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Organic chemistry 9e wade 3

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18-17

917

Use of Acetals as Protecting Groups

Proposed synthesis
incompatible
functional groups
rotate

+ BrMg CH2CH2 C H
cyclohexanone

C

H3O+

H


CH2CH2

O

C

HOCH2CH2OH

H

Br

H+

CH2CH2

HOCH2CH2OH

H

Br

H+

MgBr

MgBr

C

O

CH2CH2

C

O

O



OMgBr
C
CH2CH2

O

O

C

H

protected from
basic reagents
Mg

H

O

CH2CH2

BrMg

ether

“masked” aldehyde

OMgBr
C
CH2CH2

CH2CH2

BrMg

ether

O

O

O

Mg

H

“masked” aldehyde

+ CH2CH2 C H
O

H

target compound

O
CH2CH2

C

If the aldehyde is protected as an acetal, however, it is unreactive toward a Grignard
reagent. The “masked” aldehyde is converted to the Grignard reagent, which is allowed
to react with cyclohexanone. Dilute aqueous acid both protonates the alkoxide to give
the alcohol and hydrolyzes the acetal
O to give
O the deprotected aldehyde. O
O

O
Br CH2CH2 C
Actual synthesis

O

OH

(impossible reagent)

Actual synthesis

Br

O

OMgBr
CH2CH2

O

O

C

O

H

protected from
basic reagents

O

H

CH2CH2

O

H

O

OH

H3O+

H3O+

C

H

O
OH
target compound
CH2CH2 C

H

+ CH2CHSelective
C HAcetal Formation Because aldehydes form acetals more readily than ketones,
2

we can protect an aldehyde selectively in the presence of a ketone. This selective procompound
tection leaves the ketone available for modification under neutraltarget
or basic
conditions
without disturbing the more reactive aldehyde group. The following example shows the
reduction of a ketone in the presence of a more reactive aldehyde:

O

1 equiv

O

H

H

H

H+

O

O

C

OH

H3O+

NaBH4

OH OH

C

OH

H
O

O

C

H

C

O

O

PROBLEM 18-29
Show how you would accomplish the following syntheses. You may use whatever additional reagents you need.
O
O
(a)
(b)
O
HO
CH3

CH2OH

CHO

(c)

O

(d)

O
CH2 Br

CHO

CHO

Ph

O
CH3

CH3

O

CH3

H

OH

O

CH3

(continued)

H


918

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CHAPTER 18     Ketones and Aldehydes

(e)

(f)

O

OH

O

CH2

C

O

O

BrCH2CH2CCH3

Ph

HC

CCH2CH2CCH3

18-18 The Wittig Reaction
We have seen carbonyl groups undergo addition by a variety of carbanion-like reagents,
including Grignard reagents, organolithium reagents, and acetylide ions. In 1954, Georg
Wittig discovered a way of adding a phosphorus-stabilized carbanion to a ketone or
aldehyde. The product is not an alcohol, however, because the intermediate undergoes
elimination to an alkene. In effect, the Wittig reaction converts the carbonyl group of
a ketone or an aldehyde into a new C “ C double bond where no bond existed before.
This reaction proved so useful that Wittig received the Nobel Prize in Chemistry in
1979 for this discovery.
The Wittig reaction


R
C

+

O





C
H

ketone or aldehyde



Ph
P Ph
Ph
+

R
C



phosphorus ylide

+ Ph3P

C

O

H
alkene

The phosphorus-stabilized carbanion is an ylide (pronounced “ill´-id”)—a
molecule that bears no overall charge but has a negatively charged carbon atom
bonded to a positively charged heteroatom. Phosphorus ylides are prepared from
tri-phenylphosphine and alkyl halides in a two-step process. The first step is nucleophilic attack by triphenylphosphine on an unhindered (usually primary) alkyl halide.
The product is an alkyltriphenylphosphonium salt. The phosphonium salt is treated
with a strong base (usually butyllithium) to abstract a proton from the carbon atom
bonded to phosphorus.

H

Ph
Ph P
Ph

+ H

C

Ph +
Ph P
Ph

X

R

triphenylphosphine

C

Li

CH2CH2CH3

H

H



C

R + C4H10
butane

butyllithium

H + LiX

Ph
Ph P
Ph

R

X−

alkyl halide

δ −CH2

H

Ph
+
Ph P
Ph

δ+

phosphonium salt

C
R

phosphorus ylide

Examples

Ph3P

Ph3P

+

+ Ph

CH3

CH2

+

Ph3P

Br

Br−
Bu

CH3

Li

methyltriphenylphosphonium salt

Br

+

Ph3P

Ph

benzyltriphenylphosphonium salt

CH2

ylide

Br−
CH2



+

Ph3P

Bu

Li

+

Ph3P



CH
ylide

Ph


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18-18

The phosphorus ylide has two resonance forms: one with a double bond between
carbon and phosphorus, and another with charges on carbon and phosphorus. The double-bonded resonance form requires ten electrons in the valence shell of phosphorus,
using a d orbital. The pi bond between carbon and phosphorus is weak, and the charged
structure is the major contributor. The carbon atom actually bears a partial negative
charge, balanced by a corresponding positive charge on phosphorus.

PROBLEM 18-30
Trimethylphosphine is a stronger nucleophile than triphenylphosphine, but it is rarely used
to make ylides. Why is trimethylphosphine unsuitable for making most phosphorus ylides?

Because of its carbanion character, the ylide carbon atom is strongly nucleophilic. It attacks a carbonyl group to give a charge-separated intermediate called a
betaine (pronounced “bay´-tuh-ene”). A betaine is an unusual compound because
it contains a negatively charged oxygen and a positively charged phosphorus on
adjacent carbon atoms. Phosphorus and oxygen form strong bonds, and the attraction
of opposite charges promotes the fast formation of a four-membered oxaphosphetane
ring. (In some cases, the oxaphosphetane may be formed directly by a cycloaddition,
rather than via a betaine.)
The four-membered ring quickly collapses to give the alkene and triphenylphosphine oxide. Triphenylphosphine oxide is exceptionally stable, and the conversion
of triphenylphosphine to triphenylphosphine oxide provides the driving force for the
Wittig reaction.

MECHANISM 18-6 The Wittig Reaction
Step 1: The ylide attacks the carbonyl to form a betaine.
+

+

Ph3P

C

Ph3P



H


C

O

H



R
ylide

ketone or aldehyde

O

C

C

R





a betaine

Step 2: The betaine closes to a four-membered ring oxaphosphetane (first P ¬ O bond formed).
+

Ph3P
H



O

Ph3P


C

C

R



a betaine

H

O

C

C

R





oxaphosphetane

Step 3: The ring collapses to the products (second P ¬ O bond formed).

Ph3P
Ph3P
H

O

C

C

R



O


H


four-membered ring

C
R

C


triphenylphosphine oxide
+ alkene



The Wittig Reaction

919


920

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CHAPTER 18     Ketones and Aldehydes

The following examples show the formation of carbon–carbon double bonds using
the Wittig reaction. Mixtures of cis and trans isomers often result when geometric
isomerism is possible.

