Jones M., Fleming S.A. Organic Chemistry

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7.4 Substitution, Nucleophilic, Bimolecular: The S

2 Reaction 279

TABLE 7.3 Relative Nucleophilicities of Some Common Species

Relative

Species Name Nucleophilicity

Excellent Nucleophiles



Cyanide 126,000



Mercaptide 126,000



Iodide 80,000

Good Nucleophiles



Hydroxide 16,000



Bromide 10,000



Azide 8,000

Ammonia 8,000



Nitrite 5,000

Fair Nucleophiles



Chloride 1,000

COO



Acetate 630



Fluoride 80

OH Methyl alcohol 1

O Water 1

So we can anticipate that energy match-

ing between the orbital on the nucleo-

phile and an empty carbon orbital

will be important in determining

nucleophilicity.

Table 7.3 and Figure 7.33 show

some nucleophiles segregated into gen-

eral categories. It is not possible to do

much better than this, because nucleo-

philicity is a property that depends on

the reaction partner. A good nucleophile

with respect to carbon may or may not

be a good nucleophile with respect to

displacement on some other atom.

Energy matching is critical in the reac-

tion. If we change the reaction partner,

and thus the energy of the orbital

involved, we change the stabilization

involved as well (Fig. 7.32).

> NH

–

> SH

–

> OH

–

> OR > OH

–

––

Se > H

S > H

> R

–

> Br

–

> Cl

–

> F

–

Based on charge Based on electronegativity

FIGURE 7.33 Some relative

nucleophilicities. Be careful!

Nucleophilicity (Lewis basicity) is a

hard-to-categorize quantity. Relative

nucleophilicity depends, for example,

on the identity of the reaction

partner, which is always a Lewis acid,

as well as the nature of the solvent.

Table 7.3 and Figure 7.33 reveal some spectacular exceptions to the “a good

Brønsted base is a good nucleophile” rule. Look at the halide ions, for example.

As we would expect from the pK

values of the conjugate acids, HX (where X is

a halide), the basicity order is F



 Cl



 Br



 I



. The nucleophilicity order is

exactly opposite the basicity order (Fig. 7.34). Iodide is the weakest Brønsted base

but the strongest nucleophile. Fluoride is the strongest Brønsted base but the weakest

nucleophile. Why?

HI HBr HCl HF

–10 –9 –8 +3.2

–

Increasing pK

decreasing acidity

Increasing Brønsted

basicity in solution

Increasing nucleophilicity

in solution

FIGURE 7.34 For the halide ions in

solution, the orders of Brønsted

basicity and nucleophilicity are

opposite.

To answer this question, we need more data. It turns out that the effectiveness

of these (and other) nucleophiles depends on the so-often-neglected solvent. It is

easy to forget that most chemical reactions are run in “oceans” of solvent, and it is

280 CHAPTER 7 Substitution and Elimination Reactions

perhaps not too surprising that these oceans of other molecules are not always with-

out substantial effects on the reaction! Figure 7.35 examines rates for typical dis-

placement reactions by halides in solvents of different polarities and in the total

absence of solvent (the gas phase). Note that iodide is a particularly effective nucleo-

phile, relative to other halides, in water, a protic solvent. Recall our discussion in

Chapter 6 (p. 238): Protic solvents are polar solvents containing hydrogens (e.g., water

or alcohols). In polar solvents that do not contain hydroxylic hydrogen, such as ace-

tone, , the reactivity order is reversed. In the absence of solvent, in the

gas phase, the order is also reversed.

(CH

)

acetone

solvent

–

Br Br

–

gas phase

no solvent

Br Br

–

= halogens

–

X X

160 14 1 ---

---

15 11

<0.015 0.02 1

I Br

Relative Rates

Cl F

FIGURE 7.35 In the highly polar

and protic solvent water, iodide is

the best nucleophile (fastest

reacting). In the polar but aprotic

solvent (acetone), the

order of nucleophilicity is reversed.

In the gas phase, where there is no

solvent, fluoride emerges as the best

nucleophile! Read the table across,

not down.The dashed line means

“not measured.”

