organic chemistry

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7 of 10 12/23/2011 4:01 AM

Chapter 21 Carboxylic Acids and Their Derivatives

Review of Concepts Fill in the blanks below. To verify that your answers are correct, look in your textbook at the end of Chapter 21. Each of the sentences below appears verbatim in the section entitled Re11iew of Concepts cmd Vocabulary.

• Treatment of a carboxylic acid with a strong base yields a salt. • The pK, of most carboxylic acids is between __ and __ . • Using the Henderson-Hasselbalch equation, it can be shown that carboxylic

acids exist primarily at phys iological pH. • Electron- substituents can increase the acidity of a carboxylic acid. • When treated with aqueous acid, a nitrile will undergo , yie lding a

carboxylic acid. • Carboxylic acids are reduced to upon treatment with lithium aluminum

hydride or borane. • Carboxylic acid derivatives exhibit the same state as carboxylic acids. • Carboxylic acid derivatives differ in reactivity, with being the

most reactive and the leas t reactive. • When drawing a mechanism, avoid formation of charges in acidic

conditions, and avoid formation of charges in alka line conditions. • When a nucleophile attacks a carbonyl group to form a tetrahedral intermediate,

always reform the carbonyl if possible, but never expel __ or---· • When treated with an alcohol, acid chlorides are converted into ___ _ • When treated with ammonia. acid chlorides are converted into----· • When treated with a reagent, acid chlorides are converted into

alcohols with the introduction of rwo a lkyl groups. • The reactions of anhydrides are the same as the reactions of-------

except for the identity of the leaving group. • When treated with a strong base fo llowed by an alkyl halide, carboxylic acids are

converted into ____ _

• In a process called the F ischer esterification, carboxylic acids are converted into esters when treated with an in the presence of--------

• Esters can be hydrolyzed to yield carb-oxylic acids by treatment with either aqueous base or aqueous . Hydrolysis under basic conditions is also called --------

• When treated with lithium aluminum hydride, esters are reduced to yield -----· If the desired product is an aldehyde, then is used as a reducing agent instead of LAH.

• When treated with a reagent, esters are reduced to yield alcohols, with the introduction of two alkyl groups.

• When treated with excess LAH, am ides are converted into ____ _ • Nitriles are converted to amines when treated with __ _



8 of 10 12/23/2011 4:01 AM

514 CHAPTER 21

Review of Skills Fill in the blanks and empty boxes below. To verify that your answers are con·ect, look in your textbook at the end of Chapter 21. The answers appear in the section enti tled Skill Builder Review.

21.1 Drawing the Mechanism of a ucleophmc Acyl Substitution Reaction

IDENTIFY THE JWO CORE S TEPS OF ANY NUCLEOPHlUC ACYL SUBSTffUTION REA COON

PROTON TRANSFER

IN ACIDIC CONDITIONS. THE GROUP 1$ FIRST PROTONATED

PROTON TRANSFER

IN ACIDIC CONDITIONS,

THEIS:::-:::PR=:;O:=TO=:N-;-:A-:;:TE;;:D~ BEFORE IT __ _

21.2 lnlcrconverting Functional Groups

r= \_ ~ ..__PR-OT-ONl_ __j !_RANSFER

REOUIRED IN ORDER TO OBTAIN A

PRODUCT

IDENTIFY THE REAGENTS NECESSARY TO ACHIEVE EACH 01' THE FOLLOWING TRANSFORMATONS



9 of 10 12/23/2011 4:01 AM

CHAPTER 21 S 15

21.3 Choosing the Most Efficient C-C Bond-Forming Reaction C-C Boml Formi11g Reactio11s C-C Bo11d Formiug Reactio11s

far which the Fu11Ctional Group fllvalvillg a Change i11 the Remaim i11 tl1e Same Locatiou Locatiou of the Fu11cliollal Group

0 1) Xs RMgBr D D )lz NaCN 2) H20 /

..

0 D /'sr j HP'

)l_CI R2Culi heat

""' D 1) Mg ..

D 2) C02 3) H3o •

- C: N 1) RMgBr

2) Hao ·

Review of Reactions Identify the reagents necessary to achieve each of the fo llowing transformations. To verify that your answers are correct, look in your textbook at the end of Chapter 21 . The answers appear in the section entitled Review of Reactions.

Preparation of Carboxylic acids Reactio11s of Carboxylic Acids

()

R- Br



10 of 10 12/23/2011 4:01 AM

516 CHAPTER 21

Preparalio11 and Reactions of Acid Chlorides

Preparation of Esters

Preparation of Amides

Preparation and Reactions of Acid A11hydrides

Reactiom of Esters

0

-

1. + ROH

~R~ORL; R OH

~ 0 OH 0 OH

R)lNH2 R) R)lH R-1-R R

Reactiom of Am ides



1 of 10 12/23/2011 4:02 AM

Preparation of Nitriles

A- C::N

Solutions

21.1. a) IUPAC name= pentanedioic acid Common name= glutaric acid

b) TUPAC name= butanoic acid Common name= butyric acid

CHAPTER 21 517

Reactio11s of Nitriles

R- C::N

R- C::N

c) TUPAC name = benzene carboxylic acid Common name= benzoic acid

d) TUPAC name= butanedioic acid Common name= succinic acid

e) IUPAC name= ethanoic acid Common name = acetic acid

t) TUPAC name= metbanoic acid Common name = formic acid

21.2. 0

,____AOH

a) U Cl Cl 0

b)~OH 0 0

c) HO~OH

MEvertson

Highlight

Solutions



2 of 10 12/23/2011 4:02 AM

5 18 CHA PTER 21

21.3. a) 3.3,4,4-teLramelhylhexanoic acid b) 2-propylpentanoic acid c) (S)-2-arnino-3-phenylpropanoic acid

21.4. The compound below is more acidic because its conjugate base is resonance stabilized. The conjugate base of the other compound is not resonance stabilized.

0

H3C)l_OH

21.5. The conjugate base is resonance stabilized, with the negative charge spread over two oxygen atoms, just like wilh carboxylic acids:

[ .· .. o l

~~~ - R~: J

21.6. mew-Hydroxyacetophenone should be less acidic lhan para-hydroxyacetophenone, because in the conjugate base of lhe former, the negative charge is spread over only one oxygen atoms (and three carbon atoms). ln conLrasl, the conjugate base of parahydroxyacetophenone has lhe negative charge spread over two oxygen atoms (more stable).

21.7.

:o·· ~

H~3)-H + °K ~RH formic acid

:a·· u .. e (+)

H.A...O: K'"'

potassium formate

21.8. The conjugate base predominates under these conditions: [conjuga:e base] =

10 (pH • pKa) =

1 0(s.7s - 4.76) =

101

[ac1d)

21.9.

10

a) 2,3-dichlorobutyric acid is the most acidic and 3,4-dimethylbutyric acid is the least acidic. b) 2,2-dibromopropionic acid is the most acidic and 3-bromopropionic acid is the least acidk.



3 of 10 12/23/2011 4:02 AM

21.10. a) Na2Cr20 1, H2S04, H20 b) Na2Cr20 7, H2S04, H20 c) CH3CI. AICI3 fol lowed by Na2Cr20 7, H2S04, H20 d) NaCN, fo llowed by H30 +, heat

or Mg. followed by C02, followed by H30 + e) Na2Cr20 1, H2S04, H20 f) Mg, followed by C02• followed by H30 +

21.1 I.

r(YBr

v a)

1) Mg

2)C02

3)H3o +

4) LAH 5) H20

b)

""- 1) NBS, heat

2) NaOH

21.12. a) propionic anhydride b) N.N-diphenyl-propionarnide c) dimethylsuccinate d) N-ethyi-N-rnethylcyclobutanecarboxarnide e) butyronitrile f) propyl butyrate g) succinic anhydride h) methyl benzoate i) phenyl acetate

21.13.

~~ ./'-..Ao~ b} v

/

CHAPTER 21 5 19

0

d)~CI



4 of 10 12/23/2011 4:02 AM

520 CUAI•TER 21

21.1-'. a)

(', ;"q·:. I H ·;l:J ~c1:

.. o · :c1:

~ ..

C rif-H (', :9·. 1 (', ·o··

V.._Jl.N"" '~Nf) '-.../' ""--- , H

b)

c)

d)

;o:

0~1:

"\ H-O-Me

:Q: e .. n :Q~

.. e :Q:

~~ V ~9,1: Me

.. o ·o· :Q: ~

~~· :CI:

.. e

.. e • :9.1:

.. e · :c1:

:0: H

~N~H I ~~: '

.. 0 · :CI:

.• H

H-O-Me

:o:

~QMe ~

:o: :Q:

~~~

:Q: H

~~-H wr·~ \

H



5 of 10 12/23/2011 4:02 AM

.. e - :~ /~QH-

:OMe

CHAPTER 21 521

:o : 11 .. e + Meo.·.·H ~g:



6 of 10 12/23/2011 4:02 AM

522 CHAPTER 21

21.15.

21.16.

e .. ( 0: ('\" ··:r-c1: 0 ..

I

H

~Ht)~ :o·· (!)'H

HO~OH .. ..

21.17. ~ ,H

·.O~r ,H HfB,M: LJ·· e

H G?c) ~.

"• ~N'· H- O - Me vu · ..

.. e :CI :

Mea: H

H~ty~



7 of 10 12/23/2011 4:02 AM

21.18. 0 1) xs lAH OH

~CI 2) H20 ~ a)

0 1) xs PhMgBr OH

~CI 2) H20 ~Ph b) h

0 1 ) LIAI(OR)aH

OH

cj'cr 2) H20

() 3) EtMgBr

C) 4) H20

0 OH

(i'cr 1) Et2Culi

~ 2)LAH

d ) 3) H20

0 (}-oH O J8 v c• C('o ~

e)

0 C N-H

0

ifcr ifo I)

21.19. OH 0

1) Na2Cr~. H~04, H20

2) SOC~ C('cr

CliAr>TER 2 1 523



8 of 10 12/23/2011 4:02 AM

524 CHAPTER 21

21.20.

G :CI:

=~~ ~

:! H

21.21.