O



+

+ Ph3P

CH2

CH2
85%

C

+

+ Ph3P

O

C

H



C

H

H

C
H

(cis + trans)

PROBLEM 18-31
Like other strong nucleophiles, triphenylphosphine attacks and opens epoxides. The
initial product (a betaine) quickly cyclizes to an oxaphosphetane that collapses to an
alkene and triphenylphosphine oxide.
(a) Show each step in the reaction of trans-2,3-epoxybutane with triphenylphosphine to
give but-2-ene. What is the stereochemistry of the double bond in the product?
(b) Show how this sequence might be used to convert cis-cyclooctene to trans-cyclooctene.

Planning a Wittig Synthesis The Wittig reaction is a valuable synthetic tool
that converts a carbonyl group to a carbon–carbon double bond. A wide variety
of alkenes may be synthesized by the Wittig reaction. To determine the necessary
reagents, mentally divide the target molecule at the double bond, and decide which
of the two components should come from the carbonyl compound and which should
come from the ylide.
In general, the ylide should come from an unhindered alkyl halide.
Triphenylphosphine is a bulky reagent, reacting best with unhindered primary and
methyl halides. It occasionally reacts with unhindered secondary halides, but these
reactions are sluggish and often give poor yields. The following example and Solved
Problem show the planning of some Wittig syntheses.
Analysis

H 3C
H 3C

H 3C

CH2CH3
C

C
H

C

C
H

H

(1) Ph3P
(2) BuLi

CH2CH3

C
H

or

could come from

H 3C

Synthesis

CH2CH3

+ Ph3P



(preferred)



C

H 3C

Br

O

H 3C

+

H3C
+

Ph3P



CH2CH3

C
H

H3C

C

+

PPh3

O

+ O

CH2CH3
C
H

H 3C
H 3C

CH2CH3
C

C
H


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18-19

Oxidation of Aldehydes

921

SOLVED PROBLEM 18-2
Show how you would use a Wittig reaction to synthesize 1-phenylbuta-1,3-diene.
H
C

C

H
C

H

C
H

H
1-phenylbuta-1,3-diene

SO L UTI ON

This molecule has two double bonds that might be formed by Wittig reactions. The central double bond could be formed in either
of two ways. Both of these syntheses will probably work, and both will produce a mixture of cis and trans isomers.
Analysis

C
H
C
H

CH

+

H
Ph3P

C
CH

H

could come from

C

O

CH2

or

CH2
C

PPh3

+

H
O

H

C
CH

CH2

You should complete this solution by drawing out the syntheses indicated by this analysis (Problem 18-32).

PROBLEM 18-32

PROBLEM-SOLVING HINT

(a) Outline the syntheses indicated in Solved Problem 18-2, beginning with aldehydes and
alkyl halides.
(b) Both of these syntheses of 1-phenylbuta-1,3-diene form the central double bond. Show
how you would synthesize this target molecule by forming the terminal double bond.

Plan a Wittig synthesis so that the
less hindered end of the double
bond comes from the ylide.
Remember that the ylide is made by
SN2 attack of triphenylphosphine on
an unhindered alkyl halide, followed
by deprotonation.

PROBLEM 18-33
Show how Wittig reactions might be used to synthesize the following compounds. In
each case, start with an alkyl halide and a ketone or an aldehyde.
(a) Ph ¬ CH “ C(CH3 )2
(c) Ph ¬ CH “ CH ¬ CH “ CH ¬ Ph

(b) Ph ¬ C(CH3 ) “ CH2
H
(d)
C

CH3

18-19 Oxidation of Aldehydes
Unlike ketones, aldehydes are easily oxidized to carboxylic acids by common oxidants
such as bleach (sodium hypochlorite), chromic acid, permanganate, and peroxy acids.
Aldehydes oxidize so easily that air must be excluded from their containers to avoid slow
oxidation by atmospheric oxygen. Because aldehydes oxidize so easily, mild reagents such
as Ag2O can oxidize them selectively in the presence of other oxidizable functional groups.
O
R

C

isobutyraldehyde

O
H

[O]
(oxidizing agent)

R

C

OH

isobutyric acid (90%)


922

CHAPTER 18     Ketones and Aldehydes

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R

Examples

C

H

R

(oxidizing agent)

C

O
CH3

CH

OH
O

C

H

Na2Cr2O7

CH3

dil H2SO4

CH

CH3

C

OH

CH3
isobutyric acid (90%)

isobutyraldehyde

O

O
C

H

C

Ag2O

OH

THF/H2O

(97%)
cyclohex-3-en-1-carboxylic acid

cyclohex-3-en-1-carbaldehyde

A Tollens test is usually done on a small
scale, but it can also create a silver
mirror on a large object.

O
R

C

H

aldehyde

Silver ion, Ag + , oxidizes aldehydes selectively in a convenient functional-group
test for aldehydes. The Tollens test involves adding a solution of silver–ammonia
complex (the Tollens reagent) to the unknown compound. If an aldehyde is present,
its oxidation reduces silver ion to metallic silver in the form of a black suspension or
a silver mirror deposited on the inside of the container. Simple hydrocarbons, ethers,
ketones, and even alcohols do not react with the Tollens reagent.
O

+

+ 2 Ag(NH3)2

H2O

+ 3 − OH

2 Ag

Tollens reagent

+ R C

silver

O−

+ 4 NH3 + 2 H2O

carboxylate

PROBLEM 18-34
Predict the major products of the following reactions.
(a)

(b)

CHO

+

+

Ag2O

HO

(c)

CHO
K2Cr2O7/H2SO4

HO
CHO

+

+
Ag(NH3)2 − OH

(d)

CHO

+

KMnO4
(cold, dilute)

O

18-20 Reductions of Ketones and Aldehydes
We have discussed several reductions of ketones and aldehydes in earlier sections. Here, we
review those reactions and then cover some additional reductions that are useful in synthesis.
18-20A Hydride Reductions (Review)

Ketones and aldehydes are most commonly reduced by sodium borohydride (see
Sections 10-11 and 18-11). Sodium borohydride (NaBH4) reduces ketones to secondary
alcohols and aldehydes to primary alcohols. Lithium aluminum hydride (LiAlH4) also
accomplishes these reductions, but it is a more powerful reducing agent, and it is much
more difficult to work with. Sodium borohydride is preferred for simple reductions of
ketones and aldehydes. Sodium triacetoxyborohydride [NaBH(OAc)3] is less reactive
than NaBH4, and it selectively reduces aldehydes even in the presence of ketones.
O
OH
C

H

cyclohexanecarbaldehyde

NaBH4, CH3CH2OH

C

H
H

cyclohexylmethanol (95%)


