Jones M., Fleming S.A. Organic Chemistry

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9.3 Effects of Resonance on Regiochemistry 369

Figure 9.9 illustrates a reaction that we will return to in some detail later but can

examine quickly now. Hydrogen chloride adds to 1,3-butadiene very easily to yield

two products. One product you may be able to predict, but the other will probably

not be immediately obvious (Fig. 9.9).

1,4-Addition

(20–25%)

1,2-Addition

(75–80%)

HCl

WEB 3D

FIGURE 9.9 1,2- and 1,4-Addition

of HCl to 1,3-butadiene.

A primary carbocation

An allylic cation

H H

–

1,3-Butadiene

HCl

FIGURE 9.10 The two possible

carbocations from protonation of

1,3-butadiene.

Protonation at

the end carbon

gives a delocalized

allylic cation

Protonation at an

internal carbon

gives a localized

primary cation

FIGURE 9.11 In the allylic cation, the positive charge is shared by two carbons; in the

primary cation it is localized on a single carbon.

Look at the possibilities.There are two cations that can arise from protonation of

1,3-butadiene by hydrogen chloride (Fig.9.10). In the allylic cation,the positive charge

is on the carbon adjacent to the double bond; in the other, it is isolated on a terminal

primary carbon. The choice between these two ions becomes clear if we draw orbital

pictures of the two carbocations. In the allylic cation, the p orbital on the positively

charged carbon atom overlaps the π orbitals of the double bond (Fig. 9.11). We have

seen the system of three parallel overlapping p orbitals before (Problems 1.16c and 1.61).

Here we have an allylic cation again, and the molecular orbital description is the same

as before.The three 2p atomic orbitals can be combined to yield three molecular orbitals.

There are only two electrons in an allylic cation and they will occupy the lowest, most

stable molecular orbital (Fig. 9.12).The presence of the double bond is highly stabiliz-

ing for this cation, because the charge is delocalized over the allyl system.

Allyl cation

FIGURE 9.12 The molecular orbitals

of the allylic cation.

370 CHAPTER 9 Additions to Alkenes 1

–

Protonation to give this localized

cation will be disfavored

Protonation to give this delocalized allylic

cation will be favored

H H

–

HCl

FIGURE 9.13 Preferential formation

of the lower energy, delocalized

cation.

–

Notice the partial (dashed) bonds used

to indicate the delocalization

In an alternative

formalism, the

positions sharing the

charge are shown by

(+) placed at the

appropriate sites

H H

(+)

FIGURE 9.14 In the allylic cation, the

positive charge is shared by two

carbons. Note the summary

structures in which dashed bonds or

charges in parentheses are used to

indicate delocalization without

writing out all the resonance forms.

PROBLEM 9.3 Derive the π orbitals of the allyl system shown in Figure 9.12.

Hint: Do this task by placing a carbon 2p orbital between the two p orbitals of

ethylene. Another Hint: Watch out for “net zero,” orthogonal interactions!

For the primary carbocation in the reaction shown in Figure 9.10, the positive

charge is localized at the end of the molecule and has no resonance stabilization.

Protonation at the end carbon gives an ion (the allylic cation) that can be represent-

ed by more than one resonance structure. The choice is easy once we have looked

carefully at the structures. Formation of the lower energy delocalized allylic cation will

be favored over formation of the higher energy localized primary cation (Fig. 9.13).

The real structure of the delocalized cation is best represented as a combination

of the two resonance forms.This way of looking at the allylic cation is especially help-

ful because it points out simply and clearly which carbons share the positive charge.

When the chloride ion approaches the allylic cation it can add at either of the two

carbons that share the positive charge to give the two products shown in Figure 9.14.