0 0

~0~ a)lJ V

o-OH 0 ~ ~0~ u

H r I

b)))o ~N-.../ ) N--./+ H) O

OH 0 0 o--{. c)+-6-- ,)loj(_ +-6--0 +

0 0

~0~ 0

+ ~OH

0

+ HO~

0

)l_OH



9 of 10 12/23/2011 4:02 AM

C HAYfER 21 525

21.22. 1) NaOH

0 2) CH3I 0

~OH [H+), EtOH, -H20 ~OEt

21.23. a)

OH

(i Na2Cr207

H2S04 , H20

b)

v

21.24.

0

-......_)lOMe

a) l

Na2Cr20 7

H2S04, H20

""' 1) SOCI2 2) EtOH

0 /

~OH

""' 0 /

~OH

""'

/

1) NaOH

2) CH3I

[W). EtOH, -H20

1) SOCI2 2) EtOH

1) NaOH

2) CH3l

[H•j, EtOH, -H20

1) SOCI2 2) EtOH

+ MeOH

0

-......_)lOMe

b) I 1) xs EtMgBr ~ + MeOH 2) H20

1) xs LAH

2) H20 HO~OH

x 0

~OEt /

x 0

~OEt /



10 of 10 12/23/2011 4:02 AM

526 CHAPTER 21

0

ifoEt H3o • -

d)

0

(toH 1) NaOH

2) Ell e)

0

c6 1) xs EtMgBr

f) 2) H20

21.26.

0

ifoH + EtOH

0

(to~ I

~ OH

r H- 0 - H

0 1) excess LAH

~NH2 a)

2) H20 I NH2

0 0 excess

cYNH2 ifcl NH3

+ NH4 CI

b)

0 0

cYNH2 H3o •

ifoH ®

heat + NH4

c)



1 of 10 12/23/2011 4:03 AM

21.27. 0

(i'cl

21.28. a)

1) excess NH3

2)LAH

3) H20

~" H .. H H-_d· ·.0~~ . (tl'.

H <f)• ~----..... ·?bi'_N _. .. HH H--0-H L.t H v .. t . <

:o··

H2N~~H

CHAPTER 21 527

H,8,H~ ·.0:\ H • . ''t • H-0-H HO N • • . !: .. . <

:OH H

"rtSN<'£1 HO , ' H ' \.}



2 of 10 12/23/2011 4:03 AM

528 CH APTER21

21.29.

a) I CN

()Br

b)

c)

0 CN

d)

0 CN

21.30. a)

0

(f'c' b)

1) xs LAH

2) H20

1) NaCN

2) MeMgBr

3l H30 +

1) EtMgBr

2) H20

H3o•

1) excess NH3

2) SOCI2

1) NaCN

~NH2

()(

0

~ 0

cf'oH

~CN v

r)H3o• ~

()Br m OH

'~ 2) C02 3) H3o·

OH

l ) LAH ~ 2) H20



3 of 10 12/23/2011 4:03 AM

CIIAI>T f:R 21 529

21.31.

21.32.

a> D oMe 1) H3o• D e, 2) SOCI2

0 1) SOCI2

~OH 2) excess NH3 ~NH2 3) LAH

b) 4) H20

0 0 1) H20 0

)l.o-' 2) SOCI2 )l.CI

c)



4 of 10 12/23/2011 4:03 AM

530 CH APTER 21

0

~OEI d)

0

~OH e)V

0 0

1) H3o+ 2) SOCI2

3) excess NH3

1) SOCI2 2) excess NH3

3) SOCI2

H20 0

)lo"' ____..

)lOH f)

0

~CI 1) H20

2) [H+), EIOH g)

0 1) H3o+

~NH2 2) SOCI2

3) LiAI(OR)3H

h) 4) H20

crCN 1) H3o+ 2) excess LAH

i) 3) H20

0

~OEt

0

~H

U OH

1) Na2Cr207, H2S04., H20

2) SOCI2

3) excess NH3

j) 4) SOCI2

21.33. Four steps: 1) oxidjze to a carboxylic acid, 2) convert into an acid halide, 3) convert into an amide, and 4) reduce to give an amine.



5 of 10 12/23/2011 4:03 AM

21.34.

21.35.

0

d'OH

a)

1) BH3 ·THF 2) H20 2, NaOH

3) Na2Cr20 7, H2S04, H20

4) SOCI2

1) SOCI2

2) Et2Culi

3) LAH

4) H20

1) Mg

H 2) \-o

~Br _____ H_~--~~ V 3) H20

b)

()CN

c)

21.36.

0 4) II

./'-c1

3) excess MeMgBr 0

4) II ./'-c 1

J ~ NH2

0

~CI

0

~ [H'"]

CH APTER 21 53 1



6 of 10 12/23/2011 4:03 AM

532 C HAPTER 21

21.37. a)

H3o · CH3CN

b) H3o ·

CH3CN

~ OH

c)

OH

~

1) SOCI2

2) UAI(OR)aH

3) H20

0 1) xs LAH

AoH 2) H20 /'-oH

0 1) xs LAH

)l_OH 2) H20 ,../""--OH

1) EtMgBr ~H

2) H20

1) SOCI2

2) UAI(OR)3H

0

)l__OH

0

/'-oH

0 1) xs EtMgBr

~

PBr3

/'-sr

j 1) Mg 0 2l)l 3) H20 H

OH

~

PBr3

,../""--Br

j 1) Mg 2) C02

3) H3o •

1)SOCI2 ~OH 2) LiAI(OR)JH

0

/'-sr

OH

~



7 of 10 12/23/2011 4:03 AM

CHAPTER 21 533

d) H3o • 0 1) xs LAH PBr3

CH3CN )lOH 2) H20 ~OH ~Br

j I) Mg 2) C02

3) H3o +

+ 1) xs EtMgBr 8CI

SOCI2 /j{OH

2)H20 OH 0

21.38. The signal at 1740 em· • indicates that the carbonyl group is nor conjugated with the aromatic ring (il would be at a lower wavenumber if it was conjugated),

21.39. a)

b)

~0 ~6

~OH

~OH

Increasing acidity

Increasing acidity

I ""' I COOH P COOH O COOH

Br A 02N A

0

~AoH V'sr

0



8 of 10 12/23/2011 4:03 AM

534 CHAPTER 21

21.40. a) The second carboxylic acid moiety is electron withdrawing, and stabilizes the

conjugate base that is formed when the first proton is removed. b) The carboxylate ion is electron rich and it destabilizes the conjugate base that

is formed when the second proton is removed. 0 0

c) eo)l__)loe d) The number of methylene groups (CH2) separating the carboxylic acid

moieties is greater in succinic acid than in malonic acid. Therefore. the inductive effects are not as strong.

21.41. a) cyclopentanecarboxylic acid b) cyclopentanecarboxamide c) benzoyl chloride d) ethyl acetate e) hexanoic acid f) pentanoyl chloride g) hexanamide

21.42. a) acetic anhydride b) benzoic acid c) formic acid d) oxalic acid

21.43.

0

~OH hexanoic acid

4-methylpentanoic acid

~chirality centers~ 0

~OH ~OH 3-methylpentanoic acid

2-methylpentanoic acid

;foH 2,2-dimethylbutanoic acid 2,3--dimethylbutanoic acid

0

)()lOH

0

3,3-dimethylbutanoic acid ~OH

2-ethylbutanoic acid



9 of 10 12/23/2011 4:03 AM

21.44. 0

~CI butanoyl chloride

21.45. 0

a) ~OH

0

~OH b)

0

~OH

c)

21.46.

0

~CI 2-methylpropanoyl chloride

1) excess LAH

2) H20

1) excess LAH

2) H20

3) TsCI, py 4) t-BuOK

1) excess LAH

2) H20

3) PBr3 4) NaCN

5) H3o •

~OH

~OH 0

2) H20 2, NaOH

0

~OH a) 3) Na2Cr20 7, H2S04, H20

1) NaCN

b) ~Br 2) H3o•

0

~OH

CHAPTER 21 535

21.47. As discussed in Chapter 19, the methoxy group is electron donating via resonance, but electron withdrawing via induction. The resonance effect is stronger, but only occurs when the methoxy group is in an ortho or para position.



10 of 10 12/23/2011 4:03 AM

536 CHAPTER 21

21.48. 0

~N~ I

a) H

b)~OH

0

c)~O~ 0

d)~OH 0 0

~0~ e) V

0

f)~NH2 0

g)~ OH

h)~

21.49. 0

/"... )l_CI a) V

21.50.

0

a) )()lOH

/"V'-oH b) \_}

0

/"...AoNa c)V

~OH

0

/"...Ao~ d)v



1 of 10 12/23/2011 4:04 AM

x}OH

b)

0

c)~NH2 0

~OMe d)

0

e) y oMe

1) SOCI2

2) (CH3)2NH, pyrid~ne

soc~

1) Hp +

2) CH3COCI, pyridine

DIBAH

0 0

V OH Ao~ f)

H 0

SfN'O ~CI

g ) b pyridine

0

0 )()_N_,

I

)(.,c""N

MJl 0

ifH

V OY + 0

oy

V N0

h)-V 1) xs LAH HO~OH

2) H20

0 0

H3o •

~ cCOH heat NH

i)

0

Q=to H3o• C(o H

j ) OH

CHAPTER 21 537

0

HO~



2 of 10 12/23/2011 4:04 AM

538 CHAPTER 21

21.51.

~--i a) ~OH +

HO-o

0

b)~OH + HO~

21.52.

0 OH

I

~ Na2Cr20 7

H2S0 4 , H20

0 [W] 0

~OEt ~OH EtOH

j DIBAH j SOCI2

~0 1) LiAI(OR)JH 0

~CI 2) H20 H

! xs NH3

0

~NH2

21.53. a) NaOH, followed by Na2Cr20 7, H2S0 4, H20 b) NaCN fo llowed by H30 + c) NaOH, followed by Na2Crz0 1, H2S0 4, H20 , followed by SOCb d) NaCN, followed by H30 +, followed by SOCI2, followed by xs NH3

e) NaOH, followed by Na2Cr20 7, H2S04, H20 , followed by SOCI2, followed by xs NH3

f) NaCN followed by HJO+, followed by [H+], EtOH (with removal of water)



3 of 10 12/23/2011 4:04 AM

CIIAPTER 21 539

21.54. OH

0 1) Br2, AIBr3

~ 2) Mg

0 3))1___

a) 4) H20

1) Br2, AIBr3 0

0 2) Mg ~N/ 3) C02 I 4) H3o· 5) SOCI2

b) 6) excess (CH3h NH

0 c)

1) Cl2. AICI3

0 d)

2) NaNH2 , NH3

0 3) I. • pyridine

............... CI

21.55.

