organic chemistry

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3 of 10 12/23/2011 4:14 AM

Review of Concepts

Chapter 24 Carbohydrates

Fill in the blanks below. To verify that your answers are correct, look in your textbook at the end of Chapter 24. Each of the sentences below appears verbatim in the section entitled Review of Concepts and Vocabulary.

• Carbohydrates are polyhydroxy _____ or ketones. • Simple sugars are called and are generally classified as

aldoses and ___ _

• For all D sugars, the chirality center farthest from the carbonyl group has the _ configuration.

• Aldohexoses can form cyclic hemi ___ that exhibit a pyranose ring. • Cyclization produces two stereoisomeric hemiacetals. called . The

newly created chirality center is called the carbon. • In the a anomer , the hydroxyl group at the anomeric position is ___ to the

CH20H group. while in the p anomer , the hydroxyl group is ___ to the CH20H group.

• Anomers equilibrate by a process called , which is catalyzed by either or __ _

• Some carbohydrates. such as D-fructose, can also form five-membered rings, called rings.

• Monosaccharides are convened into their ester derivatives when treated with excess _____________ _

• Monosaccharides are convened into their ether derivatives when treated with excess and silver ox ide.

• When treated with an alcohol under acid-catalyzed conditions, monosaccharides are converted into acetals, called . Both anomers are formed.

• Upon treatment with sodlum borohydride an aldose or ketose can be reduced to yield an __ _

• When treated with a suitable oxidizing agent, an aldose can be oxidized to yield an ____ _

• When treated with HN03 , an aldose is oxidized to give a dicarboxylic acid called an ______ _

• D-Glucose and D-mannose are epimers and are inrerconverted under strongly -----conditions.

• The Kiliani-Fischer synthesis can be used to lengthen the chain of an ___ _ • The Wohl degradation can be used to shorten the chain of an ___ _ • are comprised of rwo monosaccharide units, joined together

via a glycosidic linkage. • Polysaccharides are polymers consisting of repeating monosaccharide units

linked by bonds. • When u·eated with an in the presence of an acid catalyst,

monosaccharides are converted into their corresponding N-glycosides.



4 of 10 12/23/2011 4:14 AM

640 CHAPTER 24

Review of Skills Fill in the blanks and empty boxes below. To verify that your answers are correct, look in your textbook at the end of Chapter 24. The answers appear in the section entitled Skil/Builder Review.

24.1 Drawing the Cyclic Hemiacetal of a Hydroxyaldehyde

DRAW THE CYCLIC HEMIACETAL THAT JS FORMED WHEN THE FOLLOWING HYDROXY ALDEHYDE UNDERGOES CYCLIZA TION UNDER ACIDIC CONDITIONS.

0

HO~H D 24.2: Drawing a Haworth Projection of an Aldohexose

DRAW A HAWORTH PROJECTION OF ct-D·GALACTOPYRANOSE.

;·~~H HO H

HO H

H OH

CH20H

[ H+ ]

O·GALACTOSE

24.3: Drawing the More Stable Chair Conformation of a Pyranose Ring

DRAW THE MORE STABLE CHAIR CONFORMA TfON OF a -D-GALACTOP't'RANOSE.

~HOCH2 O

OH

OH

H

a-O·GALACTOPYRANOSE ltiORE STABLE CHAIR CONFORMA TION

24.4 Identifying a Reducing Sugar

S TEP 2 · DETERMINE IF THE GROUP A TTACHED TO THE ANOMERIC POSITION IS A GROUP OR GROUP.

• IF GROUP THEN THE COMPOUND IS A REDUCING SUGAR • IF____ GROUP, THEN THE COMPOUND IS NOT A REDUCING SUGAR.



5 of 10 12/23/2011 4:14 AM

CHAPTER 24 641

24.5 Determining Whether 11 Dlsncchuridc Is 11 Reducing Sugar

STEP t - IDENTIFY rt-te ____ POSITIONS.

O~H CH20H

HO o--N:.--q OH HO~OH

OH

Review of Reactions

STEP 2 · DETERMINE IF THE GROUPS A fTACHEO TO THE ANOMERIC POSITIONS ARE HYDROXY GROUPS OR ALKOXY GROUPS.

- IF ONE 1$ A OR OVP, THEN THE COMPOUND IS A REDUCIN() SUGAR. · IF NEITHER ARE GROUPS, THEN THE COMPOUND IS NOT A REDUCING SUGAR.

Identify the reagents necessary to achieve eac h of the foll owing transformations . To verify that your answers are correct, look in your textbook at the end of Chapter 24. The answers appear in the section entitled Review of Reactions.

Hemiacetal Formation

~f~e:'l:H HO 3 H

H ' OH

H ' OH 1CH20H

~HO:H2 O OH

OH H HO H

H OH

1CH20 H

H0*2't=~ H ' OH

H ' OH

°CH20H

Chain Lengtlleni11g and Chain Shortening

HOHf : H OH

H OH CH20H

:i"c'l:H HO H

H OH

H OH

CH20H

+

H:i"c'l: HO H

H OH

H OH

CH20 H



6 of 10 12/23/2011 4:14 AM

642 CHAPTER 24

Reactio11s of Mo11osacchaddes

HO H

HtCHz~~

Solutions

24.1. a) an aldohexose d) an aldotetrose

b) an aldopentose e) a ketohexose

H OH

H OH

CH20H

c) a ketopentose

H-,-OH

HO_LH

HI OH H OH

HO 0

24.2. Roth are hexoses so both have molecular formula (C6H 120 6). Although they have the same molecular formula, they have different constitution - one is an aldehyde and the other is a ketone. Therefore, they are constitutional isomers.

24.3. All are D sugars except for (b), which is an L sugar. a) 2S, 3S, 4R, SR b) 2R, 3S, 4S c) 3R, 4R d) 2S, 3R e) 3S, 4S, SR Pay special attention to the fo llowing trend: The configuration of each chirality center is R when the OH group is on the right side of the Fischer projection, and the configuration is S when the OH group is on the left side.

24.4. H,c-:::-o

Ht OH H OH

H OH

H OH

a) CH20H

H,c-::;.o

HOtH HO H

HO H

HO H

b) CH~H

MEvertson

Highlight

Solutions



7 of 10 12/23/2011 4:14 AM

CHAPTER 24 643

24.5.

D L ~~------~ ~------~J v

enantiomers

24.6. H,c~o H,c~o H,c~o

H+OH HO+H HO_l__H

H OH HO H H+ OH CH20H CH20H CH~H

\. D

v enantiomers

L J \.

D L ----~J y

enantiomers

24.7.

L-Fructose

24.8. D-fructose and D-glucose are constitmional isomers. Both have molecular formula (Ct.H1206). Although they have the same molecular formula, they have different constitution- one is a ketone, and the other is an aldehyde.

CH20 H H ,c~0

C=O HOHf l OHH H- .- OH

HO- H H OH

H-.-OH H-1-oH

CH20 H CH20H

D-Fructose D·Giucose



8 of 10 12/23/2011 4:14 AM

644 CHAPTER 24

24.9.

-4--o, OH

a) l./

24.10.

+ ov oH

b) v

0

HO~H

24.11.

Ho~o, v-oH

a)

(OY OH

HOA._/

(0>-oH

c)Y ~0~ d) -----v-

b) The six-membered ring is expected to predominate because it has less ring strain than a fi ve-membered ring.

24.12. 6

4~0~1H ~

a) OH

6

~o"?H 4~1

d) OH 3 2

24.13. J3-D-galactopyranose

,;t-o"', 6~bH

c) OH OH

HctH2

·d:=>o~ f) OH



9 of 10 12/23/2011 4:14 AM

CHAPTER 24 645

24.14.

a·D·Mannopyranose fJ-D·Mannopyranose

24.15.

a-D·Talopyranose P·D·Talopyranose

24.16.

O~HCHOH HO OH

H a) H

HO~CHpH 0 HO H

OH b) OH

HO~CH20HI 0 HO OH

OH c) H

24.17. H, G-o:-0

HO~O\ OH ~ ~ OH H

H+OH H OH

H OH

H OH

GH20H

D·AIIose



10 of 10 12/23/2011 4:14 AM

646 CHAPTER 24

24.18.

24.19.

OH

HO_....___ < ~0 HO~OH

OH

more stable

ring flip -

OH

H OH a) OH OH b)h 24.20.

·~ H H,c~o: Hi'd: H+ OH €>\i

HO-j H

CH20H

L-THREOSE

HOP1 OH

OH '0

OH OH Jess stable

~o-r; 1-'6H

c) OH

OH

~0---1~ ~

d) OH



1 of 10 12/23/2011 4:14 AM

24.21.

C~H

:=§ D-Fructose

24.22. CH20H

~±: H=FOH

CH20H

D·Fructose

24.23.