9.3 Effects of Resonance on Regiochemistry 371

CONVENTION ALERT

Energy

⌽

C(3)

C(1)

C(2)

For allyl, the LUMO is ⌽

In this orbital, there are

only two places where

the filled orbital of

chloride can overlap, C(1)

and C(3); there is a node

at the middle carbon,

C(2), and there can be no

interaction with another

orbital there

add at C(1)

add at C(3)

–

1,4-Product

1,2-Product

C(3) C(1)

C(2)

Filled, nonbonding

orbital of chloride

HOMO

The lowest energy

unoccupied orbital

of the allylic cation

LUMO

FIGURE 9.15 In this reaction, the HOMO–LUMO

interaction is between a filled nonbonding orbital

on chloride and the lowest energy unoccupied

orbital of allyl, in this case, .

Whether one uses resonance forms or molecular orbital descriptions is largely a

matter of taste.Resonance forms are the traditional way, and they are extremely good

at letting you see where the charge resides. We will use them extensively. It’s worth

taking time now to review the whole resonance system.

Be careful not to think of this resonance-stabilized structure as spending part of its

time with the charge on carbon 1 and part of the time with the charge on carbon 3!

The cation has a single structure in which two carbons of the allylic cation share the

positive charge.

This single, summary resonance structure is often shown using dashed bonds to

represent the bonds that are double in one resonance form and single in the other.

The charge is placed at the midpoint of the dashes as shown in the structure at the

top right of Figure 9.14. Alternatively, one resonance form is drawn out in full and

the other atoms sharing the charge are shown with a () or (), as also shown in

Figure 9.14.

Now let’s ask if the molecular orbital description can predict both products. Of

course it can, or it would hardly be very useful. What happens as the chloride ion,

with its filled nonbonding orbitals,begins to interact with the allylic cation? Stabilizing

interactions occur between filled orbitals and empty orbitals. Chloride bears the filled

orbital; therefore we must look for the lowest unoccupied molecular orbital (LUMO)

of allyl. Figure 9.15 shows it, . There are two points at which chloride can add

to , and they lead to the two observed products. Note that the middle carbon,

through which the node passes,cannot be attacked. So the molecular orbital descrip-

tion also explains the regiochemistry of the addition.

372 CHAPTER 9 Additions to Alkenes 1

Summary

Addition of HX acids to alkenes really is straightforward. An initial protonation

to give the most stable carbocation possible is followed by capture by X



The only difficulty is figuring out which carbocation intermediate is the lowest

energy. Here, an analysis of the “usual suspects” is necessary. You have to consider

the substitution pattern, resonance effects, and, as we shall soon see, inductive

effects as well.

CCH

–

ClH

FIGURE 9.16 In the chlorine-

substituted carbocation, the charge

is shared by carbon and chlorine.

9.4 Brief Review of Resonance

This subject was treated in detail in Chapter 1. Here we only pick up the highlights

of that treatment. If anything seems unfamiliar or strange go back to Chapter 1! A

good way to check that this important material is under control is to see if you can

do Problems 1.18–1.21.

Remember that a molecule that is best described as a resonance hybrid, does not,

repeat not,spend part of its time as one form and part as another.That is chemical equi-

librium not resonance.The two are very different phenomena. Resonance is always

indicated by the special double headed arrow , whereas equilibrium is shown by

two arrows pointing in opposite directions .

As we pointed out as early as Chapter 1, the real structure of a resonance-

stabilized molecule is a hybrid, or combination, of the resonance forms contribut-

ing to the structure.These forms represent different electronic descriptions of the

molecule. Resonance forms differ only in the distributions of electrons and never in

the positions of atoms. If you have moved an atom, you have written a chemical

equilibrium.

It is important to be able to estimate the relative importance of resonance

forms in order to get an idea of the best way to represent a molecule. To do

so, we can assign a weighting factor, c, to each resonance form. The weighting

factor indicates the percent contribution of each resonance form to the overall

structure. Some guidelines for assigning weighting factors are summarized

below.

1. Equivalent resonance forms contribute equally.

2. The more bonds in a resonance form, the more stable the form is.

3. Resonance forms must have the same number of paired and unpaired electrons.

4. Separation of charge is bad.

5. In ions, delocalization is especially important. As we have mentioned a number of

times, small charged atoms are usually most unstable. Delocalization of electrons

that allows more than one atom to share a charge is stabilizing.The protonation of

vinyl chloride is a good example. Vinyl chloride is protonated to give the cation in

which the charge is shared by carbon and chlorine. The resonance forms show this

sharing clearly (Fig. 9.16).

