichiometry. AH®

CH,=CH—CH,

+ ‘Br —

CH,=CH—CH, + H—Br

+372

-368

(+89)

(-88)

CH=CH—CH, + Br;

—>

CH,=CH—CHBr

+192

N

...)

4

@

+ Br

247

N

(keal/mol)

) M

55

. -)

sum of AH® for chain propagation steps:

(c) Calculate the AH® for each chain propagation step and show that they add up to the observed AH® for the overall reaction. See the answer to part (b).

«l

¢ [ Cengage Solution and Answer Guide: Brown et al., Organic Chemistry

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Chapter 8: Haloalkanes, Halogenation, and Radical Reactions

Problem 8.25 Using the table of bond dissociation enthalpies (Appendix 3), estimate the BDE of each indicated bond in cyclohexene. (b)

\

H

H

H

(a)

(c)

ST

eadimal (b) —___H H

H

SRR

() 464(11D)k(keal)/mol

Estimate (a) the bond dissociation enthalpy (BDE) of this secondary site to be 414(99) kJ (kcal)/mol, (b) BDE of this allylic site to be 372(89) kJ(kcal)/mol, and (c) BDE of this vinylic site to be 464(111) kJ (kcal)/mol.

Problem 8.26 Propose a series of chain initiation, propagation, and termination steps for this reaction and estimate its heat of reaction.

Q

+ Bry m'~Br

+ HBr

(racemic)

Br *

light (racemic)

initiation:

Bry

kM

2B

~¢ ¢C Cengage «l

Solution and Answer Guide: Brown et al., Organic Chemistry

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Chapter 8: Haloalkanes, Halogenation, and Radical Reactions

Propagation: H .

+

—_—

‘Br

+

AH® kJ/mol

+372 (+89)

H-Br

(keal/meol)

-368 -88)

4 (0]

Br 4

-

O

+

Bpb

—>

+

+192 (H46)

‘Br

-247 (-59)

-55 (-13)

sum of AH® for chain propagation steps:

¢12) Termination:

‘Br + "Br

—

Bn, Br

.

e

O

‘

I a

o

Problem 8.27 The major product formed when methylenecyclohexane is treated with NBS in dichloromethane is 1-(bromomethyl)cyclohexene. Account for the formation of this product.

Recall that NBS can be considered a source of Br radicals and Bra.

cH

NBS _




CH

&

Methylene-

cyclohexane

E

1-

Qzu-l,

H

+ H-Br

%

H

H

The resulting allylic radical can be represented as the hybrid of two contributing structures. The one on the right is the major contributor because it contains the more stable trisubstituted carbon=carbon double bond.

&

:

H

H

The second propagation step completes the reaction.

Q—cu,

+ By

—>

Q—w,lr

H

+ Br

H

Problem 8.28 Draw the structural formula of the products formed

when each alkene

is treated with one

equivalent of NBS in CH.Cl, in the presence of light. NBS

light

/k/\

S

(a)

/

I‘M.

CHIC};

Br

1123118 ight

/Y\

—=

(c)

NBS light

= CH,Cly

Br

e

/\%

o

'

Br

+

3

/\\l/\

(racemic mixture)

4¢Cengage «l

Solution and Answer Guide: Brown et al., Organic Chemistry

@ 2023, 9780357451867;

Chapter 8: Haloalkanes, Halogenation, and Radical Reactions

Problem 8.29 Calculate the A4° for the following reaction step. What can you say regarding the passibility of bromination at a vinylic hydrogen? CH=CH:

+

Bre

E—

CH=CH-

+

HBr

AHP

(lucal/mol) CH,=CH,

+

Br-

—

CH,=CH*

+

+464

H-PBr

-368

(+111)

9%

(-88)

23)

This reaction step has such a high AH°, 96 kJ/mol, that it will preclude bromination at a vinylic hydrogen.

Problem 8.30 The allylic bromination of the (R) or ($) enantiomers of 3-methylcyclohexene with NBS produces 3-bromo-1-methylcyclohexene as a racemic mixture. Draw the key reactive intermediate for the reaction of each enantiomer (include stereochemical information and contributing structures). Explain the regioselectivity in the position of the double bond in

the product and identify the reason for the racemization. NBS

P

light e

0

NBS

Br,

CHyCly

X

Br

a

light —

a

P

CHqCly

a

A racemic mixture

The reactive intermediate is the carbon-centered radical obtained by hydrogen abstraction from the most substituted allylic position in the starting material. That position is the chiral center. The same sp? trigonal planar radical is formed from either one of the two

enantiomers of the starting material. The reaction prefers the formation of the product with the most stable(= most substituted) C=C double bond. The addition of Br generates a new chiral center; since the addition of the Br species can happen from either face of the intermediate radical, both enantiomers of this new chiral center are formed with equal likelihood, giving rise to the product as a racemic mixture. H g

H

H —_—

H \

—

sp’

sp

chiral center

planar less subst. C=C

Bry,,

Br

-

+

3 more subst. C=C

ol

"‘f“"’"“_"]""'-"""“" species

with

more

2 stable

(= cee sbatiiibedd) C=0

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Chapter 8: Haloalkanes, Halogenation, and Radical Reactions

ORGANIC Problem

CHEMISTRY

REACTION

ROADMAP

8.31

Use the reaction recadmap you made for Problems 6.60 and 7.34 and update it to contain

the Key Reactions listed in this chapter. Because of its highly specific nature, do not use reaction 3, autoxidation, of this chapter on your roadmap. -— ~

C-C bond forming reactions are indicared by a box surrounding

™

the reaction number

\

/4 M

geminal dhaloalkanes|

Fd

/

/

////]/[ Ila

™

-

/

//

/"J’

|(

—

48—

o~

/

\

I

s/

,

|

‘

18_~ I". f

—

\

s/

~

o

il

f

/|

([

|

{

~

\elal= VU w L e | I'. -

!

A

-.‘.

/7

I

2

t‘;";".'{,'

4

7

/

/'

= —

TR,

S

\

| [IR

l

Fl‘

Il

N\

L

|

ISR

|

Vi,

7

P

f

wa \"\“

il

e LSSN

frL

ff

\

\

CTR acid

L

|...¢-l

fll

\\

f

Nl .

vy

/

{NR

%

1 "-\

'Y

l.

"\_

\

|

/XN~ T

f

S

|

/’\_\ 4

o

A

|I

I."

)

Xe

Vi

|

o

/

/

\

/

f

A

=

\

&

/

‘-._\.\.‘

'

epoxides

s

9

|/

f

/

|—"|

aminoaicohols

«l

¢ L 4 Cengage Solution and Answer Guide: Brown et al., Organic Chemistry

@ 2023, 9780357451867;

Chapter 8: Haloalkanes, Halogenation, and Radical Reactions

Chapter 8 Roadmap Reaction Legend Reaction 8.1 X2 and light or heat

- X predominantly ends up on most substituted C atom, especially for Br; (Hammond'’s

postulate) - Radjcal chaln mechanism Reaction 8.2 NBS and light or heat - Br adds to any of the possible allylic C atoms to make the most stable alkene (most substituted alkene and trans if relevant)

- Radical chain mechanism involving an allylic radical intermedjate

Reaction 8.4 HBr and peroxides - Non-Markovnikov regiochemistry - Mixed stereochemistry of addition

- Radical chain mechanism Problem 8.32 Write the products of the following sequences of reactions. Refer to your reaction roadmap to see how the combined reactions allow you to “navigate” between the different functional groups. 1) Bry and light

@

2) —=m i Na' T

o

%) Na / NHy

1) Excess Hy / Pt /\

(b)

2) Bry and light

N

,

(c)

=

Br /k/

Br +

A

racemic mixture

1) NBS and light

O

)\/\

2) Excess Hy / Pd

Br

i

¢ Cengage '

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Solution and Answer Guide: Brown et al., Organic Chemistry

@ 2023, 9780357451867;

Chapter 8: Haloalkanes, Halogenation, and Radical Reactions

1) Na / NHgq

2) NBS and light

/ @)

—_———

M

3) Excess Hy / Pd

Br

1) 1 equivalent of NaNH,

2) CHyCH,Br (e)

=

3) Hy / Lindlar catalyst

SYNTHESIS Problem 8.33 Using your reaction roadmap as a guide, show reagents and conditions to bring about these conversions, which may require more than one step.

R

—_—

(a)

Br

)\\/\

+

HBr

——»

M

Br

() CHiCH=CHCH;

CHyCH=CHCH;

—

CH CH=CHCH,Br NBS

————» CH,Cl, light

CH,CH=CHCH,Br

«l Pl

¢ Cengage ‘

Solution and Answer Guide: Brown et al., Organic Chemistry

@ 2023, 9780357451867;

Chapter 8: Haloalkanes, Halogenation, and Radical Reactions

CH,CH=CHCH,

_":“3?“_":“(:“5 Br

)

Br

(racemic)

CH,CH=CHCH,

cHzaz

+ Br,

——————

*

*

Br

Br

CHj ct|+— (I:HCH3 (racemic or meso)

Two new chiral centers are created leading to either a meso compound or a racemic mixture depending on whether the 2-butene starting material is ¢/s or trans.

O~ —

(d)

Br +

(e

HBr

——»

O~ ) = _

CH,Cl, light

)

CH,;

—

2 CH,

(racemic)

——=——

—22 o light

(Review section 7.5.)

3 CH, CHBr

_

1. NaNH,

HC=CH

CH,Br “—"‘2_

___

e

1.NaNH,

CHyBr —‘_"‘—"“""2.

_

=

_

s

(Review section

7.5)

al

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Solution and Answer Guide: Brown et al., Organic Chemistry

@ 2023, 9780357451867;

Chapter 8: Haloalkanes, Halogenation, and Radical Reactions

M

S

/L{\

Br ®

(racemic)

=

W

/l\{\

Peroxides

Br (racemic) AUTOXIDATION

Problem 8.34 Predict the products of the following reactions. Where isomeric products are formed,

label

the major product.

Oy

H

H

initiator

(a)

O, L.

H

H

H

H M

0OH

H

H

H -

-

H

M

OOH e

M

mitiator

OOH ()2

/

-

e

—_— -

—

initiator

()

¢

i

.2 Cengage iw

Solution and Answer Guide: Brown et al., Organic Chemistry @ 2023, 8780357451861; Chapter 8: Haloalkanes, Halogenation, and Radical Reactions

Problem 8.35 Autoxidation, that is, the radical reaction of an alkene with oxygen, is not just a detrimental decomposition pathway for nutritional oils, rubbers, and plastics. Living organisms have been able to harness it for biosynthetic purposes. For instance, the synthesis of prostaglandins, lipids with hormone-like behavior, hinges on two regioselective autoxidation steps, catalyzed by cyclooxygenase enzymes: 19

20

17

s

16

16

.

WYY =

D

n

H

6

==

@

[y

b

=3

@

3

3

3

°

1

OH

H 0y l cyclooxygenase

o0 20

18

6

R

D

R

11

o

[e) T@

5

i@

3

pe

I

"OH

l cyclooxygenase

0-0 .

®

8

16

T

P

g

1

0 9

W8y

T

@y

F

1

YR

Above are the first two steps of the sequence of reactions that ultimately leads to prostaglandins. The first step involves the abstraction of a doubly allylic proton at C13 to form a hydroperoxyl radical. This radical then reacts with the C8=C9 double bond to give the ringclosed structure shown at the bottom. Draw a curved-arrow mechanism for these two steps.

4¢Cengage al

Solution and Answer Guide: Brown et al., Organic Chemistry

@ 2023, 9780357451867;

Chapter 8: Haloalkanes, Halogenation, and Radical Reactions

The mechanisms for the two steps mentioned are shown next. For further details and the complete pathway, see W.A. van der Donk, A.-L. Tsai, R.J. Kulmacz. The Cyclooxygenase Reaction Mechanism, Biochemistry 2002, 4152), 15451-15458, 19

”

15

1

i

1

==

n

9

s

B

3

==

xk‘dnif.

5

3

9

==

1 OH

I

0 19

17

15

14 .

I

11

9

i

[]

5

3

OH

20 W

e

I

+ Helnit [o] 7

u//*‘-)lu

16

Ro

:

a2 ' .

4

;

ofa) | o .6-9, £

19 17 15 ORI

1o

M E s —

Il/

TR

"N

O

7

:0-01 WM i e " 1 10 H w16 w

Problem

o -

4

2

8.36

Give the major product of the following reactions.

HBr W

(a)

—_—

peroxides BTW

AI/\

(b)

HBr

peroxides

j%A

| OH

1 "OH

+¢Cengage al

Solution and Answer Guide: Brown et al., Organic Chemistry

@ 2023, 9780357451867;

Chapter 8: Haloalkanes, Halogenation, and Radical Reactions

A

HBr ——

(c)

peroxides

O/\

.

LOOKING AHEAD Problem

8.37

Haloalkanes often represent important intermediates on the way to complex products (as

shown in Chapter 8 when talking about substitution and elimination reactions). In fact, studying the reactivity of alkenes and the reactions in this chapter equipped you with multiple reactions that add a halogen to a desired position in a molecule with good regioselectivity. The following transformations are examples of some of those reactions in use. Give reagents and conditions you would

use to carry out these three transformations

regioselectively, and briefly explain the source of their regioselectivity.

A

w

M

o

M

Br

(c)

M The indicated transformations can be achieved as follows: Key Intermediate

/l\/J\

i (a)

w

M H

H

Br

H

HBr,

)\)\

—"mxm‘

)Yk

(b)

Br

;

NBS,

CH,Ch /l\/l\

— (e)

-

M

[/LJ\

s

M]

‘ 2 Cengage

al p

Solution and Answer Guide: Brown et al., Organic Chemistry

@ 2023, 9780357451867;

Chapter 8: Haloalkanes, Halogenation, and Radical Reactions

The key to understanding the different regioselectivities lies in the nature of each transformation's reactive intermediate.

Transformation (a) goes through the formation of a carbocation by protonation of the double bond; in this case, the protonation will take place in such a position as to produce the most stable (typically most substituted) carbocation, as shown here. The Br is then added wherever the carbocation was, that is, typically at the more substituted carbon of the original double bond. Transformation (b) goes through a radical. Here the Br radical formed by initiation directly adds to the double bond to form a carbon-centered radical; once again, the more stable (more substituted) radical is formed in preference. This means that, in this case, the Br adds to the less substituted position of the original double bond. Transformation (c) also goes through a radical intermediate, but in this case an allylic radical is formed by NBS. The radical has equivalent resonance structures; the addition of BR- to either yields the same allylic bromide.

*Problem 8.38 A major use of the compound cumene is in the industrial preparation of phenol and

acetone in the two-step synthesis, shown below.

O

e

Initiator

Cumene

O

OOH

gt Oron

——

Chapter 16

Cumene hydroperoxide

—OH

+ O

Phenol

oo

Acetone

Write a mechanism for the first step. We will see how to complete the synthesis in Chapter 16 exercises.

Step 1: The reaction begins when the initiator, a radical species Xs, attacks cumene to make a highly resonance stabilized radical adjacent to the benzene ring, a species given the special name of benzylic radical.

O

O,

CH;

CH4

Cumene

-

CH,

CH,

Resonance stabilized henzylic radical

Step 2: The benzylic radical then reacts with oxygen to give a peroxy radical.

CH, ©_

+ (b)

(CHJ)JN‘FCH-‘IW}(CHS)‘N

O—(J‘Hgfir

(@

I

+

W@—cflzcfl

+ Na*CN

NaBr

= SCH;, .m(.l

()

+

{.I-lflh_l\fl

"sew

H,C

(e) CHCH.CH,CL+CH,C = C LI e —

(f)

ethanol

CHLCl + NHy —— > ethanol

CH;CH,CH,C=CCH;

+

CHzNH; CI

-

+

Naq

+

LiCl

-l

¢ L Cengage

‘

Solution and Answer Guide: Brown et al, Organic Chemistry © 2023, 9780357451861, Chapter

/\

m

O\

NH & CHY(CH,)= CH,0- @ / ethanol

(h) CH.CHCHBr Na'CN > —

[ \

N

8: Nucleophilic

Substitution and

B -Elimination

HCr

*/ N—CHp(CHy) 4 CHy

CHyCH;CHCN

+ Na'Br”

Problem 9,14 Suppose you are told that each reaction in Problem 9.13 is a substitution reaction but are not told the mechanism. Describe how you could conclude from the structure of the

haloalkane, the nucleophile, and the solvent that each reaction is an S.2 reaction. (a) A primary halide, good nucleophile/strong base in ethanol, and a moderately lonizing

solvent all favor Sw2. (b) Trimethylamine Is a moderate nucleophile. A methyl halide in acetone, a weakly lonizing solvent, all work together to favor Sy2.

(c) Cyanide Is a good nucleophile. A primary hallde in acetone, a weakly lonizing solvent, all

work together to favor Su2. (d) The alkyl chloride Is secondary, so either an Sy1 or Sy2 mechanism Is possible. Methanethiolate ion is a good nucleophile, but weak base. It, therefore, reacts by an S,2

pathway.

(e) The lithium salt of the terminal alkyne is a moderate nucleophile, but also a strong base. Because the halide Is primary, an Su2 pathway Is favored.

(f) Ammonia is a weak base and moderate nucleophile, and the halide is primary. Therefore, Sx2 Is favored. () The major factor here favoring an S,2 pathway is that the leaving group is a halide and on a primary carbon. (h) The cyanide anion is a good nucleophile and bromide Is a good leaving group on a primary carbon. Therefore, S,2 is favored.

+¢Cengage =»l

Solution and Answer Guide: Brown et al., Organic Chemistry @& 2023, 9780357451867; Chapter 2: Nucleophilic

Substitution and

8 -Elimination

Problem 9.15 Treatment of 1,3-dichloropropane with potassium cyanide results in the formation of pentanedinitrile. The rate of this reaction is about 1,000 times greater in DMSO than in

ethanol. Account for this difference in rate. HIM“

+ 2KCN

—

N(I/\V/\(‘N

1,3-Dichloropropane

+ 2 KCl

Pentanedinitrile

The hydroxyl H atom of ethanol is a hydrogen bond donor, so in ethanol the CN- is strongly solvated via hydrogen bonding. This strong solvation slows down the CN-

reaction with the haloalkane. DMSO cannot act as a hydrogen bond donor, so the CN- is not strongly solvated thereby allowing a faster reaction with the haloalkane.

Problem 9.16 Treatment of 1-aminoadamantane, CiwHnN, with methyl 2,4-dibromobutanocate in the presence of a nonnucleophilic base, R;N, involves two successive S,2 reactions and gives compound A. Propose a structural formula for compound A.

‘

NH,

(8]

+

(”Monk

RN

—

. ) HagNOs + 2 RyNH*_— Br

Br

I-Aminoadamantane

Methyl

A

2, 4-dibromobutancate

There are two successive S,2 displacement reactions to give the four-membered ring.

»

'"..-Gll,fllg(l!m |

H

Br

—_—

N

>

The stereochemistry of the product will depend on the stereochemistry of the 2,4-dibromobutanoate starting material. Racemic starting material will give a racemic

product.

al

¢ Cengage '

|

Solution and Answer Guide: Brown et al., Organic Chemistry @ 2023, 8780357451861; Chapter 9: Nucleophilic

Substitution and

8 -Elimination

Problem 9.17 Select the member of each pair that shows the greater rate of Sy2 reaction with Kl in acetone.

The relative rates of Sy2 reactions for pairs of molecules in this problem depend on two factors: (1) Bromine is a better leaving group than chlorine, and (2) a primary carbon without Fbranching is less hindered and more reactive toward S,2 substitution than a primary

carbon with one, two, or three branches on the f-carbon atom. The molecule that reacts faster Is circled. (a)

/\\/flf

or

"

'

(b)

N‘“

N“

D v O 07m

or

Br

E>—\ Br

Q/\I

O/\ (‘I

al

or

(c)

Cl or

Cl

al &

‘ ¢ Cengage Solution and Answer Guide: Brown et al., Organic Chemistry @ 2023, 8780357451861; Chapter 9: Nucleophilic

H

CHy or

E{'_g(:

‘B

;

CH, >=: —

L

CH,CH,OCH,

+

CH,CH,0=CH,

1-Chloro-2-methoxyethane (2-chloroethyl methyl ether) reacts the slowest because the carbocation produced during the reaction is somewhat destabilized by the ether oxygen atom that is two carbon atoms away. This is because oxygen is more electronegative than carbon, so there is a partial positive charge on the carbon atoms bonded to the oxygen. Thus, the carbocation produced by the departure of the chlorine atom is adjacent to this partially positive carbon atom; a destabilizing arrangement.

8 & oa a CH;-0—CH,CH,~Cl —

e

& T o8 . CH,~0-CH,CH;

Please note that in the case of 1-chloro-2-methoxyethane, there is no way to produce contributing structures with any positive charge on the oxygen atom like that just shown

for 1-chloro-1-ethoxymethane. 1-Chlorobutane does not have similar destabilization of the carbocation formed during the reaction, but it does not have any stabilization either. Its relative reaction rate is therefore intermediate between the two ether molecules.

al

¢ L 4 Cengage Solution and Answer Guide: Brown et al., Organic Chemistry @ 2023, 8780357451861; Chapter 2: Nucleophilic

Substitution and

8 -Elimination

Problem 9.27 Not all tertiary haloalkanes undergo Su1 reactions readily. For example, the bicyclic compound shown next is very unreactive under Sy1 conditions. What feature of this molecule is responsible for such a lack of reactivity? You will find it helpful to examine a model of this compound.

4g

l-lodobicyclooctane

|

fl

bridgebead

carbon atom

1-Iodobicyclooctane

| In order to form a cation, a great angle strain would have to be produced in the molecule. This is because carbocations prefer to be trigonal planar (sp? hybridized), and

loss of iodine would place a carbocation at the bridgehead position with a preferred

bond angle of 120°. However, the bicyclic structure of the molecule enforces a tetrahedral geometry at the bridgehead position (109.5° bond angles), thus preventing the formation of the carbocation.

é bridgehead carbon atom that is foreed to remain tetrahedral

«l &

‘ ¢ Cengage Solution and Answer Guide: Brown et al., Organic Chemistry @ 2023, 8780357451861; Chapter 9: Nucleophilic

Substitution and

8 -Elimination

Problem 9.28 Show how you might synthesize the following compounds from a haloalkane and a nucleophile. Q/(‘.N

@ Treatment of a halocyclohexane with sodium cyanide. Br

CN

U

+

NaCN

——

O,

+

NaBr

gm“

(b) Treatment of chloromethylcyclohexane with sodium cyanide.

O”

+ Nacn _.,O/

CH,ClI

CH,CN

v cl

op

0\“/

(e

Treatment of a halocyclohexane with the sodium salt of acetic acid.

Br

O

i 0O-C-CH,

fl)

+ CHy CONa® —»

O/

+ NaBr

al

¢ Cengage L 4

Solution and Answer Guide: Brown et al., Organic Chemistry @& 2023, 9780357451867; Chapter 2: Nucleophilic

Substitution and

8 -Elimination

(d) Treatment of a 1-halopentane with sodium hydrosulfide.

CHy(CH;)3CHo,Br + HS

(e)

“Na*

> —

CH4(CH;)3CH;SH

+ NaBr

M

Treatment of a 1-halohexane with the sodium salt of acetylene.

CHy(CH;),CH;Br+

®

/\

HC=C'Na*

——>

CHy(CH;);C=CH

+ Nabr

0/\

Treatment of a haloethane with sodium or potassium ethoxide in ethanol.

CHyCH,ONa*

+ CHCHl

—> CH;CH;

CHyCH;OCH,CH,

+ Nal

® Treatment of the appropriate frans-halocyclopentane enantiomer with sodium

hydrosulfide. 2 ..\\‘l'

+ HS "Na*

—»

+ NaBr

¢ Cengage al

L 4

Solution and Answer Guide: Brown et al., Organic Chemistry @& 2023, 9780357451867; Chapter 2: Nucleophilic

Substitution and

8 -Elimination

Problem 9.29 3-Chloro-1-butene reacts with sodium ethoxide in ethanol to produce The reaction is second order; first order in 3-chloro-1-butene and first ethoxide. In the absence of sodium ethoxide, 3-chloro-1-butene reacts produce both 3-ethoxy-1-butene and 1-ethoxy-2-butene. Explain these

3-ethoxy-1-butene. order in sodium with ethanol to results.

For the second-order reaction with sodium ethoxide, the mechanism is Sy2 so the —OCH2CHs group ends up only where the —Cl leaving group was attached. The stereochemistry of the product depends on the stereochemistry of the starting material.

J? A

CHmCH—C—CHy

+

-

82

"OCH,CH,

— >

[ L] 2CHy

CHpmCH—C—CHy

(fl

H

3-Chloro-1-butene

3-Ethoxy-1-butene

For the second reaction, the absence of a strong nucleophile allows the Sy1 mechanism to operate. The allylic carbocation intermediate can be attacked at either the1or 3 positions, a fact that is readily explained by considering the two predominant contributing structures of this intermediate. Note that both enantiomers of 3-ethoxy-1butene are formed. H I*

CHZ=CH—{i':—-CH3

Ca

S\I

—

fl

H

H

I

I

CH2=CH—§—CH3

2° > 1°,

Swl, E2, E1 (k) Order of reactivity of haloalkanes is methyl > 1° > 2° > 3°. Sn2

product.

al

¢ L 4 Cengage Solution and Answer Guide: Brown et al., Organic Chemistry @ 2023, 8780357451861; Chapter 2: Nucleophilic

Substitution and

8 -Elimination

Problem 9.46 Arrange these haloalkanes in order of the increasing ratio of E2 to S.2 products observed in the reaction of each with sodium ethoxide in ethanol.

O/

al

Me—l

A

AU

W

More E2 will occur with a more hindered site of reaction and/or a more substituted (stable) product alkene. Thus, listed in order from least to most E2 product:

Cl

ci

-

O/

3

NC;

a—f

\rl\[/

Problem 9.47 Draw a structural formula for the major organic product of each reaction and specify the most likely mechanism by which each is formed. The substitution and elimination products for the reactions are given in bold. In each case, the different parameters discussed in the chapter are considered including the

type of haloalkane (primary, secondary, tertiary, etc.) and the relative strength of the nucleophile/base.

