Ch29 Brown Organic Chemistry Solutions

8/10/2019 Ch29 Brown Organic Chemistry Solutions

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Chapter 29: Organic Polymer Chemistry

Solutions

CHAPTER 29

Solutions

t

the Problems

Problem 29.1 Given the following structure, detennine the polymer's repeat unit, redraw the structure using the simplified

parenthetical notation, and name the polymer.

~ J J ~

. c0f' c0f

Polymer

.

1

This polymer

is

derived from propylene oxide

ization

~ o t

and, therefore, named poly(propylene oxide)


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778

Solutions

Chapter29:OrganicPolymerChemistry

o

(h)

Poly(hexamethylenedecanediamide)

Problem29.6

Drawthestructure(s) ofthemonomer(s)usedtomakeeachpolymerinProblem29.5.

(a)

(b)

~

(c)

~

(d)

CF

3

-CF-CF

2

(

0.............

y

0

(e)

(g)

:::7

(OH

(0

~

OH

+

1

.HO

OH

&

. CH

2

CI

o

b H O ~ O H

+

o

Problem29.7'Drawthestructure ofthepolymer f o r ~ e d in thefollowingreactions.

o

0

+ ~

Me

eO

(a)

.,

OH

O

o

o

OMe

+

H O ~ O H

{ ~ - < : : >

~ o ~ o l

n

MeO

C

;/

Note:theremayalso be

(b)

cross-linking

KOH

(d) [ O

Problem29.8

At

onetime,arawmaterialforthe production

of

hexamethylenediaminewasthepentose-based .

polysaccharides

of

agriculturalwastes,suchasoathulls. Treatmen tofthesewasteswith

s ~ l u r i

acidorhydrochloricacid

givesfurfural. Decarbonylationoffurfuraloverazjnc-chr omium-molybdenu mcatalyst gIvesfuran. Proposereagentsand

experimentalconditionsfortheconversion offurantohexamethylenediamine.

Zn-Cr-Mo

oat hul ls , co rn H SO

catalyst

CD

CD

4

CI(CH

2

)4

C1

cobs,sugarcane - - ~ . ~

Q

Q

talks,etc H

2

0

1,4Dichloro

Furfural

Furan

Tetrahydrofuran

butane

f \ . (THF)

-

D

.

N=C(CH2)4C

=N

-

H2N(CH2)6NH2

H exaned in it ri le 1 ,6 -H exaned ia mi ne

(Adiponit ri Je) (Hexamethylenediamine)

Chapter29.: OrganicPolymerChemistry Solutions

Step1: Catalytichydrogenationusing

HZ

over

a

transition

metal

catalyst.

StepZ CleavageoftheetherusingconcentratedHCI

at

elevatedtemperature.

Step3: TreatmentofthedihalidewithNaCN,byanSNZ pathway.

Step

4:

CatalytichydrogenationofthecyanogroupsusingHZ over

a

transitionmetalcatalyst.

Problem299 Anotherrawmaterialfortheproductionofhexamethylenediamine is butadienederivedfromth

catalyticcracking

of

petroleum. Proposereagentsandexperi mentalconditionsfortheconversion

of

butadiene

hexamethylenediamine. .

CD CD

C H = C H C H = C H ~ C1CH

2

CH=CHCH

2

CI [' CCH

2

CH=CHCH

2

C =N

Butadiene

1,4-DicWoro-2-putene 3-Hexenedinitrile

H

2

N( a-l2)6NH2

1,6-Hexanediamine

(Hexameth

ylenediamine).

Step

1:

1,4- AdditionofCl

2

totheconjugateddiene.

StepZ Treatment

of

thedihalidewithNaCN"byan

SNZ

pathway.

Step3: Catalytichydrogenationofthecyanogroupsandthecarbon-carbondoublebondusing

Hz

t ransi tionmetal catalys t. .

Problem29.iO Proposereagentsandexperimentalconditionsfortheconversion ofbutadienetoadipicacid.

H O O C ~ C O O H

Hexanedioicacid

1,3Butadiene

(Adipicacid)

SeeProblem

29.9

fortheconversionofbutadieneto3hexenedinitrile. (3) Catalytichydrogenation

carboncarbon

doublebond

in

3-hexenedinitrilefollowed

by (4)

hydrolysis

of

thecyanogroupsina

acidgivesadipic

add.