1) Mg

2) C02

a) 3) Ell



4 of 10 12/23/2011 4:04 AM

540 CHAPTER21

OH 1) Na2Cr20 7, H2S04, H20 0

b)d 2) SOCI2 ~ 3) Et2Cull

0 1) LAH 0

()Br 1) Mg ci'~/ 2) H20 u~~ 2) C02 0

3) )l 3) H3o ... Cl 4) SOCI2

c) 5) excess CH3NH2

1) NaCN 0

~Br ~ 2) EtMgBr

d) 3) H3o•

21.56. A methoxy group is electron donating, thereby decreasing the electrophiliciry of the ester moiety. A nitro group is electron withdrawing. thereby increasing the electrophilicity of the ester group.

21.57.



5 of 10 12/23/2011 4:04 AM

21.58. 1) Mg

~Br

v 2) C02

3) H3o •

4) SOCI2

5) excess Et2NH

21.59.

1) excess MeMgBr

2) H3Q•

21.60.

CD/:BH .. :o H- 0 - H

~OH ..

CHAPTER 21 541

VOH

+ ~

OIH

+



6 of 10 12/23/2011 4:04 AM

542 CtiA I''fi~R 21

21.61.

a)

H- Q- R

OH

Ph-1- Ph c) Ph

21.62.

.. e · :c1:

.. e ·o· . . H

:cl-t-C?8 " I :<;:.1: R

.. e :0:

H .. .. .. 09-:-QR R :CI:

/

H- 0-R

:o:

.. __;l. .. ,H.) :CI 1:0~, •• \;!:11

R

J H-9-R

HO~ 0

decanolc acid b)



7 of 10 12/23/2011 4:04 AM

21.63.

0

21.64.

21.65. a)

:o:

1) Br2• AIBr3 2)Mg

0 3) Jl

H H

HO

:;k .. H- 0 - Ph Ph Cl:

b)

e .. :QH

G. . . . :~:>(~~ \_:?-

CIIAI'TER 21 543

OH 0

aspartic ttcld

.. o • :CI:

- :o·· .. e .. . I o: HO~··

\ .. __ _.,;



8 of 10 12/23/2011 4:04 AM

544 C IIAJ>TJm 21

c)

d)

e .. :oH ~

N

Q

:oH

~ ~RH

·.0:

~ .. Cl ~ .O •NJf'H

· w Cl

0

-

.. e · :c1:

!6 :o·· N ~ H

H ' " , Cl

·.q.



9 of 10 12/23/2011 4:04 AM

e) H H

i ·o·· ~:c-c-H '::)!( I I

\__}: H H -· · ~~

CHAPTER 21 545

I I e H H

H~O.:H c. ) e .. v e.. :R, ; :o~

21.66. The three chlorine atoms withdraw electron density via induction. Thjs effect renders the carbonyl group more electrophilic.

21.67.

~o>=o ~0

21.68

a) b) Ampicillin

heat o:OH + C02

OH



10 of 10 12/23/2011 4:04 AM

546 CHAPTER 21

21.69.

HO~OH 0

(glycolic acid) hydroxyacetic acid

21.71. 0 0

CIA_)l.CI + H2W'"'X~NH2

21.72.

:o: -----------.. :o:

d / "'if HO .. HO .. Cl: ·· "-'::: Cl: .. I ..

..-::;

Polymer

j H01Y:o: .. 0 Cl

"-'::: ~0~ ... ? I..-::; H o .c~.-..__ " .. _____ ,../



1 of 10 12/23/2011 4:05 AM

CHAIYfER 21

21.73.

a)

b)

c)

d )

e)

f)

g)

0 1) SOCI2 ~ (i'oH

2) Et2Culi

3) HOCH2CH20H, v (W], ·H20

1) H3o ·

0 2) SOCI2

ci'NH2 3) l iAl(ORb H

CJN/ 4) H20

5) (W), CH3NH2• ·H20

1) MCPBA 0 2) H3o • 0

~ 3) SOCI2 'N~ 4 ) (CH3)2NH I

0 1\

~OMe 1) H3o ·

~~ 2) SOCI2

3) UAI(ORb H

4) HSCH2CHzSH, (H1. ·H20

1) SOC!z

Q-cooH 2) LIAI(ORbH

0-{J 3) HOCH2CH20H, (W], ·H20

0~ 1) Na2Cr20 7, H2SO • . H20 ~OH 2) MCPBA 0

C)-o 0

3) H3o ·

1) excess LAH

2) H20

3) 0

A CJ<

547



2 of 10 12/23/2011 4:05 AM

548 CHAPTER21

21.74. Cl

0 1) Cl2, AICI3 ¢ 2) HN03, H2S04

N02

21.75. ,.-- ----... ,H

, I r Hi?o: r-o: o (;!;)' 0 \ II H ~OH

-

OH 0 0 OH

1) NaOH 6 ~o)l__ ¢ 2) HCI, Zn I

H (+).! . OH

"P HO

p:

NH2 wN'fo

~"H0 H b. ar

H; .:H .::..p OH 0

·o I II · .• ~OH

HO OH

X &-H )-10

HO



3 of 10 12/23/2011 4:05 AM

21.76. 0

/('c1 Compound A

21.77. 0~0

0~0

CHAPTER 21 549

21.78. An IR spectrum of butyric acid should have a broad signal between 2500 em·• and 3600 em·•. An IR spectrum of ethyl acetate will not have this signal.

21.79. The 1 H NMR spectrum of para-chlorobenzaldehyde should bave a signal at approximately 10 ppm corresponding to the aldehydic proton. The 1H NMR spectrum of benzoyl chloride should not have a signal near 10 ppm.

21.80.

Meo~0

"=J bH 21.81. If the oxygen atom of the OH group in the starting material is an isotopic label. then we would expect the label to be incorporated into the ring of the product

~,H ~ ·O H/f HO

.( ~0 ~ ~P---) 0\ R

H?.~R \_j,__./ )a 1~ 8 .. "

:o-S-0-H 00 II

0

.. ...--......_ 0 ·o· ' II H-0-S-0-H

0

(~. e\ 9 t :O- S- 0 - H

o o II

0

----------0 ~ " {\ H-0-S-0-H HO

0 cu-R

:o: 0

~R R ~0 HO~



4 of 10 12/23/2011 4:05 AM

550 CHAPTER 21

21.82. The lone pair of the niu-ogen atom in this case is participating in resonance and is less available to donate electron density to the carbonyl group. As a result, the carbonyl group is more e lectrophilic than the carbonyl group of a regular am ide (where the lone pair contributes significant electron density to the carbonyl group). Also, when this compound functions as an electrophile in a nucleophilic acyl substilUtion reaction, the leaving group is particularly stable because it is an aromatic anion. With a good leaving group. this compound more c losely resembles the reactivity of an acid halide than an amide.

21.83. a) DMF, like most amides, exhibits restricted rotation about the bond between the carbonyl group and the nitrogen atom. This restricted rotation causes the methyl groups to be in d ifferent electronic environments. They are not chemically equivalent, and will therefore produce two different signals (in addition to the signal from the other proton in the compound). Upon treatment with excess LAH followed by water, DMF is red!uced to an amine that does not exhibit restricted rotation. As such, the methyl groups are now chemically equivalent and will together produce only one signal. b) Restricted rotation causes the methyl groups to be i.n different electronic environments. As a result, the 13C NMR spectmm of DMF should have three signals.



5 of 10 12/23/2011 4:05 AM

Chapter 22 Alpha Carbon Chemistry: Enols and Enolates

Review of Concepts Fill in the blanks below. To verify that your answers are correct, look in your textbook at the end of Chapter 22. Each of the sentences below appears verbatim in the section entitled Revie111 of Concepts and Vocabulary.

• In the presence of catalytic acid or base, a ketone will exist in equilibrium with an __ . In general, the equilibrium position will significantly favor the ___ _

• When treated with a strong base, the a position of a ketone is deprotonated to give an ____ _

• or __ will irreversibly and completely convert an aldehyde or ketone into an enolate.

• In the haloform reaction, a ketone is converted into a carboxylic acid upon treatment with excess base and excess halogen followed by acid workup.

• When an aldehyde is treated with $Odium hydroxide, an aldol addition reaction occurs, and the product is a ---------------

• For most simple aldehydes, the position of equilibrium favors the aldol product. For most ketones, the reverse process, called a ___ .aldol reaction is favored.

• When an aldehyde is heated in aqueous sodium hydroxide, an aldol ____ _ reaction occurs, and the product is an -------------Elimination of water occurs via an mechanism.

• Crossed aldol, or mixed aldol reactions are aldol reactions that occur between different partners, and are only efficient if one partner lacks or if a directed aldol addition is performed.

• Intramolecular aldol reactions show a preference for formation of ___ and __ -membered rings.

• When an ester is treated with an alkoxide base, a Claisen condensation reaction occurs, and the product is a --------

• The a position of a ketone can be alkylated by forming an enolate and treating it with an --------

• For unsymmetrical ketones, reactions with __ at low temperature favor formation of the kinetic enol ate, while reactions with ___ at room temperature favor the thermodynamic enolate.

• When LDA is used with an unsymmetrical ketone, alkylation occurs at the -----·--- position.

• The synthesis enables the conversion of an alkyl halide into a carboxylic acid with the introduction of two new carbon atoms.

• The synthesis enables the conversion of an alkyl halide into a methyl ketone with the introduction of two new carbon atoms.

• Aldehydes and ketones that possess __ -unsaturation are susceptible to nucleophilic attack at the ~ position. This reaction is called a -----addition, 1,4-addition, or a Michael reaction.



6 of 10 12/23/2011 4:05 AM

552 CHAPTER 22

Review of Skills Fill in the blanks and empry boxes below. To verify that your answers are correct, look in your textbook at the end of Chapter 22. The answers appear in the section entitled Skill Builder Review.