@)

Q"""Bl'

+ CH,OH

m

|

o

(5x1)

e W CHs/c=c M

(IH3(|I(IH2(1H3 + NaOH

®

E

a cl

‘|

HCONa! /‘\/ g (c) (R)-2-Chlorobutane

CH;

s

Sl

(E1)

et

b

CH,

O+ O+ Qe N

(Sx1)

(E2)

CH, o]

———» DMSO

Q¥

)]\

R

($x2)

L

«l

‘ ¢ AR Cengage Solution and Answer Guide: Brown et al., Organic Chemistry @ 2023, 9780357451861; Chapter 9: Nucleophilic

(DA

q (:”-‘O_N&l+

e

_’d\

:

(d)

Substitution and

8 -Elimination

(E2)

methanol

“Ql

Cl +

(e)

Nal

acelone

I

(S\2)

R enantiomer

S Enantiomer

Cl |

i

)

CH,CHCH,CH,

()

w

iCH

) ll

+ HCOH

fit OFITNe

A0

Renantiomer

of

|+

(Sx1)

i

CH; ~

Guam-l CH2 CHI

Problem 9.48 Think-Pair-Share Consider the following elimination reaction:

NaOCH34

CH3OH, heat Br

(a) Draw the major expected

H

e CH,

H

CH:

R.S enantiomers

CH,CH,O Na* + CHy=CHCH,Cl ——> CH3CHOCH,CH=CH, . = b i cthanol

W

/H

/C=C\

CH,

product.

,

(Sn2)

(E1)

-l

¢ L Cengage

‘

Solution and Answer

Guide: Brown et al., Organic Chemistry © 2023, 97B0357451861; Chapter 9: Nucleophilic Substitution and 2 -Elimination

(b) Draw the transition state leading to the major product, explicitly showing stereochemistry using a Newman

I

0.

projection.

CHy “H. H

CHy “u

“Bré-

(c) Sketch a rough potential energy diagram for this transformation and indicate where the transition state from part (b) is on it.

Qe h }

i

Heaction { anrdinate

(d) What is the rate law for this reaction?

Rate

= l[)\/k(]

[ NaOCH, }

For Sy2, the rate is second order and dependent on the concentration of both the nucleophile and electrophile.

~¢ ¢ CeCengage «l

Solution and Answer Guide: Brown et al., Organic Chemistry @& 2023, 9780357451867; Chapter 2: Nucleophilic

Substitution and

8 -Elimination

(e) What simple change could be made to help induce the formation of the substitution product?

The substitution product could be encouraged if heat was removed from the reaction. Heat will help to favor the elimination reaction. Since a stable, trisubstituted alkene is formed and the electrophile is a secondary bromide, elimination can still potentially

predominate. Problem 9.49 Exposure of trans-1-bromo-4-tert-butylcyclohexane to a strong base such as sodium

methoxide does not result in elimination products in any appreciable yield. Solvolysis in methanol, however, does provide some of the alkene elimination product along with other byproducts. Explain this observation.

MeOH

NaOMe

LT

m-

T

No elimination product

B1" Other

byproducts

As shown, trans-l-bromo-4-tert-butylcyclohexane exists in chair form where the bromine and fert-butyl substituents are most stable in equatorial form. Since the tert-butyl group is so large, it locks the ring into this position. Therefore, the bromine

substituent is also locked in the equatorial position and unable to become anti-coplanar to any #~hydrogens. Elimination does not readily occur, although substitution under these conditions is possible. Conversely, elimination in methanol occurs through an EL mechanism that involves the creation of a carbocation and no

anti-coplanar arrangement is required. Therefore, El elimination occurs as well as Sy1 substitution. at W

| H

H

)
and light —

Reaction 8.1

O’ A haloolkane

o

2. KOtHuL S

Reaction 9.4

T

O An alkene

3.0, e

INTSTRES B Reaction 6.8

H

HM A dialdehyde

O

~

Br;

T

m.

Br

—_— -

Strong Base

E2

/

v

Ny

Butyronitrile

~ ~~.


NaOH

BrCH,CH=CH, —g,>

?

PCC

HOCH,CH=CH, = —

HCCH=CH,

balogenxtion)

Problem 10.43 (a) How many sterecisomers are possible for 4-methyl-1,2-cyclohexanediol?


N

OTBS

THF

1. HyC—C=C — 2. H.O* +

. N

n

H, OTBS

Hy!

Lindlar

OH

Z

OTBS

catalyst

The diol is made from nucleophilic addition to the epoxide, using propyne. To successfully achieve the addition, the alcohol functional group must be protected so it is not deprotonated. Therefore, the alcohol is first converted to the silyl ether. Subsequent addition of an acetylide ion to the less substituted epoxide carbon and an

aqueous acidic workup provides the alcohol. Reduction using Lindlar catalyst followed by removal of the silyl protecting group provides the desired diol. Problem 11.51 Vinethene, also known as divinyl ether, was sold commercially and historically used as an anesthetic much like chloroform. Using your reaction roadmap as a guide, show how to convert ethene to divinyl ether using ethene as the only source of carbon. Show all required reagents and all molecules synthesized along the way. H

H H

H

? =

fox

ol

H

Cly H,0

IBuOH

(excess)

/_/

OH

H"

—

A~

Cl

The ether is formed from the corresponding halohydrin. Chlorine and water provide 2-chloroethanol which under acid-catalyzed conditions forms the corresponding ether. Subsequent elimination with a hindered base provides divinyl ether.

=l

¢ # Cengage

‘

Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861,

Chapter 11: Ethers, Epoxides, and Sulfides

REACTIONS

IN CONTEXT

*Problem 11.52 During the synthesis of the antiasthmatic drug montelukast (Singulair), a silyl ether protecting group is used to mask the reactivity of an OH group. The silyl group chosen is

the tert-butyldimethylsilyl (TBS) group. Draw the product of the following transformation, assuming the TBS-Cl reagent reacts only once with the starting material. Briefly explain

your answer.

O

(lil Iy

HO,

HyC —C—CHyq HoC

Cl.

N Ns O

CH,

HyC

I

—Si—CHy

o

S

(l'lImidazole (base)

>

DMF/CH,C1,

Less-hindered secondary alcohol

HO

/

CH;

]

H3C‘f.;.‘—CH3 H;C

c

N

OH

CH;

H3C—Si—CH ’ 1 ci

’

\indazole (6aseT DMF/CH,0,

Tertiary alcohol CH

HiC( fc‘sj’ /™ CHy HiC 3 HaC /'™ 0 H3C c

CHy

N

Reaction will occur exclusively at the less-hindered secondary alcohol, rather than the tertiary alcohol. Note that the stereochemistry of the secondary alcohol Is not changed during the reaction.

al

.2 Cengage iw

Solution and Answer Guide: Brown et al., Organic Chemistry ©® 2023, 9780357451861; Chapter 11: Ethers, Epoxides, and Sulfides

Problem 11.53 Think-Pair-Share Tetrahydropyrans are commeon functional groups present in many organic natural products that demonstrate biological activity. Bryostatin 1, for example, shows significant promise in the treatment of Alzheimer’s disease. Ao

HO®

Ay

:

O

O

“Y

O,

8]

=

0

Aoy

en g

o W

_=

“OH l

O

0.

~

O

Bryostatin |

(a) Identify the tetrahydropyran functional groups in bryostatin.

HO °

(4]

:

i(8]

o

A

(4]

o

)(4]

/

o

Three tetrahydropyrans

OH H on

W -

T

,

-

o

|

“oH

WO

0"\

(8] Bryostatin 1

~«< Cengage «l

Solution and Answer Guide: Brown et al., Organic Chemistry & 2023, 978035T451861; Chapter 11: Ethers, Epoxides, and Sulfides

(b) The reaction below shows one method

for forming tetrahydropyrans that contain

additional functionality. Show the mechanism using curved arrow notation to transform the epoxide into the corresponding tetrahydropyran. (w

NaOCH,

O Q/\our

NaBr

+

Cl OTBS Step 1: Addition of a nucleophile. The methoxide anion attacks the least hindered side of the epoxide, opening the epoxide ring to form another alkoxide.

i;o"

OIhs

_

l)‘\/'\/\ a

—

i0:

OTBS

H;C 9 MCI

K? OCH; Step 2: Intramolecular nucleophilic substitution. Intramolecular attack of the alkoxide to

displace the chloride results in formation of the tetrahydropyran.

O

:0 C113»;

ome

-

H_,,(:o\_)\/l\/\fa OTRS

(¢) The epoxide shown in part (b) contains a TBS group. Draw out the complete structure for the epoxide, including the TBS group. Why is the TBS group used in the transformation above?

H,;C

H c\\ JCH H_,‘C—SII—CH:,

1

“SC,

o

.

«..C/H\.CJC\CI

H,

H

The TBS group is used to protect an alcohol functional group. This prevents sodium methoxide from deprotonating an alcohol, so nucleophilic addition to the epoxide can occur.

~¢Cengage al

Solution and Answer Guide: Brown et al., Organic Chemistry ©® 2023, 9780357451861; Chapter 11: Ethers, Epoxides, and Sulfides

Problem 11.54 The Sharpless epoxidation is used when a single enantiomer product is required. Predict the structure of the predominant product of the following transformation.

EU

Eu =

Q

(8]

|Q

4.5L

0.,

(= -Diethyl rarrate

OH

Ti(0-il), ——% tert Byl Byelrope roside

!

Q

[o]

iC

OH

$

Using (=)-diethyl tartrate will lead to the stereoisomer shown.

«l

¢ Cengage Solution

and Answer

Guide:

Brown

et al., Organic Chemistry

& 2023, 97B0357451861;

Chapter 12: Infrared Spectroscopy

Solution and Answer Guide BROWN ET AL., ORGANIC CHEMISTRY © 2023, 9780357451861, CHAPTER 12: INFRARED SPECTROSCOPY

TABLE OF CONTENTS IN=Chapter PrODLEIMS ...

G242

Unit Eonversions: Problem 1Y iunaiianciisiaiiasimimaiiie

Bl

Calculating IR Wavenumbers: Problem 12.2.....immsimssss s

022

Functional Group Frequencies: Problem 12.3 ... Distinguishing Different Functional Groups:

PROBIOM

sssssssssasssssssssnsses 023

Problem 12.4 ...........ccovenneiiensnnnieccsienss. 823

125 cicsiiiniiviisiniiisiniisinsiiviisiiis s b mistsuiimiiiswissnssiisin, B

PRODIBIN TRE . oo inioiiscinsimn i

PrODIBITY T2, T . ProBlemy

BB -

POl

T2

s

g i a

s

i

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¢ [ Cengage Solution and Answer Guide: Brown et al., Organic Chemistry @ 2023, 9780357451867; Chapter 12: Infrared Spectroscopy

IN-CHAPTER

PROBLEMS

UNIT CONVERSIONS: PROBLEM 12.1 Which is higher in energy? (a) Infrared radiation of 1715 cm™

or 2800 cm™'?

The higher the wavenumber, the higher the energy. As a result, 2800 cm™ is higher energy than 1705 cm™. (b) Radio-frequency radiation of 300 MHz or 60 MHz?

Energy is directly proportional to frequency, so 300 MHz is higher energy than 60 Hz. CALCULATING

IR WAVENUMBERS:

PROBLEM 12.2

Without doing the calculation, which member higher frequency?

of each pair do you expect to occur at the

(a) C=0 or C=C stretching

The atomic weight of O is slightly larger than that of C. However, the C=0 bond is much stronger than C=C and, hence, has a substantially larger force constant. Thus, C=0 stretching occurs at a higher frequency. (b) C=0

or C—O

stretching

Double bonds have higher force constants than single bonds, so the C=0 bond will have a stretching frequency that occurs at a higher frequency than C—0. (c) C=C or C=0 stretching Triple bonds have higher force constants than double bonds, so the C=C bond will have

a stretching frequency that occurs at a higher frequency than C=0. (d) C—H or C—Cl stretching

Assuming that C—H and C—Cl have similar force constants, the C—H will have an absorbance at a higher frequency because the atomic weight of H is much smaller than cL

~¢ +C Cengage «l

Solution and Answer Guide: Brown et al., Organic Chemistry @ 2023, 9780357451867; Chapter 12: Infrared Spectroscopy

FUNCTIONAL

GROUP

A compound

FREQUENCIES:

PROBLEM

12.3

shows strong, very broad IR absorption in the region 3300-3600 cm™ and

strong, sharp absorption at 1715 cm™. What functional group accounts for both of these absorptions?

The very broad IR absorption centered at 3300-3600 cm™ corresponds to O—H stretching and the strong absorption at 1715 em™ corresponds to a carbonyl C=0 stretch. The functional group responsible for these absorptions is the —COOH group.

DISTINGUISHING

DIFFERENT

FUNCTIONAL

GROUPS:

PROBLEM

12.4

Propanoic acid and methyl ethanoate are constitutional isomers. Show how to distinguish between them by IR spectroscopy.

(“a CH—C—CH,

CH,=CH—CH,—OH

Propanone

2-Propen-1-0l

(Acetone)

(Allyl alcohol)

The key difference between these constitutional isomers is the OH group of propanoic

acid that is not present in methyl ethanoate. The —OH group will show a strong, broad 0—H stretch in the IR spectrum of propanoic acid between 2500 and 3300 cm™ that will be absent in the spectrum of methyl ethanoate. Also, the strong C=0 stretching absorption will be between 1700 and 1725 cm™ for propanoic acid, but 1735-1800 cm™ for methyl ethanoate. Both spectra will have C—H stretching absorptions near 3000 cm™ and C—O stretching absorptions near 1100 cm™.

al

¢ Cengage #

Solution and Answer Guide: Brown et al., Organic Chemistry @ 2023, 9780357451867; Chapter 12: Infrared Spectroscopy

END-OF-CHAPTER

PROBLEMS

Problem 12.5 Following are the infrared spectra of methylenecyclopentane and 2,3-dimethyl-2-butene. Assign each compound its correct spectrum.

Et;N—CH,;—CH,—C—CH, Diethylamine

3-Buten-2-one (Methyl vinyl ketone)

Account for the addition of nucleophiles to the carbon-carbon double bond of an a,f-unsaturated aldehyde or ketone and the regioselectivity of the addition.

«l

~¢ & Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 16: Aldehydes and Ketones The key to this problem is realizing that the carbon-carbon double bond and adjacent

carbonyl interact with each other through their x bonds. This can be best seen by drawing the following contributing structures. The structure on the right indicates that a

partial positive charge is present on the first carbon atom, making this position slightly electrophilic. Nucleophiles such as diethylamine will therefore react at this position. Reactions of this type are referred to as a Michael reaction and are explained in detail in Section 19.8.

(iii:

| HC=CH—C—CH;

07

N+ H,C=CH—C—CH;

—l:--|IH H

mixture

An acetal

Y (L)

4 An alkane

1.Cly,hy 2. NaOH/H,0 3. BH, 1. H,0, —_— 5. PCC 6. 7.: MM“, HCL, HyO ' s a

)\

1.Cly.v

2.NaOH/H,0

Reaction 8.1

Analhane

A

Reaction 94

Al (5]

5. PCC

—_— Reaction 10.10 H

An aldehyde

6.

"

MgBe

OH

4 H,0, NaOH

Analkene

—_— 7.HCLHO

1. BH,

x

Reaction6.6 M

* i

Reaction 16.1

OH

An alcohol

Apeent o

» -

«l

~¢ & Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 16: Aldehydes and Ketones

Problem 16.78 Using your reaction roadmaps as a guide, show how to convert 1-butanol into racemic 4-octanol. You must use 1-butanol as the source of all carbon atoms in the target molecule. Show all reagents and all molecules synthesized along the way.

P

?

PS

OH

Tn—

H

1-Batanol

OGN

By

H 1 /\,Lo

AN

] e

AN

2.Hal, H0

g

Recognize that the product has eight carbon atoms, so it must be made from a combination of two four-carbon units. The key recognition element of the product is the hydroxyl group at the site of new carbon-carbon bond formation, which indicates a

Grignard reaction. The required Grignard reagent can be made by first converting the 1-butanol into 1-bromobutane followed by treatment with Mg in ether. The required butanal can be made by reacting 1-butanol with PCC.

Problem 16.79 Using your reaction roadmaps as a guide, show how to convert ethylene into 2-methyl-1,3-dioxolane. You must use ethylene as the source of all carbon atoms in the target molecule. Show all reagents and all molecules synthesized along the way.

—

5

-

O.

—

Ethylene

j

[=}

O

5

w

= N, fi/\\

A

¥ FysC

¥

NHy

NJ

Structure A

F Enamine 2

(Intermediate)

(a) Use curved arrow notation to show how Structure A is transformed into Enamine 2. Step 1: Add a proton. A proton is transferred to the alcohol to form an oxonium ion.

s

H—NH, F )

OHO:

i AN S

D

He

NH,

r F

‘f

-

N

°

N N ~SN

YR

Structure A

) 4

0 HO: NH,

He

'

¥

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~¢ 4 Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 16: Aldehydes and Ketones Step 2: Break a bond to give stable molecules or ions. Water departs.

¥ oH0! NH, “f

Nox§-

)

F

LAV

N

0 —

N

!

N

H

.

NH,

\j/\N

¥

v~

& FyC

' FyC

.

3

Step 3: Lose a proton. A proton is lost resulting in the formation of an enamine.

¥ "

N 7'\\)

B¢

o (Nm, 5

(

ZH %

7 | f ¥

& -

N

FiC

¥VN

w Z

2

.

X

¥

INHy (b) Since primary amines typically provide imines, why does the enamine product predominate?

The enamine predominates over the imine because it is lower in energy due to conjugation with the carbonyl group. (c) Since protonation of the alcohol functional group in Structure A is a necessary step proceeding to Enamine2, why does the reaction fail when run at a pH of 1? In highly acidic conditions, the ammonium ion will predominate over ammonia, and will therefore be unable to add nucleophilically to the carbonyl in Dione 1. Slightly acidic conditions provide the best yields.

(d) What reagent could be used to transform Enamine 2 into sitagliptin? Selective reduction of the enamine is possible using NaBH:CN if under acidic conditions. The enamine is converted to the imine and then subsequently reduced.

~¢ Cengage «l

Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 17: Carboxylic Acids

Solution and Answer

Guide

BROWN ET AL., ORGANIC CHEMISTRY © 2023, 9780357451861; CHAPTER 17: CARBOXYLIC ACIDS

TABLE

OF CONTENTS

INSCREPEE PrOBISIMNS oiiisssivriimiismpmssmimisimmsmismerissmismitmsissmpasiimnan B &l Naming Carboxylic Acids: Problem 17.1..

. imiimsmsinssiasorssisssssssssissnsssasssonsoosss 82T,

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Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 17: Carboxylic Acids Problamn

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~¢ Cengage «l

Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 17: Carboxylic Acids SYNREBIR. . cuunsmuimmnsm

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~¢ 4 Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 17: Carboxylic Acids

IN-CHAPTER PROBLEMS NAMING

CARBOXYLIC

ACIDS:

PROBLEM

17.1

Each of these carboxylic acids has a well-recognized common name. A derivative of glyceric acid is an intermediate in glycolysis. Maleic acid is an intermediate in the tricarboxylic acid (TCA) cycle. Mevalonic acid is an intermediate in the biosynthesis of

steroids. Write the IUPAC name for each compound. Make sure you specify the configuration.

HO™ Y @

COOH

OH Glyceric acid

(R)-2,3-Dihydroxy-propanoic acid

[COOH

®

COOH Maleic acid (2)-2-Butenedioic acid

HQ CH,

© HO

/\)’\/COOH

Mevalonic acid (R)-3,5-Dihydroxy-3-methylpentanoic acid

~¢ 4 Cengage «l

Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 17: Carboxylic Acids

PREDICTING ACID STRENGTH: PROBLEM 17.2 Which is the stronger acid in each pair?

CH3COOH

*

CH3SO3H

Acetic acid

@

Methanesulfonic acid

pK, =438

pK,=-18

[o] )‘\COOH

%

2-Oxopropanoic acid (Pyruvic acid)

)

7

COOH

Propanoic acid

pK, =25

pK, =48

In (a), the third oxygen atom on sulfur increases acidity by both inductive and resonance effects compared to the carboxylic acid. In (b), the electron-withdrawing carbonyl group in the 2 position of pyruvic acid increases acidity due to inductive effects.

SALTS OF CARBOXYLIC ACIDS: PROBLEM 17.3 Write equations for the reaction of each acid in Example 17.3 with ammonia and name the carboxylic salt formed.

N"Scoon (a)

+ NHs

>

Butanoic acid

OH

-~

= A

N"co0” NHj Ammonium butanoate

OH

cooH

(b) ($)-Lactic acid

+

NH;

—>

. ~

-

~C00 NH,

h

Ammonium (S)-lactate

LAl

~¢ Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 17: Carboxylic Acids USING THE FISCHER ESTERIFICATION REACTION: PROBLEM 17.4 Complete the equation for each Fischer esterification.

ooe [)

(o}

H*

@

(o) H

0\/\)\

OH

H’

o

=

E)SO

(b)

+ H0

(a eyclic ester)

MAKING ACID CHLORIDES: PROBLEM 17.5 Complete the equation for each reaction.

i

COOH

ccl

+socl, ®

—> GE

OCH,4

OCH3

OH

a

+50C1, —> O/ (®)

+50, + HQl

+50; + HCl

e

«l

~¢ & Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 17: Carboxylic Acids

MCAT PRACTICE QUESTIONS: PERMETHRIN AND BIFENTHRIN @

CH,

Q

— l

H

H O

[

CHy

H

O

FyC

Yon

(8]

H

|

CHy

) (o]

Permethrin

A.

cHu,

—

Bifenthrin

What is the stereochemical descriptor for the alkenes in permethrin and bifenthrin? 1.

Zfor permethrin and £ for bifenthrin.

2.

Efor permethrin and Z for bifenthrin.

3.

Neither alkene needs a stereochemical descriptor.

4.

It is not appropriate to apply a descriptor to permethrin, but it is for bifenthrin. That descriptor is £ Permethrin has two Cl atoms on the same carbon of the alkene making an £ or Z designation inappropriate. The alkene of bifenthrin has the

higher-ranked groups on opposite sides of the alkene leading to the £ designation. 5.

B.

It is not appropriate to apply a descriptor to bifenthrin, but it is for permethrin. That descriptor is Z

What is the stereochemical descriptor for the substitution pattern of the cyclopropane rings in permethrin and bifenthrin?

1.

Permethrin and bifenthrin are trans. In both cases, the higher ranking groups are on opposite sides of the cyclopropane plane.

C.

2.

Permethrin and bifenthrin are cis.

3.

Permethrin

4.

Permethrin is trans, and bifenthrin is ¢i/s.

is ¢is, and bifenthrin is trans.

Creation of the ester linkage in these two compounds is the last reaction in their syntheses. What reaction conditions could conceivably be used to create the esters? 1.

Treatment of the carboxylic acid portion on the left side of the molecules with the benzylic alcohol portion on the right side under acidic conditions.

2.

Treatment of the carboxylic acid portion on the left side of the molecules with the benzylic alcohol portion on the right side under basic conditions in water.

3.

Treatment of the carboxylic acid portion on the left of the molecule with thionyl chloride (SOCL,) followed by addition of the benzylic alcohols corresponding in structure to the right sides of the molecules.

4.

Both 1and 3. Either a Fischer esterification or conversion to an acid chloride could be considered for generating the ester.

5.

All of the above.

LAl

~¢ & Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 17: Carboxylic Acids D.

As discussed above, the natural products pyrethrins | and Il (not shown) are destroyed rapidly in the environment. One of the key changes in the structures of permethrin and bifenthrin relative to the pyrethrins was the substitution of naturally occurring methyl groups on the alkene with electron-withdrawing groups (EWGs) such as chlorine and trifluoromethyl. What reactions of the alkene would the change of methyl groups to EWGs retard? 1.

Oxidation of the double bond.

2.

Electrophilic addition reactions.

3.

Nucleophilic addition reactions.

4. Both 1and 2. Oxidation and electrophilic addition are the most common reactions of

alkenes. 5.

All of the above.

DECARBOXYLATION

REACTIONS:

PROBLEM 17.6

Account for the observation that the following f-ketoacid can be heated for extended periods at temperatures above its melting point without noticeable decarboxylation.

@

oO-@>' : ' .

OH

.

a2

~¢ & Cengage «l

Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 17: Carboxylic Acids The mechanism we have proposed for decarboxylation of a f-ketoacid involves the formation of an enol intermediate and equilibrium of the enol, via keto-enol tautomerism, with the keto form. Given the geometry of a bicycloalkane, it is not

possible to have a carbon-carbon double bond to a bridgehead carbon because of the prohibitively high angle strain. Therefore, because the enol intermediate cannot be formed, this p-ketoacid does not undergo thermal decarboxylation.

Side view

O O%,

Because of the geometry of

the bicyelie ring, it is not possible to have a double Vond o bridgehead «rho-keg

O

O

~

O o)

N

| /

o)

OH

)

N

Thisstomwosldbe

pigee END-OF-CHAPTER

—

S

R

1y

PROBLEMS

An asterisk (*) indicates an applied problem. STRUCTURE

AND

NOMENCLATURE

Problem 17.7 Write the IUPAC name of each compound, showing stereochemistry where relevant.

O

we

H

(a) 3-Cyclopentene carboxylic acid

/\r( J00™ Na* (b) Sodium 2-methylbutanoate

4

4

«l

~¢ Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 17: Carboxylic Acids

)I/w (c)

N

'

COOH

(2£,52)-8-Methyl-2,5,7-nonatrienoic acid

COOH

1-Ethylcyclobutanecarboxylic acid

)\‘A( JO0H (&

Br (R)-3-Bromo-4-methylpentanoic acid

HOOC /\[(\(:oou ® @ 3-Oxopentanedioic acid

Problem 17.8 Draw a structural formula for each compound. (a) (2)-3-Bromo-2-hexenoic acid

Br

A

@

O OH

«l

~¢ 4 Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 17: Carboxylic Acids (b) 2-Hydroxyethanoic acid (Glycolic acid)

(c) 2-Hydroxypropanoic acid (Lactic acid)

i

O

OH (d) (£)-3-Methyl-2-hexenedioic acid

COOH

(e) Dipotassium 2-hydroxybutanedioate (Potassium malate)

(o}

K

)

(¢} (6]

(f)

_IKY

OH

2-Oxobutanedioic acid (Oxalacetic acid)

O

HO

O

OH (¢]

«l

~¢ 4 Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 17: Carboxylic Acids (g Sodium benzoate

(0]

Na™

0-

(h) 2-Chloro-1-cyclohexenecarboxylic acid

(o]

OfKOH

Cl

()

(2R.35)-2,3-Dichlorobutanedioic acid (meso-2,3-Dichlorobutanedioic acid)

(‘" 1 HO

[ [e]

()

© ®

OH

Cl

Acetic acid (Ethanoic acid)

«l

~¢ 4 Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 17: Carboxylic Acids

*Problem 17.9 Megatomoic acid, the sex attractant of the female black carpet beetle, has the following structure.

| o

COOH

(a) What is its IUPAC name? Its IUPAC name is (3£, 5E)-3,5-tetradecadienoic acid. (b) State the number of stereoisomers possible for this compound. Four stereoisomers are possible; each double bond can have either an £ or Z (trans or

cis) configuration. Problem 17.10 Draw a structural formula for each salt. (a) Sodium acetate (Sodium ethanoate)

A

Na™

(b) Lithium pentanoate (Lithium valerate)

O M

(c)

Ammonium

0

Li‘

benzoate

(@] O/u\

0

NH n

-l

~¢ Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 17: Carboxylic Acids

(d) Sodium 2-hydroxybenzoate (Sodium salicylate)

OH

O _Na™

(¢}

(e) Ammonium

cyclopentanecarboxylate

(0] O

NH,

*Problem 17.1 The monopotassium salt of oxalic acid is present in certain leafy vegetables, including rhubarb. Both oxalic acid and its salts are poisonous in high concentrations. Draw the structural formula of monopotassium oxalate.

