Jones M., Fleming S.A. Organic Chemistry

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21.3 Neighboring  Systems 1099

OTs

OAc

(AcOH)

COOH

PROBLEM 21.16 Work through the reaction of the (R,R) optically active isomer of

3-phenyl-2-butyl tosylate in acetic acid (see below). This time the product is the

optically active acetate shown.

FIGURE 21.29 Two examples of carbon–carbon double bonds acting as intramolecular

nucleophiles.

21.3b Carbon–Carbon Double Bonds as Neighboring Groups

Carbon–carbon double bonds are also nucleophiles and can act as neighboring

groups in intramolecular displacement reactions. Among the examples you already

know are the sequence of intramolecular carbocation additions leading to steroids

(p. 560) and intramolecular carbene additions (p. 434). In each of these reactions, a

carbon–carbon double bond acts as a nucleophile toward an empty carbon 2p

orbital as Lewis acid (Fig. 21.29).

Neighboring carbon–carbon π bonds can influence the rate and control the stereo-

chemistry of a reaction just as a heteroatom can. For example, the tosylate of anti-

7-hydroxybicyclo[2.2.1]hept-2-ene (7-hydroxynorbornene)

reacts in acetic acid 10

The bicyclo[2.2.1]heptane cage is commonly known as the norbornyl system.This obscure name comes

from a natural product, borneol, which has the structure shown in Figure 21.30, and comes from the cam-

phor tree, common in Borneo.The prefixes syn and anti locate the direction of the OH or OTs group with

respect to the double bond, and nor means “no methyl groups.”Thus, 7-hydroxybicyclo[2.2.1]hept-2-ene

is 7-hydroxynorborn-2-ene, or, because there is only one possible place for the double bond, 7-hydroxy-

norbornene.

1100 CHAPTER 21 Intramolecular Reactions and Neighboring Group Participation

Borneol Bornane Norbornane

PROBLEM 21.17 There is only one possible place for the double bond in the bicy-

clo[2.2.1]heptyl ring system. Explain why.

times faster than the corresponding syn compound,and a staggering 10

times faster

than the tosylate of the saturated alcohol, 7-hydroxynorbornane. The product is

exclusively the anti acetate (Fig. 21.31).

Rate = 10

anti Tosylate

syn Tosylate

Rate = 10

H O

Rate = 1

COOH

OTsH

TsO

OTsH

WEB 3D

FIGURE 21.31 The anti isomer of

7-norbornenyl tosylates gives the anti

acetic acid.The anti tosylate reacts

times faster than the syn isomer

and 10

times faster than the

saturated compound!

FIGURE 21.30 Norbornane

(bicyclo[2.2.1]heptane) is related

to the natural product borneol,

endo-1,7,7-trimethylbicyclo[2.2.1]

heptan-2-ol.

A consistent mechanistic explanation for the huge difference in reaction rates

derives from the fact that only in the anti compound is the π system of the double

bond in the correct position to assist in ionization of the tosylate. For the double

bond in the syn isomer to assist in ionization would require a frontside S

2 displace-

ment, which, as we know, is an unknown process.

The rate enhancement is attributed to anchimeric assistance by the double bond.

Intramolecular displacement of the leaving group is faster than the possible inter-

molecular reactions and is therefore the dominant process. The stereochemistry of

the reaction is explained by noting that addition of acetic acid to the intermediate

ion is also an S

2 reaction and must occur from the rear of the departing leaving

21.3 Neighboring  Systems 1101

syn Tosylateanti Tosylate

anti Acetate formed

with retention of

configuration

COOH

frontside

OTs

TsO

H H

OAc

Intramolecular

Intermolecular

Deprotonation

FIGURE 21.32 The carbon–carbon double bond of the anti tosylate, but not of the syn

isomer, can react in an intramolecular S

2 reaction to displace the leaving group. Opening

of the cyclic ion must give the anti acetate, the result of net retention.

group. The net result of the two S

2 reactions (one intramolecular, one intermol-

ecular) is retention of configuration (Fig. 21.32).

