Jones M., Fleming S.A. Organic Chemistry

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7.10 What Can We Do with These Reactions? How to Do Organic Synthesis 319

PROBLEM 7.35 Predict the products of the following reactions:

(a) HO



 CH

(b) HO



 (CH

)

CBr

O  (CH

)

CBr

PROBLEM 7.36 Devise syntheses for the following molecules. You may use any

inorganic reagent, organic iodides containing up to four carbons, and the special

reagents potassium cyanide and propyne.

PROBLEM 7.37 Indicate which member of the following pairs of compounds

will react faster in (1) the S

1 reaction and (2) the S

2 reaction. Explain your

reasoning.

(a) 1-Bromobutane and 2-bromobutane

(b) 1-Chloropentane and cyclopentyl chloride

(d) tert-Butyl iodide and isopropyl iodide

(e) tert-Butyl iodide and tert-butyl chloride

PROBLEM 7.38 Devise synthetic routes that will produce the following com-

pounds as the major products from each of the indicated starting materials.

(a) (b) (c) (d) (e)

OCH

(CH

)

(CH

)

from or

(b)

C H

2 CH

from

(c)

C OH from ororC

CBrC

CCH

(a)

CCH

320 CHAPTER 7 Substitution and Elimination Reactions

PROBLEM SOLVING

Keeping track of synthetic methods can be difficult.You are right at the beginning

of synthetic chemistry now, and here is a hint on how to keep track of it. Make a

4  6-inch file card for every structural type and then keep a short list of the

available synthetic reactions on the card.To help in ordering things, and to keep as

far as possible from the dreaded mindless memorization, you might make a

mechanistic notation for each reaction as well. The reaction of acetylides with

alkyl halides to make new acetylenes might get an “S

2 with primary substrates”

beside it, for example. You can start your catalog with the reactions in Figure 7.94.

Be sure to keep up to date; synthetic ability is the kind of thing that sneaks up on

you, like whatever it was that worried Mr. Paige, as quoted at the beginning of this

chapter. A sample file card for alkyl bromides is shown in Figure 7.113.

PROBLEM 7.39 Use the following pK

data to decide whether sodium amide would

be effective at producing vinyl anions ( ). pK

: ammonia, 38; alkene, 50.

Reaction Mechanism

–

Notes

Needs good leaving group

–

tert R group, polar solvent

E1, E2 compete

Primary or secondary R

E2 competes

Alkyl Bromides

FIGURE 7.113 A sample file card for

alkyl bromides.

7.11 Summary

New Concepts

The most important idea in this chapter is the continuing

expansion of the concept of Brønsted acidity and basicity to

embrace the species known as Lewis acids and Lewis bases.

The relatively narrow notion that acids and bases compete for

a proton (Brønsted acids and bases) is extended to define all

species with reactive pairs of electrons as Lewis bases and all

molecules that will react with Lewis bases as Lewis acids. The

key word here is reactive. All bonds are reactive under some

circumstances, and therefore all molecules can act as Lewis

bases at times.

Figure 7.94 fails to show the potential complications we have talked about.

Remember: There is always a competition between the substitution reactions, S

and S

2, and the E1 and E2 elimination reactions. Under some circumstances the

major products of the reactions summarized in Figure 7.94 will be alkenes. Your file

card catalog should include the synthesis of alkenes through the E1 and E2 reac-

tions. The identity of the leaving group, so conveniently avoided by using the sym-

bol L in the figure, makes a difference,and the structure of R,also deliberately vague

in the figure, is vital. In order to design a successful synthesis, you have to think about

all the components of the reaction.

7.11 Summary 321

Another way to describe the reaction of Lewis bases with

Lewis acids is to use the graphic device showing the stabiliz-

ing interaction of a filled orbital (Lewis base) with an empty

orbital (Lewis acid) (Fig. 7.5). This idea can be generalized if

you recall that the strength of orbital interaction depends on

the energy difference between the orbitals involved. The closer

the orbitals are to each other in energy, the stronger is their

interaction and the greater is the resulting stabilization.

