Jones M., Fleming S.A. Organic Chemistry

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9.6 Inductive Effects on Addition Reactions 379

Br Cl

1Rate 10

–

FIGURE 9.28 Rates of addition to

three alkenes.

which shows typical rates of addition for a series of substituted ethylenes.The reaction

is, in fact, a hydration reaction in which HOH adds across the double bond.

Ethylene itself will protonate only very slowly. Propene will surely protonate so as

to give the more stable secondary cation,and the methoxyl group of methyl vinyl ether

will provide resonance stabilization to an adjacent positive charge. Hyperconjugative

stabilization by the methyl group of propene and resonance stabilization by the

methoxyl group of methyl vinyl ether, respectively, have a great accelerating effect on

the rate of protonation, the rate-determining step in this reaction. Figure 9.27 shows

the rates of protonation compared to that of unsubstituted ethylene itself.

PROBLEM 9.9 Why is the rate of protonation of ethylene especially slow?

Substituents can slow reactions as well, especially if they are not in a position to

provide resonance stabilization.Figure 9.28 shows typical rates for addition to some

substituted alkenes. To see why bromine and chlorine so dramatically slow the rate

of addition,we must again look more closely at the carbon–halogen bond (see p. 225

for an earlier look).

–

= Br or Cl

FIGURE 9.29 The polar bond.C

–

Notice the bad electrostatic interaction

between

and the + charge

–

FIGURE 9.30 Protonation gives an

intermediate destabilized by the

electrostatic interaction between the

newly formed full positive charge and

the partial positive charge (δ



Wouldn’t the methoxyl group of methyl vinyl ether act in the same way? Oxygen

is also more electronegative than carbon, and wouldn’t this difference in electroneg-

ativity act to discourage any addition that puts more positive charge on the already

The two atoms involved in the bond, carbon and halogen, are different, so the

bond between them must be polarized.Therefore, the electrons in the bond cannot

be equally shared. Bromine and chlorine are far more electronegative than carbon

(see Table 1.8, p. 15) and they will attract electrons strongly. The carbon–halogen

bond is polarized as shown in Figure 9.29. The carbon will bear a partial positive

charge (δ



) and the more electronegative halogen a partial negative charge (δ



For the molecules in Figure 9.28, protonation in the Markovnikov sense places

a full positive charge on the carbon adjacent to the already partially positive carbon

(Fig.9.30).This interaction will surely be destabilizing and accounts for the decrease

in rate for addition reactions to the bromine- and chlorine-substituted molecules

shown in Figure 9.28. Effects such as these, which are transmitted through the σ

bonds, are called inductive effects.

380 CHAPTER 9 Additions to Alkenes 1

partially positive carbon adjacent to oxygen? Why does addition to methyl vinyl

ether occur so rapidly (Fig. 9.27) in the Markovnikov sense? Figure 9.31 frames the

question, and answers it as well. The answer is simply that the resonance effect of

oxygen overwhelms the inductive effect.The two effects are in competition, and the

stronger resonance effect wins out. Now let’s summarize, then look at a number of

related addition reactions.

–

This intermediate is strongly

stabilized by resonance and

the stabilizing resonance

effect outweighs the

destabilizing inductive effect

The C

O σ

bond in

methyl vinyl ether is

polarized so that a

partial positive charge

is on carbon; this

makes protonation on

the adjacent carbon

more difficult

FIGURE 9.31 Stabilizing resonance

effects and destabilizing inductive

effects can “fight” each other.

Generally, resonance stabilization

is more important.

(97%)

–

FIGURE 9.32 Addition of HI to 1-octene in Markovnikov fashion.

9.7 Addition Reactions: Hydration

If and add to double bonds, it should not be surprising that other acidic

compounds undergo similar addition reactions. Hydrogen iodide is a simple

example, and addition rather closely resembles the reactions we have already

studied in this chapter. Markovnikov’s rule is followed, for example (Fig. 9.32).