H ' ' I H- 0:

0' H

HO CH20H

HO~OH c) OH

CHAPTER 24 647

CH~H

'c=P.~H H0 H

H- Qtl

H · j::!H CH~

C~!¥JH .. ~;-~~ \ @

H~

excessAc~

P'l

excess Ac~

P'l

OH /

C~I¥)H OH 0

H~H

H

CH20 Ac

AcO~ A cO

OA OAc

AcO CHzOAc

AcO~OAc OAc



2 of 10 12/23/2011 4:14 AM

648 CHAPTER 24

24.24. a)

HO~H:PHO HO

OH OH

b)

HO~CH20H 0 HO

OH OH

c)

HO CH,OH

HO~OH OH

24.25.

HO~CH20H (1

HO 0 ~H-CI OH

:QH

HO~CH20H $ ~· /~H 110 ... ....

OH H

4 ... HO CH~OH l ~({·)

HO~<l) OH

HO

Ho~ J OH

HO CH20H

I IO~O\ QCl I~

H

HO~CH20H0 HO H

OH :OEt



3 of 10 12/23/2011 4:14 AM

24.26.

24.27.

H:+'c'l:

HO H H OH H OH

CH20H

a) D·Mannose

24.28.

H;pH H OH

H OH

CH20H

0-Aitose

~it'c<~ HO H

H OH

CH20 H

0-Ta/ose

: t'C''l:H

H OH

H OH

H OH CH20 H

b) 0-A//ose

NaBH4

H20

CHAPTER 24 649

: t'c'l:H

HO H

HO H H OH

CH20H

c) D·Galactose

HO!~H HOf~H H OH HO H

H OH Rotate HO H

H OH 180°

H OH

CH20H CH20H



4 of 10 12/23/2011 4:14 AM

650 CHAPTER 24

24.29. H,c~o

~~~ H+ OH

CH20H

D-AIIose

H'C~o

:~t: HO H

HO H

CH20H

L-AIIose

NaBH4

H20

Hf~HH HOf~H H OH HO H

-H OH Rotate HO H

H OH 180°

HO H

CH20 H CH20H

24.30. The following alditols are meso compounds. and are therefore optically inactive: H,c~o

Ht OH H OH

H OH

H OH

CH20H

D-AIIose

H;'pH HO H

H OH

CH20H

D-Galactose

24.31.

a) No (an acetal)

HtCH~HH H OH

H OH

H OH

CH20 H

(meso)

HtCH~HH HO H

HO H H OH

CH20 H

(meso)

b) Yes c) Yes



5 of 10 12/23/2011 4:14 AM

CHAPTER24 65 1

24.32. HO,c70 HO,c~o HO,c~o HO,c~o

HtOH Ht OH "t~ "f" HO H HO H HO - H HO- H

HO H HO H H OH H OH

H OH H OH H OH H OH

CH20H CH20H CH20H CH20H

a) D·Galactonlc acid b) D-Galactonlc acid c) D-Giuconlc acid d) D-Giuconlc acid

24.33. This compound will not be a reducing sugar because the anomeric position is an acetal group.

24.34.

H,c~o

H$0H H OH

H OH

CH20H

Fischer-Kiliani

H,c~o H,c?o

"f" HOtH H OH H OH

"' H - OH H - OH

H OH H OH

CH20H CH20H D-Ribose

a) D-AIIose D-Aitose

H,c~o

H$0H HO H

H OH

CH20H

Fischer-Kiliani

H,c-:::-o H,c-:::-o

HtOH HOtH H OH H OH

HO H +

HO H

H OH H OH

CH20H CH~H

D-Xylose

b) D-Gulose D-ldose

H,c -:::-o

HOt HO H

H OH

H,OH

Fischer-Kiliani

H,c~o H,c~o

"tOH HOl HO H HO H

HO H +

HO H

H OH H OH

CH20H CH20H

D-Lyxose c) D-Galactose D-Talose



6 of 10 12/23/2011 4:14 AM

652 CHAPTER 24

24.35.

24.36.

H,c~o

H+ OH H-

1- 0H

CH20H

D-Erythrose

H,c~o

1)HCN :t'c~:H H OH

H OH +

CH20H

D·Ribose

H,c~o

HO-~ -H H,c~o

H OH "too H + OH Wohl Degradation "$00 H OH

+ H OH H- - OH

H OH H + OH H OH

CH20H CH20H CHpH

D-AIIose D-Aitose D·Ribose

24.37. H,c~o

"f " 1) NH20H H,c~o

H OH 2) Ac2o H+ OH

H OH H OH

CH20H 3) NaOMe

CH20H

D·Ribose D·Erythrose

24.38. H,c~o H,c~o

"t oo H,c~o

"t oo HO H Wohl ~$" Fischer- HO H

H OH H OH H OH Degradation Kiliani H OH H OH H OH

CH20H CH20H CH20H

D-Giucose D-Arablnose D-Giucose

24.39. a) Yes, one of the anomeric pos itions bears an OH group. b) No, both anomeric positions bear acetal groups. c) No, both anomeric positions bear acetal groups.

H:t'C~: H OH

H OH

Cfi:!OH

D·Arablnose

H,c~o

~t· HO H + H OH

H OH

CH20H

D-Mannose



7 of 10 12/23/2011 4:14 AM

24.40. CH20H ~ H0..--~0\ H20H

HO~O 0 OH HO

OH OH

MeOH

24.41.

b)

c) CH2<)H

HO~O, CH20H

HO~o~O, OH HO~OH

OH

d) CH20H

HO~O, CH20 H

HO~o~o, OH HO~OH

OH

py

24.42. a) a D-aldotetrose d) aD-aldohexose

b) an L-aldopentose e) a D-ketopentose

24.43.

CH20H

HO'\_~O\ HO~OCH3

OH

CHAPTER 24 653

+ HO~\ -10~

OCH3

CH20Ac

AcO~O, CH20Ac

Aco~o~N-o, OAc AcO~OAc

OAc

c) a D-aldopentose

a) D-glyceraldehyde b) L-glyceraldehyde c) D-glyceraldehyde d) L-glyceraldebyde

24.44. a) D-Glucose b) D-Mannose c) D- Galactose d) L-Glucose



8 of 10 12/23/2011 4:14 AM

654 CHAPTER 24

24.45. a) D-Ribose b) D-Arabinose

H..._CQ-0

:~±: HO=i=H

CH20H

c) L·Ribose

d) Same compound e) Diastereomers

24.46.

0 o-OH a)

24.47. 0

HO~H

24.48.

:r: H 4 OH

5CH:PH

D·rlbose

b) H,b~o

H+ OH

H 3

OH

H+OH 5CH20H

D·ribose

24.49.

)-q OH

b) \__)<

H

5

HOC~H2 4 H

H OH OH

a) epimers b) diastereomers

c) b-OH

4ro~, +

H6 ~ bH OH OH

a pyranose ring

5

4~0~1 "r--r(6H

OH OH

a furanose ring

4ro'0: H6~

OH OH

P pyranose ring

5

+ ·~=l OH OH

p furanose ring

c) enantiomers d) identical compounds



9 of 10 12/23/2011 4:14 AM

CHAPTER 24 655

24.50. H,c~o

Hi OH HO H H OH H OH

CH20H

0 -G/ucose

24.51 .

H,c~o H,c~o

H,c~o H,c~o

H~r yH20H

H* OH "*OH HO 5 H C=O

Hf OH H R OH HO 5 H H R OH HO+ H

H R OH HO 5 H H OH H R OH H R OH

a) CH20H b) CH20H c) CH20H d) CH20H e) CH20H

24.52. H,c~o H,c.yO H,~O H ,c~o

"f" lOH HOi" "f" HO H HO H HO H H OH H OH HO H H--OH H OH

H OH H OH H OH H OH CH20H CH20H CH20H CH20H

a) 0 -Giucose b) D-Ga/actose c) D-Mannose d) 0-A//ose

24.53. 6

HOC~H2 O OH

5 ~ H 2

H 14 3 CH20 H a) OH H t

HOCH2

HO~O~H

~ b) OH

24.54. HOCH2

H~H OH OH

24.55. a) a-D-allopyranose b) P-D-galactopyranose c) methyl P-D-glucopyranoside



10 of 10 12/23/2011 4:14 AM

656 CHAPTER 24

24.56. H,c~o

HtOH H OH H OH

H OH

CH20H

a) D·AIIose

24.57.

a)

CH,OH

HO~ OCH, OH

c) OH

H,c,::.o

"t" HO H HO H

H H CH20H

b) ().Galactose c)

CH20Ac

Ac0~0 OAc OAc

b) OAc

CH,OH

+ HO~O\ )bHI OH OCHs

H,c~o

Ht OH HO H H OH H OH

H20H

D·Giucose

24.58. The product is a meso compound

24.59.

;t'C~~ H OH H OH

CH20H D·AIIose

meso compound

heal

HO,C,::-0

H± OH HO H

HOT H

HT OH

HO' ""o

H- OH

HI OH

HO' C'"O

optically Inactive

meso compound



1 of 10 12/23/2011 4:15 AM

24.60.

24.61.

:~o, I~

excess CH,I

Ag20

OH

24.62. a) diastereomers b) same compound

24.63. CH20H

b:o

"$~ H OH

H OH

CH~H

24.64. H, c'lo

H OH

equatorial axial

~~ro, equatorial ~

OH OH axial axial

+

CH• O CH20CH;,

'"·0~ c~o OH

CH2 0H

~=0

"0$" H OH

H OH

C~H

CH20H

H*~=~H HO H

H OH

CH20H

CHAPTER 24 657

CH20H

HO*t:~ HO H

H OH

CH20H

H,c~o

H OH HtOH

H OH H OH

CH~H

D-AIIose

Wohl

~ation H,c'lo

H+ OH

HI OH

H I OH

Wohl

degra/

HOtH

H OH

H OH

CH~H

D-Aitose

CH20H

D-Ribose



2 of 10 12/23/2011 4:15 AM

658 CHAPTER 24

24.65. H,0~o H,0~o

H,0~o

"±00 00±" 00$" Kilian I-Fischer HO H HO H

H OH H OH + H OH

H OH H OH H OH

CH20H CH20H OH20H

D·Arablnose D-Giucose D·Mannose

24.66. CN CN

H,c,.o HCN HtOH HOt

HtOH H OH + H H

HpH ~OH H20H

D-Giyceraldehyde Diastereomers

24.67.

"f~ HO H

H OH

H OH

a) H20H

":!: HO H NaBH• H OH H20

HO H

CH20H

HfCH2~= HO H

H OH

H OH

CH20H

b) L-Gulose

24.68. D-AIIose and D-Galactose

24.69. a) This compound will not be a reducing sugar because the anomeric position is an acetal group. b) This compound will be a reducing sugar because the anomeric position bears an OH group.



3 of 10 12/23/2011 4:15 AM

24.70. a) CH30H, HCI b) CH30 H, HCI c) HN03, H20 , heat d) excess CH31. Ag20 followed by H30+

24.71. a) a-D-glucopyranose and !3-D-glucopyranose b) a-D-galactopyrano e and 13-D-galactopyranose

24.72.

24.73.

24.74.

24.75.

a) D-Arabinose b) D-Ribose and D-xylose c) D-xylose d) D-xylose

HtCH~~H HO H

H OH CH20H

D·Xylose D·Xyllto/

CH20H

HO~ HO 1

OH o,s

HO~CH2 0

HO OH

Isomaltose OH (a 1- 6-a-glycos/de)

CHAPTER 24 659



4 of 10 12/23/2011 4:15 AM

660 CHAPTER 24

24.76.

24.77.

24.78.

a) No, it is not a reducing sugar because the anomeric position has an acetal group.

b) c)

d) e)

OH

OH ~ HO~ ,0#4 OH 1-0 OH

H3o• OH _____... ~OH OH +

salicin "0~ OH

OH H

Salicin is a P-glycoside.

~QOH 0~ ~ Ac20 ~4 0" HO I A cO

H 0 py A 0 OH OAc

salicin

No. In the absence of acid catalys i , the ace tal group is not readily hydrolyzed.

OH

HO~ < _;()"./ H - H~ HO~~: -

OH 0'H

HO__.,....___(OH~O Ho~o'()

OH I ~

OH

Ho~< ~oj HO~

OH

HO~CH,OHO HO

OH ND

+

CH,OH

HO~O, 7 HO~Nv

OH I H/ ~

an a-N-Glycoside I ~ ~

a fJ-N-Giycoside



5 of 10 12/23/2011 4:15 AM

24.79.

a)

24.80.

H:*'c~~ H OH

H OH

CH20H

