Jones M., Fleming S.A. Organic Chemistry

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21.7 Additional Problems 1119

PROBLEM 21.30

Provide a detailed mechansim for the follow-

ing reaction.

PROBLEM 21.31 Now for something a little more interesting.

Explain why the regiochemical results of the following two

reactions are so different.

PROBLEM 21.32 Addition of HCl to the molecule shown

below is 10

times slower than addition of HCl to cyclopentene.

The product is exclusively (1R,2R)-1,2-dichloro-1-methyl-

cyclopentane. Provide a mechanism that accounts for the

stereospecificity observed. Why is the reaction slow?

Remember that chlorine is not a particularly large atom.

PROBLEM 21.33 Explain why compound 1 reacts much faster

than 2 (140,000 times) and 3 (190,000 times) in acetic acid.

Does this information about rates let you decide whether the

reaction proceeds through a delocalized ion?

ClCH

OH HOCH

OH Br

Br Br OH

OH Br

Br Br OH

KOH/H

Et = CH

Bz = PhCH

HCl

THF

PROBLEM 21.28 Similarly, tetrahydrofuran is the product

of this next reaction. Provide a mechanism and explain why

this process is much faster than the similar reaction of

3-chloro-1-propanol, a molecule that gives the corresponding

diol as product.

PROBLEM 21.29 Provide a detailed mechanism showing all

intermediates for the following reaction:

OBs

Bs = SO

OBs

PROBLEM 21.34 Compound 1 reacts 10

–10

times faster than

its cis isomer 2. The product is 3. Explain.

SPh

1120 CHAPTER 21 Intramolecular Reactions and Neighboring Group Participation

PROBLEM 21.37 Provide a mechanism for the following reac-

tion of 1. Be sure to explain the observed stereochemistry and

the fact that the corresponding trans compound (2) does not

react in this way.

PROBLEM 21.38 Reaction of 1 with bromine leads to a neutral

product of the formula C

BrO

. Propose a structure for the

product and a mechanism for its formation.

PROBLEM 21.39 Provide an arrow formalism for the following

reaction:

PROBLEM 21.40 Propose a mechanism for the following

change. Be sure to explain the observed stereochemistry.

PROBLEM 21.41 What do the relative rate data suggest

about the nature of the carbocationic intermediates in the

following solvolysis reactions? Hint: See Figure 21.43 and

Problem 21.23.

–

OCH

BrO

COOH

PhPh

OBs

(CH

)

COK

(CH

)

COH

OAc

HOAc

ONs

1.0

7.0

38.5

Relative

Rate

HOAc =

Ns =

PROBLEM 21.36 In Figure 21.30, we neglected to tell you

about the product of reaction of syn-7-norbornenyl tosylate (1).

The product of solvolysis in water is compound 2. Provide an

arrow formalism mechanism and an explanation for the rate

acceleration of 10

relative to 3.

PROBLEM 21.35 When Z = OCH

, about 93% of the products

of solvolysis of 1 can be attributed to neighboring phenyl group

participation. When Z = NO

, essentially none of the products

come from neighboring participation by phenyl. Explain.

HCOOH

OTs

Rate = 10

TsO

(both stereoisomers)

Rate = 1

TsO

21.7 Additional Problems 1121

NaOH/H

OCH

(a)

COOCH

(b)

hν

PROBLEM 21.42

The concerted rearrangements we saw in

Section 18.14 (p. 914) involve neighboring group participation.

With that hint, you can do this problem. Provide arrow formal-

ism mechanisms for the following reactions. Hint: See pp. 916

and 919.

PROBLEM 21.43 Compound 1 reacts about 15 times faster

than neopentyl brosylate (2). The product is isobutyraldehyde.

Provide a mechanism and explain the rate difference.

Rate

= 15

Rate

= 1

(CH

)

C (CH

)

CH CHO

OBs

(CH

)

OBs

PROBLEM 21.44 Compound 1 is the only product of the reac-

tion of compound 2 under the conditions shown. Provide an

arrow formalism mechanism, and deduce the stereochemistry of

2 from the stereochemistry of 1 shown.

PROBLEM 21.45 Treatment of the following two compounds

with HCl leads to the same chloride. Explain, mechanistically.