(

CH

2

=CHCH=CH

2

J.l.L.. CICH2CH=CHCH2C' -..m... N=CCH

2

CH=CHCH

2

C=N

utadiene 1,4DichloroZbutene

3-Hexenedinitrile

N=C(CH

2

)4C'=N ill HOOC(CH2)4COOH

H ex an edi ni tr ile He.xanedioic ac id

(Adiponitrile) .(Adipicacid)

Problem29.

I J

Polymerization

of

2-chloro-l,3-butadieneunderZiegler-Nattaconditionsgivesasyntheticelast

neoprene. Allcarbon-carbondoublebonds inthepolymerchain havethetr ns configuration. Drawtherepe

neoprene.

Problem29.

12

Poly(ethyleneterephthalate)(PET)can

be

preparedby thisreaction. Proposeamechanismfo

growthreactionin thispolymerization. .

nCH,ogO-b

3

+

n HOCH CH OH

g O - b H 2 C H ~ +

n C

Dimethylterephthalate Ethyleneglycol

Poly(ethyleneterephthalate)

Metha

Proposeadditionofahydroxylgrouptoacarbonylcarbonofdimethylterephthalatetoformatetr

carbonyladditionintermediate,followedbyitscollapsetogiveanesterbondofthepolymerplusm

Thisisanexampleoftransesterification.


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780

Solutions


Step :

Step 2:

:

0:

: 0: Proton

0'

I.:

transfer

CH

3

0C

COCH

3

.

.. 1

. -

~ ~ C H C H ~ H

H

Step

:

C H 3 ~ : ~

+ /H

.

:0

'

f

\\ II..

. . - - '

_

C ~ C H C H ? H

0

Problem 29.13 Identify the monomers required for the synthesis

of

these step-growth polymers.

o

0

"

H O C ~ _

COH

, ) + [ - { - t ~ C H 2 ~ C H 2 o t

'+

'\

Kodel

(a polyester)

HOCH2-G-CH20H

u J \ .

b) ,C(CH

2

6

C

N ~ Q i 2 ~ N H 7 ; i

+

Quiana

(a polyamide)

H2NVCH2-Q-NH2


Solutions

Problem 29.14 Nomex, another aromatic polyamide (compare ararnid) is prepared by polymerization of 1,3

benzenediamine and the diacid chloride of I ,3-benzenedicarb{)xylic acid. The physical properties of the polym

'suitable for high strength, high temperature applications such as parachute cords and jet aircraft tires. Draw a

formula for the repeating unit

of

Nomex.

polymerization

- - - - - - - - Nomex

1,3-Benzenediamine

1,3-Benzene

dicarbonyl chloride

Following is the repeat unit in Nomex.

H

I

. A

(NyyN

V 0

0

Problem 29.15 Capro lactam, the monomer from which nylon

6

is synthesized, is prepared from cyclohexanon

steps. In step I, cyclohexanone is treated with hydroxylamine to form cyclohexanone oxime. Treatment of the

concentrated sulfuric acid in Step 2 gives caprolactam by a reaction called a Beckmann rearrangement. Propos

mechanism for the conversion

of

cyclohexanone oxime to caproJactam. .

6 6

N 0H

OH

H

S0

4

Cyclohexanone

Cyclohexanone

Caprolactam

oxime

The mechanism is shown divided into six steps.

Step 1:

Proton transfer from H

3

0 to the oxygen atom of the oxime generates

an

oxonium ion, wh

conv(!rts OH, a poor leaving group, into OH

2

,

a bette r leaving group.

H

1+

o :

~ : H + H o H

Step

2: Migration

of

the.electronpair

of

an adjacent carbon-carbon bond to nitrogen accompanie

of H

2

0 . .

H

b+

6'H_


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Solutions


82

Step

3: Reactiollof thecarbocationfroinstep2withH

2

0 togivean oxoniumion.

H

\ ..

~

~ o - H

V '

Step

4:

Proton

transfertosolventgivestheenol of anamide.

H .

..

c r - H ~

).j==(

\

.. +

H-


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Solutions


(a) Draw a.structural formula for the repeat unit of this polyester

~

O

(b) Account for the regioselective reaction with the primary

y r ~ x y l

groups only.

The regioselectivity reflects the fact

that

the 1

0

hydroxyl groups

are

more accessible to reaction than

th r

hydroxyl group. .

Problem 29.20 The polyester from Problem 29. 19 be mixed with additional phthalic anhydride (0.5 mol of phthalic

anhydride for each mole of 1,2,3-propanetriol in the original polyester) to form a liquid resin. When this resin is heated,. t

forms a hard, insoluble, thermosetting polyester called glYptal.

(a) Propose a structure for the repeat unit in glyptal.