~f H OH

CH,CH

(optically active)

24.81.

"~~~ H OH

H OH

a) CH20H

N:CO ,...H (( I N

HOI___..o-. .. J NA NHz H GUANOSINE

b) OH OH

H, C;;,.O H, c--:,0

HO=tH H* OH HO H H OH

H--OH H~ CH20H CH,OH

j ~~ H20

j N•B~ H20

GH,OH GHzOH

HOT H H±OH HO--H H OH

H+ OH H f-oH

CHzQH CH,OH

(optically active) (optically inactive)

CHAPTER 24 661

H,c~o

H* OH HO H

H-1- oH

CH20H

j N~H, H20

CH2UH

H I OH

HOT H

H -l-oH

CH20H

(optically inactive)

c) Yes. The compound has chjraJiry centers, and it is not a meso compound. Therefore, it will be optjcaiJy active. d) The gluconic acid is a carboxylic acid and its IR spectrum is expected to have a broad signal between 2500 and 3600 cm·1

. The IR spectrum of the lactone will not have this broad signal.



6 of 10 12/23/2011 4:15 AM

662 CHAPTER24

24.82. In order for the CH20H group to occupy an equatorial position, all of the OH groups on the ring must occupy axial positions. The combined steric hindrance of all the OH groups is more than the steric hindrance associated with one CHzOH group. Therefore, the equilibrium will favor the form in which the CH20H group occupies an axial position. The structure of L-idose is:

24.83.

;I;: H OH

H OH

CH20H

D-AIIose (Compound A)

:~:H HO H

H OH

HO H

CH20H

L-ldose

HO--LO, OH

~ OH

~-pyranose form

HO~\ HO~OH

HOCH2 OH

Ell EtO~OEtO OEI

OEt OEt

24.84. Glucose can adopt a chair conformation in which all of the substituents on the ring occupy equatorial positions. Therefore, D-glucose can achieve a lower energy conformation than any of the other D-aldohexoses.



7 of 10 12/23/2011 4:15 AM

24.85.

Gc~~~ .. e :O :..- OH

HO H ~ H~H ~ : H

H H H OH

CH,oH

D-glucose

HqCH~= OH

HO

H OH

H,<lli

taut

CH20 H f-

"':¥\~~ CH,PH I

11H,01)

HO~CH20HOH HO H

HO H

CH20 H

taut

OH

OH

CH,oH

CH,PH

Hd-AOH HO+~ HO+ H

CH,PH

OH

taut ~

H~H~

$c~

0 H OH :~H

HO H ====""-HO H

CH,.OH

CHAPTER 24 663

HT OH

H ~OH CH,OH

H ~·H'0'H HO!?CH~ e .•'\. / oH --:~

CH,.OH

OH .El

H :) yo,H H H ====""-H H

CH,PH

OH

H~OH H--OH

HOI H HO , H

CH,PH

H,c~o

taut HO$ H H OH

HO H HO H CH,oH

L..glucose



8 of 10 12/23/2011 4:15 AM

664 CHAPTER 24

24.86. Compound X is a D-aldohexose that can adopt a p-pyranose form with only one axial substituent. Recall that D-glucose has all substituents in equatorial positions, so compound X must be epimeric with D-glucose either at C2 (D-mannose), C3 (D-allose), or C4 (D-galactose). Compound X undergoes a Wohl degradation to produce an aldopentose, which is converted into an optically active alditol when treated with sodium borohydride. Therefore, compound X cannot beD-allose, because a Wohl degradation of D-allose followed by reduction produces an optically inactive alditol. We conclude that compound X must be either D-mannose or D-galactose. The identity of compound X can be determined by treating compound X with sodium borohohydride. Reduction of D-rnannose should give an optically active alditol, while reduction of D-galactose gives an optically inactive alditol.

24.87. Compound A is a D-aldopentose. Therefore, there are four possible structures to consider (Figure 24.4). When treated with sodium borohydride, compound A is converted into an alditol that exhibits three signals in its 13C NMR spectrum. Therefore, compound A must be Dribose or D-xylose both of which are reduced to give symmetrical alditols (thus, three signals for five carbon atoms). When compound A undergoes a Kiliani-Fischer synthesis, both products can be treated with nitric acid to give optically active aldaric acids. Therefore, compound A cannot be D-ribose, because when D-ribose undergoes a Ki liani-Fischer synthesis, one of the products is D-allose, which is oxidized to give an optically inactive aldaric acid. We conclude that the structure of compound A must beD-xylose.

a) D·Xylose

b) Compound D is expected have six signals in its 13C NMR spectrum, while compound E is expected to have only three signals in its 13C NMR spectnun.

H:t'c~~H :t'c~~ H OH H OH

HO H HO H

H OH H OH HO/c~o HO/c~o

Compound D Compound E



9 of 10 12/23/2011 4:15 AM

Chapter 25 Amino Acids, Peptides, and Proteins

Review of Concepts Fill in the blanks below. To verify that your answers are correct, look in your textbook at the end of Chapter 25. Each of the sentences below appears verbatim in the section entitled Review of Concepts and Vocabulary.

• Amino acids in which the two functional groups are separated by exactly one carbon atom are called amino acids.

• Amino acids are coupled together by amide linkages called bonds. • Relatively short chains of amino acids are called -----• Only twenty amino acids are abundantly found in proteins, all of which are _

amino acids, except for which lacks a chirality center. • Amino acid exist primarily as at physiological pH • The of an amino acid is the pH at which the concentration

of the zwitterionic form reaches its maximum value. • Pep tides are comprised of amino acid joined by peptide bonds. • Peptide bonds experience restricted rotation, giving rise to two possible

confonnations, called and . The conformation is more stable.

• Cysteine residues are uniquely capable of being joined to one another via ------- bridges.

• ___ is commonly used to form peptide bonds. • In the Merrifield synthesis. a peptide chain is assembled whi le tethered to

• The primary structure of a protein is the sequence of----------• The secondary structure of a protein refers to the ----------

----------of localized regions of the protein. Two particularly stable arrangements are the _ helix and __ pleated sheet.

• The tertiary structure of a protein refers to its ------------• Under conditions of mild heating, a protein can unfold. a process called

• Quaternary structure arises when a protein consists of two or more folded polypeptide chains. called , that aggregate to form one protein complex.



10 of 10 12/23/2011 4:15 AM

666 CHAPTER 25

Review of Skills Fill in the blanks and empty boxes below. To verify that your answers are correct, look in your textbook at the end of Chapter 25. The answers appear in the section entitled Sk.ii/Builder Review.

25.1 Determining the Predominant Form of an Amino Acid at a Specific pH

CONSIDER THE POHOWINO AMINO A OlD. AND OllAW THE PORM THAT PREDOMINATES AT PHYSIOLOOICAL pH.

25.2 Using the Amidomalonate Synthesis

IDENTIFY REAGENTS THAT Wtl.t. ACHIEVE THE FOC.LOWINO TRANSFORMATION,

25.3 Drawing a Pe ptide

DRAW A BONO·LINE STRUCTURE FOR THE TRIPEPTIDE Phe·Vai· Trp.

25.4 Sequencing a Pcplidc via Enzymatic Cleavage

IDENTIFY THE FRAGMENTS OBf-AINEO FROM ENZYMATIC Ct.EAIIAGE OF t HE FOLLOWING PEPTIDE:

Ala-Phe-Lys-Pro·Met-Tyr-Giy-Arg-Ser-T rp-Leu·Hist

trypsy ~ymotrypsin

l



1 of 10 12/23/2011 4:16 AM

CHAPTER 25 667

25.5 Planning the Synthesis of a Dipeptide

IDENTIFY AU REAGENTS NECESSARY TO PREPARE THE 0/I'EPnOE A'"·G/y;

Ala Gty

! Ala + Gly Ala-Giy

Ala-Giy J

25.6 Preparing a Peptide using the Merrifield Synthesis

JOENTIFY All REAGENTS NECESSARY TO Pf!EPARE THE PENTAPEPTIOE ... a~·ltil·Alii-Pn•;

lie Gly Leu AI• Phe

J 1 --Phe -{POlYIIER) Boc- lie Gly Leu Al3 Phe

J 1 Phe Boo--' Ala Phe

Review of Reactions Identify the reagents necessary to achieve each of the following transformations. To verify that your answers are correct, look in your textbook at the end of Chapter 25. The answers appear in the section entitled Review of Reactions.

Analysis of Ami11o Acids

+ - o:f-,{o 0 0

+

H20

C02

RCHO



2 of 10 12/23/2011 4:16 AM

668 CHAPTER 25

Synthesis of Amino Acids

0

R~OH

NaCN

~OH V ~HAc

Analysis of Amino Acids 0

PEPTIDE --~~NH2 H A

Synthesis of Peptides 0

> li + H2N-i }"'"oH

0

H N~ Jl 2 I ' OH

A

0

H2N, Jl I ' OH

(H1

0

R r OH Br

0 0

EtOVOEt A N

H'll 0

~OH v ~HAc

PEPTIDE

A ~ A