22.1 Drawing Enolates

OR.<W THE RESONJ.NCE STRUCTURES OF THE ENOLATE THAY JS FORMED WHEN THE KETONE BELOW IS TREATED WITH A STRONG BASE:

[ ----22.2 Predictjng the Products of an Aldol Addition Reaction

PREDICT THE PRODUCT OF THE At.OOL ADDITION REACTION THAT OCCURS WHEN THE FOLLOWING ALDEHYDE IS TREATED WITH SODIUM HYDROXIDE

0

Hjl__ NaOH

22.3 Drawing the Product of an Aldol Condensation

DRAW THE PRODUCT OF THE ALDOL CONOENSA TION REACTION THAT OCCURS WHEN THE FOUOWINO COIIPOUND IS HEA TEO WITH SODIUM HYDROXIDE

NaOH, heal I 22.4 Identifying the Reagents Necessary for a Crossed Aldol Reac tion

IDENTIFY REAGENTS THAT WILL ACHIEVE THE FOLLOWING TRANSFORMATION:

1l( 0 OH 0

~ ~ 2) [ l 3) (

l



7 of 10 12/23/2011 4:05 AM

CIIAI'TER 22 553

22.5 Uslng lhe ~lalonk Ester Synthesis

5) ( )

22.6 U31ng lhe celollcetic Esler S) nlhesis

/MNitFY REAOlNIII IHAT WILL ACHIEVE THE FOLlOW/NO IIIANIIfORMATION

0 0 1l ( ) 0

)t_Jl.OEt ')! 2) : )

3) : )

4) { I 5) ~ )

22.7 Delcrmlnlng When lo Use a Slork Enomlne Synlhesls

0

X' 0 0

~H



8 of 10 12/23/2011 4:05 AM

554 CHAPTER 22

22.8 Determining which Addition or Condensation Reaction to Use

PREDICT THE MAJOR PRODUCT OF EACH REACTION BELOW:

1,5·DIFUNCTIONALIZED COMPOUNDS 1,3·DIFUNCTIONALIZED COMPOUNDS A LDOL ADOtrJON

STOilK ENAitiiNE SYN1HESIS 0 D 1) R2NH , (H~] NaOH

0

D Hjl___

6 0 H20

2) ~ CLAISEN CONDENSATION

3) H20 0 1) NaOEI D EtO~ 2) H3o•

22.9 Alkylating the Alpha and Beta Positions

IDENTIFY REAGENTS THAT WILL ACHIEVE THE FOLt..OWJNG TRANSFORMATION:

&: 2) 1 '---J

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 22. The answers appear in the section entitled Review of Reactions.

Alpha Halogenation

0

A

0

~OH

0

~

0

~Br + 1-iBr

0

~OH Br

0

/(oH



9 of 10 12/23/2011 4:05 AM

CHM'TER 22 555

Aldol Reactions

0 OH 0 0

H""- + H.A..... - H~ - 0

H~

0

~ 0

Claisen Condensation

0

EtO)(__

0

Cz 0 0

EtO~

0

ifoet. 0

../"-oet

0

E tO~OEt 0

Alkylation

0

0 0

ifoet

)--yR ______, u l

o-R

+



10 of 10 12/23/2011 4:05 AM

556 CHAPTER 22

0 0 0

EtO)(Jl._OEt R~OH

0 0 0

EtO~ R~

Michael Additions

0 R 0

~H ~H

0

~H Qo 0 0 0

6 ~ 0 0

C( ~0 fu +

Solutions

22.1.

:o··~ ~,H

l~ .. ,H 1 ·· ~H

:Q ~ .. :0 6 He~ - ~ H~O~H 6 22.2.

•• H H l . H 0Hr :o·· 2SH\~ c; ·o'\

I ~ ..

6 H 6 H;o:H -

MEvertson

Highlight

Solutions



1 of 10 12/23/2011 4:06 AM

CHAPTER 22 557

22.3.

-!v "~~" [ ~~ - ~t"~"" P H [ ~ - ye-r)~-,H 22.4.

a)

[ 6~ - 6] Je& - &-]

c)

[ ~~ - ~ ]

J~~ - ell [ ~ - !/ l H le H T

e)

.. H

+



2 of 10 12/23/2011 4:06 AM

558

22.5.

·o"~ . .

22.6.

CHAPTER 22

.. e :0: a

a) This anion is a doubly stabilized enolate ion. so there will not be a ubstantial amount of ketone presenr together with the enol ate at equilibrium:

[ ·oh o. 8:~q. .o~o~1 ·uv · - v - · v ··

b) This anion is a regular enolate ion (not doubly stabilized), so there will be a substantial amount of ketone present together with the enolate at equilibrium:

r ~f),o· - ~~uR. 1 c) This anion is a regular enolate ion (not doubly stabilized), so there will be a substantial amount of ketone present together with the enol ate at equilibrium:

( 0' :0:

r

. . -G 1 ~ - CJY

d) This anion is a regular enolate ion (not doubly stabilized), so there will be a substantial amount of ketone present together with the enol ate at equilibrium:

[ ~~ - t]



3 of 10 12/23/2011 4:06 AM

CHAPT ER 22 559

22.7. This anion is highly stabilized by resonance. The negative charge is spread over two oxygen atoms Uust like a doubly stabilized enolate) and three carbon atoms:

- -t

:~ e-~: \_/ }-:q.

- \';~ :cf. \:)\_/ }-:q.

22.8. a) 2,4-dimethyl-3,5-heptanedione is more acidic because its conjugate base is a doubly stabilized enolate. The other compound (4,4-dimethyl-3,5-heptanedione) cannot form a doubly stabilized enolate because there are no protons connected to the carbon atom that is in between both carbonyl groups. b) I ,3-cyclopeotanedione is more acidic because its conjugate base is a doubly stabilized enol ate. The other compound ( I ,2-cyclopentanedione) cannot form a doubly stabilized enolate because the carbonyl groups are adjacent to each other. c) Acetophenone is more acidic than benzaldehyde because the former has alpha protons and the latter does not.

22.9. a)

·o··~H~~ vv H

CN:~ :B7:

.. H c.o" ~

!\ .. :Br: . D .Br:



4 of 10 12/23/2011 4:06 AM

560 CHAPTER 22

b)

c)

I"'-' /i H

. . H-.00 HXr .H

.. ,.--------H ·o· '+'··, H' ' H \3/0...)

HJC{_ :Br:

22.10.

0 0H 1) Na2Cr20 7, H2S04, H20 cro 2) Br2, (H30 •]

a) 3) pyridine

~OH 1) PCC, CH2CI2

M H 2) Br2, [H30 +]

b) 3} pyridine



5 of 10 12/23/2011 4:06 AM

22.11.

0

ifoH 1) Br2 , PBr3

a) 2) H20

X 1) Br2 , PBr3

2) H20

b)

22.12.

~OH b)

CN

~ c)

22.13.

0

(/.;oH r

X Br

~OH

v 0

0

)()lOH

COOH

~Br

0

~OH a) V

0

~)l_OH b) Ll"-

22.14. a)

OH

r 1) Na2Cr20 7 ,

H2S04, H20

2) Br2, NaOH

a) H3o+

0

~OH

CHAPTER 22 561

Br

~OH v ~

0

~OEt



6 of 10 12/23/2011 4:06 AM

562 CHAPTER 22

b) 0 0 0

~ 1) 03 ~~ 1) Br2, NaOH

~OH SOCI2 ~CI ___..

2) OMS 2) H3o•

c)

H3o • o?J \_}

d)

N'H 0 r H3o• r -22.15.

OH 0

~H v ~ a) V

22.16. a)

dY 1) Br2, NaOH

2} H3o•

0 1} Br2, NaOH

~OH 2) H3o•

QVH b) 6

OH 0

d)~H

~COOH

0 1) SOCI2 ~NH2 2) xs NH3



7 of 10 12/23/2011 4:06 AM

CHAPTER 22 563

b)

c)

d)

22.17. The first step of an aldol addition reaction is deprotonation at the alpha position, but this compound has no alpha protons.



8 of 10 12/23/2011 4:06 AM

564 CHAPTER 22

22.18. 0 OH 0

Y H

NaOH, H20

W H

22.19. 0 OH 0 OH NaOH, H20 1) LAH

)l_H ~H 2) H20 ~OH

22.20.

a)~ b)~ 0

0

d)~ e)~ 22.21.

0

a)~H 0J b) H

0

c) H~

22.2 2. 0

~+ 22.23. a)

0

:X 0 r 1) LOA 0

~OH



9 of 10 12/23/2011 4:06 AM

CHAPTER 22 565

b)

6 1) LOA

c)

6 0 OH

1) LOA0

6XO 2) ~

d) 0 0

0 1) LOA trv 0

2) ~ H I:

3) H20

e) 0 0

if 1) LOA ()\ 0

2) H~ OH

3) H20

22.24. a)

0 H H 0

1) LOA 0 OH 1) LAH OH OH HJlH 0)(_0 Hjl___ 0 H~ 2) H20 v

[H1 v 2) Jl -H20 H H

3) H20



10 of 10 12/23/2011 4:06 AM

566 CHAPTER22

b) 0 H H

0 1) LDA 0 OH 1) LAH OH OH H)___H 0 )(0

H~ 0 H~ 2)H20 ~ [H''"] ~ 2) H~ -H20

3) H20

c) 0

0)__0 0 1) LDA 0 OH 1) LAH OH OH H~ H~ 0 H~ 2)H20

v [H1 v

2) )___ -H20 H H

3) H20

d) 0

0)__0 0 1) LDA 0 OH 1) LAH OH OH H)(_

HA__ 0 H~ 2)H20 ~ [H1 ~ 2) Jl H20 H' '

3) H20

22.25.