The polymer described in Problem 29.19 becomes cross linked as shown.

. .

This unit from a phthalic

anhydride monomer

is

the cross-linking unit.

(b) Account for the fact that glyptal is a themJOsetting plastic.

Because

of

the extensive cross linking, the individual polymer chains can no longer be made to flow and,

therefore, the polymer cannot be made to assume a liquid state.

Problem 29.21 Propose a mechanisrri for the f o ~ a t i O n

of

the following polymer.

base

+1.( 1.

U

One way

tl}

attack this problem

is

tQ first determine which rings in the product

are

present

in

the original

monomers,

and

which

are

formed during the polymerization. The monomer units

are

redrawn here to show

that

new rings

are

formed during polymerization by aldol reactions followed by dehydration.

and

by imine

formation.

hapter 29: Organic Polymer Chemistry Solutions

aldol reaction

here followed

by dehydration

~

o

0

~ ~

imine formation here

These rings formed b) the

combination of aldoVdehydrat

imine formation

Problem 29.22 Draw the structural formula of the polymer resulting from base-catalyzed polymerization of e

compound. Would you expect the polymers to be optically active? (S)-(+)-Iactide is the dilactone formed fro

molecules of (S)-(+)-Iactic acid.

o

/1

_.

H

FH J

.

H J ~ H

.

xYcJ

,

(bl A - ~ c r j

*

H J H 0 H

a

each cbiral cente r has the S (R)-Propylene oxide each dural center has

o

configuration; the polymer configuration; the po

(8)-(+)-Iactide

oS

optically active

is

optically active.

Problem 29.23 Poly(3-hydroxybutanoic acid), a biodegradable polyester,. is an insoluble, opaque material tha

proces s into shapes. In contrast, the


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Solutions


Problem 29.24 How might you determine experimentally if a particular polymerization is propagating by a step-growth or

a chain-growth mechanism?

.

nalyze the distribution of polymer molecular weights as a function

of

degree

of

polymerization. As discussed

in

the introduCtion to Section 29.5, high molecular weight polymers are

not produced

until very late in step

growth polymerization, typically past

99%

conversion of monomers to polymers. Given the mechanism of

chain-growth polymerization, high molecular weight polymer molecules are produced very early and

continuously in the polymerization process. .

Problem 29.25 Draw a structural formula for the polymer formed

in

the following reactions.

Li

~

IBN

a)

(b)

CN

~

N

Problem 29.26 Selett

the

monomer in each pair that

is

more reactive toward cationic polymerization.

The more reactive monomer in each pair is the one forming the more stable car-bocation. The first structure in

each

pair

forms the more stable carbocation.

(b)

\ or \

or

OCH

3

O ~ H

o

--:\

; /

OCH

3

This structure makes little contribution

to the hybrid because


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Solutions

788

.

.

. . .radical ol m er izati on

of

ethylenecreatesafour-car?on

'Problem29.31 Wesawhowmtramolecularcham

t r a n s f ~

m P

intramolecularchaintransferdunng.radtcal

branchonapolyethylen" chain. Whatbranch

tS

created yacompara ,

polymerization

of

styrene.

.

Ph

3P h 2

n PhCH=CH:

__

H *

Ph

*

1

Ph Ph

Ph

Ph

Asix-memberedtransition

stateleadingto

1,5-hydrogenabstraction

ThiS four-carbon

branch

is

created Ph

.

. '

.

. . I (LD PE )andhigh -densi ty po lyethy lene (HD PE )w ith

Problem29.32 Comparethedensities

o.f

l o w d e ~ s ~ t y ~ O l y e ~ y g ~ t ; o u

accountforthedifferencesbetweenthem?

thedensities

of

theliquidalkanesltstedmTable

. OW mt.

. . .

. . . t d Table

2.5

plusdensitiesforpentadecane,

Givenin tablearedensities

of

e v e r a ~ h q ~ ~ ? a } k a ~ ~ ~ : : ~ ~ ~ : a n l ; h e d a l k a n ~ s reacha m a x i m ~ intherange

eicosane andtricosane. As youcansee, ensiles or ofbothLDPEandHDPE. FromthiS data,

.

0.77.0.79g/rnL,which is significantlylessthan t h e f ~ e ~ ~ ~ i y (havegreatermassperunit volume)thantheir

concluddhat bothLDPEandHDPEpackmoreeICI

.

lowermolecularweightc o u n t e r p a r t s ~ ;. .