H20

+

0

~OH V ~Hz



3 of 10 12/23/2011 4:16 AM

Solutions

25.1. In each case, the chirality center has the R configuration.

25.2.

25.3.

HO,c~o

H+ NH2

CH3

D·Aianlne

HO,c~o

H+ NH2

CH(CH3)2

D·Vallne

a) Pro, Phe, Trp, Tyr, and His b) Phe, Trp, Tyr, and His c) Arg, His, and Lys d) Met and Cys e) Asp and Glu

0

/S~OH c) NH2

CHAPTER 25 669

f) Pro, Trp, Asn, Gin, Ser, Thr, Tyr, Cys, Asp, Glu, Arg, His, and Lys

25.4.

25.5. Arginine has a basic side chain, while asparagine does not. At a pH of II , arginine exists predominantly in a form in which the side chain is protonated. Therefore, it can serve as a proton donor.

25.6. Tyrosine possesses a phenolic proton which is more readily deprotonated because deprotonation forms a resonance-stabilized phenolate ion. In contrast, deprotonation of the OH group of serine gives an alkoxide ion that is not resonance-stabilized. As a result, the OH group of tyrosine is more acidic than the OH group of serine.

MEvertson

Highlight

Solutions



4 of 10 12/23/2011 4:16 AM

670 CHAPTER 25

25.7. a) 2.77 b) 5.98 c) 9.74 d) 6.30

25.8. a) aspartic acid b) glutamic acid

25.9. Leucine and isoleucine

25.10. The pi of Phe = 5.48, the pi of Trp = 6.11, and the pl of Leu = 6.00. a) At pH = 6.0, Phe will travel the farthest distance. b) At pH= 5.0, Trp will travel the farthest distance.

25.11.

25.12. 0

~OH 1) Br2 , PBr3

2) H20

a)

b)

c)

0

Vl-oH

3) excess NH3

1) Br2 , PBr3

2) H20

3) excess NH3

1) Br2 , PBr3

2) H20

3) excess NH3

0

n OH (racemic)

0

Y oH NH2

(racemic)

JylOH NH2

(racemic)



5 of 10 12/23/2011 4:16 AM

CHAPTER 25 671

25.13.

ILCOOH

NH2

JycooH

NH2

OL.COOH

2 H2N ............. COOH

a) Leucine b) Valine c) Phenylalanine d) Glycine

25.14.

0 0

JytOH etoVoet

t)NaOEt

2)J H~Ny NH2

Br

0 3) H,o·. heat a)

0 0 0

etoVoet

t)NaOEt

Y oH 2) CHsBr

H~Ny 3) H3o · . heat NH,

b) 0

0 0

JytOH etoV oet

1) Na0Et

2) ~ H~Ny NH2

Br

0 3) H30 ' , heat c)

25.15. 0

JytOH

0

Y oH ~OH NH, NHz NHz

a) alanine b) val ine c) leucine

25.16. Leucine can be prepared via the amidomalonate synthesis with higher yields than isoleucine, because the former requires an SN2 reaction with a primary alkyl halide, while the latter requires an SN2 reaction with a secondary (more hindered) alkyl halide.



6 of 10 12/23/2011 4:16 AM

672 CliAJ•TER 25

25. 17.

0 1) NH.CI, NaCN

's~H 2) H30'

a)

(U 1) NH.CI. NaCN

N H 2) HJO' b)

H

Wt 1) NH0CI, NaCN

H 2)H30' c)

D H

1) NH.CI. NaCN

2) H30'

d)

25.18.

0 1) NH.CI. NaCN

H3C)l_H 2)H30'

a)

D H

1) NH.CI, NaCN

2) HJO'

b)

0 1) NH.CI, NaCN

Y H 2)H30'

c)

25.19.

0

MOH

b) NHAc

0

/S~OH NH2

0

N~~ ~NH N~

0

V(oH 2

0

rrOH NH2

0

~c0~ NH2

slsnlno

0

y0~ NHz

leucine

JytOH

NHz

valine

0

,~)l~ c) T '!HAc

0

~OH N ~HAc d) H0



7 of 10 12/23/2011 4:16 AM

CHAPTER 25 673

25.20. Glycine does not possess a chirality center. so the use of a chiral catalyst is unnecessary. Also, there is no alkene that would lead to glycine upon hydrogenation.

c)

25.22. Leu-Ala-Phe-Cys-Asp or L-A-F-C-0 .

25.23. Cys-Tyr-Leu

25.24. Constitutional isomers

25.25.

0 ~ ~ C terminus

i OH

N terminus o y

25.26. Steric hindrance results from the phenyl groups:

n o,, ~OH

~NH H2N 0



8 of 10 12/23/2011 4:16 AM

674 CHAPTER 25

25.27.

b)

25.29.

0-Aspar1ie a<>d



9 of 10 12/23/2011 4:16 AM

CHAPTER 25 675

25.30. An Edman degradation will remove the amino acid residue at the N tenninus. and Ala is theN tenninus in Ala-Phe-Val. Therefore, alanine is removed, giving the following PTH derivative:

25.31. Met-Phe-Val-A Ia-Tyr-Lys-Pro-Val-Ile-Leu-Arg-Trp-His-Phe-Met -Cys-Arg-Gly-ProPhe-Aia-Val

25.32. Ala-Phe-Val-Lys

25.33. C leavage with tryps in will produce Phe-Arg, while cleavage with chymotrypsin will produce Arg-Phe. These dipeptides are not the same. They are constitutional isomers.

25-14. a)

Trp

Mel

b)

Ala

lie

(Boc),O

(H'J

(tr)

CH,OH

Boc-Trp f Boc- TIP Mot OCH3

Met OCH3

t) CF,COOH Trp - Mot

2) Na0H.H20

Boc- AJa f Boc- Ala -ne-OCH3

l le-OCH3

I) CF,COOH

2) Na0H,H20 Ala ne



10 of 10 12/23/2011 4:16 AM

676 CHAPT ER 25

c) (Bac)20

Leu

v~l (H"J

CH00H

Soc- Leu 7 Vni- OCH)

25.35.

(Boc),O lie Boc- lle

Phe CH,OH

(W] Phe- OCH,

lle-Phe-Giy 1)CF3COOH

2) NaOH, H20

Boc- letl Val •OCH3 I ) CF, COOH

2) NaOH,H20

DCC Boc-l te -t'he- OCH3j

~ NaOH, H.O

Boc- IIe-Phe

DCC ~ Gly- OCH,

Boc- lte-Pl'le-G ty-OCH3

Leu- Val



1 of 10 12/23/2011 4:17 AM

CHAPTER 25 677

25.36.

(Boc)20 l eu Boc- Leu

'\,

Val [H')

~ Vol ~""'• _/ •

Boc- Leu - Vai-Phe - lle - OCH3

1) NaOH, H20

2) Ala - OCH3

DCC

2) lie ..LoCH3

DCC

2) NaOH, HzO

Boc- Leu - Val

! Ph1>- 0CH3

DCC

Boc- LI!u - Vai-Ptoe - OCH3

Leu - Vai - Ph .. -11<~-Aia



2 of 10 12/23/2011 4:17 AM

678 CHAPTER 25

25.37. a)

Cl~ POLYMER

b)

Cl~ POLYMER

2) CF3COOH

H 0

3) Boc'N~ • DCC

,A,., 4) CF~OOH

H 0

5) Boc" N'('<>H . DCC

y 6) CF3COOH

H 0

7) Boc"N~OH , DCC

\_Ph 8) CF~OOH

9) HF

2) CF3COOH

H 0

3) Boc"'N0 0H

-y 4) CF~OOH

H 0

5) Boc" N00H

,A,., 6) CF3COOH

H 0 7) Boc, NJ OH

B) CF3COOH

9) HF

. DCC

, DCC

, DCC

Phe-Leo-Val-Pile

AJa.\lai·Leu-lle



3 of 10 12/23/2011 4:17 AM

CHAPTER 25 679

25.38. (N terminus) Val-Ala-Phe (C terminus)

25.39. The regions that contain repeating glycine and/or alanine units are the most likely regions to form ~ sheets:

25.40.

25.41. When applying the Cahn-Ingold-Prelog convention for assigning the configuration of a chirality center, the amino group generally receives the highest priority ( I ), followed by the carboxylic acid moiety (2), followed the side chain (3), and finally the H (4). Accordingly, the S configuration is assigned to L amino acids. Cysteine is the one exception because the side chain has a higher priority than the carboxylic acid moiety. As a result, the R configuration is assigned.

25.42.

25.43. a) Isoleucine and threonine b) Isoleucine= 2S,3S. Threonine = 2S,3R

25.44. - 0

~OH NH2

- 0

~OH NH2

~OH rii~

25.45. The protonated form below is highly stabilized by resonance, which spreads the positive charge over all three nitrogen atoms.

Acid -



4 of 10 12/23/2011 4:17 AM

680 CHAPTER 25

25.46. The protonated form below is aromatic. In contrast, protonation of the other nitrogen atom in the ring would result in loss of aromatic stabilization.

25.47.

25.48.

25.49.

a) 6.02

0 0

- ./"-. Jl OH Acid <±J "'--: ""' Jl OH ~~ I ' - H-rl~ I ' ~NH NH2 ~NH NH2

b) 5.41

0 0

e Jl ""' Jl e 0' ......._,- ' '0

N d) H/ 'H

0 0

e. Jl "" Jl e 0 ' ......._,- ;( '0 (j)N

H/ I ' H d) H

c) 7.58 d) 3.22

25.50. Lysozyme is likely to be comprised primarily of amino acid residues that contain basic side chains (arginine, histindine, and lysine), while pepsin is comprised primarily of amino acid residues that contain acidic side chains (aspartic acid and glutamic acid).

25.51.



5 of 10 12/23/2011 4:17 AM

CHAPTER 25 681

25.52.

{planar)

Racemic mixture

25.53. The pi of Gly = 5.97, the pi of Gin= 5.65, and the pi of Asn = 5.41. a) At pH= 6.0, Asn will travel the farthest distance. b) At pH= 5.0, Gly will travel the farthest distance.

25.54.

25.55.

25.56. 0

a) Methionine, valine, and glycine.

o:f-·{o

nR c)~H d) no reaction

Cb.) 0 0 ) The compound is highly conjugated and has a "-max that is greater than 400 nm

(see Section 17 .12)

Y'H 25.57.

25.58.

Alanine can be prepared via the amidomalonate synthesis with higher yields than valine, because the former requires an SN2 reaction with a primary a lkyl halide, while the latter requires an SN2 reaction with a secondary (more hindered) alkyl halide.

The side chain (R) of glycine is a hydrogen atom (H). needs to be installed at the a position.

0 0

etoV oet

H'NY 0

Therefore, no alkyl group



6 of 10 12/23/2011 4:17 AM

682 CHAPTER 25

25.59.

0

~OH 1) Br2, P8r3