C

C

NaBH , CH CH OH

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cyclohexanecarbaldehyde

cyclohexylmethanol (95%)
18-20

C

Reductions of Ketones and Aldehydes

OH

O
CH3

H

NaBH4, CH3OH

CH2CH3

CH3 CH CH2CH3
(±) butan-2-ol (100%)

butan-2-one

18-20B Catalytic Hydrogenation

Like alkene double bonds, carbonyl double bonds can be reduced by catalytic hydrogenation. Catalytic hydrogenation is slower with carbonyl groups than with olefinic double
bonds, however. Before sodium borohydride was available, catalytic hydrogenation
was often used to reduce aldehydes and ketones, but any olefinic double bonds were
reduced as well. In the laboratory, we prefer sodium borohydride over catalytic reduction because it reduces ketones and aldehydes faster than olefins, and no gas-handling
equipment is required. Catalytic hydrogenation is still widely used in industry, however, because H2 is much cheaper than NaBH4, and pressure equipment is more readily
available there.
The most common catalyst for catalytic hydrogenation of ketones and aldehydes is
Raney nickel. Raney nickel is a finely divided hydrogen-bearing form of nickel made
by treating a nickel–aluminum alloy with a strong sodium hydroxide solution. The
aluminum in the alloy reacts to release hydrogen, leaving behind a finely divided nickel
powder saturated with hydrogen. Pt and Rh catalysts are also used for hydrogenation
of ketones and aldehydes.
O

O

OH

C
H

C
H

C
H

Ni-H2

H

(Raney nickel)

Ni-H2

H

(90%)

18-20C Deoxygenation of Ketones and Aldehydes

A deoxygenation replaces the carbonyl oxygen atom of a ketone or aldehyde with
two hydrogen atoms, reducing the carbonyl group past the alcohol stage all the way to
a methylene group. Formally, a deoxygenation is a four-electron reduction, as shown
by the following equations. These equations use H2 to symbolize the actual reducing
agents, according to the general principle that one molecule of H2 corresponds to a
two-electron reduction. Formally, the deoxygenation requires two molecules of H2,
corresponding to a four-electron reduction.
2 H2
deoxygenation
(4-e reduction)

O
C

H2
(2-e reduction)

H

OH
C

H2
(2-e reduction)

H

H
C

+

H2O

In actual use, H2 is not a good reagent for deoxygenation of ketones and aldehydes.
Deoxygenation can be accomplished by either the Clemmensen reduction (under acidic
conditions) or the Wolff–Kishner reduction (under basic conditions).
Clemmensen Reduction (Review) The Clemmensen reduction commonly converts
acylbenzenes (from Friedel–Crafts acylation, Section 17-11B) to alkylbenzenes, but
it also works with other ketones and aldehydes that are not sensitive to acid. The
carbonyl compound is heated with an excess of amalgamated zinc (zinc treated with
mercury) and hydrochloric acid. The actual reduction occurs by a complex mechanism
on the surface of the zinc.

H
H

923


924

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CHAPTER 18     Ketones and Aldehydes

O
Ph

C

Zn(Hg)

CH2CH3

propiophenone

CH3

(CH2)5

Ph

HCl, H2O
Zn(Hg)
HCl, H2O

CHO

heptanal

O

CH2

CH2CH3

n-propylbenzene (90%)

CH3 (CH2)5

CH3

n-heptane (72%)

H
H

Zn(Hg)
HCl, H2O

cyclohexanone

cyclohexane (75%)

Wolff–Kishner Reduction Compounds that cannot survive treatment with hot acid
can be deoxygenated using the Wolff–Kishner reduction. The ketone or aldehyde is
converted to its hydrazone, which is heated with a strong base such as KOH or potassium tert-butoxide. Ethylene glycol, diethylene glycol, or another high- boiling solvent is used to facilitate the high temperature (140–200 °C) needed in the second step.
O

N

+ H2N

C

H+

NH2

NH2

H

KOH
heat

+ H2O

C

C

H

+ H2O + N

N

hydrazone

Examples

O

NNH2
N2H4

KOH, 175 ° C
HOCH2CH2OCH2CH2OH

propiophenone

(diethylene glycol)

hydrazone

O

N

N2H4

cyclohexanone

NH2

+ N2
n-propylbenzene (82%)

H
H

t-BuO− K+
O
CH3

hydrazone

S

CH3

+ N2

cyclohexane (80%)

(DMSO, a solvent)

The mechanism for formation of the hydrazone is the same as the mechanism for
imine formation (Key Mechanism 18-4 in Section 18-14). The actual reduction step
involves two tautomeric proton transfers from nitrogen to carbon (Mechanism 18-7). In
this strongly basic solution, we expect a proton transfer from N to C to occur by loss of
a proton from nitrogen, followed by reprotonation on carbon. A second deprotonation
sets up the intermediate for loss of nitrogen to form a carbanion. This carbanion is
quickly reprotonated to give the product.

MECHANISM 18-7 Wolff–Kishner Reduction
Formation of the Hydrazone: See Key Mechanism 18-4.
Reduction step 1: Proton transfer from N to C. (Basic conditions: Remove, then replace.)


R

C

hydrazone

N

N

H
H

−OH


R

remove proton from N

C

N


N

H


−C
R

N

N

replace proton on C

H

H2O


H C
R

N

+

N

H

− OH


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18-20

Reductions of Ketones and Aldehydes

925

Another deprotonation enables loss of N2
Reduction step 2: Remove second proton from N.


H C
R

N


H C
R



N

OH

H

Step 3: Lose N2.

N

N

Step 4: Protonate.


H C
R



N

N


H C
R

H2O



carbanion

H

+

− OH

product

PROBLEM 18-35
Propose a mechanism for both parts of the Wolff–Kishner reduction of cyclohexanone: the formation of the hydrazone, and then the base-catalyzed reduction with evolution of nitrogen gas.

PROBLEM 18-36
Predict the major products of the following reactions:
(a)

(b)

O

O

(1) H2NNH2

Zn(Hg)
HCl, H2O

(2) KOH, heat

(d)

(c)
O

O

O

O

(1) N2H4
(2) KOH, heat

O

Zn(Hg)
HCl, H2O

O

SUMMARY Reactions of Ketones and Aldehydes
1. Addition of organometallic reagents (Sections 9-7B and 10-9)

O− M+

O
R

C

+





M

R

(M = metal = MgX, Li, etc.)

C

OH

H3O+



R

C





alkoxide

alcohol



2. Reduction (Sections 10-12 and 18-20A)

O−

O
R

C

R´ + NaBH4 (or LiAlH4)

R

C

(or H2/Raney nickel)

ketone or
aldehyde

OH


H+

R

C

H

H

alkoxide

alcohol

R

Deoxygenation reactions
a. Clemmensen reduction (Sections 17-11B and 18-20C)

O
R

C



ketone or aldehyde

+

Zn(Hg)

HCl

H
R

H
C


(continued)


926

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CHAPTER 18     Ketones and Aldehydes

b. Wolff–Kishner reduction (Section 18-20C)

O
R

C

N
R´ + H2N

NH2

R

C

hydrazine

ketone or aldehyde

NH2

H

KOH
heat



hydrazone

+

H

R

H2O

R´ +

C

N

N

3. Hydration (Section 18-12)

O
R

HO

C

+



ketone or aldehyde

H2 O

OH

R

C



hydrate

4. Formation of cyanohydrins (Section 18-13)

O
R

C

+



HO

− CN

HCN

R

ketone or aldehyde

CN
C



cyanohydrin

5. Formation of imines (Section 18-14)

O
R

C

+



NH2



ketone or aldehyde

H+

R

primary amine

N



C



+

imine (Schiff base)