Hint: Watch the sulfur atom. What can sulfur do after an initial

protonation of the oxygen atom?

PROBLEM 21.46 Write a mechanism for the formation of the

following two products:

COOC

KOAc

OTs

COOC

CHCH

HCl

CH OH

HCl

CH Cl

CCl

PROBLEM 21.47 Heating β-hydroxyethylamide (1) with

thionyl chloride affords β-chloroethylamide (2). In turn, amide

2 yields oxazoline 3 upon treatment with NaOH.

SOCl

NaOH

(a) Write arrow formalism mechanisms for the formations of

2 and 3.Things are not quite as simple as they first appear.

For example, if 1 is treated with thionyl chloride below

25 °C, compound 4 is obtained instead of 2. Compound

4 has the same elemental composition as 2, but unlike 2,is

1122 CHAPTER 21 Intramolecular Reactions and Neighboring Group Participation

COOH

1. NaOH/H

2. Br

3. HCl (neutralize)

HOOC

1. NaOH/H

2. Br

3. HCl (neutralize)

(b) Propose a structure for 4 and show how 4 is formed and

converted into 2 and 3.

PROBLEM 21.48 Addition of iodine (I

) to an alkene does not

give a diiodide. But the reaction shown below is a standard pro-

cedure in organic synthesis. You have sufficient understanding

of alkene chemistry to be able to answer several questions about

the reaction shown. What would be the stereochemical out-

come? Is the product optically active, a meso compound, or a

racemic mixture? Draw the mechanism for this reaction. Your

mechanism should explain the stereochemical outcome you

have indicated. Why does this reaction make a five-membered

ring rather than a six-membered ring?

PROBLEM 21.49 Outline the mechanistic steps for the follow-

ing transformation:

SOCl

<25 ⬚C

ONa

soluble in water. Upon heating, 4 gives 2, and compound

4 yields 3 upon treatment with aqueous sodium carbonate.

Now explain this more complicated reaction:

21.7 Additional Problems 1123

PROBLEM 21.50

The 50-MHz

C NMR spectrum of the

2-norbornyl cation in SbF

/SO

ClF/SO

solution exhibits an

interesting temperature dependence. At -159 °C the spectrum

consists of five lines at δ 20.4 [C(5)], 21.2 [C(6)], 36.3 [C(3)

and C(7)], 37.7 [C(4)], and 124.5 [C(1) and C(2)]. However,

at -80 °C the spectrum consists of three lines at δ 30.8 [C(3),

C(5), and C(7)], 37.7 [C(4)], and 91.7 [C(1), C(2), and

C(6)] ppm. Explain. Hint: Start with the low-temperature

spectrum.

PROBLEM 21.51 2-Amino alcohols can be converted into

aziridines if the hydroxyl group is first changed into a better

leaving group. A related approach to aziridines might involve

treating a 2-amino alcohol with thionyl chloride in the presence

of base.

However, when 2-amino alcohol 1 was treated with thionyl

chloride in the presence of triethylamine, isomeric compounds 3

and 4 (C

S) were obtained rather than the anticipated

2. Propose structures for isomers 3 and 4 and explain their

formation.

Use Organic Reaction Animations (ORA) to answer the

following questions:

PROBLEM 21.52 The animation titled “Intramolecular S

2” is

a nice example of a neighboring group effect. Select that reac-

tion and observe the process. Is the starting material in a boat

or a chair conformation? Does the S

2 process occur from the

more stable chair structure? Why or why not?

PROBLEM 21.53 Observe the HOMO and LUMO of the

“Intramolecular S

2” reaction. What atom in the reaction

should have the HOMO density prior to epoxide formation?

Even more specifically, what orbital on the nucleophilic atom

do you expect to be involved in this S

2 reaction? Verify your

answer by viewing the HOMO track of the “Bimolecular

nucleophilic substitution: S

2” animation.

PROBLEM 21.54 Select the reaction titled “Baeyer-Villiger

oxidation” and observe the animation. This reaction is also an

example of a neighboring group effect. There are several steps in

the reaction. Draw out the step that involves a neighboring

group. The info page might help. What clues help you recognize

that this reaction involves assistance by a neighboring group?