9.4 Brief Review of Resonance 373

–

In this case, c

> c

FIGURE 9.17 Two resonance forms

for the enolate anion.

WEB 3D

90⬚

Rotation

–

Bridgehead

position

–

Overlap very poor

FIGURE 9.18 In this enolate-like

anion, the orbital containing the pair

of electrons does not overlap well

with the p orbitals of the .

There is no resonance, no matter

what the deceptive two-dimensional

surface says.

7. Now that we have a good deal of structure under our belts, we can look at one more

phenomenon relevant to evaluating good (low energy) resonance forms.We can tell

you to “Watch out! Geometry is important.” Is the “enolate” anion in Figure 9.18

resonance stabilized? On paper it certainly looks like it is, and there is at least a

typographical relationship to the enolate anion of Figure 9.17. If you answered

yes, however, you have been fooled by the two-dimensional surface. The three-

dimensional structure of the cage molecule prevents substantial overlap of the

orbitals. Even though the paper doesn’t protest when you draw the second reso-

nance form, in reality, it does not contribute. The negative charge is borne on only

one carbon—the bridgehead—and is not shared by any other atom. Build a model

and you will see it easily.

PROBLEM 9.4 Are the two structures shown below resonance forms? Explain why

or why not?

6. Electronegativity is important. In the enolate anion, which has two resonance

forms, the two do not contribute equally because they are not equivalent (Fig.9.17).

Each has the same number of bonds, so we cannot choose the better representation

that way. However, form A has the charge on the relatively electronegative oxygen

while form B has it on carbon. Form A is the better representation for the enolate

anion, although both forms contribute. Mathematically, we would say that the

weighting factor for A,c

, is larger than that for B,c

374 CHAPTER 9 Additions to Alkenes 1

PROBLEM 9.5 How many signals will the following species show in their

NMR spectra? In fact, not all of these species are stable enough to be detected

by NMR, but in each case it is possible to predict what would be (will be?)

seen.

(a)

(d)

(b) (c)

(e) (f)

–

Less stable

More stable

(delocalized)

–

Vinyl chloride

HCl

FIGURE 9.19 In the addition of hydrogen chloride to vinyl chloride, it is formation of the more stable,

resonance-stabilized carbocation that determines the product.

9.5 Resonance and the Stability of Carbocations

We saw in Figure 9.5 one example of how an atom capable of delocalizing electrons

can help share a charge and influence the regiochemistry of an addition reaction.

Addition of hydrogen chloride to vinyl chloride gives 1,1-dichloroethane, not

1,2-dichloroethane. Protonation produces a carbocation adjacent to the chlorine,

which is far more stable than the alternative in which the positive charge sits local-

ized on a primary carbon (Fig. 9.19).

In Section 9.2, we looked at the reaction of symmetrically substituted

2,3-dimethyl-2-butene with hydrogen chloride (Fig. 9.2). In the formation of the

carbocation, there was no choice to be made—only one cation could be formed.

When the less symmetrical alkene 2-methyl-1-butene reacts with hydrogen

9.5 Resonance and the Stability of Carbocations 375

In an unsymmetrical system, it is the more stable carbocation that leads to product:

–

Primary carbocation

(less stable)

Tertiary carbocation

(more stable)

FIGURE 9.20 The regiospecific

addition of HCl to 2-methyl-1-

butene. It is the more stable

carbocation that leads to product.

chloride, there are two possible carbocations (Fig.9.20).The only chloride formed

is the one produced from the more stable, tertiary carbocation.This product results

from the hydrogen adding to the less substituted end of the alkene and the chlo-

rine to the more substituted end (Fig. 9.20).