H2O

6. Formation of oximes and hydrazones (Section 18-15)

O
R

+



C

H2 N

ketone or aldehyde

H+

OH

+



C

H2 N

ketone or aldehyde

NH

H+



H
Ph

C



R

N

NH

C





hydrazone derivative

hydrazine reagent

R˝ =

OH

oxime

hydroxylamine

O
R

R

N

Reagent Name

Derivative Name

hydrazine
phenylhydrazine

hydrazone
phenylhydrazone

semicarbazide

semicarbazone

O
C NH2

7. Formation of acetals (Section 18-16)

O
R

C

+



2 R˝

aldehyde or ketone

H+

OH

alcohol

R˝ O
R

OR˝
C

+



H2 O

acetal (or ketal)

8. The Wittig reaction (Section 18-18)
+

Ph3P

R
C −



+
R

phosphorus ylide



R
C

O


ketone or aldehyde

C

+

C

R


alkene

Ph3P

O


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18-20

Reductions of Ketones and Aldehydes

927

9. Oxidation of aldehydes (Section 18-19)

O
R

O

C

chromic acid, bleach, Ag+ , etc.

H

R

C

aldehyde

Tollens test
R

O
C H

aldehyde

+

OH

acid

+

2 Ag(NH3)2+
Tollens reagent

O

H2O

3 − OH

+

2 Ag
silver

R

+

O−

C

carboxylate

4 NH3

+

2 H 2O

10. Reactions of ketones and aldehydes at their α positions
This large group of reactions is covered in Chapter 22.
Example
Aldol condensation

O
2 CH3

C

OH
base

H

CH3

O
CH2

C

C

H

H

S

U

M

M
N u cA
l e oR
p h iYl i c A d d i t i o n R e a c t i o n s o f A l d e h y d e s a n d K e t o n e s
In the central structure below, if R or R´ = H, the functional group is an aldehyde. If both R and R´ are alkyl or aryl
groups, the functional group is a ketone. In general, ketones are more hindered to nucleophilic attack and react
more slowly than aldehydes.

O H

H O
R

C

R



hydration—
acid- or base-catalyzed;
aldehydes form more
product than ketones;
Section 18-12

H

OH

C N

O
either:
NaBH4, ROH;
or
1. LiAlH4; 2. HOH

R

C

R

C

H2N—R3
H+

OR

R2
C–
+
P
P h P h
P h
R1

R

C

C



H



R2 Wittig reaction—

R1
H

phosphorus-stabilized ylide
is prepared in advance;
C
cis-trans isomers possible;
R´ Section 18-18
R

C

C

–H2O

H+
R O O R acetal formation—
nucleophilic addition
OR
C
followed by substitution;
R


hemiacetal
Hemiacetal formation is
reversible under acidic,
basic, or neutral conditions.
1. Ph3P
2. BuLi

followed by
elimination
R
R´ (dehydration);
Sections 18-14, 18-15
imine

H+

O R

H O
R

R1

N

R3 nucleophilic addition

N R3

H O





reduction—
both NaBH4 and LiAlH4
reduce aldehydes and ketones;
Sections 18-20A, 10-11



H

H

H

H O

C

formation—
Section 18-13

KCN
H

imine formation—

C N cyanohydrin

H O

C

Sections 18-16, 18-17
acetal
Acetal formation is
reversible under only acidic
conditions. Acetals are
stable to basic conditions.

R2
Br or I


928

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CHAPTER 18     Ketones and Aldehydes

GUIDE TO ORGANIC REACTIONS IN

CHAPTER 18

Reactions covered in Chapter 18 are shown in red. Reactions covered in earlier chapters are shown in blue.

Substitution


Addition

Nucleophilic




at sp C (SN1, SN2)
3

Nucleophilic


Ch 6, 10, 14, 22



at sp2 C (Nuc. Arom. Subst.)



at C “ O (Nuc. Acyl Subst.)







Electrophilic
at sp C (Elect. Arom. Subst.)



at sp3 C (alkane halogenation)

Ch 19




Organometallic


Gilman



Suzuki

Ch 10, 17

at C “ C (Elect. Addn.)



at C “ C (Carbene Addn.)

oxidative cleavage



Hofmann elimination



oxygen functional groups

Ch 8, 9, 11, 17, 22
Ch 11, 18, 19, 20

Acidic conditions (E1)




E1 dehydrohalogenation

Reduction


hydride reduction



hydrogenation



metals

Ch 8, 10, 11, 17, 18, 19, 20, 21

Ch 7

dehydration of alcohols

Ch 8, 9, 17, 18, 19

Ch 11

at C “ C (HBr + ROOR)



Pericyclic (Cope elimination)


Ch 9, 17, 18, 19

Ch 19

cycloaddition (Diels-Alder)



Oxidation


epoxidation
Ch 8, 10, 14

Heck
Ch 17



Ch 15

Ch 17


epoxidation

Ch 19

Pericyclic




Ch 11

Ch 8

at sp2 C (Sandmeyer rxn)

Oxidation
Ch 8, 10, 14

tosylate elimination

Radical


E2 dehydrohalogenation





Ch 4, 6, 16, 17


at C “ C (conjugate addn.)

Ch 8





Ch 7, 9

Ch 8, 9, 10

2

Radical




Electrophilic

Ch 17, 19


at C “ O (Nuc. Addn.)

Oxidation/Reduction

Basic conditions (E2)

Ch 22

Ch 10, 11, 20, 21, 22





Ch 9, 10, 18, 22

Ch 17, 19



Elimination



Reduction


hydrogenation
Ch 8, 9, 17, 18, 19

Essential Terms
acetal

A derivative of an aldehyde or ketone having two alkoxy groups in place of the carbonyl group.
The acetal of a ketone is sometimes called a ketal. (p. 911)

O
CH 3

C

H

+

CH3O

H+

2 CH3OH

CH 3

C

H

acetaldehyde
dimethyl acetal

acetaldehyde

ethylene acetal:

OCH3

+

H2O

(ethylene ketal) A cyclic acetal using ethylene glycol as the alcohol. (p. 914)

aldehyde

A compound containing a carbonyl group bonded to an alkyl (or aryl) group and a hydrogen atom.
(p. 876)

carbinolamine

(hemiaminal) An intermediate in the formation of an imine, having an amine and a hydroxy group
bonded to the same carbon atom. (p. 906)

O
R

C

HO
R

+



NH2

R

NH
C



R

carbinolamine

R

N



C

R

+

H2O

imine

carbonyl group

The C “ O functional group. (p. 876)

Clemmensen reduction

The deoxygenation of a ketone or aldehyde by treatment with zinc amalgam and dilute HCl. (p. 923)


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Essential Terms

929

condensation

A reaction that joins two or more molecules, often with the loss of a small molecule such as water
or an alcohol. (p. 906)

cyanohydrin

A compound with a hydroxy group and a cyano group on the same carbon atom. Cyanohydrins
are generally made by the reaction of a ketone or aldehyde with HCN. (p. 904)