0

HOC—CO

K*

Monopotassium oxalate

*Problem 17.12 Potassium sorbate is added as a preservative to certain foods to prevent bacteria and molds from causing food spoilage and to extend the foods’ shelf life. The IUPAC name of potassium sorbate is potassium (2£,4£)-2,4-hexadienoate. Draw a structural formula for potassium sorbate.

JrS Potassium sorbate

o]

.

«l

~¢ Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 17: Carboxylic Acids

Problem 17.13 Ringer’s lactate solution, also called “lactated Ringer’s,” is used in emergency medical treatment as an intravenous injection for fluid replacement in cases of burn injuries or extensive blood loss. The water-based solution contains sodium, potassium, and calcium

chlorides, and sodium lactate. Draw a structural formula for both stereoisomers of sodium lactate (sodium 2-hydroxypropanoate). Sodium lactate (sodium 2-hydroxypropanoate) contains one chiral center, at carbon 2 in the chain, so it exists as two stereoisomers that are mirror images of each other (that is, a pair of enantiomers):

o

o

\Hl\o

\/”\ o

OH

AH

(8)-2-Hydroxypropanoate

(®)-2-Hydroxypropanoate

*Problem 17.14 Zinc 10-undecenoate, the zinc salt of 10-undecenoic acid, is used to treat certain fungal infections, particularly 7inea pedis (athlete’s foot). Draw a structural formula for this zinc salt.

i

( CH,==CH(CH;)g co”)Z Zn?+ Zinc 10-undecenoate Problem 17.15 On a cyclohexane ring, an axial carboxyl group has a conformational energy of 5.9 kJ (1.4 kcal)/mol relative to an equatorial carboxyl group. Consider the equilibrium for the alternative chair conformations of trans-1,4-cyclohexanedicarboxylic acid. Draw

the less stable chair conformation on the left of the equilibrium arrows and the more stable chair on the right. Calculate AG® for the equilibrium as written and calculate the ratio of the more stable chair to the less stable chair at 25°C.

HOOC %

et

COOH

Hoocw

COOH

AG"=-11.8 kJ/mol

~¢ 4 Cengage «l

Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 17: Carboxylic Acids As written, the conformation on the right is (2 x 5.9) = 11.8 kJ/mol more stable than the

conformation on the left. At equilibrium the relative amounts of each form are given by the equation:

AG®=-RTInK,, Here K, refers to the ratio of the alternative chair conformations. Rearranging gives: -AG®

K= "Rr

Solving gives:

W)

K= e[ L4

Pluggingin the values for AG®, R, and 298 Kgives the answer: K

[ _~m8

™

imal)

J

= @\(831x10™ ki/mol kX208 K))

_ g476=1.1710"

PHYSICAL PROPERTIES Problem 17.16 Arrange the compounds in each set in the order of increasing boiling point. (a)

CHs(CH,)sCOOH

CH3(CH,)sCHO

CH3(CH,)CH,0H

The following are listed in order of increasing boiling point:

CH3(CHa)sCHO

CH3(CH2)sCH;0H

bp 171°C

CH3(CH2)sCOOH

bp 195°C

bp 223°C

O ®

\)k

OH

P

W

OH

N

O/\

The following are listed in order of increasing boiling point:

[o] /\O

bp 35°C

/\

/\/\OH

bp 117°C

\/l\OH

bp 141°C

«l

~¢ & Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 17: Carboxylic Acids

Problem 17.17 Dimethyl succinate (dimethyl 1,4-butanedioate, an ester) and adipic acid (1,6-hexanedioic acid) are constitutional isomers, so they have the same

molecular formula (CsH10.) and,

therefore, the same molecular weight. However, their boiling points are 197°C and 337°C, respectively. Explain this observation. (o]

/

O.

[0

Y\/L()/ (8]

HO.

MOI! (e}

Dimethyl succinate

Adipic acid

The dicarboxylic acid has two sets of excellent hydrogen bond donors (the OH groups)

and acceptors (the carbonyl C=0 groups); its molecules can form extensive intermolecular interactions with each other in the liquid phase thanks to these complementary groups. On the other hand, the succinate diester contains only hydrogen bond acceptor moieties (the C=0 and OCHs groups), so it cannot establish effective

hydrogen bonds with itself, and its liquid is only held together through much weaker dipolar and Van der Waals interactions. For any liquid to boil and the liquid-to-vapor transition to take place, interactions between its molecules have to be broken. To break the stronger cohesion interactions in the liquid adipic acid, its

molecules must possess a much higher average kinetic energy. Average molecular kinetic energy is reflected macroscopically as temperature, so a higher temperature is

required to break the intermolecular interactions among adipic acid molecules in the liquid and transition them to the vapor phase than in the case of the weaker interactions that must be broken to vaporize the succinate diester.

Problem 17.18 Acetic acid has a boiling point of 118°C, whereas its methyl ester has a boiling point of 57°C. Account for the fact that the boiling point of acetic acid is higher than that of its methyl ester even though acetic acid has a lower molecular weight.

Acetic acid can make strong hydrogen bonds, but the methyl ester lacks a hydrogen bonding hydrogen atom. Thus, acetic acid will have a much higher boiling point compared to the methyl ester. [*]

)I\0H Acetic acid bp 118°C

/

Can form hydrogen bonds with this H atom

o A

0CH3 Methyl acetate bp 57°C

«l

~¢ Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 17: Carboxylic Acids

Problem 17.19 Given here are '"H-NMR and “C-NMR spectral data for nine compounds. Each compound shows strong absorption between 1720 and 1700 cm™' and strong, broad absorption over the region 2500-3300 cm™'. Propose a structural formula for each compound. Refer to

Appendices 4, 5, and 6 for spectral correlation tables.

(a) CsHiO: 'H-NMR

BC-NMR

0.94 (t, 3H)

180.71

1.39 (m, 2H)

33.89

1.62 (m, 2H)

26.76

2.35 (t, 2H)

22.21

12.0 (s, H)

13.69

094

139

162

235

120

CH3CH,CH,CH,COOH (b) CsH0; "H-NMR

BC-NMR

1.08 (s, OH)

179.29

2.23 (s, 2H)

47.82

12.1 (s, H)

30.62

29.57

1.08

1.08

Hs 23

CcHsccl,coofi 1.08

Hs

1

~¢ Cengage «l

Solution and Answer

(€) C:HaOs TH-NMR

0.93

BC-NMR

(t, 3H)

1.80 (m,

170.94

2H)

53.28

3.10 (t, 1H)

21.90

12.7 (s, 2H)

1n.81

12.7

3.10

12.7

HOOCCHCOOH

H,CHs

180

093

(d) CsHeO: H-NMR

BC-NMR

1.29 (s, 6H)

174.01

12.8 (s, 2H)

48.77

22.56

129

HOOCCCOOH

128

H3

129

Ha

28

Guide:

Brown

et al., Organic Chemistry © 2023, 9780357451861; Chapter 17: Carboxylic Acids

Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 17: Carboxylic Acids

(e) CiHs0:

H-NMR

BC-NMR

1.91(d, 3H)

172.26

5.86 (d, TH)

147.53

7.10 (m, 1H)

122.24

12.4 (s, H)

181

CH3CH=CHCOOH (f)

C:H.CLO,

H-NMR

©C-NMR

2.34 (s, 3H)

171.82

1.3 (s, H)

79.36 34.02

Cl 2.34

113

CHj iCOOH |

~¢ Cengage )

Solution and Answer

Guide:

Brown et al, Organic

Chemistry

© 2023,

9780357451861,

Chapter 17: Carboxylic Acids

@ CiHiCLO: "H-NMR

BC-NMR

1.42 (s, 6H)

180.15

6.10 (s, H)

77.78

12.4 (s, H)

51.88 207

“I :

3

cl,CHCcooh' 6..

1

142

Hs (h) CsHsBrO, "H-NMR

BC-NMR

0.97 (t, 3H)

176.36

1.50 (m, 2H)

45.08

2.05 (m, 2H)

36.49

4.25 (t, 1H)

20.48

121 (s, 1H)

13.24

r 097

150

205|425

CHsCH,CH,

121

CHCOOH

LAl

~¢ Cengage Solution and Answer

Guide:

Brown et al., Organic

Chemistry

© 2023,

9780357451861;

Chapter 17; Carboxylic Acids

@) CiHeOs "H-NMR

BC-NMR

2.62 (t, 2H)

177.33

3.38 (s, 3H)

67.55

3.68 (s, 2H)

58.72

N5

(s, H)

338

34.75

368

262

115

CH30CH,CH,COOH PREPARATION

OF CARBOXYLIC

ACIDS

Problem 17.20 Complete each reaction.

K2Cr;07, 2V7, HS0 M9V ®

H,0, acetone

OH

CHO 1. . Ag(NHs)z* 3)2*, H H0

HO

®)

lo

N

OH

Z. R0, ACI

Ho/\r

=

iCOH

*

.

+ Ag

OH

The stereochemistry of the product will be the same as the stereochemistry of the starting material.

A

©

M

1. Cl,, KOH in water/dioxane

e =

2. HCL, H,O

———

M

S

i

+

OH

CHCI

3

~¢ Cengage LAl

Solution and Answer

Guide:

Brown et al., Organic

Chemistry

© 2023,

9780357451861;

Chapter 17: Carboxylic Acids

Br

O\\ [OH 1. Mg, ether 2. CO, e e 3. HCI, H,O

@

¢

OCHs

OCH3

Problem 17.21 Show how to bring about each conversion in good yield.

™ @

1) Mg

O

2) co,

3) HCL. H,0 Q9 B

NaClO 5, NaH ;PO

’lk/\/OH

‘—““

(Pinnick

oxidation)

o HO

[OH to CHO)

k/\/OH

°

__"_’ (see note)

HO

J\/\'r H

(®)

o [OH to CHOJ: a variety of mild and selective methods are available to oxidize alcohols to aldehydes:

= PCC, CH,Cl, (pyridinium chlorochromate) = 1) (COCl),, DMSO 2) EtsN (oxalyl chloride and dimethylsulfoxide, the Swern oxidation) - Dess-Martin periodinane (DMP)

An alternative approach:

5

Ho

~OH -

“MO"

H et

(U 0M

OH

HiG0,

o

H0 acetone

("0 OJ\/YO

OH

HCl H0

s

¢ "NT

OH

«l

~¢ 4 Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 17: Carboxylic Acids

HyCr0,

HO "o

o

——= H,0

(c)

o

HO Mcfl

acetone

1) BH; — 2) H,0,

AN (d)

HyCrOy P.

/Yo H,0

NaOH

acetone

[e]

NG

5

H

HcHps

0

R

©

OH

o

Ox]

(0x]

(sce note) O

[0X] - a variety of methods are available to oxidize aldehydes to carboxylic acids:

- H,CrO4/H,0/acetone (Jones, reagent) - Ag(D), for example, Ag;0, or AgNOs/NHs/H,0 (Tollens reagent) - NaClO,/NaH,PO, (Pinnick oxidation)

Problem 17.22 Show how to prepare cyclohexanecarboxylic acid from each compound. O/

1) Mg e

Br

O/C OOH

2) €O, 3) HCL H,0

@

O

HBr

O,

1) Mg. ether

O/COOH

2) €O,

®)

3) HCL H,0

1) BH 3

©

Br

2) H,0,

NaOH

OH

H,Cr0,

H,0

acetone

COOH

OH

0

-l

~¢ 4 Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 17: Carboxylic Acids (o}

[Ox]

H

COOH

—

(see below)

(d

OH PBr;

Br

1) Mg, ether

O/COOH

3) HCL H,0

(e)

-

OH

)

H,CrO —_— H,0 acetone

COOH

[0X] - a variety of methods are available to oxidize aldehydes to carboxylic acids:

= HaCrO4/H,0/acetone (Jones reagent)

- Ag(l), for example, Ag,0, or AgNOa/NHa/H:0 (Tollens reagent) - NaClO,/NaH,PO, (Pinnick oxidation)

Problem 17.23 Draw the structural formula of a compound with the given molecular formula that, upon oxidation by potassium dichromate in aqueous sulfuric acid, gives the carboxylic acid or dicarboxylic acid shown. (o]

a

oxidation

(@ GHiO —

/\/\)\

/\/\/\m

OH

«l

~¢ Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 17: Carboxylic Acids O (b) CeHys0 =,

OH

[o}

/\/\)I\H

CoHy,00

(o] H()\'(\/\)‘\(

oxidation

)

-

(8]

HOA

OH

Problem 17.24 Show the reagents and experimental conditions necessary to bring about each conversion in good yield.

=™ —

(a)

The most convenient way to add a carbon atom in the form of a carboxyl group is carbonation of an organolithium or organomagnesium compound.

= —_— 2) Mg, ether

—_— 2) HCl, H0

(l‘.HH

(IIHn

(IH:,(II()H —

(‘.Hfi(ll(I()()H

CHy

CHgy

®

T

Use the same set of reactions in this part as you used in part (a), except because the starting material is a tertiary alcohol, HCL is used in place of SOCl,.

CH3

|

CH; (g) CHiOH

PhCHZAOCHg

+ N

+ H.SO. (catalyst)

[o]

Il

PhCH,COH

(o]

H,S04

+ CH30H

Il

———

PhCH,COCH3

+

H0

Problem 17.37 Show how to convert trans-3-phenyl-2-propenoic acid (cinnamic acid) to each compound.

[o] @ CgHs

A N

OH

1) LiAlHg

—_—

72 H,0

CeHs

(0]

CeHs™

/\/\

N

OH

(0]

AA

OH

He —_— ~ pt 2sC

(b)

CeHs /\)I\ OH

2atm

(o]

CeHs (©)

/\)\

"o

1) LIAH4 —_—

—

H >

e

2 atm

CaHs " oH

«l

~¢ Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 17: Carboxylic Acids

Problem 17.38 Show

how to convert 3-oxobutanoic

OH

@

acid (acetoacetic acid) to each compound.

O

Mon (racemic)

[o]

o]

M

OH

OH

Bt

2t

2)H0

(0]

M

OH

(racemic mixture)

®

)Oi/\

SH

(racemic)

[o]

[¢]

M

OH

OH

— 2)H0

)‘\/\ OH (racemic mixture)

(©) /\)(])\ N OH [o]

[]

A,

e % ;

S L 5 peme LNP

OH

O

(o]

«l

~¢ 4 Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 17: Carboxylic Acids

Problem 17.39 Complete these examples of Fischer esterification.

[¢]

@

(o]

o

A, OH

*

(e

O

o

HO

on

(b)

*

0

(0]

X

J\

OH

HO

0

S

o

(excess)

s po”

+ 2H,0

o

0

™>"on

)]\

(© (excess)

0

k

0”0

+2H,0

*Problem 17.40 Benzocaine, a topical anesthetic, is prepared by treatment of 4-aminobenzoic acid with ethanol in the presence of an acid catalyst followed by neutralization. Draw a structural formula for benzocaine.

HZNQCOOH 48

+ CHiCHOH

. Aminobenzolc aci

id

i

1) H2504

—_— —o — deprotonate i e S

HZNOCOCHZCrfi

+ H0

" (a topical anesthetic) Benzocaine

Problem 17.41 Name the carboxylic acid and alcohol from which each ester is derived.

o — (a

M

O

+ ol \ (E)-2-Butenoic acid

B 2-Propanol

~¢ Cengage «l

Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 17: Carboxylic Acids

o

O (b)

(0]

O

M

()/\/

o

—

O O

*

Mofl 3-Oxobutanoic acid

-

©)

Benzoic

[e)

o

~ o 2-Propen-1-ol

HO

acid

O

Cyclohexanol

o

0O

OH

—

+

HO /\/OH

OH O @

o

(0]

Cyclopentane-1,2-dicarboxylic acid

% C

1.2-Ethanediol

() Z

=

O

HO M

©

OH

4-Hydroxybutanoic acid

(o]

O

=

YLOH

¥

H,C-OH

¥ O

(L)

2-Methyl-2-propenoic acid (Methacrylic acid)

Methanol

«l

~¢ 4 Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 17: Carboxylic Acids Problem 17.42 When 4-hydroxybutanoic acid is treated with an acid catalyst, it forms a lactone (a cyclic ester). Draw the structural formula of this lactone and propose a mechanism for

its formation. Step 1: Add a proton.

N .

:fi*—u

.

H—0—CH,CH,CH,—C—0H

H*

.

=

.

H—Q—CHzCHzCHz—C—QH

4-Hydroxybutanoic acid

Step 2: Make a bond between a nucleophile and an electrophile.

.

HO:

:?y“

OH .

Step 3: Add a proton, take a proton away. PR HO,_ :OH

="

sa

O—H

transfer

_—

.. HO

" i O=H a0

0:

Step 4: Break a bond to give stable molecules or ions.

1,

H§)

"

O-H

J

5

5"

B I+

(-H,0)

=

b

g

NG

(resonance stabilized)

«l

~¢ 4 Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 17: Carboxylic Acids Step 5: Take a proton away.

H

07 |

02

(H%) ..

y——

..

[°H

Lactone

Problem 17.43 Fischer esterification cannot be used to prepare tert-butyl esters. Instead, carboxylic acids are treated with 2-methylpropene in the presence of an acid catalyst to generate them.

oo

= H*

.

R

OH

2-Methylpropene

R

L K O

A tert-butyl ester

(Isobutylene) (a) Why does the Fischer esterification fail for the synthesis of tert-butyl esters? The Fischer esterification does not work for at least three reasons. First, the tert-butyl

alcohol is not very nucleophilic because of steric hindrance. Second, the tert-butyl alcohol dehydrates in acid. Third, any tert-butyl ester that forms falls apart to give the carboxylic acid and tert-butyl cation in acid as shown in the two-part mechanism

below. Step 1: Add a proton.

N

:o:/_\ R”

\9_—2':—c143 CH3

+O-H =

W

RrR” \g—é—cng CH3

+A

~¢ Cengage «l

Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 17: Carboxylic Acids Step 2: Break a bond to give stable molecules or ions.

:O-H

P O-H s (E RT @yDO—C—CH; —> I 3

' R"N; +

CH3

CH3

+ é"‘CHg é”; (b) Propose a mechanism for the 2-methylpropene method.

Step 1: Add a proton.

CH3

I

iy}

PN H3C

CH3

il

==

"C—CH; | CH3

CH,

+A

=

Step 2: Make a bond between a nucleophile and an electrophile.

:(l)-H

CH

o

+ é”

AL,

i

3

OH R”

§p:_ ?—CH3

3

CHy

Step 3: Take a proton away.

.

xvi"-

A

e i +

R” (ix—(rm; CH3

0t —

R7

L, & ..

\g—(IZ—CH3 CH3

+

HA

«l

~¢ 4 Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 17: Carboxylic Acids

Problem 17.44 Draw the product formed on thermal decarboxylation of each compound.

|

il

@ CoHsCCH,COOH 29

cHiCCH;

+ €O,

(l OOH ~

.

LI,

H,

) CaH:CH,CHCOOH — (o]

“ K== CCH,

©

CgHsCH,CH,COOH

+

CO,

i O™+ w

5

CH

COOH

H

Problem 17.45 When heated, carboxylic salts in which there is a good leaving group on the carbon beta to the carboxylate group undergo decarboxylation/elimination to give an alkene. Propose a mechanism for this type of decarboxylation/elimination. Compare the mechanism of these

decarboxylations with the mechanism for decarboxylation of f-ketoacids. In what way(s) are the mechanisms similar?

O Br/XLO'Na'

m»

=

X

m“

1) CO,

/Y

MgBr

[e]

—

Mon

2) HCI, H0

1) BH, PG

e

2) H0,

ZN""0H

NaOH

(¢}

H* OH

*

"S0H 7

— cat

o 07NN

«l

~¢ 4 Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 17: Carboxylic Acids

Problem 17.52 Using your reaction roadmaps as a guide, show how to convert 1-butene into 1-chloro-2-methylbutane. You must use 1-butene and carbon dioxide as the source of all carbon atoms in the target molecule. Devise a pathway that includes a carboxylic acid as

an intermediate; show all reagents needed and all molecules synthesized along the way. ?

AN

+

COq

e—

/Y\

1-Butene

Cl

1-Chloro-2-methylbutane

A possible approach is shown below:

1) Mg

HBr

P

/Y

—

NS

g

"

ether

269

3)HCi, H,0

1) LiAIH

cooH

__ether 2) H,0

COOH

—

/\r

socl, OH

—i

pyridine

Cl

~< 4 Cengage «l

Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 17: Carboxylic Acids

Problem 17.53 Using your reaction roadmaps as a guide, show how to convert 4-methyl-1-pentene and carbon dioxide into 5-methylhexanoic acid. You must use 4-methyl-1-pentene and carbon dioxide as the source of all carbon atoms in the target molecule. Show all reagents and all molecules synthesized along the way.

(o]

N

)\/\ 4-Methyl-1-pentene

+COop

? —»

ML S-Methylhexanoic acid

iy

OH

Ly 1.BH3

1.€O,

2. H;0,/NaOH

2. HCl, H0

)\/\/OH

J\/\/MQBI’

QRx

A )\/\/

Mg

B

Recognize the product as a carboxylic acid with one more carbon atom than the 4-methyl-1-pentene starting material, so propose a Grignard reaction with CO, as the last step. The Grignard reagent can be made from 4-methyl-1-pentene using the sequence of hydroboration/oxidation (non-Markovnikov regiochemistry required),

conversion to the primary bromoalkane with PBrs, and treatment with Mg in ether. Alternatively, the primary bromoalkane could have been synthesized from

4-methyl-1-pentene in one step using HBr with peroxides and light (free-radical conditions).

~¢ 4 Cengage «l

Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 17: Carboxylic Acids Problem 17.54 Using your reaction roadmaps as a guide, show how to convert cyclohexane into adipoyl dichloride. Show all reagents and all molecules synthesized along the way.

0 9

;

>

Cl

cl fo)

Adipoyl dichloride

Cyclohexane Br;

socl,

light

0

Br HO

OH 0

RO

HCro4

Strong Base

E2

0

1.03 2. J (CH3),S 3/2:

H (o)

H

Recognize the same number of carbon atoms in both starting material and product, although the ring must be opened to give the open chain adipoyl structure. The only reaction covered so far that can break a carbon-carbon bond is ozonolysis, so assume it

will be used in this synthesis. Also recognize the product as a diacid chloride that must have come from a diacid. The diacid, in turn, can be easily derived from the corresponding dialdehyde, which is the product of the ozonolysis reaction. The required cyclohexene can be synthesized from cyclohexane using the familiar sequence of free-radical halogenation followed by E2 elimination in strong base.

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~¢ 4 Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 17: Carboxylic Acids

Problem 17.55 Using your reaction roadmaps as a guide, show how to convert 5-chloro-2-pentanone and carbon dioxide into racemic tetrahydro-6-methyl-2-pyranone. You must use 5-chloro-2-pentanone and carbon dioxide as the source of all carbon atoms in the racemic target molecule. Show all reagents and all molecules synthesized along the way.

0. (0] UM“.

.

MC!

§-Chloro-2-pentancue

+

mz\,OH

Tetrakydo-6-methyl 2-pyrancae

&

HCl

'.'

MCI

H

HzS04

(catalytic amount)

_"

Mg

el

4

r\o

[e)

.

q

e

2 +CO—

"

MgCl

1. €O,

/

!

OH

i

[o]

Mk

oH

/ J/

o o

/‘\/\)k 3 OH (racemic)

14

1. NaBH, 2.H0

2.Hal, np\g

0 (\O M

[o)

0

Mo“ /HJ;

Recognize that the product has six carbon atoms, so a new carbon-carbon bond must be created between the 5-chloro-2-pentanone and CO,. Also recognize the product as a racemic lactone, that is, a cyclic ester molecule that is synthesized from the

corresponding racemic hydroxyacid. Propose that the acid function is derived from Grignard reaction with CO.. The carbonyl of the 5-chloro-2-pentanone starting material is in the proper position for a conversion to the required hydroxyl group for lactone formation using NaBH, reduction. Therefore, the required Grignard reagent can be

prepared using the reaction sequence of ketone group protection as the cyclic acetal, followed by treatment with Mg in ether. Following reaction with CO,, the cyclic acetal is removed using aqueous acid.

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~¢ 4 Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 17: Carboxylic Acids REACTIONS

IN CONTEXT

Problem 17.56 Diazomethane, CH;N,, is used in the organic chemistry laboratory despite its danger because it produces very high yields and is selective for reaction with carboxylic acids. Write the products of the following reactions. In all cases, convert the carboxylic acid functions into methyl esters. Note how many

other functional groups are compatible with this transformation

N'$" TN

O

|\

o

AL

N

H

o

oy, "of

—

4

excess CHyN,

CILON s

OH

“rny 0 H

o

OCH3

OH (a)

OCH3

O

(o]

O,

0.

WH

oH

:

CH,N,

Th

(b)

>

CILON

HY

HiCO—g7i H

[ 1™on SOloH :H OMe

[e]

(o] OH

OMe

OH

H

(o)

j/

o

20

£

OCH3

CHEH

o

5

OCH3 j/

CHEH I

0 excess CH;N,
=0

-

%:

NH

-

»—9.

NH

(o]

[¢)

N NH

o

The two contributing structures that place the negative charge on the more electronegative oxygen atoms make the greater contribution to the resonance hybrid.

*Problem 18.42 The following compound is one of a group of #-chloramines, many of which have antitumor activity. Describe a synthesis of this compound from anthranilic acid and ethylene oxide.

OO

a

NH, COOH

2-Aminobenzoic acid (Anthranilic acid)

(B‘(Il

A=) -

N O

o A B-chloramine

~¢ 4 Cengage «l

Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 18: Functional Derivatives of Carboxylic Acids Reaction of the nucleophilic amino group with two molecules of ethylene oxide gives the

diol intermediate that is converted to the dichloride by treatment with HCL The dichloride intermediate is actually a nitrogen mustard derivative that will react with the carboxylate function to give the desired lactone, presumably via the three-membered ring intermediate characteristic of nitrogen mustards (Section 9.11).

(OH NH.

j

2

O

[\

Nxo"‘

—>

I(EOM

1. Hel

R 2. Mild base

COH Il [o]

[o] a

Cl

f

Reaction with muting. via three-

Sl

N

co”

3

6O

0

o

Problem 18.43 Show how to synthesize 5-nonanone from 1-bromobutane as the only organic starting

material.