Think a bit now about the nature of the leaving group in the second, intermol-

ecular S

2 displacement.Just what is the leaving group, and what do all those dashed

bonds in Figure 21.32 mean? To answer these questions well, we must look closely

at the structure of the postulated intermediate in this reaction. When we formed

three-membered ring intermediates from heteroatoms, there was always a pair of

nonbonding electrons available to do the intramolecular displacement. An example

is the formation of a bromonium ion by displacement of water by neighboring

bromine (Fig. 21.33).

HBr

protonation

A normal,

two-electron

bond

FIGURE 21.33 In a bromonium ion, all bonds are normal two-electron bonds.

Contrast the bromonium ion of Figure 21.33 with the species formed by

intramolecular displacement of tosylate using the π bond. In the bromonium ion,

there are sufficient electrons to form the new σ bond in the three-membered ring.

In the all-carbon species there are not enough electrons. In this cyclic ion, two

1102 CHAPTER 21 Intramolecular Reactions and Neighboring Group Participation

OTs

TsO

H H H

FIGURE 21.34 Two views of the

cyclic ion from Figure 21.32 and the

resonance forms contributing to the

ion. Notice the symmetry that is

more clearly shown in the lower view.

–

FIGURE 21.35 When a

nucleophile opens a

bromonium ion, a pair of

electrons is displaced.The

same is true for the opening

of the two electron–three

carbon, cyclic ion.

PROBLEM 21.18 If three atoms share the positive charge of the ion in Figure 21.34,

why is it only the 7-position that is attacked by the nucleophile? Explain the lack

of another product. Hint: Look at the structure of the product formed by addi-

tion at the other position in the following reaction:

is a normal, two-electron σ bond that is broken.When the all-carbon three-mem-

bered ring is attacked, it is the two electrons binding the three carbons that act

as the leaving group (Fig. 21.35). The process is still an S

2 reaction, however,

and so there is no reason for softening the requirement for backside displacement.

Acceptance of the carbon–carbon double bond as a neighboring group requires you

to expand your notions of bonding,and to accept the partial bonds making up the dashed

three-center, two-electron delocalized system as a valid bonding arrangement.There are

those who resisted this notion mightily, and although they were ultimately shown to be

incorrect, their resistance was anything but foolish. The postulated intermediate in this

reaction is not a normal species and should be scrutinized with the greatest care.

–

Why is this “

?”

not formed?

electrons must suffice to bind three atoms (Fig. 21.34).The ion contains an exam-

ple of three-center, two-electron bonding, and the partial bonds are represent-

ed with dashed lines. A resonance formulation shows most clearly how the charge

is shared by the three atoms in the ring. The bromonium ion and the cyclic ion

of Figure 21.34 are very different. When the cyclic bromonium ion is opened, it

21.3 Neighboring  Systems 1103

A molecular orbital approach shows how the two electrons can be stabilized

through a symmetrical interaction with an empty carbon 2p orbital. The intellectual

antecedents for this kind of bonding relate to the cyclic H

molecule in which two

electrons are stabilized in a bonding molecular orbital through a cyclic overlap of hydro-

gen 1s orbitals. We can construct this system by allowing a 1s orbital to interact with

σ and σ* of H

as in Figure 21.36 (recall our discussion of the allyl and cyclopropenyl

π systems in Chapter 12).The result is a system of three molecular orbitals: one bond-

ing,two antibonding.Two,but only two, electrons can be accommodated in the bond-

ing orbital. The same orbital pattern will emerge for any triangular array of orbitals.

No interaction with 1s

σ, σ*

σ*

+ 1s = 1 σ – 1s = 2

= 3

Nodes

Energy

Nonbonding

Same number

of nodes

Orbital pattern for a triangular array of H

–

FIGURE 21.36 The construction of

the molecular orbitals of cyclic H

Cyclopropenium ion

PROBLEM 21.19 Work out the π molecular orbitals for the cyclopropenium ion

shown below.

Chemists resisting this notion (the “classicists”) suggested that there was no need for

this new kind of delocalized bonding, and that normal, localized two-electron models

would do as well.Let’s follow their argument.Remember as we go that they must explain

both a rate acceleration for the anti tosylate and the overall retention of stereochemistry.