HClHBr

Summary

Protonation (or addition of any Lewis acid) to an unsymmetrical double bond

leads preferentially to the more stable carbocation.The energy of carbocations is

strongly influenced by stabilizing resonance effects. Substituents bearing a lone

pair of electrons are especially effective at stabilization, but even alkyl groups can

donate electrons to the empty 2p orbital of a carbocation. In cases in which induc-

tive and resonance effects oppose each other, resonance usually wins.

Water ( ) is an reagent, but, as we saw on p. 139 it is not a strong

enough acid to undergo addition to simple alkenes at a useful rate.If one mixes water

and a typical alkene, no reaction occurs. The difference between and isHXHOH

HXHOH

9.7 HX Addition Reactions: Hydration 381

that and are strong enough acids to protonate the alkene and start things

going.Water is not.The pK

values of , ,and are 9,8,and 15.7,

respectively, which means that water is a weaker acid than and by factors

of about 10

and 10

! Acid must be added to protonate the alkene and thus gen-

erate a carbocation that can continue the reaction.The acid catalyst in this reaction

sets things going. The catalyst has no effect on the position of the final equilibri-

um,but it does lower the energy requirement for generating product.As Figure 9.33

illustrates, the catalyst provides a new mechanistic pathway for product formation.

It is able to carve a new pathway along a lower slope of the mountain we first encoun-

tered in Figure 8.20 (p. 346).

HClHBr

HOHHClHBr

HBrHCl

Alkene hydration

The lowest energy

path from A to B

A catalyst provides a lower energy path but it does not alter the

energy of starting material and product, A and B; rather, it changes

the energy of the transition state, or transition states, in the reaction

Village A Village A

Village BVillage B

FIGURE 9.33 A catalyst does not change the energy of either starting material or product but does provide

a lower energy path between them.

As can be seen in Figure 9.34, the catalyst H



is used up in the first step of the

reaction and is regenerated in the last step. A number of acids can be used in the cat-

alytic role; sulfuric (H

or ) or nitric (HNO

or ) are

commonly used to promote the addition reaction. In water containing a little sulfu-

ric acid for instance,2,3-dimethyl-2-butene is protonated to generate the tertiary car-

bocation. This intermediate is then attacked by (water, not hydroxide,



OH)

to give the oxonium ion, , which can be deprotonated by water to

regenerate the catalyst, H



,and produce a molecule of an alcohol, .Thus,

the outcome of the hydration reaction is addition of across a π bond.HOH

HOH

HONO

HOSO

WEB 3D

An alcoholAn oxonium ion

O Catalyst regenerated in this step

O Catalyst used up in this step

FIGURE 9.34 The mechanism of

hydration of 2,3-dimethyl-2-butene.

382 CHAPTER 9 Additions to Alkenes 1

The mechanism shown in Figure 9.34 predicts, indeed requires, that the

Markovnikov rule be followed, and that sets up a mechanistic test. We will exam-

ine the hydration of 2-methylpropene to probe the regiochemistry of the reaction.

Before we even start, it is worth thinking a bit about the results.What will we know

after we have done the experiment? There are only two general results possible:

Either the Markovnikov rule is followed or it is not (Fig. 9.35). If it is not, we can

say with certainty that the mechanism we have suggested for this reaction is wrong.

This compound is the

product of Markovnikov

addition, and is formed

in the reaction

Mixture

This compound is the

product of a path that does

not follow Markovnikov

addition (anti-Markovnikov

addition), and is not formed

in the reaction

path a

path b

path c

FIGURE 9.35 Three possible

regiochemical results from the

hydration of 2-methylpropene.

WEB 3D

tert-Butyl alcohol2-Methylpropene More stable, tertiary

cation intermediate

protonation addition deprotonation

FIGURE 9.36 The mechanism for Markovnikov addition of water to 2-methylpropene.The product must be tert-butyl alcohol

if this is indeed the mechanism.

A crucial step in our mechanism involves the formation of a carbocation. When

there is a choice, as there certainly is with 2-methylpropene, the more stable car-

bocation must be formed. Once formed, this cation must ultimately give the

product in which the hydroxyl group is attached to the more substituted carbon.