Y T

2 MN 5-Nonanone

T —

\/\)‘\OH

2N Br

Py 1-Bromobutane

~¢ & Cengage «l

Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 18: Functional Derivatives of Carboxylic Acids There are nine carbon atoms in the product and four in 1-bromobutane. Two

1-bromobutane molecules will be used, but that leaves one carbon atom that must be added. This is most easily accomplished through formation of a nitrile or by a Grignard reaction with CO;.

Nae/V

Y C=N

R

Hy0*

\1

(Sx2) NN

Br

[o]

\

\/\/u\

y \/\/MgBr

OH

2. HCI / H,0

1-Bromobutane is also converted to a Gilman reagent via the organolithium species. War

2uy,

cuct

2N AM

\/\/L'

’

+

(Mom2

LiBr

T

The key step in the synthesis involves reaction of pentanoyl chloride, produced from

pentanoic acid by reaction with SOCL,, with the Gilman reagent to give the product 5-nonanone. This is an example of a so-called “convergent synthesis” in which two

pieces, themselves the product of several synthetic steps, are combined in a late step in the reaction sequence. o

\/\)\

[+} OH

-

\/\)I\

Cl

_(Mw

.

\/\)ok/\/

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~¢ Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 18: Functional Derivatives of Carboxylic Acids

*Problem 18.44 Procaine (its hydrochloride is marketed as Novocaine) was one of the first local anesthetics for infiltration and regional anesthesia. See “Chemical Connections: From Cocaine to Procaine and Beyond.” According to the following retrosynthetic scheme, procaine can be synthesized from 4-aminobenzoic acid, ethylene oxide, and diethylamine as sources of carbon atoms. (o]

(0]

GAO HN

®

~~_-NEt

OH _ Lo~ HN

.

Procaine

N

NEy"

l®

4-Aminobenzoic acid

A Ethylene oxide

+

E,NH Diethylamine

Provide reagents and experimental conditions to carry out the synthesis of procaine from these three compounds. A reasonable synthetic scheme is shown below. First, the amine is reacted with

ethylene oxide to create the corresponding f-aminoalcohol:

A

+

H”

Ethylene oxide

. e

HON

g

S

Diethylamine

Next, p-aminobenzoic acid is converted to procaine by Fischer esterification.

Q*

[o]

K

OH

HoN

e N

Mild bass

———————————

He (Fischer esterification)

4-Aminobenzoic acid

K

o

HN

Q)%NNV Procaine

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~¢ & Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 18: Functional Derivatives of Carboxylic Acids

Problem 18.45 The following sequence of steps converts (R)-2-octanol to (S)-2-octanol.

OH \/\/\/\

OH PTsCI

A

CHLC00

pyridine

Na*

B

DMSO

1. LiAIH,

\/\/\)\

2. 1,0

(R)-2-Octanol

(5)-2-Octanol

Propose structural formulas for intermediates A and B, specify the configuration of each, and account for the inversion of configuration in this sequence. Formation of the tosylate ester (A) occurs with retention of configuration; reaction takes

place on oxygen of the alcohol and does not involve the tetrahedral chiral center. Nucleophilic substitution to form (B) occurs by an Sy2 pathway with inversion of

configuration at the chiral center undergoing reaction. Reduction of the ester with LAH occurs at the carbonyl carbon, not at the chiral center. These relationships are shown in

the following structural formulas. [+]

g”:

NN

P TsCl e

-

T NN

(R)-2-Octanol

g H

CH;C00™Nar 12 T wso

(a)

)l\o \/\/\/k

S

AP 2H;0

(8)

§

OH

(8)-2-Octanol

Problem 18.46 Reaction of a primary or secondary amine with diethyl carbonate under controlled conditions gives a carbamic ester. O

F.|()J\()l-‘.l Diethyl carbonate

O

+ NN Butylamine

—

rm)l\fi/\/\

+ EtOH

Ethyl N-butylcarbamate

Propose a mechanism for this reaction.

A reasonable mechanism for this reaction involves initial attack by the nucleophilic nitrogen atom of butylamine on the carbonyl carbon atom of diethyl carbonate to create a tetrahedral carbonyl addition intermediate that leads to the product carbamate after a proton transfer then elimination of ethoxide.

~¢ Cengage «l

Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 18: Functional Derivatives of Carboxylic Acids Step 1: Make a bond between a nucleophile and an electrophile.

(Id:

:

EtOCOEt

Rl + HoNCH,CH,CH,CH

—

¢oe t

EtO |

HaNCH,CH,CH,CH3 Step 2: Add a proton, take a proton away.

R

| Et0(|:OEt

0-H

&' . R

| EtO(IZOEt

HzNéHzCHZCHzCHg

HNCH,CH,CH3CH3

Step 3: Break a bond to give stable molecules or ions.

def—H 3°???

zit—H —

5

EtOCNHCH,CH,CH CH3

+ EtO

HNCH,CH,CH3CH3 Step 4: Take a proton away.

3

0:

0 + H ‘/_\

Etoe.N.HCHzCHZCHZCH3

+ Et0

-

—»

£to

l

NN\

+EtOH

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~¢ Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 18: Functional Derivatives of Carboxylic Acids

*Problem 18.47 Several sulfonylureas, a class of compounds containing RSO,NHCONHR, are useful drugs as orally active replacements for injected insulin in patients with adult-onset diabetes. These drugs decrease blood glucose concentrations by stimulating g cells of the pancreas to release insulin and by increasing the sensitivity of insulin receptors in peripheral tissues to insulin stimulation. Tolbutamide is synthesized by the reaction of the sodium salt of p-toluenesulfonamide and ethyl M-butylcarbamate (see Problem 18.46 for the synthesis of this carbamic ester). Propose a mechanism for this step.

O\s o) Q

PN

5

NH Na'0

O\s P i

+El0)kfi/\/\

Sodium salt of

—_—

Q

A carbamic ester

SN

v NN H H

Tolbutamide

p-roluenesulfonamide

(Orinase)

A reasonable mechanism for this reaction involves initial attack by the nucleophilic nitrogen atom of the sulfonamide sodium salt on the carbonyl carbon atom of the carbamic ester to create a tetrahedral carbonyl addition intermediate that eliminates ethoxide to give the final product. Step 1: Make a bond between a nucleophile and an electrophile.

:?:/_\‘ m,-@-#{unr

io: :il)'ec

+ EtOCNH(CHp)3CHy

10

—=

m,@fidml:fiu(mz),cu,

8)

:0::0:

Step 2: Break a bond to give stable molecules or ions.

CHy

L B

03

YW

P

H(CH;)3CHy

CHy’

‘3 0%

Telbataiaide

(Oramide, Orinase)

+

oo

BO:

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~¢ Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 18: Functional Derivatives of Carboxylic Acids

*Problem 18.48 Following are structural formulas for two more widely used sulfonylurea hypoglycemic agents. Show how each might be synthesized by converting an appropriate amine to a carbamic ester and then treating the carbamate with the sodium salt of a substituted

benzenesulfonamide.

O,

0 O

Q

S

N~ S

H

N H

Tolazamide

@

(Tolinase | Eene)

o o

M+ EtO

N

— EtO

o

oA ©

H

ol

oy—_—

u

oS¢ 0

S

N

H Tols

0

“NiHNa"

[o}

'OEt

‘s’

ide

(Telamide, Tobnasa)

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~¢ Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 18: Functional Derivatives of Carboxylic Acids H,

[¢)

N

S

0 O

ig;

5 N LA~ H

H

Gliclazide (b)

(Diamicron)

[}

H

H

0 Et0”

SNH N

9 “OEt

+

—

HN A=y

CHy

Et0

fog

o

%

S

—_—

N

)'\

\/

\/

Ns

0

0

H o

s,

g;

)l\NrN

H

Gliclazide

(Diamicron |

*Problem 18.49 Amantadine is effective in preventing infections caused by the influenza A virus and in treating established illnesses. It is thought to block a late stage in the assembly of the virus. Amantadine is synthesized by treating 1-bromoadamantane with acetonitrile in sulfuric acid to give N-adamantylacetamide, which is then converted to amantadine.

CH4C=N Br

in H,S0,

——@—‘*

o ll ”

NHCCHg

1-Bromoadamantane (a)

Propose a mechanism

6’

'

NH, Amantadine

for the transformation

in Step 1.

A reasonable mechanism is shown below. The last proton transfer may occur at the same time as formation of the carbonyl.

~¢ Cengage «l

Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 18: Functional Derivatives of Carboxylic Acids Step 1: Break a bond to give stable molecules or ions.

@-fiéf:

—

Q

P

Step 2: Make a bond between a nucleophile and an electrophile.

oN=CCH; 5

—

N=CCH;

Step 3: Make a bond between a nucleophile and an electrophile.

rf%ccu, +

E——

J

N=CCHj @—NH—(Z—-OEL

Phenylisocyanate

Ethanol

A urethane

Step 1: Make a bond between a nucleophile and an electrophile. Isocyanates react like other carbonyl derivatives in that a nucleophile will attack the relatively electrophilic carbon atom. The product of this step is resonance stabilized. H—0—Et

Ot

{a

M

Mo

pe

— Orbg, Co:” — O

- pe o:

Step 2: Add a proton-take a proton away. A proton transfer completes the reaction.

Ored, - O+, = Ovle H

+

)0'—5!

T

_

H

+

[if—fit

b

—

:?:

-

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~¢ 4 Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 18: Functional Derivatives of Carboxylic Acids

*Problem 18.63 Suppose that you start with a diisocyanate and a diol. Show how their reaction can lead to a polymer called a polyurethane. O=C=N.

N=C=0

U

+ ”ONO”

A diisocyanate

—

A polyurethane

Ethylene glycol

Reaction of diisocyanate with a diol such as ethylene glycol creates a urethane. The free

hydroxyl group end can react with another molecule of diisocyanate, the isocyanate end can react with another diol, and so on.

0=C=N, U

N=C=0

PO

A disocyanate

Ethylene glyeol

\l

o=c=N\O,nTo\/\°"

i

O

o n

A polyurethane

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~¢ 4 Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 18: Functional Derivatives of Carboxylic Acids

ORGANIC CHEMISTRY REACTION ROADMAP Problem 18.64 Use the reaction roadmap you made for Problems 15.22, 16.76, and 17.49 and update it to contain the reactions in this chapter. Because of their highly specific nature, do not use the Key Reactions on pages 784-785 and at the bottom of page 811in your roadmap.

o aye st

4

N [enamines] NS

——

(hemacei] -

"5

2, S

o]

15—

X

7

NaN\e. @\

Q

T W

Cy S

%

v

I\

&\

~ =

:I

\

Nz \

\ SoX Y

g

1726

A

\

] \G

8.5

2\

\

P o 5

P

\

3

[meohols] =

= f“

T

2

fd

W

N

&

o] ¢

\

e 183

ey

\%

&

1

\

1810

nitriles

Chapter 18 Roadmap Reaction Legend

Reaction 18.3 H.0 - The hydrolysis reaction occurs spontaneously and does not need acid or base catalysis - The mechanism involves direct attack of H0 at the electrophilic carbonyl carbon atom

Reaction 18.4 H.0

- The hydrolysis reaction occurs spontaneously and does not need acid or base catalysis - The mechanism involves direct attack of H,0 at the electrophilic carbonyl carbon atom

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~¢ Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 18: Functional Derivatives of Carboxylic Acids Reaction 18.5 H,0 with acid - The reaction protonation - The reaction

or in of in

base acid requires only a catalytic amount of acid and involves an initial the carbonyl group followed by water attack base requires one equivalent of HO" and involves an initial attack of

HO" on the carbonyl carbon

Reaction 18.6 H20 with acid or base - The reaction in acid requires one equivalent of acid and involves an initial protonation

of the carbonyl group followed by water attack - The reaction in base requires one equivalent of HO and involves an initial attack of HO" on the carbonyl carbon

Reaction 18.7 H;0 with acid or base - The reaction in acid requires one equivalent of acid and involves an initial protonation of the nitrile carbon atom followed by water attack. - The reaction in base requires one equivalent of HO" and involves an initial attack of

HO" on the nitrile carbon that leads to an amide.

Reaction 18.8 ROH

- The reaction does not need acid or base and involves an initial protonation attack of the carbonyl carbon atom by the alcohol O atom.

Reaction 18.9 ROH

- The reaction does not need acid or base and involves an initial protonation attack of

the carbonyl carbon atom by the alcohol O atom.

Reaction 18.11 2 R:NH, R can be H - The mechanism involves attack of the amine on the carbonyl atom. - The second equivalent of amine is protonated during the reaction.

Reaction 18.12 2 R;:NH, R can be H

- The mechanism involves attack of the amine on the carbonyl carbon atom. - The second equivalent of amine is protonated during the reaction. Reaction 18.13 RaNH, R can be H

- The mechanism involves attack of the amine on the carbonyl carbon atom

- The reaction can require heating.

Reaction 18.14 RCO;~

- The mechanism involves attack of the carboxylate on the carbonyl carbon atom.

~¢ 4 Cengage «l

Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 18: Functional Derivatives of Carboxylic Acids Reaction 18.15 2 RMgX 2. HCL, H,0

- The product tertiary alcohol has two identical R groups. - The mechanism involves formation of a ketone that is then attacked by the second equivalent of Grignard reagent. - Makes new C-C bond

Reaction 18.16 1. R,Culli 2. H,0

- This reaction stops at the ketone.

- Makes new C-C bond

Reaction 18.17a 1. LiAlH, 2. HCL, H,0

- The reaction cleaves the ester and makes two alcohols. - The mechanism involves attack of hydride anion on the carbonyl carbon atom.

Reaction 18.17b 1. DIBALH 2. HCL, H,0

- The reaction cleaves the ester but stops at the aldehyde.

Reaction 18.18 1. LiAlH, 2. H,0 - The reaction does not cleave the amide, it reduces the carbonyl group methylene.

all the way to

- The mechanism involves attack of hydride anion on the carbonyl carbon atom followed by loss of the O atom attached to the Lewis acidic Al atom.

Reaction 18.19 1. LiAlH, 2. H:0 - The reaction does not cleave the nitrile, it reduces it to a primary amine.

- The mechanism involves attack of hydride anion on the nitrile carbon atom.

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~¢ Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 18: Functional Derivatives of Carboxylic Acids

Problem 18.65 Write the roadmaps different and your

products of the following sequences of reactions. Refer to your reaction to see how the combined reactions allow you to “navigate” between the functional groups. Note that you will need both your old Chapters 6-11 roadmap new Chapters 15-18 roadmap for these.

1. BH,

2. H,0, e

e

3. HyCrOy —

SOCl,

@ An alkene » Me,NH N

Analkene

WC, 6 An acid chloride

1.BH 3

2 H20 NaOH Reaction6.6

5 Me:NH

e

.

—

Reaction 18.11

OH

R

A adsadad

3 HGo, i Reaction 105

Y

/\r OH o A carboxylic acid

/\"/N\ 0 An amide

4.50C1, Reaction 17.8

~

rin

Reaction 6.5

[o]

3pCC

)\

e

Reaction 10.10

An alcobel

A ketone

4 HON/ NaON

— >

P58

Reaction 164 A cyanohydrin

0,

5. HCI, H;0, heat

—_—

Reaction 187

HO,

>

H

An o-hydroxy

acid

1.NBS 2. Mg, cther O

s./\ 4. HCL HO

7N,

e

An alkene ¢ sy,

|

_OH

(©) 2 Mg ether 1 NBS

ii)“

Reaction 8.2

o

3

Br.

A

4.HQLH0

M‘\/\.

o

HO.

DS 7l

=———» Reaction 105 ROy

Reaction 15,3 OH Ho

6.50C1, T\/\ A carboxylic acid

3

Reaction 175

a

W 0

’

O

0.

Y\A

Reaction 185

0

An acid chloride An

ester

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~¢ Cengage Solution and Answer Guide: Brown et al. Organic Chemistry © 2023, 9780357451861; Chapter 18: Functional Derivatives of Carboxylic Acids 1. HyCrO, 2.80Cl, 3. Me,CuLi

RS

An alcohol

OH

)

i

7o

SO

5. MeoNH 6. NaBH,CN

'

1. HyCrO,

/\/\OH An alcohol

S e

B

2

2.50Q,

lE /\)\OH anll e seid

—_— Bty

/\)kg a

3. (Me),Culs e

T Reaction 18.16

Sy 5 Me:NH

/\)k

—Pb “AWN

Aketone

/\/k

Reaction 1617

Suyo” g

tmee

NN

An amine

1.0,

2. (CHy)sS N8 HGro, —_—

5

4. EtOH, catalytic HySO

A (@

ilka n atkene

5. Excess~~\~MgBr 6.N HCI,v H,O H ——

\n aldehyde

o O/R;

A carbaylic acid Ho

5.Excen ~_MgBr » _—

6 HOLHO

O)C

Reaction 18.15 An alcohol An ester

o

3 HCO, Reaction 103

1.0, 2 (CHyS Reaction 63

P An alkene

OH

EOH.caulic S0, Reaction 176

—_—

=

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~¢ & Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 18: Functional Derivatives of Carboxylic Acids MULTISTEP

SYNTHESIS

Problem 18.66 Using your reaction roadmap as a guide, show how to convert (£)-3-hexene into propyl propionate. You must use (£)-3-hexene as the source of all carbon atoms in the target molecule. Show all reagents and all molecules synthesized along the way.

? (E)-3-Hexene

0. O Propyl propionate

1.03

2. (CH3);S

w,

W“

AOH

2.H0

/YCI

We"

socl

o

HZC'CN

T

0 E

(o] Recognize that the starting alkene has six carbons in a chain, while the product is an ester with two three-carbon pieces, so a carbon-carbon bond must be broken. The only reaction covered so far that cleaves a carbon-carbon bond is ozonolysis, so propose

that the 3-hexene starting material is converted to propanal using ozonolysis as the first step. Oxidation of propanal with chromic acid to give propanoic acid followed

by reaction

with SOC, gives propionyl chloride. Reduction of propanal with NaBH, gives 1-propanol, which can be treated with propionyl chloride to give the desired propyl propionate product. Note that a Fisher esterification reaction using propanoic acid and 1-propanol

with a catalytic amount of acid would also work to create the ester product (not shown).

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~¢ 4 Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 18: Functional Derivatives of Carboxylic Acids

Problem 18.67 Using your reaction roadmap as a guide, show how to convert 1-bromopentane and sodium cyanide into N-hexylhexanamide. You must use 1-bromopentane and sodium cyanide as the source of all carbon atoms in the target molecule. Show all reagents and all molecules

synthesized along the way. o NNNEr

4

NaON

?

——»

/\/\)L

1-Bromopentane.

N

PRSI

N-Hesylhesanamide

Nacn

4

\/\/\CN

e

HyO'

2. H0

WT“!

Mu

/O:z

3

Recognize that the product amide has two six-carbon chains, while the starting 1-bromopentane contains only five carbons. Propose an initial Sy2 reaction between the starting materials to generate a six-carbon nitrile. The nitrile can be hydrolyzed in acid to give hexanoic acid followed by treatment with SOCL, to give hexanoyl chloride. The nitrile can also be reduced with LiAlH, to give 1-aminohexane. Reaction of

1-aminohexane with hexanoyl chloride gives the desired A-hexylhexanamide product.

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~¢ & Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 18: Functional Derivatives of Carboxylic Acids

Problem 18.68 Using your reaction roadmap as a guide, show how to convert 1-bromopropane and carbon dioxide into 4-propyl-4-heptanol. You must use 1-bromopropane and carbon dioxide as the source of all carbon atoms in the target molecule. Show all reagents and all molecules

synthesized along the way. OH

2 /\/B‘

+ CO,———»

1-Bromopropane Mg ether

4-Propyl-4-heptanol

AN 1.0,

hI

2

N\ MgBr Tammy

2.HQ, H0 [+]

[+]

/\)l\w

/\)km

@‘h\/\/i)k

A; Cl

Recognize that the product has ten carbons, and the two starting materials contain only three and one carbons, respectively. Therefore, propose that at least two carbon-carbon bonds need to be formed. Recognize further that the product contains the key recognition element of a tertiary alcohol with at least two identical carbon chains

attached, so propose that the last step involves a Grignard reagent of three carbons reacting twice with an ester containing four carbons in the parent chain (a butanoic acid derivative). The required propyl Grignard reagent can be produced by treating the

1-bromopropane starting material with Mg in ether. The four-carbon ester can be made by reacting the propyl Grignard reagent with CO, to give butanoic acid, which is

converted to the required ester by treatment with SOCl, then an unhindered alcohol (ROH) such as ethanol. Note that a Fisher esterification reaction using butanoic acid and the alcohol with a catalytic amount of acid would also work to create the ester (not shown).

~ “ONa*

The lactone is derived from the corresponding carboxylic acid and alcohol. Ozonolysis results in formation of an aldehyde. Hydrolysis of the cyano group affords the carboxylic acid which after Grignard addition and subsequent acid-catalyzed cyclization provides the desired lactone.

REACTIONS IN CONTEXT Problem 18.72 Think—-Pair-Share Esters are commonly found in foods and scents, and these components are significant contributors to their taste and smell. Propyl tiglate is a component of apple flavor and

aroma.

O MON

Propyl tiglate (a) Provide an IUPAC name for propyl tiglate.

Propyl (£)-2-methyl-2-butenoate

«l

~¢ 4 Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 18: Functional Derivatives of Carboxylic Acids (b) Which reaction(s) will not produce propyl tiglate? [e]

/fi)koll

+

o i CH3CHyCHoOH

NaOH ———

+

CH4CHoCHf CHyCHyCHoH

—_— — 12594

® (o]

Mon

(@

(O] (0]

N

1. SOCly OH

2. Propanol

(i) O

N

NHg

CHyCHyCHOH CH3CH2CHy

+

—R 204 (cat)

(iv) Reactions (i) and (iv) will not produce propyl tiglate. (c) Show how you could form propyl tiglate from a vinyl halide through use of a Grignard reagent addition to carbon dioxide. Show all reagents needed and all molecules synthesized along the way.

N

/\r

Br

?

—

o

/\Hko/\/ Propyl tiglate

Mg

ether

N

CHyCH,CH,OH H,S0, MEBr |

o

o

i 1]

o

th en H;O +

=

i OH

Conversion of (£)-2-bromo-2-butene to the corresponding Grignard reagent allows for addition to carbon dioxide to form a carboxylic acid after acidic workup. Esterification of the carboxylic acid using one of the methods above (ii) or (iii).

«l

~¢ 4 Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 18: Functional Derivatives of Carboxylic Acids

*Problem 18.73 Minoxidil is a molecule that causes hair growth in some people. It was originally synthesized as a vasodilator for the treatment of hypertension (high blood pressure). Most of the patients taking the drug for hypertension were seen to grow body hair. Due to other side effects, its oral use was stopped, but it became popular as a topical cream to promote hair growth.

NH,

{

N 4\o

H

N

Linezolid

T -

(Zyvox') Based on your knowledge of carboxylic acid derivatives, predict the product of the following transformation used in a synthesis of linezolid (Zyvox). The new functional group created is called a carbamate. Carbamates are often used as protecting groups for amine groups during complex syntheses.

@

=}

/—\ N =] R

Ny

S 0N

TI o

A

o

«l

~¢ & Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 18: Functional Derivatives of Carboxylic Acids

Problem 18.75 Acid anhydrides are often used in place of acid chlorides because a less acidic carboxylic acid, not the much stronger acid HCL, is the byproduct of the reaction. In the following reaction of a carbohydrate derivative, acetic anhydride is used to obtain the product in

99% yield as a single stereoisomer. Note that the stereochemistry of the starting anomeric carbon is not indicated. Draw the product of the following transformation in a chair form and show the single stereoisomer product of this transformation, which is also the most stable possible chair species. OH

anomeric

[nth-

WP

if) —

Y oy~

2\ ke 7 i =

0.

o

o

Loy

&

k

0

—

Hy

The large triethylsilyl ether will dominate the chair conformation of the molecule, remaining equatorial. This will place the methyl and methyl ether groups equatorial and axial, respectively. The anhydride will react at the only OH group on the molecule, which

is on the anomeric carbon atom. As a result, the ester group could in theory be either axial or equatorial. The equatorial ester product is the more stable, so propose it as the single sterecisomer seen in the actual reaction as shown.

Problem 18.76 The benzyl ether group (—OBn) is often used as a protecting group for OH groups during the synthesis of complex molecules. The following structure has a number of benzyl ether

groups used for this purpose. Draw the product of the following transformation. What role does the diisopropylethylamine play in this reaction? N’c‘c

M

omn

i

Mo

omn

C

H

08n

08n

"‘c“!

n -OBn mc/\O

il

°

\—“

TT N,

Rl

HiChC~0

!

w

)

08n

A“:

don

Although complex in structure, these two starting materials are really just an alcohol

and acid chloride, which combine to make an ester product. In this reaction, the diisopropylethylamine acts as a base to neutralize the HCL by-product of the reaction.