O
CH 3

HO

C

+

CH3

HCN

CH 3

acetone

CN
C

CH3

acetone cyanohydrin

deoxygenation

A four-electron reduction that replaces the carbonyl oxygen atom of a ketone or aldehyde with
two hydrogen atoms. The Clemmensen reduction and the Wolff–Kishner reduction are the two
most common methods of deoxygenation. (p. 923)

DIBAL-H

Diisobutylaluminum hydride, formula (i-Bu)2AlH. Used to reduce nitriles and esters selectively to
aldehydes. (p. 894)

enol

A vinyl alcohol. Simple enols generally tautomerize to their keto forms. (p. 891)

HO
C

C

H

O

H+ or − OH

C

enol

C

keto

hemiacetal

A derivative of an aldehyde or ketone similar to an acetal, but with just one alkoxy group and one
hydroxy group on the former carbonyl carbon atom. (p. 912)

hemiaminal

The IUPAC term for a carbinolamine, having an amine and a hydroxy group bonded to the same
carbon atom. This term is analogous to the term hemiacetal, except with an amine instead of
an alkoxy group. (p. 906)

hydrate

(of an aldehyde or ketone) The geminal diol formed by addition of water across the carbonyl
double bond. (p. 902)

O
Cl3C

H2O

H

C

HO

H+ or − OH

Cl3C

chloral

hydrazone
2,4-DNP derivative:

H

chloral hydrate

A compound containing the C “ N ¬ NH2 group, formed by the reaction of a ketone or aldehyde
with hydrazine. (p. 909)
A hydrazone made using 2,4-dinitrophenylhydrazine. (p. 910)
NO2
O2N

O 2N

NH NH2

O

N

H+

cyclopentanone

imine

OH
C

NO2

NH

2,4-DNP derivative of cyclopentanone

A compound with a carbon–nitrogen double bond, formed by the reaction of a ketone or aldehyde
with a primary amine. A substituted imine is often called a Schiff base. (p. 906)

O
CH3

C

CH3

+

acetone

CH3

NH2

H+

CH3

methylamine

N

CH3

C

CH3

+

H2O

acetone methyl imine

ketal

An acetal derived from a ketone. The IUPAC once abandoned this term, but later reinstated it as
a subclass of acetals. (p. 911)

ketone

A compound containing a carbonyl group bonded to two alkyl or aryl groups. (p. 876)

lithium dialkylcuprate

(Gilman reagent) An organometallic reagent of formula R2CuLi that couples with alkyl halides and
acyl halides (acid chlorides). (p. 896)

O
R2CuLi

+



C

O
Cl



C

R

+

R

Cu

+

LiCl

McLafferty rearrangement

In mass spectrometry, the loss of an alkene fragment by a cyclic rearrangement of a carbonyl
compound having g hydrogens. (p. 886)

nitrile

A compound containing the cyano group, C ‚ N. (p. 906)


930

CHAPTER 18     Ketones and Aldehydes

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nucleophilic addition

Addition of a reagent across a multiple bond by attack of a nucleophile at the electrophilic end of
the multiple bond. As used in this chapter, nucleophilic addition is the addition of a nucleophile
and a proton across the C “ O bond. (p. 898)

oxime

A compound containing the C “ N ¬ OH group, formed by the reaction of a ketone or aldehyde
with hydroxylamine. (p. 909)

protecting group

A group used to prevent a sensitive functional group from reacting while another part of the
molecule is being modified. The protecting group is later removed. For example, an acetal can
protect a ketone or an aldehyde from reacting under basic or neutral conditions. Dilute acid
removes the acetal. (p. 916)

Raney nickel

A finely divided, hydrogen-bearing form of nickel made by treating a nickel–aluminum alloy with
strong sodium hydroxide. The aluminum in the alloy reacts to form hydrogen, leaving a finely
divided nickel powder saturated with hydrogen. (p. 923)

semicarbazone

A compound containing the C “ N ¬ NH ¬ CONH2 group, formed by the reaction of a ketone or
aldehyde with semicarbazide. (p. 909)

Tollens test

A test for aldehydes. The Tollens reagent is a silver–ammonia complex [Ag(NH3)2+ - OH]. Tollens
reagent oxidizes an aldehyde to a carboxylate salt and deposits a silver mirror on the inside of
a glass container. (p. 922)

Wittig reaction

Reaction of an aldehyde or ketone with a phosphorus ylide to form an alkene. One of the most
versatile syntheses of alkenes. (p. 918)

R
C

O

+

R
ketone or aldehyde

ylide:
Wolff–Kishner reduction





C

+

P



Ph
Ph
Ph

phosphorus ylide



R
C

C

+

Ph3P

O



R
alkene

An uncharged molecule containing a carbon atom with a negative charge bonded to a heteroatom
with a positive charge. A phosphorus ylide is the nucleophilic species in the Wittig reaction. (p. 918)
Deoxygenation of a ketone or aldehyde by conversion to the hydrazone, followed by treatment
with a strong base. (p. 924)

Essential Problem-Solving Skills in Chapter 18
Each skill is followed by problem numbers exemplifying that particular skill.

1

Name ketones and aldehydes, and draw the structures from their names.
Identify their hydrates, acetals, imines, and other derivatives.

Problems 18-37, 38, 40, 41, and 61

2

Interpret the IR, NMR, UV, and mass spectra of ketones and aldehydes, and use
spectral information to determine the structures.

Problems 18-42, 43, 44, 45, 46, 71, 73, 74,
75, and 76

3

Propose single-step and multistep syntheses of ketones and aldehydes from
alcohols, alkenes, alkynes, carboxylic acids, nitriles, acid chlorides, esters, and
aromatic compounds.

Problems 18-55, 56, 60, and 64

4

Predict the products of reactions of ketones and aldehydes with the following
types of compounds, and give mechanisms where appropriate: (a) hydride
reducing agents; (b) Clemmensen and Wolff-Kishner reagents; (c) Grignard and
organolithium reagents; (d) phosphorus ylides; (e) water; (f) hydrogen cyanide;
(g) ammonia and primary amines; (h) hydroxylamine and hydrazine derivatives;
(i) alcohols; and (j) oxidizing agents.
Problems 18-39, 48, 49, 53, 54, 65, and 66

5

Use your knowledge of the mechanisms of ketone and aldehyde reactions to
propose mechanisms and products of similar reactions you have never seen
before.

Problems 18-50, 58, 62, 63, 68, 69, 70, 72,
73, and 75

Problem-Solving Strategy: Proposing Reaction Mechanisms

Problems 18-50, 58, 62, 63, 68, 69, 70, 72,
73, and 75

Show how to convert ketones and aldehydes to other functional groups, and
devise multistep syntheses using ketones and aldehydes as starting materials
and intermediates.