Markovnikov states in his 1870 paper (Ann. 1870, 153, 228): “wenn ein unsymmetrisch constituirter

Kohlenwasserstoff sich mit einer Haloïdwasserstoffsäure verbindet, so addirt sich das Haloïd an das weniger

hydrogenisirte Kohlenstoffatom, d.h. zu dem Kohlenstoff, welcher sich mehr unter dem Einflusse anderer

Kohlenstoffe befindet.”Which is to say “when an unsymmetrical alkene combines with a hydrohalic acid, the

halogen attaches itself to the carbon atom containing the fewer hydrogen atoms—that is to say, the carbon

that is more under the influence of other carbons.”

2-Pentene

–

(45–48%)

(52–55%)

Secondary

carbocations

FIGURE 9.21 Two products are

formed from the reaction of HCl

and 2-pentene.

These phenomena have been known a long time and were first correlated by the

Russian chemist Vladimir V. Markovnikov (1838–1904) in 1870. The observation

that the more substituted halide is formed now bears his name in Markovnikov’s

rule.

Rules are useless (or worse) unless we understand them, so let’s see why

Markovnikov’s rule works.

Unsymmetrical alkenes such as 2-pentene, from which two secondary carbocat-

ions of roughly equal stability can be formed, do give two products in comparable

amounts (Fig. 9.21).

376 CHAPTER 9 Additions to Alkenes 1

Figures 9.19–9.21 show that the mechanism of HX addition must involve for-

mation of the more substituted cation,because the less substituted cation would lead

to a product that is either not observed or is only formed in minor amounts.In turn,

a reasonable presumption is that the more substituted a carbocation is,the more sta-

ble it is (Fig. 9.22). Remember: These differences in stability are large. In practice,

reactions in which tertiary and secondary carbocations are intermediates are com-

mon, but primary and methyl carbocations cannot be formed.

Tertiary

Cation stability (R = alkyl group)

Secondary

Primary

RRR

Methyl

FIGURE 9.22 Stability order for

carbocations.

PROBLEM SOLVING

STOP

Energy

Reaction

ress

⌬G

–

Tertiary

carbocation

Secondary

carbocation

⌬⌬G

Activation energy for formation of more stable

tertiary carbocation

Activation energy for formation of less stable

secondary carbocation

Difference in energies of transition states is

the difference in activation energies

†

FIGURE 9.23 The transition states for formation of

the high-energy carbocations will contain partial

positive charges (δ



).The same order of stability

that holds for full positive charges applies to partial

positive charges. A δ



on a tertiary carbon is more

stable than a δ



on a secondary carbon, and so on.

Remember (p. 141) simple primary and methyl carbocations (no

delocalization through resonance) are mechanistic “stop signs.”

They are too high in energy to be formed, and are not known as

intermediates under “normal”conditions.

How does the difference in energy of these two carbocations, the tertiary and sec-

ondary cationic intermediates in the addition reaction, become translated into a pref-

erence for one product over the other? The relative rates of the two possible protonation

steps will not depend directly on the relative energies of the two carbocations them-

selves, but on the energy difference between the two transition states leading to them

(ΔΔG

‡

in Fig. 9.23). Both protonation steps are endothermic and it is likely that the

9.5 Resonance and the Stability of Carbocations 377

In the ethyl cation, a C

bond can overlap with the empty

2p orbital; this filled–empty

orbital interaction is stabilizing

In the methyl cation, no

such stabilization is

possible!

Methyl cation (CH

)Ethyl cation (H

CCH

)

Energy

Empty

Filled σ

bond

FIGURE 9.24 The ethyl cation is stabilized by a filled orbital overlapping with an empty orbital.

transition states will bear a strong resemblance to the cations. (Remember the

Hammond postulate, “For an endothermic reaction the transition state will resemble

the product,” p. 355.) In these two transition states, positive charge is building up on

carbon. There are partial positive charges in the transition states. A tertiary partial

positive charge is more stable than a secondary partial positive charge, just as a full positive

charge on a tertiary carbon is more stable than a full positive charge on a secondary carbon.