~¢ Cengage «l

Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 19: Enolate Anions and Enamines

Solution and Answer Guide BROWN ET AL., ORGANIC CHEMISTRY © 2023, 9780357451861; CHAPTER 19: ENOLATE ANIONS AND ENAMINES

TABLE

OF CONTENTS

INSCREPEEr PrOBLSINS uiimissivriimitismpmsismimismimmsmmismortssmismitmsissmpasiiimam Ol AldolReaction: Problem:- 19,1 iaasisaaiimenamltamnsmnisnaissmansams gt Dehydration of Aldol Products: Problem 19.2.......ccccnmcnnsinsccinisnisnssscnssnssssnseseaseess 07T Crossed Aldol Reactions: Problem 19.3.........mmmsimsmssasiansesssssessssssossssssessessensensse 78 Claisen Condensation: Problem 19.4............cccvueimmienrnnsessnsinssssssessssssssesssssssssssssessesses 918 Crossed Claisen Condensation: Problem 19.5........c.cccccueeiernrnnnsscsissinscnsiessssaessessense 979 Claisen, Hydrolysis, Decarboxylation: Problem 19.6 ........cccccuminvcinnnsenisnnsssecssensnnes 979 Reaction/of ENamines: PrOBLEM 19.7. w.sssissssmssissmsmsonsimsnsssisisesiarssasssssssssnsssn SO0 Using:Enamings’|: Problom 108 ..orsemsmsisnensssessesnsossssmasearsansanssssaspessssmasssssssassssnsoesst ST UsingEnamines; il Problem 188 ucminsaissanimamiaininnammsaimvim B3 Acetoacetic Ester Synthesis: Problem 19.10 ...........ininiicnscicninisnsnescssasinenees. 984 Variants of Acetoacetic Ester Synthesis: Problem 19.11......cccoinsinnisinsscsisasscsiense 985 Maloriic Estar Synthesis: ProblomMiIBi12 awsmumnmmssemssmesssssspmsrsasesssssssasspsorsrasseorsssess 980 Michael/Addition: ProBblerm 19018 . iurimsmmuimmiiksissimmimkis s

OO

Retrosynthetic Analysis |: Problem 19.14 ...........cccciinisnesieinssssnsssssssssssssssssssssnsasees 988 Robinson Annulation: Problem 19.15......imiismisiisssssissisesrsssisssssssssonse 988 Retrosynthetic Analysis 11 Problem 19.16 ......cccessesnsssssssnssssssonsassassnssssssassassasoasasonsessss 989 USIDELDA: PrOBISIY

DT rossccsssnsusssssnmessassssassaassssssamsesssssamssssssasarpasssonsonsasssesssssrassssasenssstssss G

MCAT Practice Questions: Ibuprofen, the Evolution of an Industrial Synthesis............. 991 ENd-of-Chaptar Probleimis wuuaaiminsssmasissainissisiisiisiisimsiisiin D0k THE AlDOL REECLION cooscosssammmmsssssessassnsssmssstonsensstusssaasessatsasesssonsionpssissosmasosssrssssasonsssassssessasntinis TOB PrODIEM 19.78 ...oveivesirscirssisssssssissssssisssssssssssssssssssssssisasssssssssssssssssssssasssssssssssssssssssssssisnsss GG PO

O

Problom

Y I AED0

s

A S

s

R

s st

B s

T

s

TR

s

s

e eSS

885 OO B

PPEBIBIMI BB 2 . cosnviimicasnsiimmsinsvssrisiismsisiissomsiivsiininsmssvinsssasnson BT FEDNOTITTIEERD.

s s o et

A

S

A

e A

PSS

s

rps st

AT

~¢ Cengage «l

Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 19: Enolate Anions and Enamines Problam

1828 ciumscouinisusmmsmmiisammmssssnninsasniasassisiio 100D

PO

IR 2T srsnserversssmrosessiomspsmesompssosasismmmsimmsinmsnsssssapmasnapiossasrmns 1000

FROBIOI IO 2B o issooismmsiomissasi s Problom 19 2T

. s

e

S e

s

i

st

N S

SR

s

s rsiasamssen NOOT szt

JOO2.

PPOBIBIT BEDD ccisuiscsvisssicssvisvmsisisissisaisssson i i issosssstsssiatiisamsisssivossivicisissaisisiiniiss 10O, ThE Claisan CONAONBRALION....osimssssssossssnsmsssessssasnosssssanssonissssorssssinsssasseosssarsssssnsassessesntiss 1007 PrOBIEM 19.29...uuueeseersesieirssinssssessssssesssssssssssssssssssssssssssssssssssssssssssssssssssssessssssssassssssssssns 1007 Problem 18,30

s

mrmmmihaaiasmassra008

PrableM AR I s sisimivsiasimsimisiassmismsaianmsimissisisissisis S PUCGBIOITE 19,321 ossrssssimmmpormesswmssssssssssssmusssssiommssismmmssommsssssssssisorssessoonsisorsn 1010 PRODI

IR, oo reremmcssmmsrmmsemenessmenmnmpssprenormpoma e

ProbIlem 19.348 it e

marsnnismiane

MY JOT

Problem 183 uuisisusamiaianmsama s st an s 1012 PYODIBITT 10,38 cinssswsseiminmimisissssmimiissmssmisnmsiiimossisisisssasiimosisisioson JOIB

Problem 19.37 TRINK=Pair=SRare ............couwewevessssesssssssessssssssssssssssssssssssasssssessasssssssssess. 1014 Problemy19:38..uuausiismsisinsimssmissmmises oy sissorsssssaassossisisosn 1010 Problem 19,39

missiiiasiisimisiwmwnisaiisisiv

ENRIMIDEHE s PIOBIGIT IO

DR

s csmuscnsmsasammmmmpapmmesmmrssusssmssosmmmmsspsssopms 1020

FroDlems TONY iR e A POO

1018

A

i

e

e

e

et

SR

A

s

10D sas s ipeos t NORT,

Problaiii T4 usmssisisnissisisinsnsisamminiisiisisisissoiii 22 PUCBIOITE 108, cssmorssrmmersommssmmmsssssmssiniossnmmssssponsississammionmsnesinivs 1000 PrOBIEM 19.45....oeeversesieirsssssssessssssssssssssssssssssssssssssssssssssssssssssssssssssassssesssssssssssssassssns IO24 PO DOy

T

0

i

e

s

st

si s

dsiinteoss O

Acetoacetic Ester and Malonic Ester Syntheses ......cc.cccincisssssssscsssscssssessens. 1026 PLOBIBITE TD. BT i mrsmsssessimsmsssessmmmommesssssinsonsssssomsasssassisssmsssipsssosiasaismossosisssssssmssiossnsssssaisessss TR0, PrOBIEM

19.48.....coeeererrrescsisississssisssssssssssssssssssssssssssssssssssssssssssssssssassssnssssssssssssssssssisnss 1028

PO

Y

O

s

A

S

e

ESRGssnTsahsssmnnw

3028

FPEBIBINY TBB .o sovsimsicamnssimmssissvssirisiisssnsorisiossomsisimsvasinimisssessvissesiasisinss 1OH BARTIENAE TR REER

IR . o oxcsvincimtp st

AR

iR

et b

eGSR

A

st L RS

~¢ Cengage «l

Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 19: Enolate Anions and Enamines Problam b

1953 s

s

inssasssisiis 1030

i o T AP

Directed Aol

—

. i

Rnd ALRVIBRION: . wssmcimssimmnrimmmnsimmsasimismtissmnimsimasismminssiini s 1039

Problom 19:55

.

s

e

e i

S

SR

szt

1O,

PPOBIBITT TEEE cvisniscsvisssisisvisivisivsississisisomRaiiiissossssisssissiisiisiisasiiisivisisissisisiiissiis 1OBD, RELOGRYIEIOEIE ANAIAIE wocmrmmrmsmaissomssmimasssaniupmssssisssmonsesmis TS PrOBIEM 19.57 ..uuoeeereeriecsieirssssssssessssssesssssssssssssssssssssssssssssssssssssssssssssssssssssessessessasssssasssnss. 14T Y818

s R

R

B

T

s

1045

Problem 19.58 ThinK=Pair—SRare.............issssissssssssissssssssissssssssssisassassssssssasss 1045 PUGBIOITE TR B D e osrsmmsimmmormesswmmssssssssssmusssnssismmssismmmssmmmssssssssssspeoonsso JOD ERODIAN 10000

s rsomsescermssrmmmammensessmesrmmnsstmesorampenemsmpeamonsmpesnrne {0,

ProbIem 19.61 it i aimmasaismian

1050

Problem 18,62 aissussmiaianssamaiaansasissasiasarsataniaaasiaiasnna 100 PYODIBITT 1088 issnssswmsesminmismisissssmimrssnsmsiaimsmosisiisssissssisiispsisison JO@

RYOBIENTIROL cmsrermpemmrmmessesemsmstssmeomssmmmemmmamanasssnmmmsninsey 00D, ProBlemmy19:65. .uuaniismisiiniussiissisimipes oy isussorsssssssssossnssaisisss 1004 Problem 19,66 .xiuvisicsinvassionss i missiusiiissiisimisimmwnissiiim iy 1056 Problom

16T ivinsusinisiiniimisiisisissniisinaimmiinimissiisiniaini 10D1

PO

1088, cosmuscnsomasammmmmpapmmesmmmrssssnssosmmmsspsssons 0D

e PO

L O

e

B

PrODIB

0

o0 PR

i

e

S

S

e

R

oo

ET

S

S

TINGon (Y OAOORTEL /.. 1 | s

sas s ipes s JOBE

IS D .csisissamesisisivsisisssinisnsissiisiissiasmsasinsissamiminiisimisisisissoii DD,

PUCBIOITE T T2 csssmorssrmsmersmmssmmossssssssmsiirsiomsssnmmssssponsississammmsosaisisssosins BOLL PrOBIEM 19.73..uuoevrrsersssieissssisssssesssssesssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssassssns 1072 OD Oy

P

i

e

e

s

ss s

dsiingiosy JON

ProBIOIT T8 vusasssiisumssiassas s s asansmisams i PLOBIBITE TDiTO

e sirsmssscssimsmssressmmmommesssssimsonsssssoasassssissssssvssipsssrosisseismossosissssssonsssmpsnssssonssessss JOTD:

Organic Chemistry Reaction PO

Oy

1004

B

T

ROAAMAP .......cccueumieisssssessnssssssssssssssssssassssssessesassessenssass 1017

A S

R

R

B

T

I

S

IR

e

A

MOLBES D SYITREBIB .csnwsmssanansmmsmasniaisimsmssabsasssiasssassin (0T T

T

TE

vsers o st

A

s

A

e A

SRS

s

s

v st

L TR

~¢ Cengage «l

Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 19: Enolate Anions and Enamines Problai

188 qiscswuiiusmmsmmiasanmmmssesinninsaaniasassiisoO09

PO

IR ER ssrsnserveprsssmrosessiomspsmessomssorsssisssmmsimmsinmsnsssssprmesnpiosioassronss (OO0,

PROBIOI

18T, issiosnmssiomisasisass i s i st miee 10T

Lroblem 19.:88 ... v

s

s

N

s

s

RBACIONSINCONEBRE oottt

1092 093

FrOBIONM 180 issrmsmmrmsmsssimpsssssissmsemssissmpppssssssmsessvimmsspasspmssssvisssemassssasnssnisnrsss 0SS PrOBIEM 19.86.......ceoevorseerieirssinssssesssssessessssssssssssssssssssssssssssssssssssssssssssssasesssssessssssssassanes 1094 Problem

1987 oisismrsininiisn

i

PrablamAQ8E s

1095, i

10

PUGBIOITE TG0 ossrssssimmmormesswmsssssssssssmusssmssiommssismmmssommssssssssssorssspesoonsaisor IO,

«l

~¢ 4 Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 19: Enolate Anions and Enamines

IN-CHAPTER PROBLEMS ALDOL

REACTION:

PROBLEM

19.1

Draw the product of the base-catalyzed aldol reaction of each compound.

(a) Phenylacetaldehyde

i

o

2 CgHsCH,CH—=

*

q

CgHsCH,CH S (racemic)

+

ON7a

H

7N

H

(racemic)

The positively charged iminium ion is attacked by water acting as a nucleophile. A proton is transferred to the nitrogen, followed by departure of the amine, and formation

of the carbon-oxygen double bond. The reaction is completed with a proton-transfer step to produce the product ketone and protonated morpholine. Under the acidic conditions in this reaction, an enamine is not formed, so hydrolysis predominates. Step 1: Make a bond between a nucleophile and an electrophile.

E

o, — L j

=

E' ‘j

@/cm

Y

H"

H +

6/@43

,\H

~¢ Cengage LAl

Solution and Answer

Guide:

Brown et al., Organic Chapter

Chemistry

© 2023, 9780357451861,

19: Enolate Anions

and

Enamines

Step 2: Add a proton, take a proton away.

E j " N.

JR—

‘.O:’H

(ti/CH;

transfer

ETD N

p—p

o

QM

CHs

Step 3: Break a bond to give stable molecules or ions.

8

H .

N

CH

[d (by m3

]o

3 *

..

N | H

Step 4: Take a proton away.

MW]

Yo

CH3

N i

0t

—J

) CH3

0

- ()

N3

H W

«l

~¢ Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 19: Enolate Anions and Enamines

UsING ENAMINES |I: PROBLEM 19.9 Show how to use alkylation or acylation of an enamine to convert acetophenone to the following compounds.

O (a

O

ph)i\)‘\

0 ”'J\/\II/ (b)

O

O

Ph)‘\/Y (©

OFEL

O

i

A

[*]

N

N

i —a

0.

R Q{z\:ipMI o

dhgd

o 1

)

q/l\

[*]

2) Ha0/HC!

OEt

o o WM

Y

na

.

2) H0/HC!

(a)

Ph)k/\l/ (b)

o)

Ph

()

o

~¢ (4 Cengage )

Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 19: Enolate Anions and Enamines

ACETOACETIC ESTER SYNTHESIS: PROBLEM 19.10 Show how the acetoacetic ester synthesis can be used to prepare these compounds. (o]

m

(a)

fl

fl

1) EtO Na*

CHiCCH,COCoHs

Lo

o

el

0o 0 sl CH3CCH COH é“ c@ 2 &il

2

ot P (

1001

«l

~¢ 4 Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 19: Enolate Anions and Enamines

Problem 18.27 Pulegone, CyH:s0, a compound from oil of pennyroyal, has a pleasant odor midway between peppermint and camphor. Treatment of pulegone with steam produces acetone and 3-methylcyclohexanone.

+H,0 |

——

0

steam

(8]

)|\ ()

(8}

Pulegone

3-Methylcyclohexanone

Acetone

(a) Natural pulegone has the configuration shown. Assign an R or S configuration to its chiral center.

i

‘0

(b) Propose a mechanism for the steam hydrolysis of pulegone to the compounds shown.

An aldol reaction is reversible, and this steam hydrolysis of pulegone is formally the reverse of an aldol reaction. Thus, a reasonable mechanism is exactly the

reverse of what would be written for an aldol reaction between acetone and 3-methylcyclohexanone.

Step 1: Hydration. This actually occurs over several steps.

|

.o

0

+

Hzo

hydration N

O 0

.o

0

1002

«l

~¢ 4 Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 19: Enolate Anions and Enamines Step 2: Break a bond to give stable molecules or ions.

+

Holg —

+

—~——

02

WP .

e

H

A

0J

Step 3: Add a proton.

HeH —_— ..-\sj./H

(a . 3

-

A

g

’ A (c) In what way does this steam hydrolysis affect the configuration of the chiral center in pulegone? Assign an R or S configuration to the 3-methylcyclohexanone formed in this reaction. This reaction mechanism does not involve the chiral center; it remains in the R configuration in the 3-methylcyclohexanone product.

f

0

1003

~¢ 4 Cengage «l

Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 19: Enolate Anions and Enamines Problem 19.28 Propose a mechanism for this acid-catalyzed aldol reaction and the dehydration of the resulting aldol product.

O

(0] ArSO,I e

o

+ H,O

©O

o (racemic)

A reasonable mechanism for this acid-catalyzed aldol reaction involves attack of the

enol on the protonated carbonyl group, followed by acid-catalyzed dehydration. Note how the first two steps could occur in either order. Step 1: Keto-enol tautomerism.

:0:

0:

ArSO;H

Z

_— s

0

9-

o

:0- -\H

.0.

Step 2: Add a proton.

i

HeH

:0: /

Qs K

e

-_———

20SH ™~ .o

:(I?:

H-cp_—fi—Ar 02

o 0—H o

0

>4

-os

{6 4

M

W0—S—Ar |

02

~¢ Cengage «l

Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 19: Enolate Anions and Enamines Step 3: Make a new bond between a nucleophile (Tt bond) and an electrophile.

0

:0:

W

, N

.

%*_H C.O.\H

O0—H

Hg ¢

oe

Step 4: Take a proton away.

0:

0. _— X

HO:

.

O’lH .

s

(0]

HO: 0.

.

0:

Tl

S ‘0t

H—0—S—Ar 01

Step 5: Keto-enol tautomerism.

HoH

:0: — —~——

HO:

0 £

HO:

0—H &

~¢ Cengage «l

Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 19: Enolate Anions and Enamines Step 6: Add a proton.

0!

¢ P

-

o—H

HO:

.

LS

o=k

H—,,? :

H el

0

0:

=

0,

H—.O.—S —Ar 1 0!

&

fi

Ar

0: 0:

Step 7: Break a bond to give stable molecules or ions.

HoH

02

0.

{ “7°

”

—

N

0—H o

.

H—ila' : H

0—H o

|

H

Step 8: Take a proton away.

0:

03 —

X

.o

QQ’-\—H

9

(x) .

S0 :0—S—Ar

0

L

P

:0:

| pd

02

Note that Steps 7 and 8 may take place more or less simultaneously. One chiral center is created, so a racemic mixture is produced in the overall reaction.

«l

~¢ 4 Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 19: Enolate Anions and Enamines

THE CLAISEN CONDENSATION Problem 19.29 Show the product of Claisen condensation of each ester. If no reaction is expected to occur, denote

with “No reaction.”

(a) Methyl 3-methylbutanoate in the presence of sodium methoxide

_ o

1. CH;0

)\/U\

OCH;

o

O

Na* OCHy

2. HCI, H,0

Methyl 3-methyl-

butanoate One chiral center is created, so a racemic mixture is produced.

(b) Ethyl benzoate in the presence of sodium ethoxide 0

1. EtO

OEt

e

Na

+

Noreaction

> — 2. HCI, H,0

Ethyl benzoate Ethyl benzoate does not contain any alpha hydrogens, so Claisen condensation cannot

occur. (c)

Methyl (R)-2-methyl-3-phenylpropanoate in the presence of sodium methoxide

o

0 : 2

¥

1. CH;0 Na* OCH;

=

2. HCL, H,0

O .

o

;

OCH

g

B

g

O

Methyl (R)-2-methyl-

3-phenylpropanoate One chiral center is created, so a mixture of diastereomers is produced.

1007

«l

~¢ Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 19: Enolate Anions and Enamines

Problem 19.30 When a 1:1 mixture of ethyl 2-methylpropanoate and ethyl 2-phenylethanoate is treated with sodium ethoxide, four Claisen condensation

products

are possible. Draw a structural

formula for each product.

For the following scheme, in the cases where one chiral center is created, a racemic mixture is produced.

0

——

EtO Na

»

Et

"

O 'Na' 'OEt

Ethyl 2-methyl‘opanoate

P

1. Ethyl 2-methyl-

L. Ethyl 2-phenylethanoate

propanoate 2. HCI, H0

0o

Y o

2. HCI, H0

o

o

|| e

o

-+

EtO Na' OCH,CHy——

O Na

H,CH.

Ethyl 2-phenylethanoate

OCH;CH; 1. Ethyl 2-methyl-

1. Ethyl 2-phenylethanoate

propanoate

2. HCI, H;0

2. HCL H0

o

o

culculo)gH/

0o

CHyCH;07

o

Y

1008

«l

~¢ 4 Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 19: Enolate Anions and Enamines

Problem 19.31 Draw structural formulas for the f-ketoesters formed by Claisen condensation of ethyl 5-methylhexanoate with each ester.

O

/‘\ @7 o

OEt o

fin

(0]

ng

®

OFEL

N 0o

o0 'OEt

N\

O OFEt

EtO (©

O o ElO

O

OEL (6]

In all three of the above reactions, one chiral center is created so racemic mixtures are

produced.

«l

~¢ 4 Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 19: Enolate Anions and Enamines

Problem 19.32 Draw a structural formula for the product of saponification, acidification, and decarboxylation of each f-ketoester formed in Problem 19.31. Following are the structures for the products of saponification and decarboxylation of

each p-ketoester from the previous problem. (o]

M

@7 Og‘:\/\/k

®

N HO. 0

(c)

1010

«l

~¢ & Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 19: Enolate Anions and Enamines

Problem 19.33 The Claisen condensation can be used as one step in the synthesis of ketones, as illustrated by this reaction sequence. (o]

M

1. EtO Na*

OEt

NaOH, H,0

5 ha o

HOLHO

S

B

e

-

Cothis®

Propose structural formulas for compounds A and B and the ketone formed in this

sequence. Compound (A) is a f-ketoester, compound (B) is the sodium salt of a f-ketoacid, and the final ketone is 5-nonanone. Compounds A and B are racemic mixtures, while the

final product is not chiral. (]

o

M

1. EtO Na*

[¢]

%

—

OEt

2. HQ, H0

OEt

‘

NaOH, H,0

2720 o heat

A [o]

[o] [o]

ONa*

\J

5

+

4o, H0

o

—

heat

NW Coll0

B

S-Nonanone

Problem 19.34 Propose a synthesis for each ketone, using as one step in the sequence a Claisen condensation and the reaction sequence illustrated in Problem 19.33.

O @

Ph/\)l\/\l’h Each target molecule is synthesized using the sequence of Claisen condensation,

saponification, and decarboxylation as outlined in the previous problem. The starting ester is ethyl 3-phenylpropanoate.

o




2-Oxepanone

H

1-Cyclopentenecarbaldehyde [o]

Q

(o]

»

H 2-Oxepanone

1-Cyclopentenecarbaldehyde i

1. LiAHg

lecular

aldol

NaOH T““‘““?.:.vd,::;:.. "

2. H,0, HCI

with

[o] PCC

HO

/W\/OH

-

H

H [0)

Recognize that the product has six carbons like the starting material, but there is an ester linkage in the ring of the starting material, and a five-member carbon-only ring of the product. Notice also that the product contains an a,f-unsaturated aldehyde, the recognition element of an aldol condensation followed by dehydration. Propose that the product is derived from an intramolecular aldol reaction followed by dehydration starting from a six-carbon dialdehyde. The six-carbon dialdehyde can be made from the starting lactone using the sequence of LiAlHs reduction to the diol followed by oxidation with PCC.

1092

8

~¢ Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 19: Enolate Anions and Enamines

REACTIONS IN CONTEXT Problem 19.85 Atorvastatin (Lipitor) is a popular treatment for high cholesterol. See “The Importance of Hydrogen Bonding in Drug-Receptor Interactions” in Chapter 10 for more information about atorvastatin. One synthesis of atorvastatin involves the following enolate reaction. Draw the predominant product of this reaction, which gives an overall yield of 90%. There are a couple of noteworthy aspects to this reaction. First, MgBr is added to exchange with Li and make Mg enolates. This is helpful for controlling stereochemistry.

Notice that the starting material is chiral and that a single enantiomer is used. The product of this reaction is a 97:3 (94% ee) mixture of two enantiomers,

not a racemic

mixture. You don’t have to be able to deduce which enantiomer is the predominant product, but be aware that being able to control the stereochemical outcome of a reaction by using a single enantiomer of a chiral starting material can save time and resources in the large-scale synthesis of chiral drugs.

HO

HO,

1.2Eq LDA. MgBr;, -78°C : .

F

HO,g/Y o

o,

2F

N N,

O "

NH

O

1093

8

~¢ 4 Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 19: Enolate Anions and Enamines

Problem 19.86 E. J. Corey used the following reaction in a synthesis of thromboxane B,. Predict the major product of the reaction. There are two possible products here. State why you think the pathway that creates the predominant product is favored under the conditions of the reaction. O

Z

I

s

. =25°C 1. LDA, THF, 2 H

s

W

c

(8]

3. H,O/THF (o] (o]

1. LDA. THF, -25°C

I

§7

N NN

|

—_—— 2 o

S

3

S snome ()°

K/v

&

N

I

oSN

SK/S\/

Z

W

0N

OH

'r-r—m

L

oL-+

H S

&

/

.

0. U

fig

&

f)

The enolate forms on the ketone as shown above. In theory, the enolate could react at

either the aldehyde carbonyl to give an aldol, or at the C=C bond to give a Michael reaction. The aldehyde carbonyl is much less hindered, so reaction primarily occurs there.

8

~¢ Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 19: Enolate Anions and Enamines

Problem 19.87 Capsaicin is a component of hot peppers that gives rise to the level of spice and “heat.” It is sometimes used medicinally as a pain reliever, for skin irritations, and as a weight loss treatment. Provide a synthetic route for capsaicin, using a malonic ester synthesis with (£)-7-bromo-2-methyl-3-heptene and a substituted benzylic amine as carbon sources. (0]

(o]

Ex( ))\/'\mt¥ HsCO,

NH,

2

Bey

8]

—— g

HCO.

N

HO

.

Capsaicin

“r/\/\/\l/ o o

o

1. NaOEt

o

EtO

o OEt

x

he: at ¢ (-_’09

H,CO.

—

i “0M

fix": -

H,CO

-

HO

HO

oH

2.HCL, H;0

2 ”M

noMon

| N.OH.H;0

o

SOCl, /3 —_—

A

i a M

o

N J\A/\/A( Capsaicin

Addition of the alkyl bromide to malonic ester under basic conditions followed by hydrolysis leads to the dicarboxylic acid. Decarboxylation leads to a carboxylic acid that can be converted to the acid chloride. Treatment of the acid chloride with the corresponding amine provides Capsaicin.

1095

8

~¢ Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 19: Enolate Anions and Enamines

Problem 19.88 The following molecule undergoes an intramolecular reaction in the presence of pyrrolidinium acetate, the protonated form of pyrrolidine. Draw the product of this reaction, assuming that a dehydration reaction takes place. %

HO_ o

o

Me

e 1\

e

S

H™)

H

B

i

9

i

’Mc

eN

(T 5

HO,

Pymolidiniss scetste

“w

Ho—rt

o

|

{

0.

o

°

o Me

Me Me

Me

Intramolecular reactions are favored that produce five- or six-membered rings, as in

this example. Problem 19.89 Organocuprates predominantly react to give 1,4-addition products with &,f-unsaturated carbonyl species, while Grignard reagents often add to the carbonyl, in a process referred to as 1,2-addition. To increase the yield of 1,4-addition products, Cul is added to convert an easily prepared Grignard reagent into an organocuprate reagent /n situ (during the reaction). Predict the major product and stereochemistry of the following reaction, assuming that the more stable chair product predominates.

L 2 N

Mghi

e,

Cul, RTHF, -30°C . 2 S 2. Equilibration to

O

more stable chair product in base

™ |

[e)

2,0 &

/

1o

?‘

x

4%

G

© A

g

2

4

ool

)

v




\@

—

2

=

Boat conformation

vb \

"

w

=

ciswcis

double bonds 9%

a1y

Goubie bonds e

The key to this problem is that the relative percentage of products is proportional to the stability of the conformations leading to their formation. Because the reaction involves a six-membered ring transition state, draw the chair-like and boat-like conformations as

107

8

~¢ 4 Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 20: Dienes, Conjugated Systems, and Pericyclic Reactions shown above. The chair-like conformation with the two methyl groups diequatorial is the most stable, so it leads to the predominant product, which is the trans-trans

product. The other chair-like conformation with the two methyl groups diaxial is the next most stable, to it leads to the intermediate product, which is the cis-cis product.

The boat-like conformation is the least stable so it leads to the smallest amount of product, which is the cis-trans product.

END-OF-CHAPTER PROBLEMS An asterisk (*) indicates an applied problem.

STRUCTURE AND STABILITY If an electron is added to 1,3-butadiene, into which molecular orbital does it go? If an

electron is removed from 1,3-butadiene, from which molecular orbital is it taken? When an electron is removed from a conjugated pi system, such as that in 1,3-butadiene, it is taken from the highest occupied molecular orbital (HOMO).

When an electron is added to a conjugated =z system, it is added to the lowest unoccupied molecular orbital (LUMO).

8

H

Electron is added to this orbital

—

—

s

Four mm

P8y g

‘molecular orbital (LUMO)

= .