Therefore the strongest interactions between orbitals are likely

to be between the highest occupied molecular orbital

(HOMO) of one molecule (A, a Lewis base) and the lowest

unoccupied molecular orbital (LUMO) of the other (B,a

Lewis acid) (Fig. 7.7). This notion is used in this chapter to

explain the overwhelming preference for backside attack in

the S

2 reaction.

Stereochemical experiments are used here for the first

time to probe the details of reaction mechanisms. The

observation of inversion of configuration in the S

reaction shows that the incoming nucleophile must approach

the leaving group from the rear, the side opposite the

leaving group. Similarly, if the stereochemistry of the E2

reaction is monitored, it can be shown that there is a

strong preference for the anti (180°) E2 reaction in acyclic

molecules.

The curved arrow formalism is used extensively in this

chapter. This bookkeeping device is extremely useful in keeping

track of the pairs of electrons involved in the making and

breaking of bonds in a reaction.

The terms transition state (the structure representing

the high-energy point separating starting material and

products) and activation energy (the height of the transition

state above the starting material) appear several times in

this chapter. Chapter 8 will continue a detailed discussion

of these terms (Fig. 7.114).

Energy

Reaction progress

Transition state

Product

Starting material

Activation energy

for starting

material product

FIGURE 7.114 A typical Energy versus Reaction progress diagram.

Key Terms

anti elimination (p. 305)

E1cB reaction (p. 309)

E1 reaction (p. 298)

E2 reaction (p. 301)

elimination reaction (p. 298)

epoxide (p. 317)

Hofmann elimination (p. 308)

inversion of configuration (p. 269)

leaving group (p. 267)

nucleophilicity (p. 278)

oxirane (p. 317)

product-determining step (p. 294)

rate-determining step (p. 294)

reaction mechanism (p. 263)

regioselective reaction (p. 307)

retention of configuration (p. 269)

Saytzeff elimination (p. 300)

1 reaction (p. 289)

2 reaction (p. 268)

solvolysis reaction (p. 290)

substitution reaction (p. 262)

sulfonium ion (p. 313)

syn elimination (p. 305)

thionyl chloride (p. 284)

Williamson ether synthesis (p. 315)

Reactions, Mechanisms, and Tools

Four exceptionally important prototypal reactions are

introduced in this chapter: the S

1, S

2, E1, and E2

reactions. The E1 and S

1 reactions involve the same

carbocationic intermediate. Addition of a nucleophile com-

pletes the S

1 reaction; removal of a proton finishes the E1

(Fig. 7.70).

The S

2 reaction and the E2 reaction are also competitive

processes. In the S

2 reaction, a nucleophile attacks the sub-

strate at the rear of a carbon–leaving group bond. In the E2

reaction, it is the adjacent hydrogen that is removed along with

the leaving group (Fig. 7.74).

For primary substrates, it is the S

2 and E2 reactions that

compete. For tertiary substrates in polar solvents with weak

bases, it is the S

1 and E1 reactions that are important. If

strong bases are employed in nonionizing solvents, E2 reac-

tions are most important. For secondary substrates all four

reactions compete.

It is absolutely vital to have these four building block reac-

tion mechanisms under complete control.

322 CHAPTER 7 Substitution and Elimination Reactions

Syntheses

5. Alkyl Sulfonates

These sulfonates are good substrates for further

conversions using S

1, S

2, E1, and E2 reactions

Cl—SO

—

2. Alkenes

polar solvent

E1 reaction. Must be a tertiary or secondary halide. Saytzeff

elimination. Competes with the S

1 reaction

CCC

strong base

E2 reaction. Competes with S

2, which dominates for

primary substrates. anti Elimination favored, regiochemistry

depends on leaving group

7. Ethers

1. Substituted Alkanes

2 reaction. The R must be methyl, primary, or secondary

Always goes with inversion. The E2 reaction is competitive

HOS

strong nucleophile

1 reaction. The R must be tertiary or secondary.

Stereochemical result is largely racemization.