2)H20

a) 3) excess NH3

vrH 1) NH4CI, NaCN

b) 2) HJ0•

0 0

E10V OE1 1)Na0E1

H ~Ny 2) CH3l

3J~o·. heat c) 0

25.60.

DOH 1) Br2 , PBr3

2) H,P

3) excess NH3

a)

0 0

E1oV oe1

1) Na0Et

2)~ H ' NY Br

0 3) H3o•, heat b)

0 1) NH4CI, NaCN

'I'H 21 H3o ·

c)

25.61. 205 = 3,200,000

25.62. 0

~OH \...-NH NHAc

0

vCOH 2

0

~ 2

0

0 oH

NH2

(racemic)

I o ~OH

0

~OH NHz



7 of 10 12/23/2011 4:17 AM

25.63. I) Leu-Met-Val, 4) Met-Leu-Val,

25.64.

N Terminus

25.68.

H2Nx=SH

0 OH

cysteine

2) Leu-Val-Met, 5) Val-Met-Leu,

"'"t 0 OH

valine

3) Met-Val-Leu, 6) Val-Leu-Met

CHAPTER 25 683



8 of 10 12/23/2011 4:17 AM

684 CHAPTER 25

25.69.

HO~ ~I tyrosine

H2N 0

serine

OH

H2N l

0 OH

glycine

HO~ ~ I H

H N O

~).J~N HOj 0~~

~

25.70. It does not react with phenyl isothiocyanate so it must not have a free N terminus. It must be a cyclic tripeptide:

Gly

Ph/ ~Ia

25.71. a) Arg + Pro-Pro-Gly-Phe-Ser-Pro-Phe-Arg

b) Arg-Pro-Pro-Gly-Phe + Ser-Pro-Phe +

25.72. Phenylalanine

25.73. Vai-Aia-Giy:

0 ~ ~N~

R: v lH~-{ s

Arg



9 of 10 12/23/2011 4:17 AM

CHAPTER 25 685

25.74. There cannot be any disulfide bridges in this peptide. because it has no cysteine residues, and only cysteine residues form disulfide bridges.

Hif,.Ser-Gln-Gly-Thr-Phe-Thr-Ser-A~p-Tyr-Ser-Ly<-Tyr-Leu-Asp-Ser-Arg-Arg-Aia-Gin-Asp-Phe-Vai

Gln-Trp-Leu-Met-Asn-Thr

25.75. Prior to acylation, the nitrogen atom of the amino group is sufficiently nucleophilic ro attack phenyl isothiocyanate. Acylation converts the amino group into an amide moiety. and the lone pair of the nitrogen atom is delocalized via resonance, rendering it much less nucleophilic.

25.76.

25.77. (Boc),O

Boc Pile >-Boc~Phe Ala~OCH,

Ala-OCH1

Phil

Ala IH"I

CH30H

l )CF, COOH Phn-Aia

2) Na0H,H20

25.78.

CH3 H 0

H2N~N~OH o CH3



10 of 10 12/23/2011 4:17 AM

686 CHAPTER 25

25.79.

Phe- l le

Val - Leu (H"J

CHoOH

25.80.

Cl~ POLYMER

Boc--,.Phe l ie

Vai - Leo- OCH2

2) CF3COOH

H 0

3) Boc" N0 0H • DCC

/-.....

4) CF3COOH

H 0

5) Boc" N0oH . DCC

~

6) CF3COOH

7) HF

25.81. y

DCC

Leu-Val-Ala

Boc- Phe-lle-Vai-Leu--OCH3

J 1) CF, COOH

2) NaOH, H20

Pile lie- Val-Leu

25.82. A proline residue cannot be part of an u helix, because it lacks an N -H proton and does not participate in hydrogen bonding. (The amino acid proline does indeed have an N-H group, but when incorporated into a peptide, the proline residue does not have an N-H group)



1 of 10 12/23/2011 4:18 AM

25.83.

HO~Br 25.84. --v-

CHAPTER 25 687

25.85. The stabilized enolate ion (fonned in the first step) can function as a base, rather than a nucleophile, giving an E2 reaction:

o r~s EtO~OEt A

H'NI( 0

25.86. The lone pair on that nitrogen atom is highly delocalized via resonance and is participating in aromaticity. Accordingly, the lone pair is not free to function as a base.



2 of 10 12/23/2011 4:18 AM

688 CHAPTER 25

25.87. 0

OH

a) HO

Br~0

I ~ OH

HO b

Br

OH

HO

b)

25.88. At low temperature, the barrier to rotation keeps the two methyl groups in different elecu·onic environments (one is cis to the C=O bond and the other is trans to the C=O bond), and they therefore give rise to separate signals. At high temperature, there is sufficient energy to overcome the energy barrier, and the protons change electronic environments on a timescale that is faster than the timescale of the NMR spectrometer. The result is an averaging effect which gives rise to only one signal.

25.89.

25.90.

a) The COOH group does not readily w1dergo nucleophilic acyl substitution because the OH group is not a good leaving group. By converting the COOH group into an activated ester, the compound can now undergo nucleophilic acyl substitution because it has a good leaving group.

b) The nitro group stabilizes the leaving group via resonance. As described in Chapter 19, the nitro group serves as a reservoir for electron density.

c) The nitro group must be in the ortho or para position in order to stabilize the negative charge via resonance. Jf the nitro group is in the meta position, the negative charge cru1not be pushed onto the nitro group.

heat

0

H;2N00H

~OH J:!OH NH2

th reonine



3 of 10 12/23/2011 4:18 AM

Review of Concepts

Chapter 26 Lipids

Fill in the blanks below. To verify that your answers are correct, look in your textbook at the end of Chapter 26. Each of the sentences below appears verbatim in the section entitled Review of Co11cepts and Vocabulary .

• Lipids are naturally occurring compounds that are extracted from cells using -------solvents.

• Complex lip ids readily undergo , while simple lipids do not. • are high molecular weight esters that are constructed from carboxylic

acids and alcohols. • are the triesters formed from glycerol and three long-

chain carboxylic acids, called fatty acids. The resulting triglyceride is said to contain three fatty acid _____ _

• For saturated fatty acids, the melting point increases with increasing ____ _ ------· The presence of a __ double bond causes a decrease in the melting point.

• Triglycerides that are solids at room temperature are called ___ , while those that are liquids at room temperature are called __ _

• Triglycerides containing unsaturated fatty acid residues will undergo hydrogenation. During the hydrogenation process, some of the double bonds can isomerizes to give 1t bonds

• To the presence of molecular oxygen, triglycerides are particularly susceptible to oxidation at the position to produce hydroperoxides.

• Transesterification of triglycerides can be achieved either via __ catalysis or ___ catalysis to produce biodiesel.

• are similar in structure to triglycerides except that one of the three fatty acid residues is replaced by a phosphoester group.

• The structures of steroids are based on a tetracyclic ring system, involving three six-membered rings and one ___ -membered ring.

• The ring fusions are all in most steroids, giving steroids their rigid geometry.

• AU steroids, including cholesterol, are biosynthesized from _____ _ • Prostaglandins contain twenry carbon atoms and are characterized by a ___ •

membered ring with two side chains. • Terpeoes are a class of naturally occurring compounds that can be thought of as

being assembled from units. • A terpene with I 0 carbon atoms is called a , while a terpene

with 20 carbon atoms is called a--------



4 of 10 12/23/2011 4:18 AM

690 CHAPTER 26

Review of Skills Fill in the blanks and empty boxes below. To verify that your answers are correct, look in your textbook at the end of Chapter 26. The answers appear in the section entitled Skill Builder Review.

26.1 Comparing Molecular Properties of Triglycerides

CIRCLE THE TRIGLYCERIDE BELOW THAT IS EXPECTECD TO HAVE A HIGHER MEL TING POINT.

0

~ to~ 0~

0

26.2 Identifying the Product~ of Triglyceride Hydrolysis

DRAW THE PRODUCTS OBTAINED WHEN THE FOLLOWING TRJQL YCERIDE IS TREATED WITH AOUEOUS SODIUM HYDROXIDE.

NaOH

26.3 Drawing a Mechanism for Transesterilication of a Triglyceride

PROTON NUCLEOPHILIC TRANSFER ATTACK

A THE THE RESUL TJNG IS FUNCTIONS AS A

PROTONATED NUCLEOPHILE AND INTERMEDIATE ATTACKS THE IS DEPROTOANTED. PROTOANTEO THEREBY REMOVING

CARBONYL GROUP THE POSITIVE CHARGE

26.4 Identifying Isoprene Units in a Terpene

IDENTIFY THE iSOPRENE UNITS IN THE FOLLOWING TERPENE

LOSS OFA LEAVING GROUP

THE THE ---IS IS

PROTONATED REGENERATED VIA EXPULSION OF THE

LEAVING GROUP

PROTON TRANSFER

DEPROTONATJON YIELDS THE PRODUCT



5 of 10 12/23/2011 4:18 AM

CHAPTER 26 691

Review of Reactions Identify the reagents necessary to achieve each of the following transformations. To verify that your answers are correct, look in your textbook attlle end of Chapter 26. The answers appear in the section entitled Review of Reactions.

0