Problems 18-47, 51, 52, 59, and 60

6


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Study Problems

931

Study Problems
18-37

18-38

Draw structures of the following derivatives.
(a) the 2,4-dinitrophenylhydrazone of benzaldehyde
(c) cyclopropanone oxime
(e) acetaldehyde dimethyl acetal
(g) the (E) isomer of the ethyl imine of propiophenone

the semicarbazone of cyclobutanone
the ethylene acetal of hexan-3-one
the methyl hemiacetal of formaldehyde
the hemiacetal form of 5-hydroxypentanal

Name the following ketones and aldehydes. When possible, give both a common name and an IUPAC name.
(b) CH3(CH2)2CO(CH2)2CH3
(c) CH3(CH2)5CHO
(a) CH3CO(CH2)4CH3
(d) PhCOPh
(e) CH3CH2CH2CHO
(f) CH3COCH3
(g) CH3CH2CHBrCH2CH(CH3)CHO
(h) Ph ¬ CH “ CH ¬ CHO
(i) CH3CH “ CH ¬ CH “ CH ¬ CHO
O
(l)
O
O
(k)
(a)
OH

(j)

(k)

CH3

CHO

Predict the major products of the following reactions.
(a)
O
PhNHNH

(b)

O

2

CN

(2) H2O
(d)

(1) DIBAL-H

Cl

(2) H3O+
(e)

OCH3

(f)

H3O+

Cl

(h)

COOH (1) 2 CH CH Li
3
2

O

H
(j)

H3O+

+

O

O
CH3

H+, heat

NH2

(lose H2O)

Rank the following carbonyl compounds in order of increasing equilibrium constant for hydration:
CH3COCH2Cl

18-41

PPh3

C

O
18-40

LiAlH(O-t-Bu)3

O

(2) H3O+
(i)

(CH2CH)2CuLi

O

O
(g)

(1) DIBAL-H
(− 78 °C)

OCH3

H+
(c)

CH3

(l)

O

18-39

(b)
(d)
(f)
(h)

ClCH2CHO

CH2O

CH3COCH3

CH3CHO

Acetals can serve as protecting groups for 1,2-diols, as well as for aldehydes and ketones. When the acetal is formed from
acetone plus the diol, the acetal is called an acetonide. Show the acetonides formed from these diols with acetone under
acid catalysis.
O
HO

OH
OH

HO

OH

OH

HO

18-42

Sketch the expected proton NMR spectrum of 3,3-dimethylbutanal.

18-43

A compound of formula C6H10O2 shows only two absorptions in the proton NMR: a singlet at 2.67 ppm and a singlet at
2.15 ppm. These absorptions have areas in the ratio 2:3. The IR spectrum shows a strong absorption at 1708 cm-1. Propose
a structure for this compound.


932

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CHAPTER 18     Ketones and Aldehydes

18-44

The proton NMR spectrum of a compound of formula C10H12O follows. This compound reacts with an acidic solution of
2,4-dinitrophenylhydrazine to give a crystalline derivative, but it gives a negative Tollens test. Propose a structure for this
compound and give peak assignments to account for the signals in the spectrum.
200
180
Offset: 40 ppm

160

140

120

8

7

6

100

80

60

40

20

0

5

4

3

2

1

0

C10H12O

10

9

δ (ppm)

18-45

The following compounds undergo McLafferty rearrangement in the mass spectrometer. Predict the masses of the
resulting charged fragments.
(a) pentanal
(b) 3-methylhexan-2-one
(c) 4-methylhexan-2-one
An unknown compound gives a molecular ion of m>z 70 in the mass spectrum. It reacts with semicarbazide hydrochloride
to give a crystalline derivative, but it gives a negative Tollens test. The NMR and IR spectra follow. Propose a structure
for this compound, and give peak assignments to account for the absorptions in the spectra. Explain why the signal at
1790 cm-1 in the IR spectrum appears at an unusual frequency.

18-46

3

2.5
100
80

3.5

4

4.5

wavelength (μ m)
5
5.5
6

7

8

9

10

11

12

13

14 15 16

Problem
18-46

60
40
20
1790
0
4000

3500

3000

2500

2000 1800 1600
wavenumber (cm−1)

1400

1200

1000

800

600

(continued)


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200
180
Offset: 40 ppm

160

140

120

8

7

6

Study Problems

933

100

80

60

40

20

0

5

4

3

2

1

0

Problem
18-46

10

9

δ (ppm)

18-47

Show how you would accomplish the following synthetic conversions efficiently and in good yield. You may use any
necessary additional reagents and solvents.
(a)

(b)

O

OCH3

OCH3

CHO
(c)

O
CH2Br

(d)

CH2Br

H

(e)

O

(f)

O

O

COOH

H

O

OH
(g)

O
COOH

(h)

CH2OH

CHO

Ph

CH
HO
18-48

C
H2C

CH3

CH3

The following road-map problem centers on the structure and properties of A, a key intermediate in these reactions. Give
structures for compounds A through J.
OH

H

NaBH4

G

H+
warm gently

(1) J
(2) H3O+

I

(1) CH3MgI
(2) H3O+

OH
CH2OH

PCC
(excess)

B

OH OH (1 equiv)
TsOH

A

PhNHNH2
(excess)
dil H2SO4

C

Tollens reagent

D

H3O+ , then excess NaOCl/HOAc

E

Zn (Hg)
HCl

F


934

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CHAPTER 18     Ketones and Aldehydes

18-49

For each compound,
1. name the functional group.
2. show what compound(s) result from complete hydrolysis.
(a)

CH3O
CH3CH2CH2

(e)

OCH3
C

(b) HO

CH3

O

(f)

O

(g)

O

O
18-50

(d)

(c)

OCH2CH3

N

C

CH3OH, H+

H

(c)

O

CH3O
Ph

Ph3P

(b)

OCH3
C

CH3

H
(d)

CH2

CH2

PhNHNH2, H+

O
C

H

O

(e)
N

H

H+
H2O

(f)
CHO
+NH
3

NNH2

OCH3
OCH3

O

+

N

NHPh

C

H

CH2

CH2

OH

OH

+

NCH2CH3

CH3CH2NH3 Cl−
CH3CH2NH2

Show how you would accomplish the following syntheses efficiently and in good yield. You may use any necessary reagents.
(a) acetaldehyde ¡ lactic acid, CH3CH(OH)COOH
O
O
O
O
(d)
CHPh
(b)
(c)
H OH

(e)

O

CHO

CH2OH

O

O

CHO

CHO

CHCH2CH3

(e)

(f)
CHO

(g)

OH

O

CHO

Show how you would synthesize the following derivatives from appropriate carbonyl compounds.
(c)

(b)

(a)
N

N

N

OH

(e)

(d)
O

18-54

CH3
H+
H2O

O

18-53

(h)

N

O

O
Ph

18-52

O

O

Propose mechanisms for the following reactions.
(a)

18-51

O

N

(f) CH3O

OCH3

O
Predict the products formed when cyclohexanone reacts with the following reagents.
(b) excess CH3OH, H +
(a) CH3NH2, H +
(c) hydroxylamine and weak acid
(d) ethylene glycol and p-toluenesulfonic acid
(e) phenylhydrazine and weak acid
(f) PhMgBr and then mild H3O+
(g) Tollens reagent
(h) sodium acetylide, then mild H3O+
(i) hydrazine, then hot, fused KOH
(j) Ph3P “ CH2
(k) sodium cyanide
(l) acidic hydrolysis of the product from (k)