8

%

ul m

Electron is removed from this orbital

x molecular orbitals

of 1.3-butadiene

1108

8

~¢ 4 Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 20: Dienes, Conjugated Systems, and Pericyclic Reactions Problem 20.14 Draw all the important contributing structures for the following allylic carbocations, then

rank the structures in order of relative contributions to each resonance hybrid. >+

(a)


>~,/

H ©

CgH;—=—=——"

H, )‘\q

,

+*

CGHS>>-./

¢

a®

",

H

(D

Double dehydrohalogenation of product (d) using sodium amide as the base.

cl CgHs’

L

*

+2NaNH;

—»

CgHg—=——

+

2 NaBr+

2 NH;3

Cl

CeHs fl (U] Catalytic reduction of the alkyne using the Lindlar catalyst to reduce the alkyne of part (e) to the cis alkene. Lindlar

CeHs—=—o

+H,

—tuint

CGHS/fi

~¢ Cengage 8

Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 21: Benzene and the Concept of Aromaticity

@ Cotls X The trans alkene will be the primary product generated upon dehydrohalogenation of

the 1-bromo-1-phenylpropane produced in (a).

Br *

CSN5/|\/

+KOH

——

CeH

N

Alternatively, the alkyne from part (e) can be treated with Na in ammonia to give the trans alkene.

CeHs—=—

+Na

NH3(1) ——

CHs" AN S

OH

Gl l_-,)\‘/ OH (h)

(racemic)

Treatment of the £ alkene from part (g) with 0s0./H,0, will give the desired product because the syn stereoselectivity of the reaction will place both OH groups on the same

face of the double bond. Note that the product of this reaction will be a racemic mixture, only one enantiomer of which is shown above.

OH

0s0,4

CeHe

N

0% o H,0;

OH H

cohg OH

+ CsHs/Y OH (racemic)

OH

(II;H.',/'\:/

OH [0)

(racemic) Oxidation of the alkene from part (g) by first creating an epoxide, followed by opening in

acid or base gives the glycol with the trans configuration because the anti stereoselectivity of the epoxide ring opening will place the OH groups on opposite faces

1227

8

~¢ 4 Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 21: Benzene and the Concept of Aromaticity of the double bond. Note that a racemic mixture is formed, only one enantiomer of which is shown above.

OH

1.RCOH

H

Y

YT+ ool

e

L

2. HOor H30

CHs N

OH

/'\/ OH

OH (racemic)

Problem 21.59 Carbinoxamine is a histamine antagonist, specifically, an Hi-antagonist. The maleic acid salt of the levorotatory isomer is sold as the prescription drug Rotoxamine.

a ()\/\N/

«l

Me

5

™

|

l

Ny

CHO

7"_

N|

i (ll/\/‘\\

Z>

;

N

Me

Me

Br

Z Carbinoxamine (a) Propose a synthesis of carbinoxamine. (Note: Aryl bromides form Grignard reagents much more readily than do aryl chlorides.) CHO

a

a m

Mg e

Br

ether

cl

7 G

. _OH

e

\O\

MgBr

2.HCI 7 H0

|

N

c

1. NaH —_—n i

A

5 0\/\N/M° Me pl;

NN

""Me

|

Me

| Carbinoxamine

1228

8

~¢ Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 21: Benzene and the Concept of Aromaticity (b)

Is carbinoxamine chiral? If so, how

many

stereoisomers

are possible? Which of the

possible sterecisomers are formed in this synthesis? Yes, carbinoxamine is chiral with one chiral center. Both enantiomers are formed in this synthesis as a racemic mixture.

*Problem 21.60 Cromolyn sodium, developed in the 1960s, has been used to prevent allergic reactions

primarily affecting the lungs, as, for example, exercise-induced emphysema. It is thought to block the release of histamine, which prevents the sequence of events leading to swelling, itching, and constriction of bronchial tubes. Cromolyn sodium is synthesized in the following series of steps. Treatment of one mole of epichlorohydrin (Section 11.10) with two moles of 2,6-dihydroxyacetophenone in the presence of base gives I. Treatment of | with two moles of diethyl oxalate in the presence of sodium ethoxide gives a diester II. Saponification of the diester with aqueous NaOH gives cromolyn sodium.

HO.

HO. l

base

¥ ‘b\/ O

Epichlorohydrin

FiONa' Z OH ()\)\/()

[8)

2,6-Dihydroxy-

11

FO—C—C—OF:t

|

OH

acetophenone

0o

OH

N

(8]

I

(e}

(e} O,

NaOM, 1,0

p MOLHO,

Na' "0

).

|

| OH (Wt)

[8)

O™Na*

8]

Cromolyn sodium

(a) Propose a mechanism for the formation of compound

I.

Step 1: Take a proton away.

HO?

HOS >

p

+ H-Base

o_.r\:sm 0

3

_ o

o

1229

~¢ 4 Cengage 8

Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 21: Benzene and the Concept of Aromaticity Step 2: Make a bond between a nucleophile and an electrophile and simultaneously break a bond to give stable molecules or ions.

HO:

5

Ho

o

*

NO;

NO,

NO; SO3H

COOH

HNO,

_—

H.S0,

©

NO, COOH

COOH

\m m-directing

/

NO,

HNO3

“Hesor™ O;N

NO,

NUCLEOPHILIC AROMATIC SUBSTITUTION: PROBLEM 22.6 In S\2 reactions of haloalkanes, the order of reactivity is Rl > RBr > RC| > RF. Alkyl iodides are considerably more reactive than alkyl fluorides, often by factors as great as 10°. All 1-halo-2,4-dinitrobenzenes, however, react at approximately the same rate in nucleophilic aromatic substitutions. Account for this difference in relative reactivities. Recall that the overall rate of a reaction is determined by the rate-determining (slow)

step. In an Sy2 reaction, departure of the leaving group is involved in the ratedetermining step, thus the nature of the leaving group influences the rate of the overall reaction. As shown in the text, the mechanism of nucleophilic substitution with 1-halo-

2,4-dinitrobenzenes involves addition to the ring to form a Meisenheimer complex. Because formation of this complex is the rate-determining step and departure of the halide is not involved in Meisenheimer complex formation, the nature of the halogen has

little influence on the overall rate of the process.

1267

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~¢ Cengage Solution and Answer Guide: Brown et al.,, Organic Chemistry ® 2023, 9780357451861; Chapter 22: Reactions of Benzene and Its Derivatives

END-OF-CHAPTER PROBLEMS An asterisk (*) indicates an applied problem.

ELECTROPHILIC AROMATIC SUBSTITUTION:

MONOSUBSTITUTION

Problem 22.7 Predict the monoalkylated products of the following reactions with benzene.

|

a0

®)

x

Sl

O

(“W

©/k/

—_—T

AICl3

_— AlCly K“

©)
%0

Reaction coordinate

O/

SOH

(c) Which step in the reaction mechanism is highest in energy? Explain.

The highest energy step is the electrophilic addition of the protonated sulfur trioxide to benzene since aromaticity is broken in that step.

(d) Which of the following reaction conditions could be used in an electrophilic aromatic substitution with benzene to provide substituted phenyl derivatives?

a /Y

T

0

reaction

= HNO; —’H SO

(if)

reaction

20874

1270

-l

~¢ Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861;

Chapter 22: Reactions of Benzene and Its Derivatives

KoCryO; (i

™

e

o

HoSOy

T

reaction

HyPO,

NaOH O

)

w

(wil)

No reaction

. Na*

-~C=C—CHg

—————————

—

Clo

AlClg

it

fl

(wilh)

No reaction

gur—

H3POy

reaction

reaction

No reaction

~¢ 4 Cengage 8

Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 22: Reactions of Benzene and Its Derivatives Problem 22.9 Write a stepwise mechanism for each of the following reactions. Use curved arrows to show the flow of electrons in each step.

Q

+

AlCly

(ii/Y

—_—

(a) Step1: Make a bond between a Lewis acid and Lewis base. This forms a complex between 1-chloro-2-methylpropane a (Lewis base) and aluminum chloride (a Lewis acid).

Y\g:\

All_c.

ap—

Y\’CI_ Alf—Cl

1

o]

Step 2: 1,2 Shift and break a bond to give stable molecules or ions. Rearrangement to form tert-butyl cation.

Le

@

\{>\¢(,:,'_AII:_C|

—

YB

l

%

a

Cl—‘AlI—CI é!

Step 3: Make a new bond between a nucleophile (pi bond) and an electrophile. The pi bond from benzene attacks the carbocation.

/—\

. Y+

Step 4: Take a proton away. Proton transfer to regenerate the aromatic ring.

H /\

¢ Cl—-l?l—Cl

—

+ HCl + AICl

(8}

1272

~¢ Cengage -l

Solution and Answer

Guide: Brown et al., Organic Chapter

O

O/\()ll

HyPOy

Chemistry

22: Reactions

© 2023, 9780357451861,

of Benzene and

Its Derivatives

O/\O

. —

+

() Step 1: Add a proton. An oxonium lon is generated by protonation with phosphoric acid.

Q”?!e

s

H—0-F-on

77

.

"

om

O H

0

4 :5-p-om * OH

Step 2: Break a bond to give stable molecules or lons. Loss of water results In formation of the benzylic carbocation. I“).

©/\.gfl

-

—

\©

+ H0

Step 3: Make a new bond between a nucleophile (pi bond) and an electrophile. The pi

bond from benzene attacks the carbocation.

0

//_\\.

Q0O—

’

Step 4: Take a proton away. Proton transfer to regenerate the aromatic ring.

+

O

N\ O

N

B

:ob-om “da

e

e

~¢ & Cengage 8

Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 22: Reactions of Benzene and Its Derivatives

©

[y

o,

O

[\

SOy

o~

SOsH

Step 1: Add a proton. Sulfur trioxide is protonated using sulfuric acid. o o U /N Ol’s‘\O A H—?—fi—ofl .~

9 O”S‘\O'H

—

o

o

+

w9 :‘0_'5'_0'{

+

T

)

Step 2: Make a new bond between a nucleophile (pi bond) and an electrophile. The pi bond from furan (nucleophile) attacks the protonated sulfur trioxide (electrophile) to

form a resonance-stabilized cation.

Y

Ny

0"

—

o070,

Olyo0

0. 4 ou

—

L) 2

9+ 4ou

Step 3: Take a proton away. Proton transfer to regenerate the aromatic ring.

v ¥s 0

o“

\OH

&3 0-S-om —»

‘oo .

(II)

I\ g®o [N o

(l’, \ofl

21] . no-b-on 6

1274

~¢ 4 Cengage 8

Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 22: Reactions of Benzene and Its Derivatives (o]

+

2

AlCly

Q

_—

(d Friedel-Crafts acylation involves electrophilic attack by an acylium ion. Step 1: Make a bond between a Lewis acid and Lewis base. This forms a resonance-

stabilized acylium ion.

B~ T a

Al

.

%

G

-—

Sk

+

ba

Cl—AlI&}

Step 2: Make a new bond between a nucleophile (pi bond) and an electrophile. The pi bond from benzene attacks the acylium ion.

i e — g /

\

HO

Step 3: Take a proton away. Proton transfer to regenerate the aromatic ring.

fs

s w

e

= N atara

o —_—

+ HC + AICl

«

1275

8

~¢ 4 Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 22: Reactions of Benzene and Its Derivatives

Problem 22.10 Pyridine undergoes electrophilic aromatic substitution preferentially at the 3 position, as illustrated by the synthesis of 3-nitropyridine.

N

NO,

Mg

Oi O 5 + HNO, —23%, 00°C N

Y8

+ HO 2

N

Pyridine

3-Nitropyridine

Under these acidic conditions, the species undergoing nitration is not pyridine, but its conjugate acid. Write resonance-contributing structures for the intermediate formed by attack of NO,* at the 2, 3, and 4 positions of the conjugate acid of pyridine. From examination of these intermediates, offer an explanation for preferential nitration at the

3 position. Pyridine is a base, and in the presence of a nitric acid-sulfuric acid mixture, it is protonated. It is the protonated form (that is, conjugate acid) that must be attacked by the electrophile *NO,. For nitration at the 3-position, the additional positive charge in the cation

intermediate may be delocalized on three carbon atoms of the pyridine ring. None of the contributing structures, however, places both positive charges on the same atom.

R H

H

N

‘

i

e

N

H

H

H

H

For nitration at the 4-position or the 2-position, the additional positive charge in the cation intermediate is also delocalized on three atoms of the pyridine ring, but one of these contributing structures has a charge of +2 on nitrogen. This situation is thus less stable than that which occurs for nitration at the 3-position.

H_

NO;

H_

NO;

H_

+

NO,

+

NN+ H

2 " H

N'+ H

PCober | ssrirmrom | charge of +2 on nitrogen

H

+ NO;

-— PI! + H

+

y NO,

'|4 + H

-—

y

| NO,

7b H

Prrubid] iseliog jspn charge of +2 on nitrogen

1276

8

~¢ 4 Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 22: Reactions of Benzene and Its Derivatives

Problem 22.11 Pyrrole undergoes electrophilic aromatic substitution preferentially at the 2 position, as illustrated by the synthesis of 2-nitropyrrole.

N,

HNO, -hC00H Q

N

.

5%

4+ HO

N

|

NO,

b

|

H

H

Pyrrole

2-Nitropyrrole

Write resonance-contributing structures for the intermediate formed by attack of NO,* at the 2 and 3 positions of pyrrole. From examination of these intermediates, offer an explanation for preferential nitration at the 2 position. Pyrrole is nitrated under considerably milder conditions than pyridine. For nitration at

the 2-position, the positive charge on the cation intermediate is delocalized over three atoms of the pyrrole ring whereas for nitration at the 3-position, it is delocalized over

only two atoms. The intermediate with the greater degree of delocalization of charge has a lower activation energy for its formation and hence is formed at a faster rate. .

H

H -

) T

+

N7

N02

T

H

(/

§’N°2 H

N02

H H

g No; H

Q N"

H H

(f

H -

N02

8

~¢ 4 Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 22: Reactions of Benzene and Its Derivatives

Problem 22.12 Addition of m-xylene to the strongly acidic solvent HF/SbF; at —-45°C gives a new species, which

shows 'H-NMR

resonances

at &2.88 (3H), 3.00 (3H), 4.67 (2H), 7.93 (1H), 7.83 (1H),

and 8.68 (1H). Assign a structure to the species giving this spectrum. The strong acid results in protonation of the aromatic ringto create a positively charged

species as shown below.

CHy

CH;

A H

&

+

H-F/?‘SbFs

H

—

H3C

A

+ SbFg

-

H3C

Problem 22.13 Addition of tert-butylbenzene to the strongly acidic solvent HF/SbF; followed by aqueous

workup gives benzene. Propose a mechanism for this dealkylation reaction. What is the other product of the reaction?

A reasonable mechanism for this reaction involves protonation of the aromatic ring, followed by heterolytic bond cleavage and release of the tert-butyl cation. After addition

of water, tert-butyl alcohol will be the other product of the reaction. Step 1: Make a new bond between a nucleophile (x bond) and an electrophile—add a proton.

f"; HyC—C—CH;

Hs3C e H

— -~

+

CH

e C\ K

CH3

Step 2: Break a bond to give stable molecules or ions.

H3C "

4

CH

3

H

\013

+

—_—

+

+

CH3—C—CH3 |

CH3

1278

~¢ 4 Cengage 8

Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 22: Reactions of Benzene and Its Derivatives Step 3: Make a bond between a nucleophile and an electrophile. H

c"'a‘(l*‘c";

+

:+—H

.A

+ H=0—H

—»

(:yq]—(l:—c”3

CH3

CH3

Step 4: Take a proton away.

i

0 a“

CH3

_éJ

I—Gig

+

i

0:

y

H=0—H

H

I

—»

(:Hz—(l:—ct-l;

CH3

I+

+

H—O—H

CH3

Problem 22.14 What product do you predict from the reaction of SCl, with benzene in the presence of AlCL;? What product results if diphenyl ether is treated with SCl, and AlCL,? The Lewis acid, AlCL,, facilitates departure of one of the chlorine atoms from SCl, and

the resulting electrophile takes part in an electrophilic aromatic substitution reaction to create the —SCl derivative that then reacts again to create diphenyl sulfide. The

diphenyl sulfide is activated compared to benzene, so this will react further to generate a polymeric product as shown.

SCI

=2

O

ALCI,

S

|2~ AICI

U

AICly

—-

S

n

If diphenyl ether were treated in a similar manner, the resulting polymeric species will

have alternating ether and thioether functions.

(0}

(o} AlC!

O

©

'

SCI:

a

s

O

s

n

1279

8

~¢ 4 Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 22: Reactions of Benzene and Its Derivatives

Problem 22.15 Other groups besides H* can act as leaving groups in electrophilic aromatic substitution. One of the best is the trimethylsilyl group, Me;Si—.

with CF,COOD

For example, treatment of Me;SiC¢Hs

rapidly forms CsHsD. What properties of a silicon-carbon bond allow you to

predict this kind of reactivity? Based on simple electronegativities, the C—Si bond is polarized such that a partial

positive charge is on the Si atom. Furthermore, the heterolytic bond cleavage is facilitated because the (CH,)sSi* cation is so stable.

DISUBSTITUTION AND POLYSUBSTITUTION Problem 22.16 The following groups are ortho-para directors. Draw a contributing structure for the resonance-stabilized cation formed during electrophilic aromatic substitution that shows the role of each group in stabilizing the intermediate by further delocalizing its positive charge.

@

——C=CHy

H

%

E

-

H

(b) —Br E

. Br+

HG ©

——OCH:; E

-

3 ¥,

l_l>':o;cn,

1280

8

~¢ 4 Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 22: Reactions of Benzene and Its Derivatives e

=—iN

-

@

X0

H

@

+\—

NH: E

+

NH.

HG

:

Problem 22.17 Predict the major product or products from treatment of each compound with Cl,/FeCl.. When there is more than one substituent on a ring, the predominant product is derived

from the orientation preference of the more activating substituent.

Q/k

Cl

d\

/@)\

e

(a

+

Fecls

OH

CU

—

s

- X +

NO,

cl

NO,

CHO

Ci") C

@

¢

OH

L CHO

()

a

(e}

eC! FeCly

NOs»

(b)

=2

a

=

FeCl

I

~CN

(e}

L0

a C12 FeCly

oN C

1281

~¢ 4 Cengage 8

Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 22: Reactions of Benzene and Its Derivatives Problem 22.18 How do you account for the fact that phenyl acetate is less reactive toward electrophilic aromatic substitution than anisole?

O

0.

Y

“CH,

(¢} Phenyl acetate

Anisole

The unshared pair of electrons on the oxygen atom of phenyl acetate is involved in a

resonance interaction with the carbonyl group of the ester, and therefore, less available for stabilization of an aryl cation intermediate compared to anisole.

Problem 22.19 Propose an explanation for the fact that the trifluoromethyl group is almost exclusively meta directing.

CFy

G

CFy

+ HNO, 12504, Q

+ Hy0 NO,

Shown below are contributing structures for meta and para attack of the electrophile.

For meta attack, three contributing structures can be drawn and all make approximately equal contributions to the hybrid. Three contributing structures can also be drawn for ortho/para attack, one of which places a positive charge on carbon bearing the

trifluoromethyl group; this structure makes only a negligible contribution to the hybrid. Thus, for meta attack, the positive charge on the aryl cation intermediate can be

delocalized almost equally over three atoms of the ring giving this cation’s formation a lower activation energy. For ortho/para attack, the positive charge on the aryl cation intermediate is delocalized over only two carbons of the ring, giving this cation's

formation a higher activation energy. meta attack:

[ CF3

5 CFs 3+ +

> CF3

+ =

NO,

gy 2 —

NO,

H

+

NO;

1282

~¢ 4 Cengage 8

Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 22: Reactions of Benzene and Its Derivatives ortho/para attack:

&,

ON

&,

OzN

H

&,

ON

H

H

+

'+ Adjacent positive charges

[ CF3 3

[ CF3 418+ s

H

'NO,

5 CF3 3+

-

-

+

H NO; Adjacent positive

H®

'NO,

charges

Problem 22.20 Suggest a reason why the nitroso group, —N=0, is ortho-para directing whereas the nitro group, —NO,, is meta directing. Like other ortho-para directing groups, the —N=0 group has a lone pair of electrons

that can stabilize an adjacent positive charge of ortho or para attack via the resonance structures shown below. Without a lone pair of electrons on nitrogen, the nitro group

cannot take part in similar stabilization. In fact, an adjacent positive charge is destabilized by the electron-withdrawing nature of the nitro group and the partial

positive charge on the nitro N atom.

HeH

in E H

0 G

E

-

i*

]

!+

08

0

< :

H

—

or Ortho attack

H

E

H

E

Para attack

1283

8

~¢ Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 22: Reactions of Benzene and Its Derivatives

Problem 22.21 Arrange the compounds in each set in order of decreasing reactivity (fastest to slowest) toward electrophilic aromatic substitution.

(@)

Q

Oi

Q()(:H3 (b)

o

Q(IN

@

CH:‘

—(:H3

Oroan> Or-on » O

OL

O

©

O

Orom L

oo >

o

- O

1284

8

~¢ 4 Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 22: Reactions of Benzene and Its Derivatives

(d

(e)

:

Q

-

QB].

-

QF

Problem 22.22 For each compound, indicate which group on the ring is more strongly activating and then draw a structural formula of the major product formed by nitration of the compound.

In the following structures, the more strongly activating group is circled and arrows show the position(s) of nitration. Where both ortho and para nitration is possible, two

arrows are shown. A broken arrow shows a product formed in only negligible amounts. Cl

@

i

CHg Cl

h t.

1285

~¢ Cengage «l

Solution and Answer Guide: Brown et al.,, Organic Chemistry ® 2023, 9780357451861; Chapter 22: Reactions of Benzene and Its Derivatives

Br

)

CHO

CHO

CHO

©

CHs CHO

(:[ocn3 @

COsH

o

i

Lo,"

~¢ Cengage «l

Solution and Answer Guide: Brown et al.,, Organic Chemistry ® 2023, 9780357451861, Chapter 22: Reactions of Benzene and Its Derivatives

(@

NHCOCHy

"

l

NOy

SOsH

NHCOCH

®

3

SOsH

i Q/CN

®

CH:

3

CN

/

‘

1287

8

~¢ 4 Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 22: Reactions of Benzene and Its Derivatives

(h)

Cl

Pl i;(\ Cl Problem 22.23 The following molecules each contain two aromatic rings. Which ring in each undergoes electrophilic aromatic substitution more readily? Draw the major product formed on nitration.

(@

OO

¥

F F

OZN

F

"

Phenyl groups are activating at the para position and, therefore, ortho/para directing. Note that nitration takes place on the ring that does not contain the electron-

withdrawing fluorine groups. Furthermore, nitration does not take place at the ortho position because of steric interactions between the rings.

1288

~¢ 4 Cengage 8

Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 22: Reactions of Benzene and Its Derivatives

HyC, S

(b)

\ N

O H,C

The phenyl ring directly bearing the nitrogen is more highly activated and therefore nitrated. The para-substituted product is favored since the nitrogen is ortho/para directing and the ortho position is blocked due to steric hindrance.

OZN

The phenyl ring conjugated with the alkene group is more prone to nitration since the cation intermediate formed is stabilized through resonance. Furthermore, nitration does

not take place at the ortho position because of steric interactions between the rings. *Problem 22.24 Reaction of phenol with acetone in the presence of an acid catalyst gives a compound known as bisphenol A, which is used in the production of epoxy and polycarbonate resins. Propose a mechanism for the formation of bisphenol A.

OH ©/

o +

A

Lo

O

O

HO

Phenol

Acetone

+ H,0 OH

Bisphenol A

The reaction begins by proton transfer from phosphoric acid to acetone to form its

conjugate acid which may be written as a hybrid of two contributing structures. The conjugate acid of acetone is an electrophile and reacts with phenol at the para position by electrophilic aromatic substitution to give 2-(4-hydroxyphenyl)-2-propanol.

Protonation of the tertiary alcohol in this molecule and departure of water gives a resonance-stabilized cation that reacts with a second molecule of phenol to give bisphenol A.

1289

~¢ & Cengage 8

Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 22: Reactions of Benzene and Its Derivatives Step 1: Add a proton.

«fa/N

P

6

ey H

H

fis

CHy;—C—CHy + “Eb—go“

~==

it

CHi—C—CH;

.-,°i H

HeH Step 6: Take a proton away.

g w?"

08

&H

CH "

sa o

o

==

PR A o

08

oi“

oo-H

NUCLEOPHILIC AROMATIC SUBSTITUTION Problem 22.30 Predict all possible products formed from the following nucleophilic substitution reactions.

O @

OuN

.

Cl

1. NaOH 2. HCL, HO

O 0O,N

OH

.

1297

~¢ 4 Cengage 8

Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 22: Reactions of Benzene and Its Derivatives

a + NaNH,

NH.

NH,

NHy (1 i—

-38°C

+

e

.

NH,

(b)

Cl

O

OH

O

1. NaOH 2. HCI, HyO

(©) The presence of a strong electron-withdrawing group positioned ortho or para to the leaving group on the ring means that the nucleophilic aromatic substitution mechanism will be addition-elimination instead of being through a benzyne intermediate.

*Problem 22.31 Following are the final steps in the synthesis of trifluralin B, a preemergent herbicide.

¢

l O,N

NO, /l\/'\ ON " —= 5

HNO, —,

@

CFy

/L\J\ NO,

)]

CFy

CFy Trifluralin B

(a) Account for the orientation of nitration in Step 1. The trifluoromethyl group is strongly deactivating and meta directing (Problem 22.17).

Chlorine is weakly deactivating and ortho/para directing. The combination directs the incoming nitro groups ortho to the chlorine atom.

1208

~¢ Cengage 8

Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 22: Reactions of Benzene and Its Derivatives (b) Propose a mechanism for the substitution reaction in Step 2.

The second step of this transformation is an example of nucleophilic aromatic substitution. In the following structural formulas, the propyl group of dipropylamine is abbreviated R. Step 1: Make a bond between a nucleophile and an electrophile (aromatic).

e

Rama. ON

Co:

H

&

A 0: o

R,

O

ON

Nosso *o:o

>

CF3

CF3 A Meisenheimer complex

Step 2: Take a proton away.

ON

iy

)

INR, O: NG .-

QM

()

—

ON

1Ci_

NR, O Noes

R

CF3

CF3

Step 3: Break a bond to give stable molecules or ions.

1CI)

NR, (j); N...-

ON

0!