~~~ 0 0 to .. ~~ 0~

0

0

~

~0~ 0~

0

0

0~

to~/'-/'-/'-/'-. /'-/'o~

0

OH

0

o~'-.../'-/'-./~-~-, 0

o"·~ I

o.lr~~ 0

+

0 Na@ 0 U . . . . . .

0~

0 Na<£1 e u . . . . . .

0~

0

... o~ 0

U!>f'l_.)l~



6 of 10 12/23/2011 4:18 AM

692 CH APTER 26

Solutions

26.1 0

hydrolysis )lOH CH3(CH,),.

26.2 0

0

26.3. a) trimyristin° b) triarachadin c) triolein d) tristearin

26.4. tripalmitolein, tripaJmjtin, and tristearin

26.5. The fatty acid residues in lriarachadin have more carbon atoms than the fatty acid residues in tristearin. Therefore, triarachadin is expected to have a higher melting point. It should be a solid at room temperature, and should therefore be classified as a fat. rather than an oiJ. Therefore, triglycerides made from lauric acid will also have a low melting point.

26.6. a) All three fatty acid residues are saturated, with either 16 or 18 carbon awms, so the triglyceride is expected to have a high melting point. It should be a solid at room temperature, so it is a fat. b) All three fatty acid residues are unsaturated, so the triglyceride is expected to have a

low melting point. It should be a liquid at room temperature, so it is an oil.

26.7. 0

0~~ < 0

I o

0~~

a) o b) Tristearin c) The melting point of tristearin is higher than triolein. d) Stearic acid

MEvertson

Highlight

Solutions



7 of 10 12/23/2011 4:18 AM

26.8. 0

o·'~-......_,....-......./ .......... ./ ............. ,...~ ..... .......,.... ....... ..../'.,.

)-o.Jl...__~--../"''~ \ 0~""''-./",/~'-... /',/'

0

0

t~/~/'-...,~~ ( o)l....../'--./'-,,/~./''-...,-""'-.~,""'-...

o . .........,,__.,-./-......_.,-··~ ...... - .... ~"'"'-·-.... / 0

26.9. 0

CHAPTER 26 693

0

0)'..,/''-.,/''-./'-./""~-/'-.-/''/"' ' 0 \ )... "" }-o ' '-~./"-..,/".,/""::~.~ .....

< 0.~'-.../'-/'~/

0

0

o~/'......_/,/'""'/"'--./'-./'--../"''

\ J'-./'~""~'~'/' ,. 0

< o,~~/,..,.,...·,~---....../'-~"' .. ........_,.""

0

(:) G II Na 70~

26.10.

26.11

~OH OH

(!)8 ~ NaO~

0

0

0~ , .. /too chiral itycen ler ~

0



8 of 10 12/23/2011 4:18 AM

694 CHAPTER 26

26.12. Each of the three ester moieties is hydrolyzed via the following mechanism:

. . { --.... {\ IH ~ H Me ~ ·0· H--:0@ ;o"' ··~. HO' ·1()_H'" •. ;. HO: :O- Me Jl ~ .,Me,_ '-VMe- 9 - H \!'1f5 Me-~-H \/

!-v 'i !-R/ 'i 1-~1 ~ ~~~ ~ 1~;l:"

,------...H@

Me-0-H {;q: ~·· j-·· Me, ~Jl !

9H .. o:- 'i ;::::

26.13.