RNH

SOCl

base

RNH

(CH

)

S)(C

hexane, 25 ⬚C

SOCl

(CH

)

C(CH

)

Special Topic

Carbohydrates

1124

22.1 Preview

22.2 Nomenclature and Structure

of Carbohydrates

22.3 Formation of Carbohydrates

22.4 Reactions of Carbohydrates

22.5 The Fischer Determination of

the Structure of

D-Glucose

(and the 15 Other

Aldohexoses)

22.6 Special Topic: An Introduction

to Di- and Polysaccharides

22.7 Summary

22.8 Additional Problems

FIELD OF CARBOHYDRATES Cellulose is a polymer of the monosaccharide glucose found

in all plant matter. It is the most common organic compound found in Nature. Converting

all that cellulose into biofuel is an active area of chemical research.

22.1 Preview 1125

There exists no better thing in the world than beer to blur class and social

differences and to make men equal.

—EMIL FISCHER

22.1 Preview

In the preceding chapter, we encountered neighboring group effects, the influence of

one functional group on the chemistry of another. In this chapter, we discuss mol-

ecules that bristle with functionality, and for which intramolecular neighboring group

effects play most important roles, influencing both structure and reactivity.These are

the carbohydrates, which are molecules that have the formula of C

.The name

carbohydrate comes from the formulas of these compounds, which can be factored

into C

.These molecules appear to be hydrates of carbon. Of course they are

not literally hydrates,which are compounds having water molecules clustered around

an atom,but the classic demonstration experiment in which a carbohydrate sample—

usually a sugar cube—is treated with sulfuric acid does yield a spectacular sight as

exothermic dehydration reactions produce steam and a cone of carbon.

Carbohydrates

are also called sugars or saccharides.

Carbohydrates serve as the storage depots for the energy harvested from the

sun through photosynthesis by green plants. Moreover, carbohydrates are present in

nucleic acids (Chapter 23) and are thus vital in controlling protein synthesis and in

transmission of genetic information.Table sugar is a carbohydrate.More than 100 mil-

lion tons of this sugar are produced yearly, and each American consumes about 64 lb

of table sugar per year (down from over 100 lb just a few years ago). Carbohydrates

are vital to human survival,and it is somewhat surprising that their study took so long

to emerge as a glamorous area of research. Perhaps the development of carbohydrate

chemistry was delayed by the simple technical difficulties of working with goopy syrups

rather than nice, well-behaved solids. A full scale renaissance is underway now, how-

ever, and carbohydrate chemistry is very much a hot topic.

The carbohydrate chemistry we will see in this chapter involves only a few new

reactions, but the polyfunctionality of these compounds can make for some com-

plications. To understand carbohydrate transformations you will have to be able to

apply old reactions in new settings, and work with a good deal of stereochemical

complexity as well. It’s not a bad idea to review the concepts in Chapter 4 before

you start on this chapter.That’s a hard thing to do—time is probably pressing,and it’s

always more interesting to move on to new stuff—but at least reread the summary

sections of Chapter 4 and be prepared to look back if things get confusing. Dust off

your models. There will be moments when they will be essential.

In this chapter, we will also meet Emil Fischer (1852–1919) again. He is surely

one of the greatest of organic chemists and a person of heroic theoretical and exper-

imental achievements. Carbohydrates can be hideously frustrating molecules to work

with experimentally. They form syrups and are notoriously difficult to crystallize.

Emil Fischer (1852–1919) was perhaps the greatest of all organic chemists. He won the 1902 Nobel Prize

in Chemistry and made extraordinary contributions to many fields, including the chemistry of carbohydrates

and proteins.

DO NOT try this experiment unsupervised! The reaction is very exothermic and sometimes violent. Sulfuric

acid can be thrown about, and sulfuric acid in the eyes is no picnic!

Although commercial sugars are mostly harvested from plants (mainly sugar beets or sugarcane), there are some

marvelously exotic laboratory syntheses. One favorite is that of Philip Shevlin (b. 1939) of Auburn University,

who fired bare carbon atoms into water and obtained carbohydrates. Shevlin has suggested that this reaction

may well have occurred on the icy surfaces of comets, transforming them into giant orbiting sugar cones.