(CH

)

–

BrR

slower

––

Cl R

–

faster

–

I R

FIGURE 7.36 In a hydroxylated solvent, the highly solvated chloride ion reacts more slowly with than

the less encumbered iodide ion does.

In a protic solvent, a halide can act as a nucleophile (Lewis base) and react with

the substrate through interaction with σ

of the bond. Alternatively, a halide

can behave as a Brønsted base by interacting with one of the protonic solvent mol-

ecules. Chloride and fluoride are by far the better bases and thus hydrogen bond to

protonic solvents much more strongly than iodide does. The halides become highly

solvated in protonic solvents and their relative sizes increase dramatically. They

become more encumbered, and therefore less effective nucleophiles. Iodide, the poorer

base, is less hydrogen bonded to protonic solvents and thus more free to do the S

reaction (Fig. 7.36).

7.4 Substitution, Nucleophilic, Bimolecular: The S

2 Reaction 281

As the polarity of the solvent decreases, fluoride becomes more competitive with

iodide. What’s the limit of this effect? A reaction with no solvent, of course. Pure

nucleophilicity, with no competing effects of solvent, can be examined only in the

absence of other bulk molecules. We can do that by running the reaction in the gas

phase.It is no trivial matter to examine reactions of ions in the gas phase, as the sta-

bility of ions depends heavily on solvation. Yet, since the 1990s such experiments

have become common. In the gas phase, fluoride, the better base, is also the better

nucleophile (Fig. 7.35).

7.4e Effect of the Leaving Group Many leaving groups depart as anions.

It seems obvious that the more stable is, the easier it will be to displace it.

The stability of the anion is related to how easily its conjugate acid, ,

dissociates and thus to the pK

of . Low-pK

acids (strong acids) are related

to good leaving groups, (weak bases). High pK

acids (weak acids) are related

to poor leaving groups (strong bases). Table 7.4 shows some good and poor leaving

groups and the pK

values of their conjugate acids, .

L + H

+ L

+ R

R + L

TABLE 7.4 Some Good and Poor Leaving Groups and Their Conjugate Acids

Acid pK

Leaving Group Name

Good Leaving Groups

HI 10



I Iodide

HBr  9



Br Bromide

HCl  8



Cl Chloride

HOSO

R  3



OSO

R Sulfonate



 1.7 OH

Water

Poor Leaving Groups

HF  3.2



F Fluoride

S  7.0



SH Thiolate

HCN  9.4



CN Cyanide

O 15.7



OH Hydroxide

HOCH

15.9



OCH

Ethoxide

HOR 16–18



OR Alkoxide

7.4f How to Play “Change the Leaving Group” Many times it would be

convenient to convert an alcohol ( ), which is generally cheap and available

(in the chemical industry, those two terms are usually synonymous), into another

compound, . One naturally considers an S

2 substitution reaction of

(Fig. 7.37) with some nucleophile Nu .

–

FIGURE 7.37 A hypothetical S

reaction converting an alcohol,

, into a new compound,

.Nu

The problem is that



OH is an extraordinarily poor leaving group (Table 7.4).

The pK

of water is 15.7. Water is a rather weak acid, and its conjugate base, hydroxide

ion, is a strong base.There are several ways to circumvent the problem, the simplest

of which is to transfer a proton to the alcohol (protonate the alcohol) with a strong

282 CHAPTER 7 Substitution and Elimination Reactions

acid to give the conjugate acid of the alcohol, . Protonation converts the

leaving group from a very poor leaving group, hydroxide, into an excellent leaving

group, water (Fig. 7.38).

OH BRRB H

–

FIGURE 7.38 The first step in the

reaction of an alcohol with the acid

is protonation of the Lewis

base oxygen.

A variant of the “change the leaving group” game is very closely related to the

reactions of alcohols in halogenated acids to give halides. It is called ether cleavage,

and it follows essentially the same mechanism. In strong halogenated acids, such as

HI or HBr, ethers can be cleaved to the corresponding alcohol and halide. In the

presence of excess HBr, the alcohols formed in this reaction are often converted into

the bromides (Fig. 7.40).