~OH OH

+

(three equivalents)

26.14. a) Hydroxide functions as a catalyst by establishing an equilibrium in which some ethoxide ions are present.

ethoxide

Then, each ester moiety undergoes transesterification via the following mechanism:

i'f,-;--§?oEt ~),(0~1 ~~H-(9';_H :o·· t---~4 ~ t-~~ $ - t-~~ Et':-p_: l ~ t-~H Et':.o.~

b) Hydroxide could function as a nucleophile and triglyceride would undergo hydrolysis rather than transesterification.

26.15. 0

0~

to~ o, p ' /

Y .... o~N, ... ,0 Q)

b) '-?

c) No. The C2 position would no longer be a chirality center.



9 of 10 12/23/2011 4:18 AM

CHAPTER 26 695

26.16. 0

a)

b) Yes. The C2 position would still be a chirality center.

26.17.

26.18. Octanol has a longer hydrophobic tai l than hexanol and is therefore more efficient at crossing the nonpolar environment of the cell membrane.

26.19. No. Glycerol has three OH groups (hydrophilic) and no hydrophobic tail. It cannot cross the nonpolar environment of the cell membrane.

26.20. A ring-flip is not possible for trans-decal in because one of the rings would have to achieve a geometry that resembles a six-membered ring with a trans-alkene, which is not possible. The ring fusions of cholesterol all resemble the ring fusion in trans-decalin, so cholesterol cannot undergo ring-flipping.

Hypothetical ring flip



10 of 10 12/23/2011 4:18 AM

696 C I-I AI'TEK 26

26.21. axial

equatoriai~Me Me Me

equatorial Me

a) axial

axial Me Me axial

"'"""'~'~ b) axial

axial axial CH

3

~0 rP2 H OH H H

c) axial

26.22. OH

HO

0 H oxymetholone

26.23. 0

0~

0

norgestrel



1 of 10 12/23/2011 4:19 AM

CHAPTER 26 697

26.24. a) PGE1 b) PGFra

26.25.

rl-·"~OH 0

OH il p ;);·'' ~

'!! a) menthol b) grand/sol c) carvone

26.26. a) Yes, it has I 0 carbon atoms. which are comprised by the joining of two isoprene units. b) No, it has II carbon atoms. c) No, it has ll carbon atoms. d) No. It has 10 carbon atoms. but the branching pattern cannot be achieved by joining two isoprene units.

26.27.

·OPP --

----OPP OPP s·



2 of 10 12/23/2011 4:19 AM

698 CHAPTER 26

26.28.

I ~ ~OPP

26.29.

26.30.

a) steroid b) terpene c) triglyceride d) phospholipid e) prostaglanctin f) wax

0

0 ~ 0

J ( 0~ o, r ......-~"-..../'~

a) o

() -OPP

I ... 1 ~OPP

~t1 OPP geranyl

pyrophosphate

a-fsrnesene



3 of 10 12/23/2011 4:19 AM

OH

t OH

b) OH

+

26.31. Both compounds are chiral:

0

0~

to~ o, /? (i}

_P,0~NH3 Go

CHAPTER 26 699

(three equivalents)

26.32. The fatty acid residues in this triglyceride are saturated, and will not react with molecular hydrogen.

26.33.

26.34. 0

HO

a) not a lipid b) a Lipid c) a lipid d) a lipid e) a lipid f) not a lipid g) a lipid h) not a lipid

0

o/"-....--"".....,_/'"-/',/'-.../"~,

\_o...-~..-........,~ ............. ~/"-/~ ( 0,~/'-~

0

26.35. The fatty acid residues of tristearin are saturated and are therefore less susceptible to auto-oxidation than the unsaturated fatty acid residues in triolein.

26.36. a < b < c



4 of 10 12/23/2011 4:19 AM

700 CHAPTER 26

26.37. Water would not be appropriate because it is a polar solvent, and terpenes are nonpolar compounds. Hexane is a nonpolar solvent and would be suitable.

26.38. a) saturated b) saturated c) unsaturated d) saturated e) unsaturated f) unsaturated

26.39. Arachidonic acid

26.40. a) No. It is an o il. b) No. It is reactive towards molecular hydrogen in the presence of Ni. c) Yes. It undergoes hydrolysis to produce unsaturated fany acids. d) Yes. It is a complex lipid because it undergoes hydrolysis. e) No. It is not a wax. f) No. It does not have a phosphate group.

26.41. a) Yes. It is a fat. b) Yes. It is unreactive towards molecular hydrogen in the presence of Ni. c) No. It undergoes hydrolysis to produce fany acids that are saturated. d) Yes. It is a complex lipid because it undergoes hydrolysis. e) No. It is not a wax. f) No. It does not have a phosphate group.

26.42. 20 CARBON ..t1 TOMS 30 CARBONA TOMS

26.43. Trimyrislin is ex peered ro have a lower melting point than tripalmitin because the former is comprised of fatty acid residues with fewer carbon atoms ( 14 instead of 16).

0

0~

~~ 0~

0 trim yrlstln tripalmitin



5 of 10 12/23/2011 4:19 AM

CHAPTER 26 701

26.45. See the solution to Problem 26.1 4.

26.47.

0

~0 0

0

26.48.



6 of 10 12/23/2011 4:19 AM

702 CHAPTER 26

26.49.

a) bisabolene b) flexlbl/ene

d ) VItamin A e)

26.50. Hydrophobic tails

~

a)

geraniol

OH

c) humulene

f) sablnene

Polar Head

e o,? \I

p /'... _N o' 'o- '--"' (£)'

b) Yes, they have one polar head and two hydrophobic tails.

26.51.

HO a)

H H

b) The methyl group (CI9) provides s1eric hindrance thai blocks one side of the 11: bond. and only the following is obtained:

H

HO



7 of 10 12/23/2011 4:19 AM

CHAPTER 26 703

26.52. OH

Br

HO

a) Br

0

b)HO~ OEt

c)E~~ 0

o_f(_

d)~O~ 26.53. The compound is chiral.

R

0~ 0

~Jl, \ _

0/1)

0 o

0 o..._ CH

20

r >=o yo ro R R R

0 0 o"'{ )=o

R R

26.54. a) H2, Ni b) H2, Ni, followed by NaOH, followed by Etl. c) H2, Ni, followed by LAH, followed by H20 d) 0 3, followed by OMS, followed by Na2Cr20 7 and H2S0 4

e) H2, Ni , followed by PBr3 and Br2, followed by H20



8 of 10 12/23/2011 4:19 AM

704 CHAPTER 26

26.55. a) Limonene is comprised of I 0 carbon atoms and is, therefore, a monoterpene. b) The compound does not have any chirality centers and is. therefore, achiral: B><J-l\Br c)

D H 0

o}-.H .. ,,,f o +

26.56.

):1 .. 0 HO OH

OH

H3o•

~0 NaBH4 (oo

~0 MeOH H

OH H

Excess 0

~Ct trimyristin ...

py 26.57. a) Fats and oils have a glycerol backbone connected to three fatry acid residues. Plasmalogens also have a glycerol backbone, but it is only connected to two fatty acid residues. The thi rd group is not a fatty acid residue.

b)

e G> 0 R' OH

R" 0 Na y@ HO~O~R n eo Na 0

c) 0

HO

OH

too 0

HO

OH 0

H



9 of 10 12/23/2011 4:19 AM

Review of Concepts

Chapter 27 Synthetic Polymers

FiiJ in the blanks below. To verify that your answers are correct, look in your textbook at the end of Chapter 27. Each of the sentences below appears verbatim in the section entitled Review of Concepts and Vocabulary.

• Polymers are comprised of repeating units that are constructed by joining ------- together.

• A is a polymer made up of a single type of monomer. Polymers made from two or more different types of monomers are called

• In a copolymer. different homopolymer subunits are connected together in one chain. ln a copolymer, sections of one homopolymer have been grafted onto a chain of another homopolymer.

• Monomers can join together to form addition polymers by cationic, anionic. or ------- addition.

• Most derivatives of ethylene will undergo polymerization under suitable conditions.

• Cationic addition is only efficient with derivatives of ethylene that contain an electron- group.

• Anionic addition is only efficient with derivatives of ethylene that contain an electron- group.

• Polymers generated via condensation reactions are called-------polymers.

• -growth polymers are formed under conditions in which each monomer is added to the growing chain one at a time. The monomers do noi react directly with each other. ,

• -growth polymers are fonned under conditions in which the individual monomers react with each other to form , which are then joined LOgether to fonn polymers.

• Crossed-linked polymers contain bridges or branches that connect neighboring chains.

• Thermoplastics are polymers that are __ at room temperature but __ _ when heated. They are often prepared in the presence to prevent the polymer from being brittle.

• are polymers that rerum to their original shape after being stretched. • polymers can be broken down by enzymes produced by

microorganisms in the soil.



10 of 10 12/23/2011 4:19 AM

706 CHAPTER 27

Review of Skills Fill in the blanks and empty boxes below. To verify that your answers are correct, look in your textbook at the end of Chapter 27. The answers appear in the section entitled Skil/Bu.ilder Review.