Once the carbocation intermediate is formed, the structure of the product is deter-

mined. Capture of the tertiary carbocation leads to the observed product, in which X

is attached to the more substituted carbon of the starting alkene (the site of the terti-

ary positive charge in the cation). Similarly, the secondary carbocation would lead

inevitably to a product in which X is attached to the less substituted carbon (Fig. 9.23).

In Chapters 3 and 7, we avoided the issue, but it is now time to deal with the

question of why alkyl groups stabilize carbocations. Our mechanistic hypothesis is

consistent with the idea that the more substituted halide is produced because it

is the result of the initial formation of the more substituted (and hence more sta-

ble) carbocation (Fig. 9.20). But this consistency doesn’t answer anything; we have

just pushed the question further back.

Why are more substituted carbocations more stable than less substituted ones?

To answer such questions,we start with structure. Carbocations are flat, and the cen-

tral carbon is sp

hybridized and therefore trigonal. Figure 9.24 compares the methyl

cation (



) with the ethyl cation (



).Note that in the ethyl cation there

is a filled carbon–hydrogen bond that can overlap with the empty p orbital. This

phenomenon has the absolutely grotesque name of hyperconjugation, but there is

nothing particularly “hyper” about it. It is a simple effect and can be rationalized

in familiar molecular orbital or resonance terms. The orbital mixing is stabilizing

(Fig. 9.24) and nothing like it is available to the methyl cation. The more interac-

tions of this kind that are possible,the more stable the ion.Thus the tert-butyl cation

is most stable, the isopropyl cation next, and so on.

Alternatively, we could write the following kind of resonance form, which shows

the charge delocalization from carbon to hydrogen especially well (Fig. 9.25).

It’s important to see that the second carbon–carbon bond in the newly drawn

FIGURE 9.25 A resonance description

of the ethyl cation (



378 CHAPTER 9 Additions to Alkenes 1

In the “hyperconjugative” resonance

form for the ethyl cation, the “double”

part of the double bond is not made up

of 2p /2p overlap, but of 2p /sp

overlap

The π bond in ethylene is

made up of 2p /2p overlap

FIGURE 9.26 A detailed look at

hyperconjugative stabilization of the

ethyl cation.

resonance form of Figure 9.25 is not identical to a π bond of an alkene.It is not formed

from 2p/2p overlap but rather from 2p/sp

overlap (Fig. 9.26). Be sure you see the

difference between these two fundamentally different,but similar-looking structures.

PROBLEM 9.6 It is often claimed that the tert-butyl cation has 10 resonance

forms, 1 in which the carbon bears the positive charge and 9 in which each hydro-

gen bears the positive charge. Is this claim true?

PROBLEM 9.7 On page 115 we described the stabilization of alkenes by attached

alkyl groups in different terms, focusing on the σ bond system. We could con-

struct a similar treatment here. Provide a justification for the relative stability of

differently substituted carbocations using an analysis of the σ system.

PROBLEM 9.8 Do you think the π part of the double bond in the resonance form

in Figure 9.25 is stronger or weaker than a normal π bond? Explain why.

We have just seen that carbocations are stabilized by delocalization through the

overlap of filled and empty orbitals. One example of this effect is fairly straightfor-

ward; a lone pair of electrons on an adjacent atom interacts with an empty orbital.

Less obvious is the last phenomenon, the overlap of filled σ orbitals of

carbon–hydrogen bonds with empty orbitals, called hyperconjugation. Now we pro-

ceed to other factors influencing the stability of carbocations.

Relative rate of

addition at 25 ⬚C

Ethylene

Propene

Methyl vinyl ether

H X1

–

H X2 ⫻ 10

OCH

5 ⫻ 10

H X

FIGURE 9.27 Addition reaction to

three different alkenes.

in this case.

HX = HOH

9.6 Inductive Effects on Addition Reactions

Polar bonds in a molecule can have a major effect on both the rate and the regiochem-

istry of an addition reaction. Let’s look first at the series of ethylenes in Figure 9.27,