1126 CHAPTER 22 Carbohydrates

These properties were especially troublesome in Emil Fischer’s time because there

were no fancy chromatographs and spectrometers in his basement, and separation

was routinely achieved through crystallization. One of Fischer’s major contributions

was the discovery that phenylhydrazone derivatives of many carbohydrates were

rather easily crystallized.That point may seem simple now, but it certainly was not

in 1875! One of the great logical triumphs of organic chemistry was the deciphering

of the structures of the aldohexoses (we’ll see just what this means later, but glucose,

a vital source of human energy, is one of the aldohexoses), and Emil Fischer was at

the heart of this work for which he won the Nobel prize.

ESSENTIAL SKILLS AND DETAILS

1. Carbohydrate chemistry uses an important structural convention, the Fischer

projection. To understand this chapter, you have to be able to move easily from

Fischer projections to conventional structures. The directions are right there in

Section 22.2.

2. Understanding the logic of structure determination in carbohydrate chemistry is an

essential skill.

3. Though neither a skill nor a detail, the Fischer proof of the structure of glucose is

something you should know, at least in general, because it is one of the supreme

intellectual achievements of human history!

4. Understanding the differences between the anomeric hydroxyl group in an aldose,

which is attached to C(1) in the cyclic form, and all the other hydroxyl groups in the

molecule is critical to “getting”carbohydrate chemistry.

5. The reactions of carbohydrates can be controlled by use of protecting groups.

22.2 Nomenclature and Structure of Carbohydrates

Carbohydrates could all be named systematically, but their history has led to the

retention of common names throughout the discipline. There is also a special con-

vention, named after Emil Fischer, that is used to simplify representations of these

stereochemically complex molecules.

The simplest carbohydrate is 2,3-dihydroxypropanal, which has the common

name glyceraldehyde (Fig. 22.1). It contains a single stereogenic carbon and thus can

exist in either the (R) or the (S) enantiomeric form.

Glyceraldehyde is an aldotriose. The tri tells you it is a three-carbon molecule,

and the aldo tells you that the molecule contains an aldehyde.The ose tells us that the

molecule is a carbohydrate. The term aldose is a more general name that means the

molecule has an aldehyde (CHO) at one end,a primary alcohol (CH

OH) at the other

end,and these ends are attached through a series of groups (Fig.22.2).

In glyceraldehyde, of course, there is only one linking group.H

(R)-Glycer-

aldehyde

(S)-Glycer-

aldehyde

Mirror

HOCH

CHO

FIGURE 22.1 The two enantiomers

of the simplest carbohydrate,

glyceraldehyde.

Aldehyde group = aldotriose

Glyceraldehyde

Three carbons = triose

WEB 3D

FIGURE 22.2 Glyceraldehyde is an

aldotriose because it has a three-

carbon backbone and an aldehyde

group. Other aldoses have longer

carbon backbones.

WORKED PROBLEM 22.1 Generalize from the definition of an aldotriose to gener-

ate the structures of an aldose with four carbons and an aldose with five carbons.

ANSWER The aldose means the sugar has an aldehyde group. The aldehyde of an

aldose and the primary alcohol at the end of the chain are attached through a

series of groups. In the four-carbon sugar there are twoH

(continued)

22.2 Nomenclature and Structure of Carbohydrates 1127

CHO

OHH

CHO

OHH

If glyceraldehyde

is a triose, …

In Fischer Projections

…then this compound

is a tetrose...

...and this one

is a pentose

CHO

OHH

CHO

OHH

CHO

OHH

CHO

OHH

D-Glyceraldehyde (R)-Glyceraldehyde=

on right

Configurational

carbon

on left

CHO

L-Glyceraldehyde

CHO

(S)-Glyceraldehyde

HHO

CHO

FIGURE 22.3 Aldoses are written

with the aldehyde group at the top

and the primary alcohol at the

bottom in the Fischer projection.

Horizontal bonds are taken as

coming toward the viewer and

vertical bonds as going away from

the viewer. If the OH group on the

configurational carbon is on the

right, the carbohydrate is a member

of the

D series. If it is on the left, the

carbohydrate is in the

L series. Notice

that to get the designation right, the

carbohydrate must be drawn with the

aldehyde at the top and the primary

alcohol at the bottom.