CF3

N —

NO;

ON

+

g

:Q:

CF3 Trifluralin B

1299

8

~¢ 4 Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 22: Reactions of Benzene and Its Derivatives

*Problem 22.32 A problem in dyeing fabrics is the degree of fastness of the dye to the fabric. Many of the early dyes were surface dyes; that is, they did not bond to the fabric, with the result that they tended to wash off after repeated laundering. Indigo, for example, which gives the

blue color to blue jeans, is a surface dye. Color fastness can be obtained by bonding a dye to the fabric. The first such dyes were the so-called reactive dyes, developed in the 1930s for covalently bonding dyes containing —NH: groups to cotton, wool, and silk fabrics. In the first stage of the first-developed method for reactive dyeing, the dye is treated with cyanuric chloride, which links to the fabric through the amino group of the dye. The remaining chlorines are then displaced by the —OH groups of cotton (cellulose) or the

—NH; groups of wool or silk (both proteins). i

h"

J:

SN

(:1)\1\)\(:1 Cyanuric chloride

Dye—NH,

\|

SN

)()\—('uuun

HO—Cotton

l)‘u-—\'llJ\_\;;\(:I A reactive dye

\I

SN

l)\r-—\ll)\.\)\()—(lmmn Dye covalently bonded to cotton

Propose a mechanism for the displacement of a chlorine from cyanuric chloride by (a) the NH: group of a dye and (b) by an —OH group of cotton.

In each reaction, the chlorine atoms are replaced by nucleophiles, either the amine groups of the dye or the hydroxyl groups of the cotton fibers. The mechanism will be

nucleophilic aromatic substitution, analogous to that presented in Problem 22.28 above, involving a Meisenheimer complex intermediate. Note that the chloro groups here can be

considered the nitrogen analog of an acid chloride.

1300

8

~¢ Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 22: Reactions of Benzene and Its Derivatives

ORGANIC CHEMISTRY REACTION ROADMAP Problem 22.33 Use the reaction roadmap you made for Problems 20.61 and 21.55 and update it to contain the reactions in this chapter. In anticipation of reactions in Chapter 23, on this roadmap, you should include the H./Ni and 1) Fe, HCL, 2) NaOH reactions that convert aryl nitro compounds to aryl amines. These two reactions were covered in the chapter (Section

22.1B). 03]

C-C bond forming reactions are

N

lmfimrd’hya box surrounding

the reaction number

[a|

)

|

P I

.

[aryliodides

=] Pr3

1]

\

/

a 4 [carbompteacs | =~ I\—[Ponae] “—*— [auinones]

i

-e

my? /

/

\ /y

RS ¥/ &,

Chapter 22 Roadmap Reaction Legend

Reaction 22.1 Xa, FeXa

- The X; reacts with the Lewis acid FeX; to give the electrophileX*

1301

~¢ 4 Cengage 8

Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 22: Reactions of Benzene and Its Derivatives Reaction 22.2 HNO;, HzS04 - The electrophile is NO;* formed by the reaction of HNOs with H2SO4

Reaction 22.3 S0 - The electrophile is either SO; or HSO,* depending on conditions

Reaction 22.4 R-X, AlCls - Friedel-Crafts alkylation

- The haloalkane reacts with the Lewis acid AlCLs to give the carbocation electrophile - Rearrangements can be a problem

- Makes a new C—C bond

Reaction 22.5

X

R™"al, alcl, - Friedel-Crafts acylation

~ The acid chloride reacts with AlCL to give the acylium ion electrophile - Makes a new C—C bond

Reaction 22.6 Alkene, Brgnsted or Lewis acid

- The alkene reacts with the Brensted or Lewis acid to give a carbocation electrophile - Makes a new C—C bond Reaction 22.7 Alcohol, Bronsted or Lewis acid - The alcohol reacts with the Bronsted or Lewis acid to give a carbocation electrophile

- Makes a new C—C bond

Reaction 22.8 NaNHa, NHx()

- The strong base causesan elimination reaction to give a benzyne intermediate followed by addition of NHz

Reaction 22.10 Ha, Ni or other transition metal - This converts a nitro group into a primary aryl amine making it a very useful reaction - These conditions also reduce alkenes or other susceptible groups

Reaction 22.11 1. HCL, FeCls, 2. NaOH

- This converts a nitro group into a primary aryl amine making it a very useful reaction - The initially formed salt is converted to the amine usinga strong base

treatmentin the

second step

1302

8

~¢ Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 22: Reactions of Benzene and Its Derivatives

Problem 22.34 Write the products of the following sequences of reactions. Refer to your reaction roadmaps to see how the combined reactions allow you to “navigate” between the different functional groups. Note that you will need your old Chapters 6-11, Chapters 15— 18, and Chapter 19 roadmaps along with your new Chapters 20-22 reaction roadmaps for these.

1. HNO,, H,S0, 2. Cly, FeCly —_—

3. Hy, Ni

[0

An aryl ring

i

NO, R

1 HNO, H,50,

O/

——

NO, 2.0, FeQly

B

Reaction 22.2

O

—

Reaction 22.1

An ‘ aryl ring

3 Hy Ni

B

NH, Q

Reaction 22.10

e A nitrobenzene

&

&

A halobenzene

An aniline

1. Cly, FeCly 2. HNOy, H,S0, —_—

3. Hy Ni

®) An aryl ring

1.Qy, FeQly

At

—

’”"InflhnJ

.

2 HNO;, H:SO,

—

Reaction 222

X

X

Ci

-

NO; 3 Hy.Ni

o

>

Cl

NHy ol

Reaction 2210

A halobenzene ON

A nitrobenzene

HN

An aniline

1303

~

+

Wer

Br

1306

8

~¢ 4 Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 22: Reactions of Benzene and Its Derivatives o

(b)

e

0

©)l\/

fiBr.

Br. \O)‘\/

+

HBr

Problem 22.37 Show how to convert ethyl benzene to (a) 2,5-dichlorobenzoic acid and (b) 2,4-dichlorobenzoic acid. o

cl, FeCly ethylbenzene

H,CrO,

o

i, FeCly

al para-substituted isomer will also

i

NOz

d, FeCly

NHz

1. Fe/HCl 2 NaOH >

(=] K2Cr207,

fi

fl

SOCI, 22207 M504 o oy ehpCon —2%

CHyCHpCH0H

N

H,SO.

[«]

2

N \'/\

(fl

+

cHyeH,Cal

CHyCHCO

——

]

(_

[§]

J) -11—}|’—n

¢

o

5

Gusnine ()

base pairs showing the locations of planar —NH,

to the aromatic bases as well as the specific patterns of hydrogen

bonds responsible for recognition between complementary strands of DNA. E.

In the structures of T—A and C—G base pairs, there are three amino groups specifically labeled as “sp; hybridized and planar.” What is the primary difference between these structures and that of aniline that lead to their planarity? 1.

In contrast to aniline, the amino groups on the DNA bases are necessary to make the heterocyclic rings aromatic.

2.

In contrast to aniline, the contributing structures that delocalize the nitrogen lone pairs onto the rings creates partial negative charges on electronegative atoms.

3.

In contrast to aniline, the hydrogen bond accepting ability of the lone pairs on the —NH; groups of the DNA bases is better when these amino groups are sg? hybridized.

4.

Both 2 and 3.

1358

«l

¢ Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451867; Chapter 23: Amines

F.

In the structures of T—A

and C—G

base pairing, four nitrogens are circled. Given your

knowledge of organic functional group names, which of the following is the most appropriate descriptor for the kind of functional group that these nitrogens are part of?

1.

An N-heterocyclic ester.

2.

An N-acetal.

3.

An imide.

4.

An imine.

Chemists have studied base pairings analogous to those found in DNA in order to shed light on the strength of the hydrogen bonds. For example, the strength of the association of the following three base pairs increases in the order given (as an abbreviation, the pattern can be written with D = hydrogen bond Donor and A =

hydrogen bond Acceptor).

X,

2r

0”? NAM) ' 1 1

woyo }

H”

N

|

N

=

N-H

:

1

H

N

N

[8]

oo

H

N

t'J”l\N’ [ '

~H

o

H”

1

Y

N,

=

=

N

MN“!

R,NJ

o0

Ar” SNCSNESNT ' [l '

uou

“I

H

|

'

N

N

[ ]

N

Ar H

Or

Q N

A

RO

A—D—A

A—A—D

A—A—A

D—A—D

D—D—A

D—D—D

x“

O

-_— Increasing strength of association

G.

Which of the following is the most likely explanation for the order of association found experimentally?

1.

A trend is not expected, and hence the result is random.

2.

The overall number of hydrogen

bond donating and hydrogen

bond accepting

interactions increases from left to right. 3.

The hydrogen bonds are increasingly more linear in the complexes from the left to the right.

4,

By decreasing the alternation of hydrogen bond donors and acceptors on the same molecule, the hydrogen bonds become stronger due to less repulsive interactions between neighboring hydrogen bonds.

1359

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¢ [ Cengage Solution and Answer Guide: Brown et al., Organic Chemistry @ 2023, 9780357451867; Chapter 23: Amines

RELATIVE AMINE

BASICITY:

PROBLEM

23.6

Select the stronger acid from each pair of compounds.

OzNONH3+

of

CHaONH3+

(A)

(@

(B)

4-Nitroaniline (pAg 13.0) is a weaker base than 4-methylaniline (pA; 8.92). The decreased basicity of 4-nitroaniline is due to the electron-withdrawing effect of the para nitro

group. Because 4-nitroaniline is the weaker base, its conjugate acid (A) is the stronger acid.

\ (b)

.

or

NH

NH3*

(C)

(D)

Pyridine (p/t 8.75) is a much weaker base than cyclohexanamine (pA; 3.34). The lone pair of electrons in the sp? orbital on the nitrogen atom of pyridine (C) has more s

character, so these electrons are less available for bonding to a proton. Because pyridine is a weaker base, its conjugate acid, (C), is the stronger acid. AMINE ACID-BASE

REACTIONS:

PROBLEM

23.7

Complete each acid-base reaction and name the salt formed.

(Et)3N

+

HC———3=

(a)

Triethylammonium chloride

N—H (b)

(Et)3NH* CI

+

CHsCOOH —»

,H

N;H

CH;C00~

Piperidinium acetate

1360

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¢ Cengage '

|

Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451867; Chapter 23: Amines

AMINO ACID

PROTONATION:

PROBLEM

23.8

Following are structural formulas for propanoic acid and the conjugate acids of isopropylamine

and alanine, along with pA, values for each functional group.

pk, 10.78

i,

(IH,(I_.H(,HR

O

pk, 478

I Rl

CH,CH,COH «

INT-13+.) Conjugate acid of isopropylamine

0

R

pk, 235

(_.Hfltl,H(;oH

it pK,

9.87

NH.,.*‘.__./ Propanoic acid

Conjugate acid of alanine

(a) How do you account for the fact that the —NH" group of the conjugate acid of alanine is a stronger acid than the —NH;' group of the conjugate acid of isopropylamine? The electron-withdrawing properties of the carboxyl group adjacent to the amine of alanine make the conjugate acid of alanine more acidic than the conjugate acid of isopropylamine. (b) How do you account for the fact that the —COOH group of the conjugate acid of alanine is a stronger acid than the —COOH group of propancic acid? The —NH;* group is electron-withdrawing, so the adjacent —COOH group is made more

acidic by an inductive effect. This situation is analogous to the electron-withdrawing effects of halogens adjacent to carboxylic acids in molecules such as chloroacetic acid, which has a p& of 2.86. In addition, deprotonation of the carboxylic acid function of alanine results in formation of an overall neutral zwitterion. Thus, the carboxylate form

of alanine can be thought of as being neutralized by the adjacent positively charged

ammonium ion. SEPARATIONS

BY AQUEOUS

EXTRACTIONS:

PROBLEM

23.9

In what way(s) might the results of the separation and purification procedure outlined in Example 23.9 be different if the following conditions exist? (a) Aqueous NaOH is used in place of aqueous NaHCO:. If NaOH is used in place of aqueous NaHCOs, then the phenol will be deprotonated along

with the carboxylic acid, so they will be isolated together in fraction A. (b) The starting mixture contains an aromatic amine, ArNHs, rather than an aliphatic amine, RNH2.

If the starting mixture contains an aromatic amine, ArNH,, rather than an aliphatic amine, RNHy, then the results will be the same. The aromatic amine will still be protonated by the HCl wash, and deprotonated by the NaOH treatment.

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Solution and Answer Guide: Brown et al., Organic Chemistry @ 2023, 9780357451861; Chapter 23: Amines

AMINE

SYNTHESIS:

PROBLEM

23.10

Show how to bring about each conversion in good yield. In addition to the given starting material, use any other reagents as necessary.

OH “MIG—NH!

—-

Lien—ON

(a)

OH OH MeO

Q

NH;

2A

—————®

MeO

Q

a OH

OH

O

(b)

(racemic)

.

{racemic)

«

OH

0 Et;NH .

(racemic)

g

NEt;

(racenmc)

The stereochemistry of the product will be determined by the stereochemistry of the

starting material; if the product is racemic, the starting material must have been racemic as well.

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TIFFENEAU~-DEMJANOV

REACTION:

PROBLEM

23.11

How might you bring about this conversion?

OroOxim=Cr 00—

CHyNH,

-

O 22Ol ethanal

Aieaveaions

5

Ol

CH;NO;

CH,NH;

Go

The synthesis begins with an aldol reaction using nitromethane, followed by reduction to give a f-aminoalcohol that then undergoes the ring expansion reaction.

MULTI-STEP SYNTHESIS |: PROBLEM 23.12 Show how to convert toluene to 3-hydroxybenzoic acid using the same set of reactions as in Example 23.12 but changing the order in which two or more of the steps are carried out,

CHy

(j

COOH

iy

B

O

2)

e

COOH

COOH

Qua

ij\m

e NO;

COOH

e NH;

Q OH

The key to this question is that the methyl group is converted to a meta-directing carboxyl group before the nitration reaction. This leads to the desired product with the hydroxy group in the 3 position. Step 1: Oxidation at a benzylic carbon (Section 20.6A) can be brought about using chromic acid to give benzoic acid.

Step 2: Nitration of the aromatic ring using HNOs in H2S04. The meta-directing carboxyl group gives predominantly the desired 3-nitrobenzoic acid product. Step 3: Reduction of the nitro group to 3-aminobenzoic acid can be brought about using Hz in the presence of Ni or other transition metal catalyst. Alternatively, it can be

brought about using Zn, Sn, or Fe metal in aqueous HCL Step 4: Reaction of the aromatic amine with HNO; followed by heating in water gives 3-hydroxybenzoic acid.

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Solution and Answer Guide: Brown et al., Organic Chemistry @ 2023, 9780357451867; Chapter 23: Amines

MULTI-STEP

SYNTHESIS

Il: PROBLEM

23.13

Starting with 3-nitroaniline, show how to prepare the following compounds. (a) 3-Nitrophenol

NH,

Ny*

OH

Namfi,uzso‘

)

20

50, heat

NO;

NO;

NO;

(b) 3-Bromoaniline

NH;

Q

N,*

NaNO,, Hal

r

@\

Br

CuBF, heat

H, Ni

NH;

NO;

NO;

NO;

Q

(c) 1,3-Dihydroxybenzene (resorcinol) NH;

O

NH;

Ha, Ni

—_—

Q

NO;

N;*

NaNO,, HCl

OH

Q

———

H;0, heat

Q

——

NH;

Nz*

OH

(d) 3-Fluoroaniline NH;

a

N2*BFy

1) NaNOy, HCI

F

On...

2) HBF, NO;

Q

F

Hy, Ni

: NO;

Q '

NO;

NH;

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Solution and Answer Guide: Brown et al., Organic Chemistry @ 2023, 9780357451867; Chapter 23: Amines

(e) 3-Fluorophenol

CL—-L: o O\—-“' Q—+ : QL’* NOp NO; NO; NH;

Oz, F

Nz*

(f)

F

OH

3-Hydroxybenzonitrile

NH;

Nz*

OH

OH

Qe Q- O Q= NO;

OH

NO;

NO;

NH;

OH

CL N,*o

—_—

(Ll

CN

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Solution and Answer Guide: Brown et al., Organic Chemistry @ 2023, 9780357451861; Chapter 23: Amines

HOFMANN

ELIMINATION:

PROBLEM

23.14

The procedure of methylation of amines and thermal decomposition of quaternary ammonium

hydroxides was first reported by Hofmann

in 1851, but its value as a means of

structure determination was not appreciated until 1881, when he published a report of its use to determine the structure of piperidine. Following are the results obtained Hofmann. 1. CHgl

(excess), KaCOy

CsHyN Fo000,

L CHyl

(excess), RaCOy

Hy N 208000,

Piperidine

(H,=CHCH,CH=CH,

(A)

(a) Show that these

by

1,4-Pentadiene

results are consistent with the structure of piperidine (Section 23.1).

As shown below, the structure of piperidine is consistent with the formulas given, as

well as the final product. 1. CH;l (excess)

4. CH3l (excess)

2. Ag20, H,0

|

—_— 3. heat

N

5. Ag;0, H:0

—————»

N

6. heat

PAS

I'l'i

CH;

CH;=CHCH,CH=CH;

14-Pentadiene

CH3

C7H1sN

CsHiN

@)

Piperidine

(b) Propose two additional structural formulas (excluding stereoisomers) for C:HyN that are also consistent with the results obtained by Hofmann.

The following two molecules also have structures that are consistent with the formulas given, as well as the final product.

Remember that in Hofmann eliminations, the least

substituted alkene is formed predominantly.

QC“; +*

N H

CHs ——CH3 *

*

N H

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CoPE

ELIMINATION:

PROBLEM

23.15

In Example 23.15, you considered the product of Cope elimination from the 2R,3S stereoisomer of 2-dimethylamino-3-phenylbutane.

What is the product of a Cope

elimination from the following sterecisomers? What is the product of a Hofmann elimination from each

stereoisomer?

(a) 25,3/ stereoisomer

The Cope elimination gives predominantly (£)-2-phenyl-2-butene.

3® CH3>

*_CH-_;‘

CGHS"}C—C{‘H

H”

# “N(CHy),

-

&-.

cH

3o

2l

CeHY

2 G)

3

H

+

(CH3),NOH

=

1 E)=2-Phenyl-2-butene

The Hofmann elimination gives predominantly (&)-3-phenyl-1-butene.

3®

2 ®

cH;)

CosmicZ H

CH;

£

oM

(

N(CH3),

1. CHl (excess)

2. Ag;0,

CH3>

H;0

L

o

3. heat

CeHsetZ cu—cH, H

+ (CHN

(R1-3-Phenyl-1-butene

2(8)

(b) 25,35 sterecisomer The Cope elimination gives predominantly (2)-2-phenyl-2-butene.

3® C5H5>

.CH3

CHys 4 L

oM

£

1. H;02 Loth02

2 @

H CMsaemct™

CH

4 (cHy)NOH

(Z)-2-Phenyl-2-butene

The Hofmann elimination gives predominantly (S)-3-phenyl-1-butene.

36

o) g S

CH;-;‘C_C':"'

H

(

N(CH3),

2 @

.

e

=

=

2

2®

N2

"

o

CHimeZ_CcH=CH,

N

+

(CH3)sN

15)1-3-Phenyl-1-butene

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Solution and Answer Guide: Brown et al., Organic Chemistry @ 2023, 9780357451867; Chapter 23: Amines

END-OF-CHAPTER

PROBLEMS

An asterisk (*) indicates an applied problem.

STRUCTURE

AND

NOMENCLATURE

Problem 23.16 Draw a structural formula for each amine and amine derivative.

(a) N-Ethylpiperidine

iEt (b) M-Methylpyrrole

I\

NCH, ()

N-Ethyl-N-methylpropanamine

I (d) (A-4-Amino-3-hydroxy-2-butanone (0]

MN“: OH (e) (5)-3-Amino-1-pentene

NH, “\\/‘\/

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Solution and Answer Guide: Brown et al., Organic Chemistry @ 2023, 9780357451867; Chapter 23: Amines

(f)

9,70-Diaminoanthracene NH,

NH,

(g) MAN-Diethylaniline

o

Et

N

(h) Triethylammonium

chloride

EtaNH*Cl"

(i)

(E)-N-Ethyl-1-amino-2-butene

MN“ [ H

()

3-Phenylpyridine

l

S N

~

(k) N-Ethylcyclohexylamine H

\/N\O

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Solution and Answer Guide: Brown et al., Organic Chemistry @ 2023, 9780357451867; Chapter 23: Amines

Problem 23.17 Give an acceptable name for these compounds.

@ A NH; Isopropylamine

H

(b) Benzylethylamine

®

+N”

@

=

I Cl H

2

2-Ethylpyridinium chloride

HN

(d) é Diphenylamine

o-Benzenediamine

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Solution and Answer Guide: Brown et al., Organic Chemistry @ 2023, 9780357451861; Chapter 23: Amines

®

“w

NH;

2-Amino-3-chloronaphthalene

N. al @

H

H

Piperidinium chloride

O

iEt

(h)

Et

N, N-Diethylbenzamide

Problem 23.18 Classify each amine as primary, secondary, or tertiary and as aliphatic or aromatic. MeO

Torus-2; a:.lph.lfic

o I /

H

N-CHy

0 (a)

Oxycodone

137

+

Br

Nitrobenzene

Br A-Bromophenol

(c) Toluene to p-cyanobenzoic acid

CH,

CH;

COOH

COOH

NO;

NOz

NHz

Q-2 Faad

|

COOH

== )

Toluene

Np*

COOH

... heat

CN p-Cyanobenzoic acid

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Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451867; Chapter 23: Amines

(d) Phenol to p-iodoanisole

OH

OCH;3

G

1) NaOH

O

OCH,

HNO,

2) CHyl

OCH3

¢

Ha, Ni

I¢

OCH,

NaNO, HCI

H;50,

¢ B

Plienol

NO;

NH,

N2*

OCH3

KI

[ p-lodoanisole

Note that dimethyl sulfate could be used in place of CHal to give the methyl ether in the first step of the synthesis.

(e) Acetanilide to p-aminobenzylamine NHCOCH;

NHCOCH,

HNO,

NH;

NaOH

Nz* CuCN

NaNO, HO

H2504

a

heat

Acetanilide

NO;

CN H,, Ni

CN

NO,

NO;

NO;

CH,NH, 1. LiAlH4

-

B

2. H;0

NH,

NH; p-Aminobenzylamine

1402

4

~'

TP Ko, o "

0

N-"ocH,

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Solution and Answer Guide: Brown et al., Organic Chemistry & 2023, 978035T7451861; Chapter 24: Catalytic Carbon-Carbon Bond Formation

HECK REACTION |I: PROBLEM 24.2 Give reagents and conditions for the following reaction. Ph Ph Br

|

Br

Br

Ph —

Br

Br

Ph

Br

‘

Ph IP’h

Ph Ph

Br Br.

Br

Br

Br Br

B

l

PA(OAG)z 2 PhaP, _._._-' (CH3CHz)sN

Ph

Ph

Styrene Ph Ph

Reaction of hexabromobenzene with 6 equivalents of styrene under the normal Heck conditions gives the desired compound.

ALLYLIC ALKYLATION: PROBLEM 24.3 Write the product(s) of the following reaction. How many stereoisomers of the product are formed?

OH

OH

Y\\\/'\

ol L E—

ALl

PhOS™

=

"NO,

I“

PhO,S

NO;

This is the only

stereoisomer formed

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Suzukl COUPLING: PROBLEM 24.4 Show how the following compound eight carbons or less. MeO

can be prepared from starting materials containing

: (8]

()Xi‘l W

MeO

(o]

Gamma

| Br

MeO

&_Q_om or

2 PdL;

"1

o

O I

(Hoyza—@—om

Because both pieces are aryl, there are two combinations of reagents that would work for a Suzuki coupling to make the biaryl product. Note that a borane or boric ester could also have been used.

STILLE COUPLING:

PROBLEM

24.5

What is the product of the following reaction?

L

0

L

; o

+ oTf

nBusSn

JI’

COEt

Pa(O),

o)

; 3

Fav

CO;Et

The triflate (—OTf) group is the leaving group under these conditions.

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SONOGASHIRA COUPLING: PROBLEM 24.6 What sequence of reactions will produce the following product if starting with trimethylsilylacetylene and the appropriate two aryl iodides?

0—

0o—

9 £y

o

Cul, NEts

=—SiMey

L@v 00 o

%%m

W

BuN F

0 —

Cul, NEts

I

a

o

-

O—=

o

0—

Sonogashira coupling conditions using trimethylsilyl acetylene give the substituted

alkyne after deprotection using fluoride. Another such coupling using the appropriate phenyl iodide reactant shown gives the product.

RING-CLOSING METATHESIS: PROBLEM 24.7 Show the product of the following reaction.

Bf, g

10

&

'

LU

1

J

_—

catalyst

)

>

4

Kfi

04

0

nucleophilic carbene

2

A)

U

i 4 ©

\

2+

3

L

H,C=CH,

0]

Numbers were added to help you keep track of the different atoms as the ring is

formed. The alkene metathesis reaction will generate an eleven-membered ring by forming a bond between carbon atoms labeled as 2 and 12. Carbons labeled as 1 and 13 will be found in the ethene product.

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Solution and Answer Guide: Brown et al., Organic Chemistry ® 2023, 9780357451861, Chapter 24: Catalytic Carbon-Carbon Bond Formation

Cu(l) CLick CHEMISTRY: PROBLEM 24.8 Show how the following compound of an alkyne and an aryl azide.

can be synthesized from starting materials consisting

If =N

OH

V4

HO\QN

OH

HO

N=N

N,

cu(l) +

=

OH

L OH 3\-/3_@‘

HO

EE——

Ascorbic acid OH OH

END-OF-CHAPTER

PROBLEMS

An asterisk (*) indicates an applied problem.

THE HECK

REACTION

Problem 24.9 As has been demonstrated in the text, when the starting alkene has CH: as its terminal group, the Heck reaction is highly stereoselective for formation of the £isomer. Show how the mechanism proposed in the text leads to this stereoselectivity. /\O v

+

f\©

GgHsl R

Pd(OAc)s,2 2 PheP —-l{l:;N i

L,PACHCl

I’h’/\\/

Ph

cl, PdL, CeHs

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In the second step of the mechanism, the Pd complex adds to the alkene via a syn addition from either face of the double bond. This leads to the creation of a mixture of intermediate enantiomers as shown below. l

l\

PdL, Cél'lfi

2

l\

PdL,

:

CeHjs

—_

PdL,

..IH

+

H H

CeHjs

.

H H

.H

©

In either case, the central bond of each intermediate enantiomer rotates to minimize torsional strain between the large phenyl groups and still allow for subsequent syn elimination. l\

CGHS

PdL,

it

Lak H

~

H

m—

HH

PdL, _ah H

-

CeHsH

+ l\

+ l'.

PdL,

6511531,

PdL,

ragga

H

H H

H

HCH

Syn elimination of the most stable palladium complex conformers (bulky phenyl groups held apart from each other) leads to the £ alkene in both cases, I\

/,- PdL,

H/ Csfis

QH \

H

! Ho

+ PdL,

®

N

Q

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¢ Cengage Solution and Answer Guide: Brown et al., Organic Chemistry & 2023, 978035T7451861; Chapter 24: Catalytic Carbon-Carbon Bond Formation

Problem 24.10 The following reaction involves two sequential intramolecular Heck reactions. Draw structural formulas for each organopalladium intermediate formed in the sequence and show how the final product is formed.