The E1 reaction competes. HOS is

a polar, protic solvent

polar solvent

–

3. Alkoxides

+ –

OHR O

–

OHR O

–

6. Amines

4. Alkyl Halides

This reaction can be S

1 or

2, but the intermediate is

always the protonated alcohol

This reaction can be S

1 or

2, but the intermediate is

a chlorosulfite ester

Inversion is the usual result

SOCl

CCl

(Ph)

PCl

PBr

+ CH

Primary

amine

–

N CH

–

NCH

HN(CH

)

Restrictions on S

2 reaction apply;

overalkylation is a serious problem

+ CH

Secondary

amine

N(CH

)

(CH

)

(CH

)

+ CH

Tertiary

amine

–

R XR OR R

Williamson ether synthesis.

R must be able to undergo

the S

2 reaction—it cannot

be tertiary

The S

1 and S

2 reactions allow the conversion of alkyl

halides into a host of new structures. These reactions were

summarized earlier in Figure 7.94. Remember that the S

reaction works only for primary and secondary halides and

the S

1 reaction works only for tertiary and secondary

molecules.

Alcohols can also be used as starting materials in the S

and S

2 reactions, but the OH group is a very poor leaving

group and must be transformed either by protonation into

water or by other reactions into some other good leaving

group.

7.12 Additional Problems 323

Common Errors

Students of organic chemistry long for certainty. They have not

yet learned to love the messiness of the subject or at least to

accommodate to it. That the answer to the question, What

happens if we treat A with B? might be “All sorts of things” can

be profoundly unsettling. In this chapter, we meet four basic

building blocks: the S

1, S

2, E1, and E2 reactions. It is a sad fact

that in many instances they compete with one another. We can usu-

ally adjust reaction conditions to favor the reaction we desire, but it

will be a rare joyous occasion when we achieve perfect selectivity.

There is much detail in this chapter, but that is not usually

a problem for most students. Good nucleophiles can be distin-

guished from good bases, good leaving groups can be identified,

and so on. What is hard is to learn to adjust reaction conditions

(reagents, temperature, and solvent) so as to achieve the desired

selectivity, to favor the product of just one of these four reac-

tions. That’s hard in practice as well as in answering a question

on an exam or problem set. To a certain extent, we all have to

learn to live with this problem!

7.12 Additional Problems

PROBLEM 7.42 Explain with crystal clarity why treatment

of propyl alcohol with bromide ion fails to give propyl bromide,

but treatment with HBr succeeds.

PROBLEM 7.43 Propose a synthesis of each of the following

compounds starting with 1-butene:

(a) 2-chlorobutane

(b) 2-methoxybutane

(d) 2-butene (mixture of cis and trans)

(e) butane

PROBLEM 7.44 Propose a synthesis of each of the following

compounds starting with 2-butanol:

(a) 2-chlorobutane

(b) 2-methoxybutane

(d) 2-butene (mixture of cis and trans)

(e) 2-butanethiol

HBr

–

This chapter expands the topic of organic synthesis. We still

have very few reactions for making molecules at our disposal,

basically just additions and the E1, E2, S

1, and S

2 reactions.

Nonetheless, it is possible to devise simple routes to target

molecules and even to craft a few multistep sequences. Organic

synthesis is one of the most creative (and fun) parts of this

science, and so it is important to start early. Here we group

several synthesis-related problems before going to more

mechanistic material. In just a few chapters, we will have many

more reactions under control, and be able to do much fancier

work. Stay tuned.

PROBLEM 7.40 The S

2 reaction occurs between an alkyl halide

and a nucleophile. It is a very important process in organic chem-

istry. For this reason it is necessary to recognize the nucleophilicity

of various reagents. In the following pairs of compounds, indicate

the more nucleophilic reagent. Explain your reasoning.

(a) (CH

)

N (CH

)

(b) (CH

)

N (CH

)



SNa CH

PROBLEM 7.41 Which nucleophiles would serve to effect the

following conversions of 1-iodopropane?