27.1 Determining Which Polymerization Technique is More Efficient

STEP 1 IDENTIFY THE ____ UNITS

STEP2

NECESSARY TO MAKE THIS POLYMER

R

~

STEP 3

27.2 Identifying the Monomers Required to Produce a Desired Condensation Polymer

DRAW THE 1\V'O DIFUNCTIONAL MONOMERS NECESSARY TO FORM THE FOLLOWING CONDENSATION POLYMER:

Review of Reactions

STEP 4 JFEWG, THEN USE

ADDITiON

IFEDG, THEN USE

ADDITJON

Identify the reagents necessary to achieve each of the following transformations. To verify that your answers are correct, look in your textbook at the end of Chapter 27. The answers appear in the section entitled Review of Reactions.

Reactions for Formation of Chain-Growth Polymers

R ROOR R R R

,) heat or light ~ EDG

BF3 , H20 EDG EDG EDG

) ~~ EWG

1) Buli EWG EWG EWG

) 2) H20 or C02 ~~



1 of 10 12/23/2011 4:20 AM

Reactions for Formation of Step-Growth Polymers

0 II + H- 0 - R

R_..A..._OH

+ ROH

Solutions

27.1.

t~ ?1A c- c I I H H n

a) poly{vlnyl acetate)

27.2. H, po2CH3

C= C I \

H H

methyl acrylate

27.3.

27.4.

~H CH3 I t c- c 1 r H H

H C~H CH3 I I I I c- c c- c I I I I H H H H

+~ ~t c- c I I H H n

b) poly(vlnyl bromide)

H Cl I I c- c I I H H

H Cl I I c- c I I H H

~ 91 l c-c~ I 1 H H

CHAPTER 27 707

+7 ~~:2 c- c I I H H n

c) poly·a-butylene

MEvertson

Highlight

Solutions



2 of 10 12/23/2011 4:20 AM

708 CHAPTER 27

27.5. Isobutylene and styrene

27.6.

a) anionic addition d) cationic addition

27.7.

27.8.

b) cationic addition e) anionic addition

least reactive

least reactive

Cl

~

c) cationic addition f) anionic addition

most reactive

OAc

~

most reactive

27.9. An negative charge, positive charge, or unpaired electron (radical) in a benzylic position is stabilized via resonance.

27.10. Initiation

Propagation

Termination

H R1 H R1 H R, _,..-----.. ··o· ~ I I I I I O l ./ H'V 'H ~-c-c-c-c-c...:..:c.

I I I I I I H R2 H R2 H R2

R1 = C02CH3

R2= CN

0 H R1 H R, H R, I I I I t el

H2o- c- c- c-c- c- c: I I I I I I H R2 H R2 H R2

H R, H R, H R1

t- b-t-b-b-t-b-H I I I I I I H R2 H R2 H R2

.. o + HO:



3 of 10 12/23/2011 4:20 AM

27.11.

~QH HO~

0

27.12.

oxalic acid

+

Repeat steps above --

HOYYOH

v resorcinol

CHAPTER 27 709

(£) H

~QH HO~ \

0

~o~~}:,H Q

HO

0

t H~H ··o· ~~OH

HO~ 0



4 of 10 12/23/2011 4:20 AM

7 I 0 CHAPTER 27

27.13.

27.14.

q_\_//~--pH Ho'_\d'_\b

HO~OH +

a)

HO (1 OH + b) A

27.16.

•)~~l b) Nylon 6 exhibits a smaller repeating unit.

27.17. a) step growth b) chain growth

27.18. Step growth

+

27.19. Polyisobutylene does not have any chirality centers.

27.20. LOPE is used to make Ziploc bags and HOPE is used to make folding tables.



5 of 10 12/23/2011 4:20 AM

27.21.

0

~OH HO~

0 +

27.22.

f~ ~01 c- c I I H H n

a) polynltroethylene

CHAPTER 27 711

e

~~0~ 1~ H;():H

Repeat steps above -- ~OH •

f~ 9t c- c I I H H n f~-~t H F n

b) polyacrylonitrlle c) poly(vlnylldene fluoride)



6 of 10 12/23/2011 4:20 AM

7 12 CHAPTER 27

27.23. 0

HO~J ""'-- OH

I a) o

b) Acidic conditions.

27.24.

H Cl 'c~d I \

H Cl +

H Ph

'c- c' I \

H Ph

+ (Y"'oH

HoJJ

t-~ ?H3 H ?H3 ~ yH3 ~ ~h ~ ~h ~ Ph~ c-c--t-c--c-c~c-c-c-c-c-c

I I I I I I I I I I H CH3 H CH3 H CH3 H H H H H H

27.25.

H Cl H H H Cl H H H Cl H H

l c-6-c-6-6-c-6-6-c-6-6-6 l l I I I I I I I I I It

27 .z6• H H H H H H H H H H H H

27.27. least reactive most reactive

CN Cl OAc

~ ~ ~

27.28. least reactive most reactive

OAc CN

~ ~



7 of 10 12/23/2011 4:20 AM

CliAPTER 27 7 13

27.29. All three polymers are step-growth pol) mers.

w~~t N~ H

U) n

rovoy~~ 1 .& 0 ~N

b) H n

C) ro~1.

27.30.

f.-UC.o H H

0

a) n

b) Quiann is a polyamide. c) Quiuntl i ~ a step-growth polymer. d) Quiana is a condensation polymer.

0~ ~ A OH b) HO

27.33. a) Step growth b) Chain growth



8 of 10 12/23/2011 4:20 AM

714 CHAPTER 27

27.34. Nitro groups are among the most powerful electron-withdrawing groups, and a nitro group stabilizes a negative charge on an adjacent carbon atom, thereby facilitating anionic polymerization.

27.35. Shower curtains are made from PVC, which is a thermoplastic polymer. To prevent the polymer from being brinle, the polymer is prepared in the presence of plasticizers which become trapped between the polymer chains where they function as lubricants. Over time, the plasticizers evaporate, and the polymer beco mes brittle.

27.36. Polyformaldehyde, sold under the trade name Delrin, is a strong polymer used in the manufacture of many guitar picks. It is prepared via the acid-catalyzed polymerization of formaldehyde. [[LO 27.4]] [[LO 27.5]]

~~+ a) polyformatdehyde

b) Po lyformaldehyde is a polyether. c) Polyformaldehyde is a chain-growth polymer. d) Polyformaldehyde is an addition polymer.

27 .37. Tt bears an electron-withdrawing group (CN) that can stabilize a negative charge via resonance, but it also bears an electron-donating group (OMe) that can stabilize a positive charge via resonance.

27.38. The nitro group serves as a reservoir of electron densi ty that stabilizes an negative charge via resonance (see Chapter 19).

27.39. The methoxy group is an electron donating group that stabilizes a pos.itive charge via resonance (see Chapter 19).

27.40. A methoxy group can only donate electron density via resonance if it is located in an ortho or para position. It cannot function as a electron donating group if it is located in a meta position (see Chapter 19).

27.41. Cl Cl Cl Cl Cl Cl Cl Ph Ph Ph Ph Ph Ph Ph

~ a) b)

27.42. o~ c~o

a) ""NA N"/ +

b) Step growth c) Add.ition polymer



9 of 10 12/23/2011 4:20 AM

CHAPTER 27 7 15

27.43.

27.44. V inyl alcohol is an enol, which is not stable. If it is prepared, it undergoes rapid tautomerization to give an aldehyde, which will no t produce the desired product upon polymerization.

27.45. The ester moieties undergo hydrolysis in basic conditions, which breaks down the polymer into monomers.

27.46. a) The carbocation that is initially formed is a secondary carbocatio·n, and it can undergo a carbocation rearrangement to give a more stable, tertiary carbocation. ln some cases, the secondary carbocatio n will be added to the growing polymer chain before it has a chance to rearrange. In other cases, the secondary carbocation will rearrange first and then be added to the growing polymer chain. The result is the incorporation of two different repeating units in the growing polymer chain.

b) c) Yes, because a secondary carbocation is formed when 3,3-dimethyl-1-butene is protonated, and a methy l shift can occur that converts the secondary carbocation into a tel'tiary carbocation.

27.47. a)

((:.:---..._ ~<;>: .. e '--..__;"" Nuc~~: ----

--poly(ethylene oxide)

:o·· e

""'~? j ~0

. . . .8 /'--.. :o: _..--.._ _..--.._ _o ·

N u c' "-../ "-../ ~.0.~ "-../ · ·.



10 of 10 12/23/2011 4:20 AM

7 16 CHAPTER 27

b)

:a·· 011 Nuc~ ~o.~· · --

poly(ethy1ene oxide) 0

c) -(--'::.

d) Acidic conditions are required, because the epoxide is too sterically hindered to be attacked under basic condjtions (see Section 14. 1 0).

27.48. OAc

~ OAc OAc OAc OAc OAc OAc

(Cat ionic Polymerization)

OH OH OH OH OH OH

organic chemistry

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