OH CH

Hydroxide

(a very poor leaving group)

Ethyl alcohol

Ethyl bromide

Water

(an excellent leaving group)

H Br Br

–

FIGURE 7.39 Playing “change the

leaving group” with ethyl alcohol.

PROBLEM 7.10 What will happen when an alcohol is treated with a strong base?

Think “simple.”

Water is a weak base. Its conjugate acid, H



, is a strong acid (pK

−1.7).

Therefore, water is easily displaced by a good nucleophile. In the reaction of an alco-

hol with HBr, the strongest available nucleophile is Br



, the conjugate base of the

HBr used to protonate the alcohol. So, if we want to convert an alcohol into a bro-

mide we need only treat it with HBr. The reaction and its mechanism are shown for

the conversion of ethyl alcohol into ethyl bromide in Figure 7.39. Note especially

the technique of changing the leaving group from the very poor hydroxide to the

excellent water.

HBr

excess 47% HBr

8 h

–

HBr

FIGURE 7.40 The mechanism of the

cleavage of ethers by strong haloacids.

7.4 Substitution, Nucleophilic, Bimolecular: The S

2 Reaction 283

In this reaction, the ether oxygen is first protonated, turning the potential leav-

ing group in an S

2 reaction from alkoxide,



OR, into the much more easily dis-

placed alcohol, HOR. The displacement itself is accomplished by the halide ion

formed when the ether is protonated. This reaction is quite analogous to the reac-

tion of alcohols with haloacids.The halide must be a nucleophile strong enough to

do the displacement, and the reaction is subject to the limitations of any S

2 reac-

tion (Fig. 7.40).

PROBLEM 7.11 Clearly, an unsymmetrical ether can cleave in two ways. Yet often

only one path is followed. Explain the specificity shown in the following reaction.

PROBLEM 7.12 Provide a mechanism for the following reaction:

excess

H I

OH +OCH

H I

Another technique for changing the leaving group uses a two-step process in

which the alcohol is first converted into a sulfonate. Sulfonates are excellent leav-

ing groups (Table 7.4) and can be displaced by all manner of nucleophiles to give

desirable products (Fig. 7.41).

THE GENERAL CASE

A SPECIFIC EXAMPLE

An ethyl sulfonate

Ethyl alcohol

Ethyl tosylate

(an ethyl sulfonate)

Tosylate

(an excellent leavin

g group)

–

Cl SO

O SO

OSO

–

WEB 3D

FIGURE 7.41 The conversion of ethyl alcohol into another ethyl derivative. In this sequence of reactions the poor leaving

group OH is first converted into the good leaving group sulfonate.The leaving group is then displaced through an S

reaction with a nucleophile, here CH



284 CHAPTER 7 Substitution and Elimination Reactions

PROBLEM SOLVING

The role of the tosylate group is essentially always to leave. It is a generic “good

leaving group.” Every time you see it in a problem, you are likely to have to break

the bond. There should be a little pull-down menu in your head that

says “leaving group!” when OTs appears.

There are other common “GO” signs, or clues, in many S

2 problems. Every

time you see “S

2” you must think “inversion”—all S

2 reactions go with

inversion. There are also good nucleophiles that appear often in S

2 problems.

Cyanide (NC



), mercaptan (HS



), and thiolate (RS



) are good examples of ions

that should make “good nucleophile” appear in your mental pull-down menu.

OTs

THE GENERAL CASE

A SPECIFIC EXAMPLE

HCl

100 ⬚C, 2 h

SOCl

CHCl

(80%)

HCl

Thionyl chloride

FIGURE 7.42 Treatment of an alcohol

with thionyl chloride results in the

formation of an alkyl chloride.

SOCl

–

A chlorosulfite

ester

–

(a)

(b)

FIGURE 7.43 Mechanisms of chloride

formation from reaction of an alcohol

with thionyl chloride. (a) The S

pathway. (b) Direct decomposition.

A variation on the sulfonate reaction of Figure 7.41 involves the treatment of an

alcohol with thionyl chloride (SOCl

).The end result is the conversion of an alco-

hol into a chloride (Fig. 7.42).