(a)

(b)

(c)

(d)

(e)

(f)

–

324 CHAPTER 7 Substitution and Elimination Reactions

PROBLEM 7.45 Predict the product(s) for each of the reactions

shown. Indicate the pathway (S

1, S

2, E1, or E2) for forma-

tion of each product.

PROBLEM 7.46 The contents of four flasks containing the

compounds shown in (a), (b), (c), and (d) are treated with

. In two cases, a product C

is formed that shows only five signals in its

C NMR spectrum.

In two other cases, a product of the same formula, C

,is

formed that shows seven signals in the

C NMR spectrum.

What are the products, and which of the compounds, (a), (b),

(c), and (d), leads to which product? Explain.

PROBLEM 7.47 Why are sulfonates (p. 283) such good partici-

pants in the S

1 and S

2 reactions?

PROBLEM 7.48 Write a mechanism for the formation of alkyl

bromides from the reaction of alcohols and PBr

PROBLEM 7.49 You may be familiar with the compound

tert-butyl methyl ether. Its common acronym is MTBE. It has

been an additive for gasoline since 1979, although its use has

declined in the United States since 2003. How would you

make MTBE if you have bottles of tert-butyl alcohol and

methyl alcohol? You may use any other inorganic reagents

needed. Your proposed route should avoid mixtures containing

undesired products.

(a)

(b)

(d)

(c)

(CH

)

/(CH

)

(d)

OTs



(b)

catalytic



(c)

ONa

(e)

HO OTs

(a)

NaSCH

THF

PROBLEM 7.50 The reaction shown below was used to help clarify

the S

2 nature of the alkylation reactions of S-adenosylmethionine.

Explain what the results tell us about the mechanism.

COO

–

= Tritium (T)

= Deuterium (D)

OHHO

PROBLEM 7.51 (a) Reaction of 1,2-dimethylpyrrolidine

(1) with ethyl iodide leads to two salts, C

IN. Explain.

(b) However, when 2-methylpyrrolidine (2) undergoes a similar

reaction with ethyl iodide, followed by treatment with weak

base, analysis of the product by

H NMR spectroscopy

reveals only one set of signals for C

N. Explain.

PROBLEM 7.52 Treatment of 1,2-dibromoethane with the dithio-

late shown below leads to two products, C

and C

Write structures for these products and explain how they are

formed.

A dithiolate

–

(b)

I1.

weak base2.

One

(a)

Two

7.12 Additional Problems 325

PROBLEM 7.53 Devise a synthetic route that will convert

(R)-sec-butyl alcohol into (S)-sec-butyl alcohol.

PROBLEM 7.54 Provide reagents that will effect the following

changes:

PROBLEM 7.55 Devise S

2 reactions that would lead to the

following products:

PROBLEM 7.56 Predict the products of the following S

reactions:

(a)

–

(a) (b)

NH(CH

)

(c)

(b)

(a)

(R)(S)

OHH

PROBLEM 7.57 Treatment of butyl ethyl ether with HI

leads to both butyl iodide and ethyl iodide, along with the

related alcohols. By contrast, when tert-butyl ethyl ether is

treated the same way, only ethyl iodide and tert-butyl alcohol

are formed. tert-Butyl iodide and ethyl alcohol are not

produced. Explain.

PROBLEM 7.58 Write all the possible products of the reaction

of alcohol 1 with HCl/H

HCl

R'R

I CH

+CH

OH CH

and

(CH

)

C +(CH

)

COH CH

O CH

–

(c)

(d)

(e)

(b)

–

326 CHAPTER 7 Substitution and Elimination Reactions

PROBLEM 7.59 Predict the products of the following S

reactions:

PROBLEM 7.60 You are given a supply of ethyl iodide, tert-

butyl iodide, sodium ethoxide, and sodium tert-butoxide. Your

task is to use the S

2 reaction to make as many different ethers

as you can. In principle, how many are possible? In practice,

how many can you make?

PROBLEM 7.61 The following molecules can form conjugate

bases by loss of a proton from more than one position (arrows).

Start by drawing a good Lewis structure (electron dots are

deliberately left out) and then choose which proton will be lost

more easily. Explain your choice.