Predict the products formed when cyclohexanecarbaldehyde reacts with the following reagents.
(b) Tollens reagent
(c) semicarbazide and weak acid
(a) PhMgBr, then H3O+
(f) zinc amalgam and dilute hydrochloric acid
(d) excess ethanol and acid
(e) propane-1,3-diol, H +


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Study Problems

935

18-55

Show how you would synthesize octan-2-one from each compound. You may use any necessary reagents.
(a) heptanal
(b) oct-1-yne
(c) 2,3-dimethylnon-2-ene
(d) octan-2-ol
(e) heptanoic acid
(f) CH3(CH2)5CN

18-56

Show how you would synthesize octanal from each compound. You may use any necessary reagents.
(a) octan-1-ol
(b) non-1-ene
(c) oct-1-yne
(d) 1-bromoheptane
(e) 1-bromohexane
(f) octanoic acid
(g) ethyl octanoate

18-57

Both NaBH4 and NaBD4 are commercially available, and D2O is common and inexpensive. Show how you would synthesize
the following labeled compounds, starting with butan-2-one.
(a)

(b)

OH
CH3

C

CH2

CH3

(c)

OD
CH3

C

D

CH2

CH3

OD
CH3

D

C

CH2

CH3

H

18-58

When LiAlH4 reduces 3-methylcyclopentanone, the product mixture contains 60% cis-3-methylcyclopentanol and 40%
trans-3-methylcyclopentanol. Use your models, and make three-dimensional drawings to explain this preference for the
cis isomer.

18-59

The Wittig reaction is useful for placing double bonds in less stable positions. For example, the following transformation
is easily accomplished using a Wittig reaction.
O
cyclohexanone

CH2
methylenecyclohexane

(a) Show how you would use a Wittig reaction to do this.
(b) Show how you might do this without using a Wittig reaction, and then explain why the Wittig reaction is a much
better synthesis.
18-60

Show how you would accomplish the following syntheses. (d)
(a) benzene ¡ n@butylbenzene
(b) benzonitrile ¡ propiophenone
(d) Ph
(c) benzene ¡ p@methoxybenzaldehyde

O
(CH2)4

OH
tetralone

18-61

There are three dioxane isomers: 1,2-dioxane, 1,3-dioxane, and 1,4-dioxane. One of these acts like an ether and is
an excellent solvent for Grignard reactions. Another one is potentially explosive when heated. The third one quickly
hydrolyzes in dilute acid. Show which isomer acts like a simple ether, and then explain why one of them is potentially
explosive. Propose a mechanism for the acid hydrolysis of the third isomer.
O
O
O
O
1,2-dioxane

O

O

1,3-dioxane

1,4-dioxane

18-62

Two structures for the sugar glucose are shown on page 914. Interconversion of the open-chain and cyclic hemiacetal
forms is catalyzed by either acid or base.
(a) Propose a mechanism for the cyclization, assuming a trace of acid is present.
(b) The cyclic hemiacetal is more stable than the open-chain form, so very little of the open-chain form is present at
equilibrium. Will an aqueous solution of glucose reduce Tollens reagent and give a positive Tollens test? Explain.

18-63

Two structures of the sugar fructose are shown next. The cyclic structure predominates in aqueous solution.
1
(a) Number the carbon atoms in the cyclic
CH2OH
structure. What is the functional group at C2
2
in the cyclic form?
C O
O
(b) Propose a mechanism for the cyclization,
HOH2 C
3
HO
H
assuming a trace of acid is present.
H

4

H

5

OH
OH

6

CH2OH

fructose

H+ or −OH

H

H
OH

HO

OH
C

H

fructose
(cyclic form)

CH2OH


936

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CHAPTER 18     Ketones and Aldehydes

18-64

Hydration of alkynes (via oxymercuration) gives good yields of single compounds only with symmetrical or terminal
alkynes. Show what the products would be from hydration of each compound.
(a) hex-3-yne
(b) hex-2-yne
(c) hex-1-yne
(d) cyclodecyne
(e) 3-methylcyclodecyne

18-65

Which of the following compounds would give a positive Tollens test? (Remember that the Tollens test involves mild
basic aqueous conditions.)
(b) CH3 CH2 CH2 CH2 CHO
(c) CH3 CH “ CHCH “ CHOH
(a) CH3CH2CH2COCH3
(d) CH3 CH2 CH2 CH2CH(OH)OCH3

(f)
(f)

(e) CH3 CH2 CH2 CH2CH(OCH3)2

O

18-66

OH

Solving the following road-map problem depends on determining the structure of A, the key intermediate. Give structures
for compounds A through K.
hept-1-yne

D
SOCI2

F

E

G

(1) O3, – 78 ° C

(CH3)2CuLi

(2) (CH3)2S

CH3(CH2)4MgBr

(1) B
(2) H3O+

NaOCl
HOAc

C

A

(1) K
(2) H3O+

HCN
(1) J
(2) H3O+

H
H3O+

CH3
OH

I

Ph

18-67

OH

O

CH3

Within each set of structures, indicate which will react fastest, and which slowest, toward nucleophilic addition in basic
conditions.
(a)

O

O

O
CH3

(b)

H

O

CH3

O

O
H
O

(c)
F3C
18-68

O

O
CF3

H3C

O
CF3

H3C

CH3

One of these reacts with dilute aqueous acid and the other does not. Give a mechanism for the one that reacts, and show
why this mechanism does not work for the other one.
(a)

O
O

+

H3O

?

(b)

O
O

+

H3O

?


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18-69

937

Study Problems

Show a complete mechanism for this equilibrium established in diethyl ether with HCl gas as catalyst.
OH
O
HCl(g)
Et2O

N

18-70

NH

Show a complete mechanism for this reaction.

O

H
N
18-71

+

CH3OH

OCH3

N

HCl

+

H2O

The UV spectrum of an unknown compound shows values of lmax at 225 nm (e = 10,000) and at 318 nm (e = 40).
The mass spectrum shows a molecular ion at m>z 96 and a prominent base peak at m>z 68. The IR and NMR spectra
follow. Propose a structure, and show how your structure corresponds to the observed absorptions. Propose a favorable
fragmentation to account for the MS base peak at m>z 68 (loss of C2H4).