__

..

i r l : . ~ r v _

Density

Fornlllla (g1mL)

Alkane

CsH

12

0.626

Pentane

C,H

16

0.684

Heptane

C a

H

22

0:730

Decane

C

sH

32

0;769

Pentadecane

Eicosane

Czo

H

42

0.789

Tricosane

C:3o

H

62

0.779

LDPE

+a-t

2

1n

0.91 0.94

HOPE

-{-a-t

2

-+nO.96

Problem29.33Naturalrubberistheall

is

polymerof2-methyl-1,3-butadiene(isoprene).

Poly(2-methyl.l,3-butadiene)

(Polyisoprene).

C )

Ornw' ruetu,,1

frumol' foe th '

p ,'nn;'

nf ; )1

'";') ,

. tru

t

al

formula

of

he product

of

oxidation

of

naturalrubberbyozone followedbyaworkup

in

thepresence

(b)

~ ~ a ~ ~ ~ ; h ~

u ~ a m e eachfunctionalgrouppresent in thisproduct. ,

o

A l d h Y : ~ O Koto",

4-0xopentanal

Chapter29:OrganicPolymerChemistry Solutions

(c) Thesmogprevalentin manymajorm etropolitanareascontainsoxidizingagents,includingozone. A,ccou

thatthistype

of

smogattacksnaturalrubber(automobiletiresandthelike)butdoesnotattackpolyethyle

polyvinylchloride.

Polyethyleneandpoly(vinylchloride)donotcontaincarbon-carbondoublebonds,whichare susc

attack by oxidizingagentssuchasozone.

(d) Accountforthefactthatnaturalrub beris an elastOmerbutthesyntheticall trans isomeris not.

Thenaturalcisisomer

is

kinkedbyvirtueofthe

cis

bondgeometrywhiletheall

trans

syntheticru

moreuniformstaggeredpolymerchain. The

tra ns

syntheticrubber chainscanthuspack ogether

makingitmorerigidcomparedtonaturalrubber.

Problem29.34 Radicalpolymerization

of

styrenegivesalinearpolymer. Radicalpolymerization

of

amixtu

and1,4-di vinylbenzeneg ivesacrosslinkednetwork polymer

of

thetypeshowninFigure29.1. Showbydra

structuralformulashowincorporation

of

afewpercentI ,4-divinylbenzeneinthepolymerizationmixture give

linkedpolymer.

- - - i ~

.

acopolymerof styreneanddivinylbenzene

+

Styrene

1,4-Divinylbenzene

Drawnhere

is

asectionofthecopolymershowingcrosslinkingbyonemoleculeof 1,4-divinylbenz

Benzeneringsderivedfrom,PhCH=CH

z

,areshownasPh,

Fromthecarbon-carbondouble

bonds

of

1,4-divinylbenzene

n

Acopolymerof styreneand1,4-divinylbenzene

Problem29.35Onecommontypeof cationexchangeresin is preparedbypolymerization of amixtureconta

and1,4-divinylbenzene(Problem29.34). Thepolymer

is

thentreatedwithconcentratedsulfuricacidtosulfo

majority

of

thearomaticringsinthepolymer.

(a) Showtheproduct

of

sulfonation

of

eachbenzenering.

Thefollowingis astructural formulaforasectionof thepolymer. Structura lformulasforonlyth

ringsare writteninfull;unsulfonatedbenzeneringsare shownasPh.

SOJH


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Solutions


90

(b) Explain how this sulfonated polymer can act asa 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 -803- groups. . .

Problem 29.36

The

most widely used synthetic rubber is a copolymer.of styrene and butadiene called

58

rubber. Ratios

of

butadiene to styrene used in polymerization vary depending on the end use

of

the polymer. Th e ratio used most

commonly in the preparation

of 58

rubber for use in automobile tires is I mole styrene to 3 moles butadiene, Draw a

structural formula

of

a section

of

the polymer formed

fwm

this ratio

of

reactants, Assume that all carbon-carbon double

bonds

in

the polymer chain are in the

is

configuration, .

. derived from 1,3-butadiene

d,dvedf om

tyrene,.

- - - - .

h

*

.

, .

-

-

-

Problem ?9,37 From what two monomer units is the following polymer made?

c=N

c =N

The section of polymer drawn here is derived six 1,3-butadiene monomer units and two acrylonitrile monomer

units.

, C=N

. Problem 29.38 Draw the structure

of

the polymer formed from ring-opening metathesis polymerization (ROMP)

of

each

monomer.

b )

a ~

. H H n

e)

d) .

n

. -M--t

Ch29 Brown Organic Chemistry Solutions

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