The initial product in this reaction is a chlorosulfite ester. Mechanisms of break-

down of the intermediate chlorosulfite ester involve both S

2 displacement by chlo-

ride ion to give the alkyl chloride, sulfur dioxide (SO

), and a chloride ion (Fig. 7.43a),

as well as direct decomposition to SO

and a pair of ions that recombine (Fig. 7.43b).

7.4 Substitution, Nucleophilic, Bimolecular: The S

2 Reaction 285

PROBLEM 7.13 Treatment of an alkoxide with dimethyl sulfate leads to methyl

ethers. Write a mechanism for this reaction.

–

RCH

OCH

Dimethyl sulfate Meth

l etherAlkoxide

–

Halide formation

Several related reactions involve the use of phosphorus reagents. Alkyl halides

can be made from the treatment of an alcohol with phosphorus halides,PX

or PX

as shown in Figure 7.44. Many of these phosphorus-containing reagents cause

rearrangement, especially in secondary systems, and the stereochemical outcome,

retention or inversion,depends on solvent and other reaction conditions.Accordingly,

modifications have been worked out in recent years in order to avoid such stereo-

chemical problems.

OH Cl

(CH

)

CHCH

OH (CH

)

CHCH

Br + P(OH)

(58%)

PBr

–10 ⬚C, 4 h

PCl

CaCO

ether

0 ⬚C, 3 min

(100%)

FIGURE 7.44 Formations of alkyl

halides with phosphorus reagents.

For example, alcohols can be converted into the related bromides and chlorides

through treatment with triphenylphosphine and a carbon tetrahalide. The alkyl

halide is generally formed with inversion, and a rough mechanism involving a series

of S

2 reactions is sketched out in Figure 7.45.

THE GENERAL CASE

A SPECIFIC EXAMPLE

Cl CCl

CCl

(Ph)

Triphenylphosphine

(Ph)

CCl

25 ⬚C, 24 h

–

ROH

(Ph)

P (Ph)

–

O RCl

(~85%)

FIGURE 7.45 Halides are formed from the reaction of alcohols with the intermediate

produced from triphenylphosphine and a carbon tetrahalide.

In either case, the end product is the chloride. Once again, the strategy in this

reaction has been to convert a poor leaving group (



OH) into a good one (SO

and Cl



286 CHAPTER 7 Substitution and Elimination Reactions

(CH

)

NCH

(CH

)

–

(a)

Rate increases as solvent polarity increases

–

S(CH

)

S(CH

)

(b)

Rate decreases as solvent polarity increases

FIGURE 7.46 Opposite effects of a

change in solvent polarity on the

rates of two S

2 reactions. Reaction a

goes faster when the solvent is made

more polar, whereas reaction b goes

more slowly.

7.4g Effect of Solvent At first, the behavior of the S

2 reaction as the

solvent polarity is changed is perplexing. As the solvent polarity is increased,

some S

2 reactions go faster, some slower. Two examples are shown in Figure

7.46. In the first reaction (a), a change to a more polar solvent results in a faster

reaction, but in the second reaction (b), the rate decreases when the solvent is

made more polar.

(a)

(CH

)

N (CH

)

N ICH

I (CH

)

–

In this transition state, a

partial positive charge is

developed on nitrogen and

a partial negative charge

on iodine

(b)

HO HO S(CH

)

S(CH

)

HO CH

S(CH

)

.. ..

–

In this transition state,

charge is dispersed. The

transition state is less polar

than the starting material

In the starting material

two full charges exist

Uncharged

starting materials

Charged products

Uncharged products

FIGURE 7.47 The transition state for reaction a is more polar than the starting material.The transition state for

reaction b is less polar than the starting material.

There is a way to attack almost every problem that poses a question having to

do with rates. The rate of a reaction is determined by the energy of the transition

state—the high point in energy as the reaction proceeds from starting material to

products. In order to solve a question involving rates, we must know the structure

of the transition state of the reaction.

Figure 7.47 shows the structures of the transition states for the two S

reactions of Figure 7.46. In reaction a, a more polar solvent will act to stabilize the

charged products and the polar, charge-separated transition state.There will be very

PROBLEM 7.14 Explain why sulfonates should give rather stable anions (weak

nucleophiles) and are therefore good leaving groups.