2.5
100
80
60
40
20

3

%
T
R
A
N
S
M
I
T
T
A
N
C
E

3.5

4

wavelength (μm)
5
5.5
6

4.5

7

8

9

10

11

12

13

14 15 16

Problem
18-71

0
4000

3500

3000

2500

200

180

160

140

2000 1800 1600
wavenumber (cm−1)
120

100

1400

1200

80

199

1000

800

600

60

40

20

0

3

2

1

0

CDCl3

Problem
18-71

10

*18-72

9

8

7

6

5
δ (ppm)

4

(a) Simple aminoacetals hydrolyze quickly and easily in dilute acid. Propose a mechanism for hydrolysis of the following
aminoacetal:
N(CH3)2
OH H
O
H3O+

H

O

+

+

(CH3)2NH2

(continued)


938

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CHAPTER 18     Ketones and Aldehydes

(b) The nucleosides that make up DNA have heterocyclic rings linked to deoxyribose by an aminoacetal functional
group. Point out the aminoacetal linkages in deoxycytidine and deoxyadenosine.
NH2
NH
2

N

N
N

O

HOH2C
H

H

H

OH

H

O

HOH2C

H

H

deoxycytidine

N

N

O
H

H

OH

H

N

H

deoxyadenosine

(c) The stability of our genetic code depends on the stability of DNA. We are fortunate that the aminoacetal linkages of
DNA are not easily cleaved. Show why your mechanism for part (a) does not work so well with deoxycytidine and
deoxyadenosine.
18-73

2.5
100
80
60
40
20

The mass spectrum of unknown compound A shows a molecular ion at m > z 116 and prominent peaks at m > z 87 and
m > z 101. Its UV spectrum shows no maximum above 200 nm. The IR and NMR spectra of A follow. When A is
washed with dilute aqueous acid, extracted into dichloromethane, and the solvent evaporated, it gives a product B. B
shows a strong carbonyl signal at 1715 cm-1 in the IR spectrum and a weak maximum at 274 nm (e = 16) in the UV
spectrum. The mass spectrum of B shows a molecular ion of m > z 72. Determine the structures of A and B, and show
the fragmentation to account for the peaks at m > z 87 and 101.
3

3.5

4

%
T
R
A
N
S
M
I
T
T
A
N
C
E

wavelength (μm)
5
5.5
6

4.5

7

8

9

10

11

12

13

14 15 16

Compound A

0
4000

3500

3000

2500

200

180

160

140

120

100

80

60

40

20

0

8

7

6

5
δ (ppm)

4

3

2

1

0

2000 1800 1600
wavenumber (cm−1)

1400

1200

1000

800

600

Compound A

10

9


www.downloadslide.net
*18-74

60
40
20

939

(A true story.) The chemistry department custodian was cleaning the organic lab when an unmarked bottle fell off a shelf
and smashed on the floor, leaving a puddle of volatile liquid. The custodian began to wipe up the puddle, but he was
overcome with burning in his eyes and a feeling of having an electric drill thrust up his nose. He left the room and called
the fire department, who used breathing equipment to go in and clean up the chemical. Three students were asked to
identify the chemical quickly so that the custodian could be treated and the chemical could be handled properly. The
students took IR and NMR spectra, which follow. The UV spectrum showed lmax at 220 nm (e = 16,000). The mass
spectrometer was down, so no molecular weight was available. Determine the structure of this nasty compound, and show
how your structure fits the spectra.

2.5
100
80

Study Problems

3

%
T
R
A
N
S
M
I
T
T
A
N
C
E

3.5

4

wavelength (μm)
5
5.5
6

4.5

7

8

9

10

11

12

13

14 15 16

Problem
18-74

0
4000

3500

3000

2500

2000 1800 1600
wavenumber (cm−1)

1400

1200

1000

800

600

Hz
600

500

400

300

200

100

60 MHz NMR spectrum

0

3

Problem
18-74

1

1
1

10.0

9.0

8.0

7.0

6.0

5.0
δ (ppm)

4.0

3.0

2.0

1.0

0


940

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CHAPTER 18     Ketones and Aldehydes

*18-75

In the absence of water, o-phthalaldehyde has the structure shown. Its strongest IR absorption is at 1687 cm-1; the proton
NMR data are shown by the structure. In the presence of water, a new compound is formed that has a strong IR absorption
around 3400 cm-1 and no absorption in the C “ O region. The proton NMR data are shown. Propose a structure of X
consistent with this information, and suggest how X was formed.

O O

singlet at
H
δ 10.5

HNMR data
broad singlet at δ 5.0 (2H)
sharp singlet at δ 5.9 (2H)
doublet at δ 7.2 (2H)
doublet at δ 7.3 (2H)

H

doublet at
H
δ 8.0

X

+ 1 H2O

C8H8O3

doublet at
H
δ 7.8
o-phthalaldehyde
C8H6O2

*18-76

Assume you are a research physiologist trying to unravel a serious metabolic disorder. You have fed your lab animal Igor
a deuterium-labeled substrate and now need to analyze the urinary metabolites. Show how you would differentiate these
four deuterated aldehydes using mass spectrometry. Remember that deuterium has mass 2. (Hint: Predict the important
fragmentations, and show how the different compounds give unique peaks.)

O

D

O

H

O
H

H

D

D

D
W

18-77

O

X

Y

Z

The family of macrolide antibiotics all have large rings (macrocycle) in which an ester is what makes the ring; a cyclic
ester is termed a lactone. One example is amphotericin B, used as an anti-fungal treatment of last resort because of
its liver and heart toxicity. Professor Martin Burke of the University of Illinois has been making analogs to retain the
antifungal properties but without the toxicity, including this structure published in 2015. (Nature Chemical Biology (2015)
doi:10.1038/nchembio.1821) The carboxylate of amphotericin B has been replaced with the urea group (shown in red).

OH

OH

O
HO

O

OH

OH

OH

OH

analog of amphotericin B

OH
O

O

O

N
H
O
NH
OH + 3

(a) Where is the lactone group that forms the ring?
(b) Two groups are circled. What type of functional group are they? Explain.

N
H

CH3

OH


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19

Amines
Goals for Chapter 19
1 Draw and name amines, and use
spectral information to determine
their structures.

2 Compare the basicity of amines
with other common bases, and
explain how their basicity varies
with hybridization, aromaticity,
resonance, and induction.
3 Describe the trends in the physical
properties of amines, and contrast
their physical properties with those
of their salts.
4 Predict the products and propose
mechanisms for the reactions of
amines with ketones, aldehydes,
acid chlorides, nitrous acid, alkyl
halides, and oxidizing agents.
5 Propose single-step and multistep
syntheses of amines from compounds
containing other functional groups.
O–
N+

H3C

CH3
CH3

trimethylamine oxide
(TMAO)

N

H3C

CH3
CH3

+

1
2

O2

trimethylamine

▲ Amines

are well-known for their fishy odors. Live fish use amine derivatives such as
trimethylamine oxide (TMAO) to control the osmotic pressure in their cells. Once a fish dies,
enzymes and bacteria begin to convert these compounds to trimethylamine and other degraded
amines. In fish that are not fresh, these amines give the fishy odor. Some people suffer from
the metabolic disease trimethylaminuria (TMAU), in which a defect in an enzyme prevents the
oxidation of trimethylamine (from food digestion) to trimethylamine oxide. Trimethylamine builds
up, and gives a strong fishy odor to the person’s breath, sweat, and urine.

19-1 Introduction
Amines are organic derivatives of ammonia with one or more alkyl or aryl groups
bonded to the nitrogen atom. As a class, amines include some of the most important
biological compounds. Amines serve many functions in living organisms, such
as bioregulation, neurotransmission, and defense against predators. Because of their
high degree of biological activity, many amines are used as drugs and medicines.
The structures and uses of some important biologically active amines are shown in
Figure 19-1.

941


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