If our mechanism postulated in Figure 9.36 is correct, there is no other possible

result. So if we get either the “wrong” alcohol or a mixture of alcohols (paths b

and c in Fig. 9.35), we know the mechanism is wrong. It may need only a minor

adjustment, or it could require complete scrapping, but it cannot be correct as

written.

More interesting, and less obvious, is what we know if our prediction turns

out to be correct. Have we proved the mechanism? Absolutely not. Our mecha-

nistic hypothesis remains just a hypothesis: a guess at how the reaction goes that

9.7 HX Addition Reactions: Hydration 383

is consistent with all experiments that have been done so far.There may be some

other test experiment that would not work out as well. The sad fact is that no

number of experiments we can do will suffice to prove our mechanism. We are

doomed to be forever in doubt, and our mechanism to be forever suspect. There

is no way out. Indeed, all the mechanisms in this book are “wrong,” at least in

the sense that they are broad-brush treatments that do not look closely enough

at detail.

Organic chemistry has come a long way since the days of the rectangular

hypothesis, which was little more than a bookkeeping device. In olden times it was

common practice to write a reaction such as the S

1 reaction of tert-butyl iodide

with water in the way shown in Figure 9.37. To use the rectangular hypothesis,

one simply surrounds the appropriate groups with a rectangle, which somehow

removes the captured atoms and reconstitutes them, in this case as hydrogen iodide.

C I OHH HI

COH

FIGURE 9.37 The rectangular

hypothesis applied to the reaction

of tert-butyl iodide with water.

This archaic device is scarcely a reasonable description of how the reaction occurs

and has almost no predictive value. If you know one rectangle, you do not know

them all! But let’s not get too smug. We do know more than chemists of other gen-

erations. We know, for example, that carbocations are involved in the reactions we

are currently studying,and molecular orbital theory gives us additional insight into

the causes of these reactions. But we are still very ignorant. We are certain that

much of what we “know” is wrong, at least in detail. Chemists of the future will

regard us as primitives, and they will be quite right. It doesn’t matter, of course.

Our job is to do the best we can; to add what we can to the fabric of knowledge

and to provide some shoulders for those future chemists to stand on, as our earlier

colleagues did for us.

In this case, things turn out well for the mechanistic hypothesis shown in

Figure 9.36, and the Markovnikov rule is followed in hydration reactions.

There are variations on this reaction. Ethylene, for example, is hydrated only

under more forcing conditions—concentrated sulfuric acid must be used—and the

initial product is not the alcohol, but ethyl hydrogen sulfate, which is converted in

a second step into the alcohol (Fig. 9.38). It is not easy to be sure of the mechanism

of this reaction.The concentrated sulfuric acid may be required to produce the very

unstable primary carbocation, or a different, perhaps cyclic cation may be involved

(see Problem 9.9).

OSO

Ethyl hydrogen sulfate Ethyl alcohol

sulfuric acid

O , 0 ⬚C

HO SO

C CH

FIGURE 9.38 Ethyl hydrogen sulfate

is an intermediate in the hydration of

ethylene.

384 CHAPTER 9 Additions to Alkenes 1

Now we have a structure that contains eight carbons, as do the two new prod-

ucts. It looks as though we are on the right track.The only remaining trick is to see

how to lose the “extra” proton and form the new alkenes. Losing H

as shown in

HOSO

CCH

FIGURE 9.39 Dimeric products

formed from 2-methylpropene.

9.8 Dimerization and Polymerization of Alkenes

Let’s look again at the hydration of 2-methylpropene. What if we change the con-

ditions a bit from what they were in Figure 9.36? What if both the temperature and

concentration of sulfuric acid are increased? Figure 9.39 shows that two new prod-

ucts,each containing eight carbons,are formed.These are dimeric products (C

formed somehow by the combining of two molecules of 2-methylpropene (C

Brønsted acid

The nucleophilic

alkene acts as

a Brønsted base

Lewis acid

The nucleophilic

alkene acts as

a Lewis base

CCH

HOH

FIGURE 9.40 Many different acids add to 2-methylpropene (and other alkenes). Here we see H



and the tert-butyl

cation adding to 2-methylpropene to give tertiary carbocations. In each case, addition occurs in the Markovnikov sense, as

the more stable carbocation is formed preferentially.