7.4 Substitution, Nucleophilic, Bimolecular: The S

2 Reaction 287

Reaction b is different. Here the starting materials bear full charges.This charge

is dispersed as the reaction goes on. A change to a more polar solvent will stabilize

the fully charged starting materials more than the transition state or products.

Therefore, the activation energy increases and the reaction slows (Fig. 7.49).

Energy

Reaction progress

(CH

)

N ICH

–

(CH

)

NCH

(CH

)

–

Activation

energy in

more polar

solvent

Activation

energy in

less polar

solvent

FIGURE 7.48 The difference between

the energy of the starting materials

and the energy of the transition state

is smaller when the reaction is run in

a more polar solvent (red curve) and

larger when the reaction is run in a

less polar solvent (blue curve).The

result is a decreased activation energy

for the reaction and a faster rate.

little effect on the uncharged starting materials in reaction a. The energy diagram

in Figure 7.48 shows what this differential solvation does to the rate of this reac-

tion. The height of the transition state is decreased relative to the starting material

and the reaction goes faster. It takes less energy for the molecules to traverse the

barrier when the polarity of the solvent increases.

Energy

Reaction progress

HO S(CH

)

–

HO CH

S(CH

)

....

S(CH

)

HO CH

Activation

energy in more

polar solvent

Activation

energy in less

polar solvent

FIGURE 7.49 The difference between

the energy of the starting materials

and the energy of the transition state

is larger when the reaction is run in a

more polar solvent (red curve) and

smaller when the reaction is run in a

less polar solvent (blue curve).The

result is an increased activation energy

for the reaction and a slower rate.

288 CHAPTER 7 Substitution and Elimination Reactions

Summary

The S

2 reaction—Substitution, Nucleophilic, bimolecular—is common for

methyl, primary, and secondary substrates. It is disfavored by steric hindrance in

either the substrate (the “R” group) or the entering nucleophile. It is favored by

a strong Lewis base (nucleophile) and a good leaving group.The response of the

2 reaction to changes in solvent polarity varies with the charge type of the reac-

tion. Each case must be examined individually. The S

2 reaction always involves

inversion of configuration at the center of substitution.

FIGURE 7.50 Replacement of the

hydrogens by an alkyl group

would hinder—or even eliminate—

hydrogen bonding. Thus, proper

association of the bases would no

longer be possible.

7.5 The S

2 Reaction in Biochemistry

As mentioned in Section 7.1, your vital biomolecules—DNA, for example—are

covered with nucleophiles, each of them ready to do the S

2 reaction if presented

with a suitable alkylating agent. Once that alkylation takes place, two

important changes are induced. First of all, the shape of the mol-

ecule is changed, and the critical fit, on which proper bioaction

depends, is altered. Second, alkylation often removes the possibility

for hydrogen bonding, and in turn that changes the ability of DNA

to associate properly with other molecules. As we will see in Chapter

23, DNA exists as pairs of associated “nitrogenous bases” and that

association is achieved through hydrogen bonding (Fig. 7.50).

Alkylation of one of the nucleophilic groups in DNA can disrupt base

pairing and lead to destructive mutations.

Remember the methyl transfer from S-adenosylmethionine (p. 142)? Heinz Floss

(b. 1934) and his co-workers have demonstrated that a similar transfer occurs through

an S

2 displacement. They did this experiment by showing that transfer takes place

with complete inversion of configuration—the hallmark of the S

2 reaction. How can

one show inversion in a simple methyl group, CH

? That’s not easy! Floss and his co-

workers used two isotopes of hydrogen, deuterium (D) and tritium (T), so that their

methyl was not CH

but CHDT.Use of a single enantiomer revealed the diagnostic inver-

sion in the transfer shown in Figure 7.51.The work involved in synthesizing that single

enantiomer and in analyzing the product was prodigious,but these prototypal techniques

have been borrowed several times in similar studies of biochemical transformations.

Note inversion

Modified

S-adenosylmethionine

COO

“methyl”

transfer

FIGURE 7.51 Methyl transfer involves

an S

2 reaction with inversion.

Deoxyribose

C(1')

Deoxyribose

C(1')