The stereochemical convention called a Fischer projection translates the drawings

of Figure 22.1 and Problem 22.1 into three-dimensional representations (Fig. 22.3).

In a Fischer projection, all horizontal lines are taken as coming out of the plane of

the paper toward you, and vertical lines as going away from you.The carbons link-

ing the aldehyde and the primary alcohol ends of the molecule are not explicitly

written out. By convention, Fischer projections are written with the most oxidized

carbon of the molecule, which in an aldose is the aldehyde carbon, at the top and

the primary alcohol at the bottom.

CONVENTION ALERT

Fischer projections do not attempt to represent energy minimum conformations.

Instead, they are designed to help distinguish the stereoisomers. An analogy would be

the representation of substituted ethanes in an eclipsed conformation,called the sawhorse

structure (p. 65). Fischer projections are used because they are simple to draw and read.

We will see soon how to translate them into more accurate representations of molecules.

groups between the aldehyde and primary alcohol, and the five-

carbon sugar has three groups between the aldehyde and the

primary alcohol.

1128 CHAPTER 22 Carbohydrates

For any sugar shown in Fischer projection, the stereogenic carbon closest to the

primary alcohol (the stereogenic carbon closest to the bottom of the Fischer projec-

tion) is called the configurational carbon. If the OH group of the configurational

carbon is on the right in the Fischer projection then the molecule is designated as

a “

D” sugar; if this OH is on the left, the sugar is a member of the “L” series.

The

D and L terminology for carbohydrates is based on the optical activity of glyc-

eraldehyde.

D-glyceraldehyde is the enantiomer that rotates plane-polarized light clock-

wise or dextrorotatory (p. 156) and that means the enantiomeric

L-glyceraldehyde

rotates plane-polarized light by the same amount counterclockwise or levorotatory. For

all other carbohydrates, the D and L terminology only relates to the stereochemistry

at the configurational carbon of glyceraldehyde. If the OH bonded to the configura-

tional carbon is shown on the right (in other words,on the same side as in the Fischer

projection of

D-glyceraldehyde), that carbohydrate is a D-carbohydrate. There is

absolutely no correlation between

D or L in the name of a carbohydrate and the direc-

tion in which the carbohydrate rotates plane-polarized light.Any

D-carbohydrate other

than glyceraldehyde can rotate plane-polarized light either clockwise or counterclockwise.

The convention for indicating the direction of rotation of plane-polarized light is the

addition of () or () in the name. Clockwise rotation is indicated by prefacing the

name of the sugar with (), and counterclockwise rotation with (). So we might

write

D-()-glyceraldehyde or D-()-erythrose (Fig.22.4). Both sugars are members

of the

D series, but one is dextrorotatory () and the other is levorotatory (). It is

important that you recognize the difference between a

D or L carbohydrate, particu-

larly if you will continue on in the biological sciences. They are enantiomers of each

other. The D-carbohydrates are the isomers most frequently encountered in nature.

OHC

(a)

OHC

(b)

OHC

(c)

OHC

(d)

OHH

CHO

HHO

CHO

D-(+) -Glycer-

aldehyde

L-(–)-Glycer-

aldehyde

OHH

CHO

Mirror

CHO

HHO

D-(–) -Erythrose L-(+) -Erythrose

Mirror

OHH

CHO

HHO

CHO

FIGURE 22.4 Other than for

glyceraldehyde, there is no relation

between

D and L in a carbohydrate

name and the sign of optical rotation,

() and ().

PROBLEM 22.3 Write Fischer projections for the following molecules:

OHHOCH

FIGURE 22.5 1,3-Dihydroxyacetone,

a ketotriose.

PROBLEM 22.2 Use solid-wedge, dashed-wedge notation to write three-

dimensional representations for the following molecules shown in Fischer

projection. Label each stereogenic carbon as either (R) or (S).

There are also carbohydrates that contain no aldehyde group,but rather a ketone.

This kind of sugar is called a ketose. Figure 22.5 shows 1,3-dihydroxyacetone,which

is a three-carbon ketose, a ketotriose.