Surely, increasing the acid concentration can only make protonation of the alkene

more probable, so the first step of the reaction is again formation of the tertiary car-

bocation intermediate shown in Figure 9.36. Now, however, the initially formed car-

bocation finds itself in the presence of fewer water molecules than it did in the reaction

we looked at before. Accordingly,formation of the alcohol through reaction with water

is less likely.We might ask what other Lewis bases are available for the strongly Lewis

acidic carbocation.The most obvious one is the starting alkene itself. The tert-butyl

cation intermediate can react with it, just as the proton did. Both the proton and the

carbocation are Lewis acids and can add to the double bond of 2-methylpropene to

give a new cation (Fig. 9.40). There is no essential difference in the two reactions,

and once more it is the more stable, tertiary carbocation intermediate that is formed.

Summary

Water is not a strong enough acid to protonate a simple alkene.However, if a trace

of acid catalyst is present, an initial protonation will be followed by addition of

water to produce an oxonium ion.A final deprotonation gives the alcohol.As usual,

you have to be careful to analyze which carbocation is formed, and to consider

structural, resonance, and inductive effects in making that determination.

9.8 Dimerization and Polymerization of Alkenes 385

–H

deprotonation

protonation

–H

deprotonation

protonation

C CH

CCH

FIGURE 9.41 The two products of

Figure 9.39 are formed by

deprotonation of the intermediate

carbocation.

Figure 9.41, will lead to the first product and losing H

gives the second. We have

seen similar deprotonations before in our discussion of the E1 reaction (p. 298), and

we have also seen the reverse reaction. Loss of these protons is just the reverse of

protonation of an alkene (Fig. 9.41).

Alkene polymerization

The proton can’t just leave, however. It must be removed by a base. In concentrated

sulfuric acid, strong bases do not abound.There are some bases present, however, and

the carbocation is a high-energy species, and thus rather easily deprotonated. Under

these conditions, both water and bisulfate ion (HSO



) are capable of assisting the

deprotonation.

Note the exact correspondence of the steps involved in alkene protonation and

deprotonation.The two reactions are just the two sides of an equilibrium (Fig. 9.42).

protonation

loss of a proton

(deprotonation)

FIGURE 9.42 Protonation and

deprotonation are just two sides

of an equilibrium.

If the concentration of acid is even higher, the base concentration will be even

lower and both alcohol formation and deprotonation will be slowed (both require water

to act as Lewis base). There is no reason why the dimerization process cannot con-

tinue indefinitely under such conditions, leading first to a trimer (a molecule formed

from three molecules of 2-methylpropene) and, eventually, to a polymer (Fig. 9.43).

Repeat

FIGURE 9.43 The beginning stages of the cationic polymerization of 2-methylpropene.

386 CHAPTER 9 Additions to Alkenes 1

PROBLEM 9.10 Treating 2-ethyl-1-butene with H



O leads to three prod-

ucts each having the formula C

. Draw the three structures and provide a

mechanism for their formation.

HCl

CCH

CH CH

2-Chloro-3-methylbutane

(40%)

3-Methyl-1-butene 2-Chloro-2-methylbutane

(60%)

FIGURE 9.44 Reaction of HCl with

3-methyl-1-butene gives a new

product in addition to the expected

one.

The obvious question is, Why? Will our mechanism for addition have to be

severely changed to accommodate these new data? In fact, we will be able to graft

a small modification onto our general mechanism for addition reactions and avoid

drastic changes. The way to attack a problem of this kind—explaining the for-

mation of an unusual product—is to write the mechanism for the reaction as we

have developed it so far and then see if we can find a way to rationalize the new

product. In this case, protonation of 3-methyl-1-butene can give either a secondary

Teflon is the polymer formed from

polymerization of tetrafluoroethene.