~~OH 0

0

~.0 ~0: "'-----1

0 0

Cz - ¢1: :OH



1 of 10 12/23/2011 4:07 AM

0

~ 0

22.27.

o o ~OH ~

0

Q~-

22.28. a) NaOEt b) t-BuOK

~OM• b) 0 0

e .. :OH

CHAPTER 22 567

0

~.e ·.0:/



2 of 10 12/23/2011 4:07 AM

568 CHAPTER 21

22.30. a)

b) 0

~OEt

e)

6

1)LDA

2) ~0 \~OEt

3) H3o•

1) LOA

Ko

2)

OEt

1)LDA

1) LOA

2) ~pMe

~0 3) H3o·

1) LOA

CMO

2)

OMe

0 0

~OEI

0 0

ftoet

~ 0

0 0

v1:)



3 of 10 12/23/2011 4:07 AM

22.31.

cOO OEt

0 a)

22.32.

gO ~OEt ~

b)

0 Eto):) • ~OEt

c)

0

CHAPTER 22 569

0

Eto)j-

c) 0

0

~

~ 0-c~ 0

-~



4 of 10 12/23/2011 4:07 AM

570 CIIAI>'f'ER 22

22.34. OH

~ 1) Na2Cr20 1. H2so • • H20

22.35. a)

0 0

EIO~OEt

b)

0 0

EtO~OEt

c)

0 0

EtO~OEt

d)

0 0

EtO~OEI

2) LOA

3) Ell

4) LAH

5) H20

1) NaOEI / EIOH

2)lfer

3) H30 ', heat

1) NaOEt l EtOH

2) V Br

3) NaOEt I EIOH

4) CH3Br

5) H30 ' heat

1) NaOEt EtOH

2)~Br

3) NaOEt EIOH

4) Br

5) H30'. heat

1) NaOEt I EtOH

2) CH3Br

3) NaOEI / EtOH

4) ~Br

5) H30 ', heat

0

~OH

0

~H

COOH

~

~OH 0

OH



5 of 10 12/23/2011 4:07 AM

CHAPTER 2l 571

e) 1) NaOEt f EtOH 0

0 0 2) EtBr

Y:(oH EtO)(_jlOEt 3) NaOEt f EtOH

4)JBr

5) H30 +, heat

22.36. a)

0 0 1) NaOEI f EtOH 0

EtO)(_jlOEt 2) ~Br ~0~

3) H3o•, heat

4) £H•), EtOH, (·H20)

b) 0 0 1) NaOEt I EtOH

~OH EtO~OEt 2)Vsr

3) H3o•. heat

4) excess LAH

5) H20

c) 0

0 0 1) NaOEI/ EtOH ~NH2 EtO)(_jlOEt

2) Vsr

3) ~o·. heat

4) SOCI2

5) excess NH3

22.37. Preparation of the desired compound requires the installation of three alkyl groups at the alpha position. The malonic ester synthesis can only be used to install two alkyl groups because the starting material (diethyl malonate) has onJy two alpha protons.



6 of 10 12/23/2011 4:07 AM

572 C IIAfl'fER 22

0

22.38. lfoH

22.39. a)

0

0 0 1) NaOEt I EtOH ~ ~OEI 2) u ar

3) H3o ·. heat

b) 0 1) NaOEt I EtOH

0 0 2) Ell v(' ~OEI 3) NaOEI / EtOH

4)V8r

5) H3o•. heat

c) 1) NaOEt l E10H

0 0 2)~1

~ ~OEI 3) NaOEt l EtOH

4) II 5) H3o •. heat

d) 1) NaOEI I EtOH

0 0 2) CH3I

~ ~OEI 3) NaOEt I EtOH

4)~1

5) H3o• , heat



7 of 10 12/23/2011 4:07 AM

22.40.

22.41. a)

0 0

~OEt

b)

0 0

~OEt

c)

0 0

~OEt

1) NaOEt I EtOH

2) ~Br

3) H3o •, heat

4) HOCH2CH20H, [H+], (-H20)

1) NaOEt/ EtOH

2) ~Br v 3) H3o•, heat

4)LAH 5)H20

1) NaOEt I EtOH

2) U Br

3) H3o•, heat

4) [W]. NH3, (-H20)

CHAPTER 22 573

OH

~

NH

~

22.42. Preparation of the desired compound requires the installation of three alkyl groups at the alpha position. The acetoacetic ester synthesis can only be used to install two alkyl groups because the starting material (diethyl malonate) has only two alpha protons.

22.43. 0 0

HO~OH 1) Na2Cr207, H2S04, H20

2) [H+], EtOH, (-H20 ) EtO~OEt

0 0 1) NaOEtl EtOH ~OEt



8 of 10 12/23/2011 4:07 AM

574 CHAPTER 22

22.44.

•>~ b)(r 22.45.

0 0 0 ..

~/:~H

22.46. a)

0 0

EIO~OEI 1) NaOEI / EtOH

2)M 3) H3o•, heat

0

;--)lOEt

c)~

~ \/ ~ HO~

0 0

~



9 of 10 12/23/2011 4:07 AM

b) 0 0 1) NaOEt I EtOH

EtO~OEt 2)~N02

22.47. a)

0

d'

b)

0

~

c) 0

H~

3) H30'", heat

1) R2NH, (H•), (-H20)

0

2) ~ 3) H3o•

1) R2NH, [H'], (-H20)

2)00

3) H30 +

1) R2NH, [W], (-H20 )

0 2)~

CHAPTER 22 575

0

HO~N02

~

~



10 of 10 12/23/2011 4:07 AM

576 C IIAI'TEK 22

22.48. 0

~ ~NaOH, H20 , heat (Aldol

Condensation)

R2NH

[H'J

0 ~

R . R ~ N ,~

~ ~ H30' v -----22.49.

6( 0

Cha "8

J ,H

q.H

mo 0 .. :OH ~

0 HQI

22.50.

0

~ + ~ 0

NaOH, heat

0 m



1 of 10 12/23/2011 4:08 AM

22.51. a)

0

6

b)

0

6

c) 0

6

22.52. a)

OH

0

CHAPTER 22 577

w 1) R2NH, (H"], (-H20)

2)~ 3) H3o •

&to 1) LDA

0

2) H~ 3) CH3I

OH 1) R2NH, [H1. (-H20) ~OH

0 2)~ ~ OEI

3) H3o•

4) excess LAH

5) H20

~H 0

~H NaOH



2 of 10 12/23/2011 4:08 AM

578 CHAIYf'ER 22

b)

1) [H3o•1. Br2

12) pyrid ine

( PCC H~ R;!NH

CH2CI2 [W) OH 0

(-H20)

c)

( OH

H~ 0

PCC NaOH H~ 0 OH

HO~OH

H~ 0 0

l [H+)

NH3

(-H20 }

H~ H'N N,H



3 of 10 12/23/2011 4:08 AM

22.53. a)

~OH

PCC CH2CI2

0

)lH

b)

,............._OH

22.54. a)

0

Na2Cr20 7

H2S04 , H20

~H 0 0

NaOH , H20 , heat

0

)l_H

6 2)rV

CHAPTER22 579

0 [W) 0

)lOH EtOH )lOEt

l' ) NaOEI/ E10H 2) H3o•

1) NaOEt Y OE1

2)~H

3) H,o · ~

DJ H H

HO~OH



4 of 10 12/23/2011 4:08 AM

580 CHAPTER 22

1) Me2Cull

2) Mel

3) LAH

4) H20

c) 0

~H 1) Me2Cull

2) Mel

d) 1\

0 e)

3) Na2Cr20r

H2S04, H20

4) SOCI2

1) H3o•

2) Et2CuU

3) Ell

4) HOCH2CH20H

[W). (-H20)

1) PCC, CH2CI2

~OH 2) Me2Cull

3) Mel

f)

4) MeNH2 [H•], (-H20 )

(ToH

Jytc,

1) PCC. CH2CI2

2) Me2Cull

3) Mel

4) HCI, Zn[Hg], heat



5 of 10 12/23/2011 4:08 AM

22.55.

22.56.

0 2) OMS

0

H~H 0

22.57.

0

~H v

1) Me2Culi

2) Mel

CHAPTER 22 581

11) Et2Culi

~ 2) Mel

0

ct"

22.58. The conjugate base of this compound is a doubly stabilized enolate.

o o ·oH .o~o. :o~o. .oyyo: UH~ [ .. r:e .. e.~ {'0.·· • • ..el ·.. 'lt v ·- ·· v · - · v ··

22.59.

[~ 1 .. OEt a) v

- e .. . . c:o: ·o~

~OEt -. . ..e ]

U OEI

-e .... c:.o: ·o· CfOEI -



6 of 10 12/23/2011 4:08 AM

582 CHAPTER 22

22.60. Increasing acidity

0 0 0

~ ~OH ~ 0

22.61. (YOH

a) This enol does not exhibit a significant presence at equilibrium: V OH 0

b) This enol does exhibit a significant presence at equilibrium: AA 0 OH

c) This enol does not exhibit a significant presence at equilibrium: r OH

~ d) This enol does exhibit a significant presence at equilibrium: Ul)

22.62. OH 0

EtO~

22.63.

0 OH

EtO~

e 0

0 OH

EtO~

e 0 n 98

a) ~H b)~ e

0

c)~OEI d)~



7 of 10 12/23/2011 4:08 AM

CHAPTER 22 583

22.64. Deprotonation at the following y-position results in an anion that has three resonance strucrures. The negative charge is spread over one oxygen atom and two carbon atoms:

:o··

~ r: . ~ ve

t e .. :o:

~

22.65. Deprotonation at the a carbon changes the hybridization state of the a carbon from sp3 (tetrahedral) to sl (planar). When the a position is protonated once again, the proton can be placed on either side of the planar a carbon, resulting in racemization:

22.67. HO H 0

A)lH HO H 0

~H



8 of 10 12/23/2011 4:08 AM

584 CHAPTER 22

22.68. HO H 0

~ 22.69. The carbonyl group and the aromatic ring are conjugated in the product. but are

not conjugated in the starting material. Formation of conjugation serves as a driving force in formation of the product.

OH , ... ~. HO H~· ·

/

... H'O~H ~ H~O~H 0 ~ I .o .o""/

. ' H e

0

ciY-22.70.

HOH~ :~H

~ .

OH

~ u bH /



9 of 10 12/23/2011 4:08 AM

22.71.

b)~ 22.72. Trimelhylocetaldehyde does nol have any a protons.

0

/('H

22.73. a)

0

if' +

b)

0 0

~

c)

0

0 NaOH

H)lH H20

heat

0

NaOH Q H20

heat

H~H NaOH

0

d)

0 + A

0

~ ~

CHAPTER 22 585



10 of 10 12/23/2011 4:08 AM

586 CHAPTER 22

22.74.

r

~ H ·o~ x H -

OH 0

~H

. . ~----;;;, H~O~H I.±I,..H

OH :0 ' ~~

22.75. 0 0

EtoV oEt a) Br

b)

.. ... H :o 0

EtO~OEt

c) The product should be more acidic than diethyl malonate because of the inductive effect of the bromine atom.



1 of 10 12/23/2011 4:09 AM

CHAJYf ER 22 587

22.76.

0-{ NaOEt / EtOH

22.77.