The two compounds in question here, the syn and anti tosylates, are different, and

there must be some rate difference between them.One can scarcely argue with this point,

and one is reduced to relying on the magnitude and direction of the rate acceleration as

evidence for the delocalized ion. A classicist might argue that in cases such as this, the

rate increase could easily come from the formation

of a normal, localized or “classical,” carbocationic

intermediate. The anti tosylate has the leaving

group in an excellent orientation to be attacked by

the intramolecular nucleophile, the filled π orbital

of the carbon–carbon double bond (Fig. 21.37).

The syn tosylate does not.

no frontside

2 possible

H OTs TsO H

synanti

FIGURE 21.37 A “classical”

displacement reaction by the π bond

of the anti isomer would give a

localized ion that could not be

formed from the syn isomer.

1104 CHAPTER 21 Intramolecular Reactions and Neighboring Group Participation

As in the symmetrical displacement to form the delocalized ion, the required

attack on the leaving group from the rear could occur only in the anti compound.

Of course,there are two equivalent intermediates possible (A and B,Fig. 21.38),and

it was suggested that they were in equilibrium. The compound with the charge at

the 7-position (C) might also be a partner in the equilibrium. Be sure you see the

difference between the three equilibrating structures of Figure 21.38 and the three

resonance forms contributing to the single structure of Figure 21.34.

rapid

equilibrium

A racemic pair of ions

ABC

7-Position

FIGURE 21.38 The first-formed ion

A equilibrates with B and, perhaps, C.

Addition of acetic acid to ions A and B of Figure 21.38 must give the anti acetate,

because the opening is an S

2 reaction.Attack on ion C is more problematic,because

it is difficult to explain the stereospecificity of the reaction. Still, it must be admit-

ted that the syn and anti faces of the ion are different, and in principle, this differ-

ence could lead to a stereospecific reaction (Fig. 21.39).

1. S

2. deprotonation

anti Acetate

syn Acetate

1. S

2. deprotonation

1. add

anti Face

syn Face

2. deprotonate

If ion C is involved,

why only anti acetate—

why not syn as well?

OAc

anti Acetate

OAc

AcO

FIGURE 21.39 Addition of acetic acid

to the equilibrating ions A and B

must lead to anti acetate. Attack at

the 7-position (ion C) to give only

the anti acetate presents some

difficulties.

21.3 Neighboring  Systems 1105

Stop right now and compare the two mechanistic hypotheses (Fig. 21.40). In one,

there is a symmetrical displacement of the leaving group to give a resonance-stabilized

delocalized intermediate. In the other, either the leaving group is displaced to give a

carbocationic intermediate containing a three-membered ring (A and B) in equilibrium

with at least one other ion, C, or a simple S

1 ionization to C takes place.

Symmetrical, delocalized

intermediate

H OTs

Unsymmetrical, normal localized

pair of equilibrating carbocations

A and B

H H H

FIGURE 21.40 A comparison of the symmetrical, delocalized intermediate with the localized ions A, B, and C.

The whole difference between the two mechanisms is the formulation of the

structure of the intermediate! The structures in the two versions of the mechanism

for this reaction are only very subtly different. In one, the constituent structures are

contributing resonance forms to a single intermediate delocalized species; in the

other they are equilibrating molecules, each with its own separate existence.

Are we now down to questions of chemical trivia? Are we counting angels on

pin heads? Many thought so, but we maintain that they were quite misguided in

their criticisms of those engaged in the argument. All questions of mechanism come

down to the existence and structures of intermediates, and to the number of barri-

ers separating energy minima.

This argument has been recast many times in related reactions. Not all questions

have been resolved, but those that have been have generally involved two kinds of

evidence. Sometimes a direct observation of the ion in question was made, usually

under superacid conditions (p. 679) at low temperature.In other reactions,an exper-

iment like the following one provides evidence for symmetrical participation to pro-

duce a delocalized intermediate.

The reactions in acetic acid of the monomethyl and dimethyl derivatives of

the p-nitrobenzoate of anti-bicyclo[2.2.1]hepten-7-ol have been examined. The

1106 CHAPTER 21 Intramolecular Reactions and Neighboring Group Participation

rate = 1.0

rate = 13.3

A more stable bridged ion—

one resonance form is a

tertiary carbocation

COOH

FIGURE 21.41 Formation of a

symmetrical, delocalized ion from the

methyl-substituted molecule would

be faster than ion formation from

the parent.