PROBLEM 7.62 The following molecules can accept a proton

to form their conjugate acids by addition of a proton at more

than one possible place (arrows). Choose the position at which

protonation will be preferred and explain your choice. Once

again, it makes sense to start by drawing a good Lewis structure

for the molecule in question.

(a) (b) (c)

(CH

)

(a) (b) (c)

(a)

(b)

(c)

(d)

PROBLEM 7.63 Write arrow formalisms for the following

reactions. Charges and electron dots are deliberately left out of

the reagents and products, so you must start by putting them in.

PROBLEM 7.64 The S

2 reactions at the allyl position

are especially facile. For example, allyl compounds are even

more reactive than methyl compounds (Table 7.1, p. 275).

Explain this rate difference. Why are allyl compounds

especially fast? Hint: This, like all “rate” questions, can be

answered by examining the structures of the transition

states involved.

PROBLEM 7.65 In the following pairs of compounds, rank

the molecules in order of rate of reaction in the S

2 process.

Explain your reasoning.

(a)

(b)

(c)

(d)

CH CH

C CH CH

–

(a)

(b)

(c)

(CH

)

(CH

)

–

HOH

OHH

7.12 Additional Problems 327

PROBLEM 7.66 In the following pairs of compounds, rank the

molecules in order of rate of reaction in the S

1 process.

Explain your reasoning.

PROBLEM 7.67 Each of the following molecules contains two

halogens of different kinds. They are either in different posi-

tions in the molecule or are different halogens entirely. In each

case, determine which of the two halogens will be the more

reactive in the S

2 reaction. Explain.

PROBLEM 7.68 Each of the following molecules contains two

halogens at different positions in the molecule. In each case,

determine which of the two halogens will be the more reactive

in the S

1 reaction. Explain.

(a)

(b)

(c)

Cl Br

(c)

(b)

(a)

PROBLEM 7.69 You have two bottles containing the two

possible diastereomers of the compound shown below

). When the salts of these two carboxylic acids are

made, only one of them is converted into a new compound,

. What is the structure of the new compound, and

how does its formation allow you to determine the stereo-

chemistry of the original two compounds? The squiggly bond

indicates both stereoisomers.

PROBLEM 7.70 You have two bottles containing the two

possible diastereomers of the compound shown below

IO). When the molecules are treated with a strong

base, such as sodium hydride, only one of them is

converted into a new compound, C

O. What is the

new compound, and how does its formation allow you

to determine the stereochemistry of the original two

compounds?

NaH

–

NaHCO

One remaining C

Two isomers, each

–

(c)

(b)

328 CHAPTER 7 Substitution and Elimination Reactions

PROBLEM 7.71 Predict the products of the following E2

reactions. If more than one product is expected, indicate which

will be the major compound formed.

PROBLEM 7.72 Predict the products of the following E1 reac-

tions. If more than one product is expected, indicate which will

be the major compound formed.

(b)

(a)

(e)

–

S(CH

)

(d)

–

(c)

–

(b)

–

(a)

–

PROBLEM 7.73 (a) Write the “obvious” arrow formalism for

the following transformation. Of course it is going to be wrong;

you are being set up here. But don’t mind, it’s all a learning

experience.

(b) OK, what you wrote is almost certainly wrong. To find the

correct mechanism, you measure the kinetics of this process

and discover that the reaction is bimolecular. Now write

another arrow formalism.

nal arrow formalism? Hint: Think about the geometry of

the transition state for the S

2 reaction.

PROBLEM 7.74 (a) Menthyl chloride reacts with sodium

ethoxide in ethyl alcohol to give a single product as shown.

Analyze this reaction mechanistically to explain why

this is the only alkene produced. Hint: Think in three

dimensions.

By contrast, neomenthyl chloride treated in the same

fashion gives an additional, major, product. Moreover,

menthyl chloride reacts much more slowly than

neomenthyl chloride. Explain these observations

mechanistically.

Menthyl chloride

CH(CH

)

CH(CH

)

–

OTs

–

OTs

(d)

S(CH

)

(c)