.

1 mol % Pd{OAc)s

4 mol % PhgP

CH4CN

CysHasl

o

CisHyy

q

4R l

1% mol Pd(OAc),

|N

R

x4

it

4% mol PhyP¥

A

=

s 'i:P o

T

{1/1T

CHCN

10

3 et

«”\si A AU

s

C sl 1L, Pd T 8

1)

13

l!_,“\_~ i

a

o/~7

. 1

s

I,Pd~ S—

- B

fl\

73

it

S0 ‘f’:.'\:n/ 9

13

1L-'

1

“’4 ~ 8

\F —-

3

:

SO \ - “in 7

(J!

"

Cysllyy Problem 24.11 Think-Pair-Share Answer the following questions as they pertain to the Heck reaction.

(a) Indicate which of the following are good solvents for the Heck reaction: DMSO, diethyl ether, tetrahydrofuran, ethanol, acetonitrile. DMSO and acetonitrile are good solvents because they are polar protic. Polar protic solvents have the polarity needed to dissolve the palladium acetate but do not have protons to interfere with the reaction conditions. Some Heck reactions can occur in

ethanol.

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Solution and Answer Guide: Brown et al, Organic Chemistry © 2023, 9780357451861, Chapter

24: Catalytic Carbon-Carbon

Bond Formation

(b) Which substrates work well in a Heck reaction, meaning they provide only one major product? Explain your reasoning for those that do not work well.

o

CHyCHoCH,CH, CH,CHy

an

=

CHy

» ("

Br

CHy M

/=/

HyC

)

)

CgHs Substrates (ii), (Iv), and (v) work well in the reaction and provide a single major product. Substrate (J) does not work well because the two substituents on the alkene (ethyl and butyl) are not substantially different in size to preferentially afford the £vs Z product

Substrate (i) does not work well because It contains a halogen that can act as a leaving group and lead to an elimination product under the Heck reaction conditions. The

resulting diene may not be selective enough In a Heck reaction.

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(c) Predict the product from the following Heck reaction. Explain the reasoning behind

the

regioselective carbon-carbon bond formation.

CH;

I i

+

Pd(OAC)s,Clo, 22 PhyPPhyP Q e

CHy

\

EtsN

The syn addition of the palladium complex to the alkene occurs with the palladium preferentially adding to the more substituted (sterically hindered) side of the double bond. Because of the long Pd—C bond, the palladium is sterically less demanding than

the organic group, therefore, it ends up on the more hindered carbon. This leads to the diene product shown. Problem 24.12 Rank the following substrates in terms of increasing reaction

rate in a Heck

reaction. Draw

the palladium-containing intermediate that results in each case after syn addition with vinyl iodide.

Least reactive

Most reactive

CsHz

cae

Corresponding

poladiem

intermediates

L;Pd’

1

“A CgHg

—

=

™




~

A

Py

05

R

o

"if“,

“J

e

500

Bottom lace

Fsc” %

J".:':

7

PO

; [s)

wgn

-

J

Reactive double hond

N

S

et

._J S

Because of the "top face” orientation, the syn addition step occurs primarily on the top face, leadingto the observed cis ring junction.

R

Lo

&

y, Uta-S, F C’ 2\

e

0

>

PA(OAC);, PPhs Ag,CO3, CH3CN

I

H

g

5 H

!

HE

Cix ring

junction

tot

face of double bond

.

£ H

PdL,OAc

Syn elimination of this complex gives the predominant product.

R

5

8 syn elimination

Only H atom capable / of syn elimination

H

A

——

Cg

£

=

H

H

H

H

PdL,0Ac

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Solution and Answer Guide: Brown et al., Organic Chemistry & 2023, 9780357451861, Chapter 24: Catalytic Carbon-Carbon Bond Formation

Problem 24.17 The aryl diene undergoes sequential Heck reactions to give a product with the molecular formula CisH:. Propose a structural formula for this product. 1

1%

mol Pd(OAc),

4% mol PhyP

1

—_—

O H

CH,CN

(CysH gl

1 a

5

9

6

1

3 4

2/

1% mol Pd(OAc).

PdL,l

4% mol PhsP ————

8

CH3CN

9

7

5 3

1

3 4

V

7

(CysHygl)

] 3 4

—_—

(o

+

-

"

—

PdL,| 8

s 9

1

1

PdL,l

a

—

s )

CysHie

"

(racemic)

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Solution and Answer Guide: Brown et al., Organic Chemistry & 2023, 978035T7451861; Chapter 24: Catalytic Carbon-Carbon Bond Formation

Problem 24.18 Heck reactions take place with alkynes as well as alkenes. The following conversion involves an intramolecular Heck reaction followed by an intermolecular Heck. Propose structural formulas for the palladium-containing intermediates involved in this reaction. |i

. . COOMe

—

Heck reaction

|

Me Me

\ Me OO

5

j 5

4

N

3

7

Heck reaction —_—

3

'™ e —_—

2

PdLzf

Me

(D

== [

a

N, Me PdL,l

_b

(D

\e,Me

s

(D

N, Me

i \ b)—PdL,l

MeOOC ¢

b MeOOC e

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Solution and Answer Guide: Brown et al., Organic Chemistry & 2023, 978035T7451861; Chapter 24: Catalytic Carbon-Carbon Bond Formation

Problem 24.19 The following conversion involves sequential Heck reactions. Propose structural formulas for the palladium-containing intermediates involved in this reaction. —

Heck reaction

: EiOOC

=

—

F100C

———————————

ey

F1OOC

|

EtOOC

SiMe, 4

6 Et00C—J8

7

EtOOC

3

5 9

2 1 10

i

A

1 4

~ EtOOC

4

EtOOC

EtO0C—|8

SiMes 1

PdLeI

—

EtOOC

n

EtOOC

EtOOC

9

2 1 10

PdLsl

7

6

8

5

9

11

—

SiMes

a

9

5 ‘

1

n SIM33

76

EtOOC

5

2

SiMe;

—

7

EtOOC

2

3

‘-.‘_

B

_

6

Heck —

3

1

o

n

SiMe;

PdLz'

R

——

al

.2 Cengage iw

Solution and Answer Guide: Brown et al., Organic Chemistry & 2023, 978035T7451861; Chapter 24: Catalytic Carbon-Carbon Bond Formation

Problem 24.20 The following transformation involves a series of four consecutive Heck reactions and the formation of the four-ring steroid nucleus as a racemic mixture. Propose structural formulas for the palladium-containing intermediates involved in this reaction. P a

Fi00C

S

EtOOC

»

Heck reaction

E1O0C

I

EtOOC

F1O0C

a8

4

75

Et00C—g

1

5. ~\2

|

3

Bk

Et0OC

g

—————r

Et0OOC—Y5

10

7

1

5 5~

PdL,|

5

10 12

1

14

13

18

15

16

EtOOC & § Ew00C a 2

1

3

PdL|

—

e

m

15

1t

15

18

1%

1

17

3

i

5

13

14

13

13

4

5

12

n

19

17 S

—

rl

PdL,| | _ Y,

BOOC 3 el Etooc @Q P

"

QL 17

g

1

4

s

EtO0C

EtOOC

8

3

s

7.8k

15

“PdL;l

J

18

17

EtOOC

——p

EtOOC

19

11

8

>

s

,5

!

W

Q

Y

s

-

PdLyl

Y 1

-

EtOQC

8

7

3

i

6,

&

18

19

1

1461

~¢ ¢ CelCengage al

Solution and Answer Guide: Brown et al., Organic Chemistry & 2023, 978035T7451861; Chapter 24: Catalytic Carbon-Carbon Bond Formation

Problem 24.21 Show the sequence of Heck reactions by which the following conversion takes place. Note from the molecular formula given under each structural formula that this conversion corresponds to a loss of H and | from the starting material.

w

1% mol Pd{OAc) 4%

M

mol Py

l;;\/‘\\ (G H:D

(CryHyg) 1 I

7

8

5

4

1% mol. Pd(DAc); 4% mol PhyP —

56

! qr"

fi

Q

Hé’ 32 \ (CyqHy7l)

(Cy4H15)

1

I

PdL;l 7

1%

5

8

mol.

4%

4

6

NmAC);

L2

T

mol PhiP

—

5

>

8

4

N

6§

—=

§

H (CrqHy71)

3

2

H

3

2

1

1

1

_ %, ‘

TN Q

g

TR

3 2

1

|

3 :

e

Pl

. "% ’ ‘-._‘ 8 flk:

(Crathe)

In the following sequence, note how the stereochemistry at the chiral center labeled as 4 controls the stereochemistry at the carbon labeled as 5. A common theme in modern

organic synthesis is that the presence of one chiral center is used to control the stereochemistry of chiral centers produced in later synthetic steps. Although a complete

explanation for the stereospecificity of these reactions can be quite complex, one contributing factor is likely to be the conformational preference of the starting material. Rotation about the carbon-4 carbon-5 bond yields two conformational isomers that would give rise to the two stereocisomer products as shown, The conformer on the right is the more stable, and this leads to the observed predominant product sterecisomer.

1462

=»l Pl

¢ Cengage [

Solution and Answer Guide: Brown et al., Organic Chemistry ® 2023, 9780357451861, Chapter 24: Catalytic Carbon-Carbon Bond Formation

Less stable conlormer

Meore stahle conformer

¥

(@]

_

:

(I?

.

i

I

.

PdL;|

®

:

"

5

PdL,l

rotation

g

o

A

H

a3

H

s

o

6 C ®)

~

L

&

A ¢

58

245 O

37,

1

1

Ao

e

More stable tin-formatio-

Less stable Thu-M0.

7

TdLzi

v g&;

s

7

3

&

GO1S

OO

0

C4-CS bond

[

Y

$2.

=

(D

6

— Z

PdL,|

3

Predominant product 1

>

—

stereoisomer

CATALYTIC ALLYLIC ALKYLATION Problem 24.22 Draw the products of the following reactions.

O

Pd(PPhy),

I

o)

(a)

: o

O (b)

O

o

A e

Pd(PPhy),

&

H

°

O

H

-

CN

1463

«l

¢ Cengage '

|

Solution and Answer Guide: Brown et al., Organic Chemistry & 2023, 978035T7451861; Chapter 24: Catalytic Carbon-Carbon Bond Formation

Cl

PdCly

0O

X ()

NOo

H

W

NO,

Pd(PPhy)y

/\—\_/5|:.Bu3

+

/@

—

or

>—\_/l

1

™ W

+

—

Q B’Il;Sn

—

(b) Possible Suzuki Cross-Coupling Reactions BY:

I

—

x

or

+

x

NS

+ CH3 BY;

or

/=
:

TP —r CHyCly, 40°C, 30 min

(b)

ti] TBS

f\!/\" (c)

(8)

CgHs,

o ARG T CHyCly, 40°C, 3) min

+

7N\

CHoCly, 40°C, 30 min

S mole % Ru catalyst

1

AH— ~ /

—

0A

-

=t

O

5 mole % Ru catalyst (8]

on

CHyCls, 40°C, 30 min

5 mole % Ru catalyst {self-reaction )

(d)

—

HO'

p [-

o

e

= _2_\

Ha=C] e

H:

-~0\[/C6H.'-

,k,o

+ CH:=CH:

mo_/_

O‘ +

Cl

c

'H:=CH;

1479

al

¢ Cengage Solution and Answer Guide: Brown et al., Organic Chemistry & 2023, 978035T7451861; Chapter 24: Catalytic Carbon-Carbon Bond Formation

Problem 24.40 The following transformation can be accomplished by reacticns we have studied in this chapter and Chapter 20. Name the type of reaction used in each step.

MeOOC m MeOO(C

:

—

(1) MeOQOC

:

MeOOC

MeOOC

MeQOC (2 3

o

MeQOC

A

MeooC”

(3 4

5

i

\1o 1112

2

3

1

Heck reaction R ——

(0 g

Intramolecnla Diels-Alder —,-

B g

Me0QC

7

[

8 7

MeQOC

A

G

Me0OC

U 55

2

=

)

4

2 .

A

3

|

Me0OC

‘

1

(3)

This molecule is made from an intramolecular Heck reaction followed by an intramolecular Diels-Alder reaction. The Heck reaction creates intermediate (2) and makes a bond between the carbons labeled as 7 and 11.

MeOOC

mecoc”

8

N 10

6

¥

!

=

11

5

.

1

2

1

8

Me0OC

Heck rencion _ MeOOC”

12

MeOOC

Me0OC

8.7

a

¥

N\

—_—

4

.

12

-

a“

5

02!

10

6

[

1

=

e

.

3

2

1

.

12

1

-

1480

al

¢ Cengage Solution and Answer Guide: Brown et al., Organic Chemistry & 2023, 978035T7451861; Chapter 24: Catalytic Carbon-Carbon Bond Formation

The Diels-Alder reaction takes place between the diene and terminal alkene so that new

bonds are made between the carbons labeled as 6 and 2 as well as between 12 and 1. &

5

6

4

2

-)

MeQOC

MeOOC

£V

2, 2

e

4

5 =

i

a

@

:]

MeOOC

i 6

Za

(;

3

A|

MeOOC

l Intramolecnlar Diels-Alder

;

MeOOC

4

g

5

MeOOC

1

Problem 24.41 Compound A below represents a structural intermediate in the synthesis of ingenol. Ingenol has interesting biological activity and represents the core structure for a topical skin medication used to treat precancerous skin lesions. What reaction outlined in this

chapter could be used to form the seven-membered ring containing the unsaturated ketone in Compound A? What was the starting material? Note that the PMB group represents p-methoxybenzyl and is an alcohol protecting group.

(e

=

(8]

S

OPMB

HO HO HO

Compound A

OH Ingenol

PMB = p-methoxybenasyl

Alkene metathesis

[3

e

[03“.

OPMB

of

-0

4

5 mole % Ru catalyst

.v oa,

(8]

‘E

H

OPMB

1481

+¢ Cengage i

Solution and Answer Guide: Brown et al., Organic Chemistry & 2023, 978035T7451861; Chapter 24: Catalytic Carbon-Carbon Bond Formation

CLick CHEMISTRY *Problem 24.42 The following two structures have been synthesized for testing as an antibiotic and an antiviral, respectively. Given that the structures contain triazole rings, show precursors to both products as well as the proper metal and conditions for their generation. O

Xo 0\\8//0

8]

Y CO,H

b J?

X

HO

N,

.N

N

W

/ N N

OH

Tazobactam

an antiviral

O\\ pe S

0\\ 0 S

Cu(l) = ;

8]

N

=

— Ascorbic acid

N;

i

CO-H

(8]

o m N. .N

i

N

CO-H

ethvne (acetylene)

Tazobactam )

H,N ?N

”

HO 0

3

(0]

"

H,N

-

E uf 1 j

.

HO

W

/

Ascorbic acid

(]

= N

5

OH OH

an antiviral

1482

~¢Cengage «l

Solution and Answer Guide: Brown et al., Organic Chemistry ©® 2023, 9780357451861; Chapter 24: Catalytic Carbon-Carbon Bond Formation

SYNTHESIS *Problem 24.43 Following is an outline of the stereospecific synthesis of the “Corey lactone.” Professor E. J. Corey (Harvard University) describes it this way. “The first general synthetic route to all

the known prostaglandins was developed by way of bicycloheptene intermediates. The design was guided by the requirements that the route be versatile enough to allow the synthesis of many analogs and also allow early resolution. This synthesis has been used on a large scale and in laboratories throughout the world; it has been applied to the production of countless prostaglandin analogs.” Corey was awarded the 1990 Nobel Prize in Chemistry for the development of retrosynthetic analysis for synthetic production of complex molecules. See E. J. Corey and Xue-Min Cheng, The Logic of Chemical Synthesis, John Wiley & Sons, New York, 1989, p. 255. Note: The wavy lines in compound C indicate that the stereochemistry of —Cl and —CN groups was not determined. [The conversion of (D) to (E) involves an oxidation of the ketone group to a lactone by the Baeyer-Villiger reaction, which we have not studied in this text.]

9

OCH, 1. Nabl.

HyCO

:
&

mme

Figaut oy

™y A~~~ 8|

—

o

(I;

o

Ccon

o

0o

0=

0=

——

(1’

]

J

N, Se

CrL on om

o

on

Step 4: Termination. The reaction of the polymer radical with another polymer radical leads to the final polymer. Nac

v

S0

OH

OH

OH

e

VNN OH

OH

OH

Nac

"

e

220NN

OH

OH

QC

EH

OH

OH

(c) Formation of materials using this process allows for simple addition of additives. For acrylic nails, additives include UV absorbers, crosslinking agents, pigments, and an

inhibitor in the monomer solution. i.

What value is gained by adding crosslinking agents?

Crosslinking agents increase the durability and strength of polymers by linking polymer chains to one another. ii.

Why does the manufacturer add a radical inhibitor to the monomer solution?

An inhibitor is needed in the monomer agent to ensure polymerization does not occur

before it is intended. This increases the shelf life of the material prior to use.

8

~¢ 4 Cengage Solution and Answer Guide: Brown et al., Organic Chemistry ® 2023, 9780357451861; Chapter 26: Organic Polymer Chemistry

Problem 26.34 The dental industry commonly uses polymers in corrective applications. Like in the acrylic

nail process, new fillings are now made with a mixture of polymerizable monomer and particles that act as a filler material. Historically, silicon dioxide has been used as an inorganic particle additive in dental fillings (about 50% by volume), although nanomaterials are increasingly being used due to their enhanced strength and polishability. It is important for the new material to blend in appearance with natural teeth, be durable, and have minimal material shrinkage after the reaction (curing) is complete. (a) Bis-GMA has been commonly used as the monomer in filling material for dental applications. Draw the linear polymer formed from a chain-growth polymerization of this monomer.

%()fi/\”

(‘Y(’)kr

Bis-GMA (Bisphenol A-glycidylmethacrylate)

ix o

LT Y

0

U

CY

I

OJ\(

(b) The term “bis” relates to two identical, but separate complex groups in one molecule.

How does the “bis” nature of the monomer bring about crosslinking in the polymer? Likewise, how does crosslinking affect the durability and strength of the polymer? The bis-GMA has two alkene groups, which allows for a significant amount of crosslinking in the polymer. This increases the durability and strength of the polymer,

which is important in a dental application. Problem 26.35 Low-density polyethylene (LDPE) has a higher degree of chain branching than high-density polyethylene (HDPE). Explain the relationship between chain branching and density. Unbranched polyethylene packs more efficiently into compact structures which have

more mass per unit volume than structures formed by packing of branched polyethylene chains. Therefore, unbranched polyethylene has a higher density than branched-chain polyethylene.

1582

8

~¢ 4 Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 26: Organic Polymer Chemistry Problem 26.36 We saw how intramolecular chain transfer in radical polymerization of ethylene creates a four-carbon branch on a polyethylene chain. What branch is created by a comparable intramolecular chain transfer during radical polymerization of styrene?

&

T Ph

~

5 Phs

T

S

4

Ph

Ph

. Ph 8 Ph

S

nPhCH=CH,

Ph

A six-membered transition

1

. "Emu-f" to Dhkpieopes Svetrnetion

This four-carbon branch is created

b

Ph

N

s

M

“my

Ph

Ph

Problem 26.37 Compare the densities of low-density polyethylene (LDPE) and high-density polyethylene (HDPE) with the densities of the liquid alkanes listed in Table 2.5. How might you account

for the differences between them? Given in the table are densities of several liquid alkanes reported in Table 2.5, plus

densities for pentadecane, eicosane, and tricosane. As you can see, densities for these unbranched alkanes reach a maximum in the range 0.77-0.79 g/mL, which is significantly less than the density of both LDPE and HDPE. From this data, we conclude that both

LDPE and HDPE pack more efficiently (have greater mass per unit volume) than their lower molecular weight counterparts. Alkane

Formula

Density (g/mL)

Pentane

CsHi

0.626

Heptane

CiHw

0.684

Decane

CioH2z2

0.730

Pentadecane

CisHaz

0.769

Eicosane

CaoHaz

0.789

Tricosane

CaoHez

0.779

LDPE

{CH,},

0.91-0.94

HDPE

{CH,},

0.96

1583

8

~¢ 4 Cengage Solution and Answer Guide: Brown et al., Organic Chemistry ® 2023, 9780357451861; Chapter 26: Organic Polymer Chemistry *Problem 26.38 Natural rubber is the all-c¢/is polymer of 2-methyl-1,3-butadiene (isoprene).

Poly(2-methyl-1,3-butadiene)

(Polyisoprene)

(a) Draw a structural formula for the repeat unit of natural rubber.

(b) Draw a structural formula of the product of oxidation of natural rubber by ozone followed by a workup in the presence of (CH,),S. Name each functional group present in this product.

(o]

Aldehyde

O

H

Ketone

4-Oxopentanal (c) The smog prevalent in many major metropolitan areas contains oxidizing agents, including ozone. Account for the fact that this type of smog attacks natural rubber (automobile tires and the like) but does not attack polyethylene or polyvinyl chloride.

Polyethylene and poly(vinyl chloride) do not contain carbon-carbon double bonds, which are susceptible to attack by oxidizing agents such as ozone. (d) Account for the fact that natural rubber is an elastomer but the synthetic all-trans isomer is not. The natural cis isomer is kinked by virtue of the cis bond geometry while the all trans

synthetic rubber has a more uniform staggered polymer chain. The trans synthetic rubber chains can thus pack together better making it more rigid compared to natural rubber.

1584

8

~¢ Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451867; Chapter 26: Organic Polymer Chemistry Problem 26.39 Radical polymerization of styrene gives a linear polymer. Radical polymerization of a mixture of styrene and 1,4-divinylbenzene gives a cross-linked network polymer of the type shown in Figure 26.1. Show by drawing structural formulas how incorporation of a few percent of 1,4-divinylbenzene in the polymerization mixture gives a cross-linked polymer.

N

X +

—

acopolymer of styrene and divinylbenzene

[

~

Styrene

1,4-Divinylbenzene

Drawn here is a section of the copolymer showing crosslinking by one molecule of 1,4-divinylbenzene. Benzene rings derived from, PhCH=CH,, are shown as Ph. From the carbon-carbon double bonds of 1 4-divinylbenzene

*

*

Ph

*

Ph

Ph

Ph *

W |

h

Ph

Ph

Ph

Ph

*

*

*

Ph

n

A copolymer of styrene and 1 4-divinylbenzene

Ph

Ph

n

8

~¢ Cengage Solution and Answer Guide: Brown et al., Organic Chemistry ® 2023, 9780357451861; Chapter 26: Organic Polymer Chemistry Problem 26.40 lon exchange resins are insoluble small beads or powders created from linear or crosslinked polymers that possess like charges along their backbone or side groups. The charges must be paired with a counterion. For example, negative charges along the polymer are paired with cations. Because the counterions are free (not attached to the polymer), they can migrate within the otherwise insoluble polymeric structure and thereby exchange with different like-charged counterions. The charge of the counterion gives rise to the name of the exchange resin. For instance, those resins with cationic counterions are called cation exchange resins. One common cation exchange resin is prepared by polymerization of a mixture containing styrene and 1,4-divinylbenzene (Problem 26.39). The polymer is then treated with concentrated sulfuric acid to sulfonate a majority of the

aromatic rings in the polymer. (a) Show the product of sulfonation of each benzene ring.

The following is a structural formula for a section of the polymer. Structural formulas for only the sulfonated rings are written in full; unsulfonated benzene rings are shown

as Ph. *

*

*

*

AL

*

A n

OsH

0sH

0sH

0sH

(b) After sulfonation, the resin can be treated with sodium acetate and then washed with water to generate a cation exchange resin. What is the role of the sodium acetate, and what counterion now exists within the resin? The role of the sodium acetate is to displace protons in the resin and replace them with

sodium counterions. The resin will then contain sodium sulfonate salts instead of sulfonic acid groups. (c) Explain how this sulfonated polymer can act as a cation exchange resin.

The resin is shown in the acid or protonated form. When functioning as a cation exchange resin, cations displace H* and become bound to the negatively charged

—S0,™ groups.

1586

8

~¢ Cengage Solution and Answer Guide: Brown et al., Organic Chemistry ® 2023, 9780357451861; Chapter 26: Organic Polymer Chemistry Problem 26.41 The most widely used synthetic rubber is a copolymer of styrene and butadiene called SB rubber. Ratios of butadiene to styrene used in polymerization vary, depending on the end use of the polymer. The ratio used most commonly in the preparation of SB rubber for use in automobile tires is one mole styrene to three moles butadiene. Draw a structural formula of a section of the polymer formed from this ratio of reactants. Assume that all carbon-carbon double bonds in the polymer chain are in the cis configuration. derived from 1,3-butadiene

derived from

\

9

Problem 26.42 From what two monomer units is the following polymer made?

W‘— The section of polymer drawn here is derived from two 2,3-dimethyl-1,3-butadiene monomer units and three methoxyethene monomer units

A

OMe

A

OMe

A

OMe

Problem 26.43 Draw the structure of the polymer formed from ring-opening metathesis polymerization (ROMP) of each monomer.

(@ m

®) j

:

f*/_/_\—\_%.

{WW

1587

«l

» 2 Cengage ie

Solution and Answer

Guide: Brown

et al., Ovganlc

Chemistry

© 2023, 9780357451861;

Chapter 26: Organic Polymer Chemistry

©

M

(d)

1588

~¢ Cengage 8

Solution and Answer

Guide: Brown

et al., Organic

Chemistry

© 2023, 9780357451861;

Chapter 27: Lipids

Solution and Answer Guide BROWN ET AL., ORGANIC CHEMISTRY © 2023, 9780357451861; CHAPTER 27: LIPIDS

TABLE OF CONTENTS In=Chapter Problems.iiuiiiiimiiss i 1600 Triglyceride Structure: Problem 27.1........cciicisssnssasisssssassssssesssssessasssassssssasassssssssassassess 1991

MCAT Practice Questions: Vitamin K, Blood Clotting, and BasiCity........umsmsnnnens 1591 End=ofsCHuDEor

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8

~¢ Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 27: Lipids PIOBIBITE

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2728 sisiisessisusuasisns ottt

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8

~¢ 4 Cengage Solution and Answer Guide: Brown et al., Organic Chemistry © 2023, 9780357451861; Chapter 27: Lipids

IN-CHAPTER

PROBLEMS

TRIGLYCERIDE STRUCTURE:

PROBLEM

27.1

(a) How many constitutional isomers are possible for a triglyceride containing one molecule

each of palmitic acid, oleic acid, and stearic acid?

There are three constitutional isomers possible, the difference being which fatty acid is in the middle of the molecule:

i

_O—C-paimitic

HoC

o

W| ~O—C-stearic n

HoC

O—fi-oleic (