COOH

rate = 1.0

rate = 13.3

Secondary carbocation

Tertiary carbocation

FIGURE 21.42 There would also be

a rate increase in the formation of

the unsymmetrical, localized ion

postulated by the classicists.

relative to that of the parent, which can produce only a less stable secondary carbo-

cation (Fig. 21.42).The observed rate increase of 13.3 for the monomethyl analogue

is accommodated by both mechanisms.

monomethyl compound reacts faster than the parent (Fig. 21.41), as predicted by either

mechanism. A delocalized ion would be stabilized because one of its contributing

resonance forms is tertiary.Another way of putting this is to note that the methyl group

makes the double bond a stronger nucleophile. In the localized mechanism, a terti-

ary carbocation is formed and this would surely accelerate the rate of the reaction

PROBLEM 21.20 Draw resonance forms for the monomethyl intermediate of

Figure 21.41.

21.3 Neighboring  Systems 1107

Relative rate

= 13.3

Relative rate = 1.0

Predicted rate (13.3)

= 177

Both methyl groups stabilize

the cation at the same time

COOH

FIGURE 21.43 Reaction of the

dimethyl compound to give a

delocalized ion should be still faster

than the reaction of the monomethyl

compound.The two methyl groups

simultaneously stabilize the cation.

Predicted rate = 2  13.3 = 26.6

COOH

FIGURE 21.44 In the localized, unbridged ions, the two methyl groups can never stabilize the ion at the same time.

Now consider the dimethyl compound. If a delocalized intermediate is formed

through a symmetrical displacement, both methyl groups exert their influence at the

same time.If one methyl group induces a rate effect of 13.3,two should give an effect

of 13.3 × 13.3

= 177 (Fig. 21.43).

On the other hand, if it is the formation of a localized tertiary ion that is impor-

tant, the second methyl increases the possibilities by a factor of 2, but the two methyl

groups never exert their influence at the same time. The rate change should approxi-

mate a factor of 2 (Fig. 21.44).

1108 CHAPTER 21 Intramolecular Reactions and Neighboring Group Participation

The fundamental difference between a mechanism involving a symmetrical delo-

calized ion and the traditional mechanism, which postulates a localized, less sym-

metrical ion, is that in the delocalized case both methyl groups exert their stabilizing

influence at the same time.In the pair of equilibrating open, localized ions they can-

not both act simultaneously.

The observed rate for the dimethyl compound is 148, much closer to that

expected for the delocalized intermediate than for the localized ion.

Summary

We have progressed from a discussion of obvious neighboring groups,heteroatoms

with nonbonding electrons, to less obvious internal nucleophiles, the filled π

orbitals of double bonds. In every case, a pair of internal electrons does an initial

intramolecular S

2 displacement, which is followed by a second, external (inter-

molecular) S

2 reaction. The result is overall retention of configuration.

21.4 Single Bonds as Neighboring Groups

Surely,one of the weakest nucleophiles of all would be the electrons occupying the low-

energy bonding σ orbitals. Yet there is persuasive evidence that even these tightly held

electrons can assist in the ionization of a nearby leaving group. For example, when the

tosylate of 3-methyl-2-butanol is treated with acetic acid the result is formation of the

rearranged tertiary acetate, not the “expected” secondary acetate (Fig. 21.45).

(97%) (3%)

OTs OAc

OAc

(AcOH)

COOH

25 ⬚C

FIGURE 21.45 The reaction of the

tosylate of 3-methyl-2-butanol in

acetic acid involves a rearrangement.

This result is easily rationalized through a hydride shift to form the relatively

stable tertiary carbocation.The interesting mechanistic question is whether the sec-

ondary carbocation is an intermediate, or whether the hydride moves as the leaving

group departs (Fig. 21.46).Thus, the timing of the steps becomes an issue. Is formation

of the tertiary carbocation a two- or one-step process? This example recapitulates

OTs

1. CH

COOH

2. deprotonation

–

OTs

Secondary

carbocation

Tertiary

carbocation

–

H migration

–

OTs

leaves

Ionization

H shift

FIGURE 21.46 Two possible

mechanisms for the acetolysis of

3-methyl-2-butyl tosylate.The green

route is a two-step process. The red

path is a one-step process.