0

~OEt a) 0

0~oet 0

(~ 0 b) -

22.78. a)

0 0 1) NaOEt I EtOH m OH EtO~OEt

2) 0 Br

3) H3o• heat

b) 1) NaOEt I EtOH

0 0 2) CH3Br 0

EtO~OEt 3) NaOEt I EtOH ~OH 4) CH3Br

5) H3o·. heat

c) 1) NaOEt I EtOH

0 0 2) Ph...._......Br COOH

EtO~OEt 3) NaOEt I EtOH Ph~Ph

4) Ph ...._......Br

5) H3o ·. heat



2 of 10 12/23/2011 4:09 AM

588 CHAPTER 22

22.79. a)

0 0

~OEI

b)

0 0

)l__)l.OEt

c)

22.80.

1) NaOEI/ EIOH

2) Q Br

3) H30 +, heat

1) NaOEt I EtOH

2) Mel

3) NaOEt / EtOH

4) Mel

5) H3o•. heat

1) NaOEt I EtOH

2) Eli

3) NaOEt I EtOH

4)Vsr

5) H3o•. heat

®. ~H j &



3 of 10 12/23/2011 4:09 AM

22.81.

LJd 22.82.

0

vv

22.83.

0

~OEt

0

1) [H30+), Br2

2) pyridine 3) Me2Culi 4) H3o+

LOA

~ NaOH, H20

0

H)l_H

heat

22.84.

heat

0

~

0 0

~OEt

0

~ i) Et2Culi

2) H3o+

CHAPTER 22 589

~OH a)V

0

~Br

b)v o I ~

c) V V



4 of 10 12/23/2011 4:09 AM

590 C HAPTER 22

22.85.

0 + A

22.86. a)

¢ + 4 EtOH + 4 C02

0

b)

1) NaOEt

2) H3o+ heat CO=o



5 of 10 12/23/2011 4:09 AM

c) 0 0 0

)ly [W] , Br2

~r pyridine

/y d)

0 0

Cr 1)LOA ~ 2)EII

22.87. a)

0 0 1) NaOEt

EtO~OEt

b)

c) 0

~OEI

d) 0

uY'

NaOH

heat

2) SOCI2

1) R2NH, [W], (-H20 )

0 2)~

0

6

0

~

CHAPTER 22 591

1) Et2Culi 0

~ 2) Mel

1) LOA

2) Mel

3) LOA

4) Mel



6 of 10 12/23/2011 4:09 AM

592 CHAPTER 22

22.88. 0 0

~H HO)l_)

~ H04

22.89. a)

0

6 b)

0

6 c)

0

6 d )

0

6 e)

0

6

1) LOA

2) Ell

2) NaOH, H:P. heat, 0

A 3) H2, Pt

4) TBAF

0

1) Br2. [H30+) Q 2) pyridine

3) Et2Culi

4)H3o •

0

1) Br2. [H30 •1 Cc 2) pyrid1ne

3) Et2Culi

4) Mel

0

1) R2NH, (W), (·H20 ) 6:l 0 2)~

3) H3o ·

0 NaOH, H20, heat vDv X$~0

..-;:;

- H



7 of 10 12/23/2011 4:09 AM

CHAPTER 22 593

f) 0

~" 6 1) R2NH, [H•], (·H20)

0 2)~0Et 3) H3o •

g) 0 1) LDA 0

6 2) Mer LY 3)LDA 4) Mei

22.90. T he driving force for this reaction is formation of a doubly stabilized eno late. After the reaction is complete, an acid is requLred to protonate this anion.

e

6'~-y .8

~ r- j .1.8 ~0

EtO: l¥D _ _ .. _ I /

0 0

ifo



8 of 10 12/23/2011 4:09 AM

594 CHAPTER 22

.e JS4Et u OEt

0 0

~OEI

J ~~ 0

fe ~ Et~: u OEI

1 ~kH 0 0

~OEI

22.92.

O ~ ~~y· O [ __ :k_·. _EtO~~ Eto~ __ 1 __ EtO~· e T · )

1 e67 H~··'H

EIO~



9 of 10 12/23/2011 4:09 AM

22.93.

b)

22.94.

a) Michael Donor

0

~.(3 CN

b) Michael Donor

0 0

~ EtO •• OEt e

c) Michael Donor

0 0 )lJl

•• OEt 8

d) Michael Donor

1) NaCN

2) LOA

3) CH31

4) LDA

5) CH3I 6) H3o•

~OH ug

0

~ Michael Acceptor

0

~OEt Michael Acceptor

Michael Acceptor

Michael Acceptor

CHAPTER 22 595

~CN R



10 of 10 12/23/2011 4:09 AM

596 C HAPTER 22

e 02N- CH2

e) Michael Donor

22.95.

0 0

a)

EtO~OEt

0 0

EtO~OEt

d) 0

0 0

~N02 Michael Acceptor

0 0

EtO~B b) OH

0 0

Et0\3 1-...::::

e) .&

0 0

EtO~EI

0 0

EtO~ o I -...::::

c) 0

0 0

EtOX OEt

t) CN

EtOX OEt

g) 0 h) N02

22.96. 0

\ 0 mo Cc NaOH NaOH, heat

Michael Aldol Addition Condensation

22.97.

NaOH, heat

(&0



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

CHAPTER 22 597

22.98.

-

22.99. 0

1) 0 3 0 0

H~ 6 2) OMS heat

1) Me2Culi

2) H3o+

22.100. A ketone generally produces a strong signal at approximately 1720 cm' 1 (C=O stretching), while an alcohol produces a broad signal between 3200 and 3600 cm' 1 (0-H stretching). These regions of an IR spectrum can be inspected to determine whether the ketone or the enol predominates.



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

598 CHAPTER 22

22.101. H

~ \I . : . """' ,0~ :OH H e H

HO~OH

22.102. 0

roCN H3o• h heat

22.103. 0 0

1) NaOEt EtO)(__)lOEt

2)~ 3) H3o •

heat

acrolein

0 0 0

~OH H3o• C) + C02 heat

C10H1oO

000



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

CHAIYfER22 599

22.104. n 0 0 0 0 0

cX(oH 00 roC02Et H3o + H3o + ± COe

.& heat heat

C10H1o0

22. 105. When treated with aqueous acid, both compound A and compound B undergo racemization at the Ct position (via the enol as an intermediate, see problem 22.65). Each of these compounds establishes an equilibrium between cis and tr.ans isomers . .But the position of equiLibrium is very different for compound A than it is for compound B. The equilibrium for compound A favors a cis configuration, because that is the configuration for which the compound can adopt a chair conformation in which both groups occupy equatorial positions. The equilibrium for compound B favo•·s a fran.~ configuration, because that is the configuration for which that compound can adopt a chair conformation in which both groups occupy equatorial positions.

22.106 .

•• ~ H ~· \\6~

Ph u H ·· H [

®

Ph~-

0 OH

AifPh



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

600 CHAPTER 22

22.107.

Gr----._ o

~ e .~ o

:qH ~ .... > 0 0



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

CHAPTER 22 60 1

22.108.

e.C.b5. L 7 o: :oH · ------

1f .. ~ Q.

"5'~ t9<55



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

602 CHAPTER 22

b)

- r:_.e o ·

/;

:q.

.. ·

22.109. Direct alkylation would require performing an SN2 reaction on a tertiary substrate, which will not occur. Instead the enolate would function as a base and E2 elimination would be observed instead of SN2. The desired transformation can be achieved via a crossed aldol condensation, followed by a Michael addition:

6° if~ if 1) LOA 1) Me2Culi

2) ~ 2)H30 +

3) H20, heat



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

Review of Concepts

Chapter 23 A mines

Fill in the blanks below. To verify that your answers are correct, look in your textbook at the end of Chapter 23. Each of the sentences below appears verbatim in the section entitled Revie111 of Concepts and Vocabulary.

• Amines are , or _____ , depending on the number of groups attached to the nitrogen atom.

• The lone pair on the nitrogen atom of an amine can function as a or

• The basicity of an amine can be quantified by measuring the pK. of the corresponding----------

• Aryl amines are less basic than alkyl amines, because the lone pair is ____ _ • Pyridine is a stronger base than pyrrole, because the lone pair in pyrrole

participates in------• An amine moiety exists primarily as --------------- at

physiological pH. • The azide synthesis involves treati ng an with soclium azide,

followed by-------• The synthesis generates primary amines upon treatment of

potassium phthaJimide with an alkyl halide, followed by hydrolysis or reaction with Nv·k

• Amines can be prepared via reductive amination, in which a ketone or aldehyde is converted into an imine in the presence of a agent, such as sodium cyanoborohydride (NaBH3CN).

• A mines react with acyl halides to produce -----• In the Hofmann elimination, and am ino group is converted into a better leaving

group which is expelled in an __ process to form an ____ _ • Primary amines react with a nitrosonium ion to yield a salt in a

process called diazotization. • Sandmeyer r eactions utilize copper salts (CuX), enabling the installation of a

haiogen or a group. • In the Schiemann r eaction, an aryl diazonium salt is converted into a

fluorobenzene by treatment with --------• Aryldiazonium salts react with activated aromatic rings in a process called __

coupling, to produce colored compounds called __ dyes. • A cycle is a ring that contains atoms of more than one element. • Pyrrole undergoes electrophilic aromatic substitution reactions, which occur

primarily at c_ .



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

604 CHAPTER 23

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 23. The answers appear in the section entitled Skit/Builder Review.

23.1 Naming an Amine

PROVIDE A SYSTE•MTIC NAME FOR THE F:OLLOWING COMPOUND

I} IDENTIFY THE PARENT

2) IDENTIFY AND NAME SUBSTITUENTS

31 ASSIGN LOCANTS TO EACH SUBSTITUENT

4) ALPHABETIZE

5} ASSIGN CONFIGURATION

23.2 J>rcparing a Primary Amine via tbe Gabriel Reaction

IOENTIFY REAOENTS THAT WILL ACHIEVE THE FOLLOWING TRANSFORt.fA TJON,

1) [ )

2)~ 0

o=;N- H

0 3l ._[ --.J

23.3 Prepa ring an Amine via n Reductive Amination

IDENTIFY REMJENTS THAT W"L ACHIEVE EACH OF THE FOLlOWING TRANSFORMATIONS

0

~

0 R/"oo..N_....CH3

I H

0 I

)



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

CHAPTER 23 605

23.4 Synthesis Strategies

IDENTIFY THE REACTANT THAT IS USED AS A NUClEOPHU IN EACH OF THE THREE PROCESSES SHOWN BELOW:

DOD AZIDE

SYNTHESIS REDUCTIVE AM/NATION

GABRIEL SYNTHESIS

PRIMARY AMINE

REDUCTIVE

AM.WAriON

SECONDMY AMINE

REDUCTIVE

11MINATION

23.5 Predicting the Product of a Hofmnnn Elimination

PREDICT THE WoJOR PRODUCT OF THE FOUOWING REACTJON

11oxcess CH3I

21 Ag2o. H2o. heal

23.6 Determining the Reactants for Preparing an Azo Dye

JDENTIFY REAGENTS THAT Wlli ACHIEVE THE FOLLOWING TRANS FOR~ TION-

I

TERTIARY AMINE

OUATERNAR't' AMMONIUM ION

ALKrtA TION 1



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

606 CHAPTER 23

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 23. The answers appear in the section entitled Review of Reactions.