Here are several useful applications of

this wonderful material.

The cation-induced polymerization is called, logically enough, cationic poly-

merization and ends only when the alkene concentration decreases sufficiently

to slow the addition step.

9.9 Rearrangements during Addition

to Alkenes

When adds to isopropylethylene (3-methyl-1-butene), a quite surprising thing

happens. We do get some of the product we expect, 2-chloro-3-methylbutane, but

the major compound formed is 2-chloro-2-methylbutane (Fig. 9.44).

HCl

9.9 Rearrangements during HX Addition to Alkenes 387

Less stable primary cation is not formed

3-Methyl-1-butene

More stable secondary cation is formed

–

HCl

FIGURE 9.45 No real choice here—

the more stable secondary

carbocation is formed.

–

Secondary

carbocation

Minor product

shift of

hydride,

Major product

Tertiary

carbocation

chloride

adds

path a

path b

chloride

adds

FIGURE 9.46 In the reaction between

3-methyl-1-butene and HCl the

major product is formed through a

rearrangement mechanism (path b)

involving the migration of hydrogen

with its pair of electrons (hydride,



) to give the more stable tertiary

carbocation from the less stable

secondary ion.

or primary carbocation (Fig. 9.45). There is no real choice here, because the

secondary carbocation is of far lower energy than the hideously unstable primary

carbocation and will surely be the one formed.

If the chloride ion adds to the positively charged carbon of the secondary carbo-

cation, we get the minor product of the reaction,and the one we surely expected (path

a, Fig. 9.46). To get the major product, we must have a rearrangement, which is a

word describing the relocation of an atom (or atoms) in a molecule as a result of elec-

trons shifting. In this case,a hydrogen must move, with its pair of electrons, from the

carbon adjacent to the cation to the positively charged carbon itself (path b,Fig. 9.46).

Such migrations of hydrogen with a pair of electrons are called hydride shifts.

Why should this rearrangement happen? Note that in such a process we have

generated a lower-energy tertiary carbocation from a higher-energy secondary car-

bocation. The cation is certainly “better off” (lower in energy) thermodynamically

if the hydride moves. In practice, such rearrangements are quite common. Hydride

shifts to produce more stable cations from less stable ones are easy. Indeed, not only

are hydride shifts common, but so are alkyl shifts, called Wagner–Meerwein

rearrangements (Georg Wagner, 1849–1903; Hans Lebrecht Meerwein,1879–1965).

388 CHAPTER 9 Additions to Alkenes 1

For example, 3,3-dimethyl-1-butene adds hydrogen chloride to give not only

2-chloro-3,3-dimethylbutane but also 2-chloro-2,3-dimethylbutane. The new,

rearranged compound results from an alkyl shift—in this case the migration of a

methyl group with its pair of electrons (Fig. 9.47). Why is this reaction called an

“alkyl shift,”when by analogy to “hydride shift” it should be “alkide shift”? We don’t

know, but it is!

PROBLEM 9.11 Write mechanisms that account for the products in the following

reaction:

PROBLEM 9.12 A hydride shift to a positively charged carbon is just another

example of a Lewis base–Lewis acid reaction. Although it looks very different

from the more obvious electrophile–nucleophile reactions we have studied, it

really is quite similar. In the reaction shown in the previous problem, there is a

very strong and strategically located electrophile and a weak, but nonetheless

reactive enough nucleophile. Identify the nucleophile and electrophile in this

reaction.

HCl

–

Secondary carbocation

–

shift

Tertiary carbocation

path a

path b

2-Chloro-2,3-dimethylbutane

(60–75%)

2-Chloro-3,3-dimethylbutane

(25–40%)

3,3-Dimethyl-1-butene

–

The minor, unrearranged,

product is formed from the

secondary carbocation

(path a)

The major product is formed

from the rearranged, tertiary

carbocation (path b)

FIGURE 9.47 Formation of the major product involves a rearrangement mechanism in which a shift of a methyl

group with its pair of electrons generates a more stable tertiary carbocation from a less stable secondary

carbocation.