Preparatioll of Amines

~Br

0

~OH

0 ~X

0

~N-H 0

0

)(

0

0

~NH2

~ 0 ~

~ ~N3

~

OINH2 I

NH2

A HW'R

A R,N/R

A

~NH2

~NH2



1 of 10 12/23/2011 4:11 AM

CHAPTER 23 607

Reactions of Amiltes 0

H R 'N)l__ R-N:

\ I H H

NH2

~ ~

H (f) c1° R- N' R- N:::N

'H

R R 0 ' \

,, N- H N- N ' r:l R

Reactions of Aryldiazonium Salts

Q-cN

~OH

0

0



2 of 10 12/23/2011 4:11 AM

608 CHAPTER 23

Reactions of Nitrogen Heterocycles

H H . I 'N

0 . I

o--Br

N

Q 0 Br

Solutions

23.1. a) 3,3-dimethyl-1-butanamine b) cyclopentylamine c) N,N-dimethylcyclopentylamine d) triethylamine e) (I S,JR)-3-isopropylcyclohexanamine t) (1 S, JS)-3-aminocyclohexanol

23.2.

? b)CTU

23.3.

I I /N'-.. ~N,H

trimethylamine ethyl methyl amine

23.4. Increasing boiling point

o-{

H I

~N,H

propylamine isopropylamine

MEvertson

Highlight

Solutions



3 of 10 12/23/2011 4:11 AM

CHAPTER 23 609

23.5. a) No. This compound has eight carbon atoms and only one functional group. b) Yes. c) Yes.

23.6.

a) ~N--< b) ON- co d) N

23.7. Increasing Basicity

23.8. In the reactant, the lone pair of the amino group is delocalized via resonance. ln the product, the lone pair of the amino group is localized.

o H o

HzN~ Pt2

H2N~

23.9.

~ Q H CI)=K_J-r-CH3

a) Cl 23.10. a)

~Br

0

~OH

1) NaCN

2) LAH

3)Hp

1) SOCI2 , py 2) NH3 /

3) LAH 4) H20

OH

vJ:. c) <D



4 of 10 12/23/2011 4:11 AM

610 CHAPTER 23

1) NaCN

1) SOCI2, py

OH err 2) NH3 /

c)

aBr

3) LAH 4)H20

1) NaCN

2) LAH "' 3) H20 (J'NH2

1} SOCI2, py 0

ci'OH 2) NH3 / 3) LAH

4) H20

23.11. This compound cannot be prepared from an alkyl halide or a carboxylic acid, using the methods described in this section, because there are two methyl groups at the alpha position (the carbon atom connected to the amino group). These two methyl groups cannot be installed with either of the synthet ic methods above, because both methods produce an amine with two alpha protons.

23.12. 1) KOH

0 (JBr o:l-N 2) (JNHz 3) H3o+

a) o

1)KOH 0 Br

dO ~N-N 2) PhAPh

3) H3o+ ~

b) 0



5 of 10 12/23/2011 4:11 AM

0

cQ-" c) o

0

~·-" d) 0

23.13. a)

1)KOH Br

2)~

1) KOH

2) ~Br

>-< 2) t-BuOK

3) HBr, ROOR

b)

~OH

v 0

23.14. a)

0

d'H 0 ctH

1) excess LAH

2) H20 3) PBr3

[ H+ ] , Na8H3CN

CHAPTER 23 611



6 of 10 12/23/2011 4:11 AM

612 CHAPTER 23

b)

~0

H

0

HA H

C) H I

Q ¢ d)

0

Co

~ 0

I w ). NaBH:JCN

I H"N~

I w I • NaBH,CN I H I

/""-.../N~

I H' J, NaBHJCN

\ (H'),Na~CN I

I f-1) . NaBH3_cN_ I

I /""-.../N~

C(

cc



7 of 10 12/23/2011 4:11 AM

CHAPTER 23 613

e)

q_ [ W I , NaBH3CN

0 \ A Q~ t¢ I I

I W ] , NaBH3CN

23.15.

~ 23.16. The last step of reductive ami nation is the reduction of a C=N bond. That step introduces a proton on the alpha position (the carbon atom that is connected to the nitrogen atom in the product):

H,N/R

I _....c..._ I H

As a result, the product of a reductive amination must have at least one proton at the alpha position. In the case of tri-tert-butyl amine, there are three alpha positions, and none of them bears a proton. Each of the alpha positions has three alkyl groups and no protons.

Therefore, this compound cannot be made with a reductive ami nation.

23.17.

cY 2) NaOEt 2) OMS



8 of 10 12/23/2011 4:11 AM

614 CHAPTER 23

23.18.

0

H~

b) Q=o

c)

d) 0 [ H+ ] , NaBH3CN [ H+ ] , NaBH3CN

)lH NH3 ~NH2 0 ~N~

H~ H

e) 0 [ W J , NaBH3CN [ H' ] , NaBH3CN

)lH NH3 ~NH2 0 ~N~

H~ H

j ,,.~~('"' ~N~

)



9 of 10 12/23/2011 4:11 AM

CHAPTER 23 615

1)

Q=o 0

H)(_ o-{_

0

A

o-C 23.19. The first alkyl group is installed via a Gabriel synthesis, and the remaining alkyl

groups are installed via reductive amination processes. For most of the following syntheses, there is a choice regarding which group to attach via the initial Gabriel synthesis. ln such cases, the least sterically hindered group is chosen (the group whose installation involves the least hindered alkyl halide):

a)

cr$/ 0

l ) Br /"'.....

b)

0 $

~- ·' 0

t) Q-ar

c)

~ K$ t) CH3Br

~~:e ---'2J-H-30- .--.....

0 Q=o

I W 1. NaBH3CN

0

H)lH

o-\



10 of 10 12/23/2011 4:11 AM

616 CHAPTER 23

d) 0 ®

~·' 0

l)Br~

1) Br~

2) Hso•

/'N~

)

o-N~ 1)Br~

Q=o

o-t 23.20. The first alkyl group is instaUed via an azide synthesis, and the remaining alkyl

groups are installed via reductive amination processes. For most of the following syntheses, there is a choice regarding which group to auach via the initial azide synthesis. In such cases, the least sterically hindered group is chosen (the group whose installation tnvolves the least hindered alkyl halide):

a) [ H+ ] , NaBH3CN

Br~ 2) H2, Pt

xJLH



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

b)

Q-er 2) H2. PI

c) 1)NaN3

CH3Br

d) l) NaN3

Br~

e)

Br~ 1) Na~

2)H2,. PI

[ H+ I . NaBHsCN

Q=o

I H+ I , NaBH,CN

I H· J ' NaBH, CN

0

)l_H

CHAPTER 23 6 17

[ W J , NaBH3CN

/'.N~ H

/'.N ............... H

/'.N ...............

)



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

618 CHAPTER 23

f) 1) NaN3 [ W I , NaBHaCN

o-N~ Br~ 2) H2, Pt H~~

C)=o [ H' I , NaBH3CN

0

A

o-t 23.2 1.

H~ 1) 03 0 0 0

2) OMS H~H ! I H+ J , xs NaBH3CN

NH3

co 23.22.

0 0

NH2 0 HNA._ HN)l__

Q ' CIA 6 ¢ 0 HN03 NaOH

~o. or

H30'

N~ N~

23.23. 0 0

N02 NH2 0 HN~ HN)l__ NH2

0 HCI 6 CIA 6 Cl2 ¢ NaOH ¢ -Zn or

H3o• Cl Cl



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

23.24.

0

)l_OH SOCI2

23.25.

b);y n ,~/

~ c)

23.26.

~Br

23.27.

Compound A

1) excess NH3

2) LAH 3)H20

0

)l_CI

0

1) NaCN

1) excess CH3!

2) Ag20, H20

heal

l /'NH2

CHAPTER23

0

./'--N~ I H

1) excess CH3 1

2) Ag20 , H20 heal

j 1) 03

2) OMS

~H+ 0

0

H~

619



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

620 CHAPTER 23

23.28.

1} excess CH31

2) Ag20 . H20 , heat

23.29.

N"o <D -;.N I N' ON' j} cle a) "'o b) "'o

23.30. a)

0

o-NH2

1) )l_CI >-o-NH2

2) (CH3)2CHCI, AICI3

3) H3o+

b) V N02 1) Br2, FeBr3 BrvNH2 --:;:. 2) Fe, H3o+

I""' --:;:.

c) 0 0

0 ~CI ~

0 II

c)

o-N

1) NaN0 2, HCI

2) CuCN

1) NaN02, HCI

2) CuBr

0 I I N'N

d) b

)-0--cN

sr"Vs' 1.--::-

I HCI, Zn[Hg]

1 heat

~OH

v 1)NaN02 ,HCI ~NH2

2) H20, heat V



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

d)

0 e)

0 f)

V CI

23.31. a)

b)

1) (CH3)3CI, AICI3 ~N02

2) HN03. H2S04

1 ) Cl2, AICI3

Cl-o-N02 2) HN03, H2S04

U Br 1) NaNH2 -..::::,

2) NaN02, HCI ..?

3) CuBr

I) NaN02 , HCI

2) o-NH2

I ) NaN02 , HCI

2) o-OH 0

c)

O:!N-o-NH2 I ) NaNO, , HCI

2) g-'\( j - \

~ /;

CHAPTER 23 621

1} Fe, H30~ 2} NaN02, HCI ~:H 3) CuCN

4)H3o +

1} Fe, H3o~

Cl-o-F 2) NaN02, HCI

3) HBF4



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

622 CHAPTER 23

23.32. Cl

0 0 ~ 1) Fuming H2S04 &N~ AICI3 2) HN03, H2S04 -&

j 1) HNO,. H,so, 3) Dilute H2S04 l ''· H,o• 2} Fe, H3o+

0 &NH, 0 h-

1} Ac1 02Nv-=:N .. 2) HN03, H2S04 c1° 3) H3o+

4) NaN02, HCJ

23.33.

a) o-;-Q-NH2



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

CHAPTER 23 623

23.34. Attack at either C2 or C4 generates an intermediate that exhibits a resonance s tructure with a nitrogen atom that lacks an octet (highlighted below). Attack at C3 generates a more stable intermediate:

N

(~f

y E(±)

23.35. H

··~ 0

23.36.

E<D

C2 ATTACK

C3 ATTACK

C4 ATTACK

o·c

r-----------------------------------------~

0-E @ H

0 (±)

H E

•. H

f N'f-E -CD~

06-- :::,.... E H

(±) I N

e) -H E

I

-

a - E H

Q - 0 H E

a) The second compound will have an N-H stretching signal between 3300 and 3500 em·•. The first compound will not have such a signal.

b) When treated with HCJ, the first compound will be protonated to form an ammonium salt that will produce an IR signal between 2200 and 3000 cm· 1

• The second compound is not an amine and will not exhibit the same behavior.

23.37. a) The 1H NMR spectrum of the first compound will have a singlet resulting from theNmethyl group. 1H NMR spectrum of the second compound w ill not have any singlets.

b) The 1 H NMR spectrum of the first compound will have six signals, while the 1H NMR spectmm of the second compound will have only three signals.

ENGINEERING

Highlight



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

624 CHAPTER 23

23.38. secondary

I \ ~t ~ H,N~N~N~N'H

~~ primary/

23.39. a) The lone pair that is farthest away from the rings is the most basic, because its lone pair is localized. The lone pair of the other nitrogen atom is delocalized via resonance.

23.40.

23.41.

0 b) N

23.42.

~N~

a) 6 c)



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

CHAPTER 23 625

23.43. Only one of the nitrogen atoms has a localized lone pair (highlighted in the following structure). The other rwo nitrogen atoms have delocalized lone pairs.

23.44. a) two

23.45.

b) two c) one

a) 2,2,3,3-tetrarnethyl-1-hexanarnine b) (S)-4-arnino-2,2-dimethylcyclohexanone c) dicyclobutylrnethylamine d) 3-bromo-2,6-dimethylaniUne e) N,N-dimethyl-3-propylaniline t) 2,5-diethyl-N-rnethyl pyrrole

23.46.

I -......._,N'-.

'] ~N,H

ethyldimethylamine diethylamine

H H I I

~N,H ~N,H

1-butanamlne 2-butanamine

I ./"'...../N' H

methylpropylamine

A,tH

2-methyl-1-propanamine

23.47. None of these compounds are chiral.

I ~N....._

I I N'H

isopropylmelhylamine

2-melhyl· 2-propanamine

dimethy~ropylarrine dielhylmethylamine

isopropyldimethylamine



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

626 CHAPTER 23

a) Base Aetd

" :vo H t ':p_. ~

Acid A

+ Q .. ~o-<:::: ·R I

A b) Base

23.49.

&i• JCO CH3 te ~ CN HN

H,C cS" l A 6 0 6 n) r b) c) d) e)

23.50. a)

~OH 1) PBr3 ~NH:! 2) NaN3

3) H2. Pt

b)

~OH 1) PBr3 ~NH2

2) NaCN

3)LAH

4)H~

c)

~OH 1) PBr3

~NH, 2) 1-BuOK

3)03

4) OMS

5) (WJ, NaB~CN, NH3



1 of 10 12/23/2011 4:13 AM

CHAPTER 23 627

23.51. a)

1) NaN3 ~Br

2) H2, Pt ~NH2

b)

~Br 1) NaCN

2)LAH ~NH2

3) H20

c)

0 1) SOCI2, py

~OH 2) XS NH3 ~NH,

3) LAH 4) H20

d)

~CN 1) LAH

2) H20 ~NH2

23.52. Aziridine has s ignificant ring strain, wbich would increase significantly during pyramidal inversion. This provides a significant energy barrier that prevents pyramidal inversion at room temperature.

23.53.

~ 0 H

~N~~~

23.54.

I H



2 of 10 12/23/2011 4:13 AM

628 CHAPTER 23

23.55. ln acidic conditions, the amino group is protonated to give an ammonium ion. The ammonium group is a powerful deactivator and meta-director.

23.56. a) The presence of the nitro group in the para position helps stabilize tbe conjugate base via resonance. As seen in chapter 19, this effect only occurs when the nitro group is in the ortho and para positions. b) The basicity of ortho-nitroaniline should be closer in value to para-nitroaniline.

23.57.

23.58. Protonation of the oxygen atom gives a resonance stabilized cation (as een in chapter 20). In contrast, protonation of the nitrogen atom gives a cation that is not resonance stabilized.

23.59. a)

0 1) CH3CI, AICI3 ()Br 1) NaN3 ()NH2

2) NBS, heat 2) H2 , Pt ! (W), N;BH3CN

()')__ 0

H)lH ~a ()NH I I

b)

0 l)Q-ct

AJCI3

1) Fe, H30 *

2 ) NaN02, HC I

3 ) CuCN

2) HN03 , H2S04 4) HaO*



3 of 10 12/23/2011 4:13 AM

CHAPTER 23 629

23.60. e .. C : :OH

~,~, -·.0 :

e H

s jx-7 OH

-

23.61. 0 0 ex,_, H2N- NH2 c¢~" RNH2 +

NH

0 0

23.62.

0/H 1) excess Mel I a)

2) Ag20 , H20, heat ~N""-

0

CX·-" 1) KOH ,............._NH2 2)EtBr

b) 0 3) H2NNH2



4 of 10 12/23/2011 4:13 AM

630 CHAPTER 23

()Br ------'---'-'.....;...;.....~ __1_!_ NaCN. OMSO 2) 1-\,0'. heal 3) SOC!o. py

c)

d) 0

23.63.

4) xs NH,

1) HNO:! I H2S04

2)Fe,H3o• 3)NaN~. HCI

4) CuCN

[W) 1) excess Mel

11) 03

2) 0MS

23.64. The conjugate base of pyrrole is highly s tabilized because it is an aromatic anion and it is resonance stabilized, spreading the negative charge over a.ll five atoms of the ring:

N 0 .N l ( /"i> - 0 ~ 0 ·

23.65.

~ 0



5 of 10 12/23/2011 4:13 AM

23.66.

~NH 23.67.

a)

<D , N

U N" e Cl

A

23.68. G ~N

O N'Cie +

~N

23.70. N02

Q, a)

Fe

I

o:;OH 0

b) co A

CHAIYfER 23 631



6 of 10 12/23/2011 4:13 AM

632 CHAPTER 23

rYCN

c) V

23.71. OH

•l 6_,, 23.72.

0

0

~

F

Q b) Br

L~ ( HzSO,)

NaBH3CN

23.73. a)

[ WJ

NaBH3CN

CN

Q c) h Br

~ NaNH2, NH3

0

~ d) Br

~ H2N-o

0 Br2 V Br 1) Mg ci'H -FeBr3 h 0 [Hj 2) ).__ H H NaBH3CN

3) H20

4) PCC, CH2CI2

Br

Q e) h- Br

~ CJN H h



7 of 10 12/23/2011 4:13 AM

b)

0

c)

0

d)

0

0 1) HN0 3, H2S0 4 xs Cl2

2) Fe, H3o•

~Br _1_:_)_M-=9_.,_

V 2)C02

3) H3o• 4) SOCI2

0

1) ~CI AICI3

2) HN03, H2S04

3) HCI, Zn[Hg], heat

1) Cl2, AI%

2) NaNH2, NH3

CHAPTER 23 633

ova ti 1) NaN02, HCI Cl Cl "":

I ~ 2) CuCN _,;:;

Cl 3) H3o• Cl

j NaN02 HCI



8 of 10 12/23/2011 4:13 AM

634 CHAPTER23

23.74. TheIR data ind.icates that we are looking for structures that lack an N-H bond (i.e tertiary amines):

I ~N"

I ~N~

23.75. H

o ~ I

/V~t"

23.76.

H ;~~

6~1"CI ,_

cf> H

$~

0 +

e - Cl

H 0

if

[

H 0

if. $

ec_{_ ~ - \J'H'

! _o_~ -~~

23.77. The compound is a tertiary amine with the appropriate symmetry that provides for only three signals :

I ~N~

23.78.

Q ... ,/~ H

Coniine

or

1) excess CH31



9 of 10 12/23/2011 4:13 AM

23.80.

0 1) cone. H2S04, heat

2) MCPBA

l H' I . NaBH3CN

23.81.

0

23.82.

1) Cl2, AICI3

2) NaOH, heat

3)Eti

a\ • \ ><:'. .,:.N '

~"e heat

·P:

1) HN03, H2S04

2)Fe,H3o+ 3) 0

)lCI

o~~~ benzyne~

(see Chapter 19)

CHAPTER 23 635

0 0 r(Y-....../ )l)J

N I H

0

of?



10 of 10 12/23/2011 4:13 AM

636 CHAPTHR 23

23.83. /"N~

l__

23.85.

0 23.86.

Na, NH3 0 CH30H

0 1) Br2, hv

2) NaOEt

3)03 4) OMS

1) 0 3 0 0 ( H+ I HNf\H H~H 2) OMS NaBH3CN \...._-)

NH3

0 0 ( H+ I (v H~ NaBH:1CN .,__) CH3NH2



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

23.87. u·· ~f\ .. II '.fc) ,., H-Q : .N, .. - f.t\ • • Q l Na~

H \ ..

. p

-!()" J ~?

H' ··' H

H,~

~N~~ -H,,

CHAPTER 23 637

N ..

~ ~ '<::: . N• ~ A

~

Xa ./ :±a ·o· H' 'H 0 '<:::

lA



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

638 CHAPTER 23

23.88. Protonation of the nitrogen highHghted below results in a cation that is highly resonance stabilized. Protonation of either of the other nitrogen atoms would not result in a resonance stabilized cation:



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.

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