Jones M., Fleming S.A. Organic Chemistry

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12.10 Thermodynamic and Kinetic Control of Addition Reactions 539

The 1,2-addition product is favored under conditions of kinetic control

(short time, lower temperature), and the 1,4-product is favored under con-

ditions of thermodynamic control (long time, higher temperature). The

1,2-product is the kinetic product, and results from taking a pathway over a

lower-energy transition state.The 1,4-product results from conditions that favor

formation of the more stable product; it is the thermodynamic product in

this case.

Now we have a fourth question, and it’s a nice one. For both Cl

and HCl

addition, we know why thermodynamics favors the 1,4-product; a disubstituted

alkene is more stable than a monosubstituted alkene. But why does kinetic con-

trol favor the product of 1,2-addition?

Let’s look at HCl addition because the best experiments have been done in

this area.The simplest explanation focuses on distance. Where is the chloride ion

after the initial protonation at C(1)? At the moment of its birth, it is much closer

to C(2) than to C(4). Perhaps addition of chloride is faster at the 2-position than

at the 4-position by virtue of simple proximity (Fig. 12.44).

A second possibility looks at the two resonance forms for the allylic cation

(Fig. 12.44), which are not equivalent. The one on the left is a secondary carbo-

cation, the one on the right is a primary carbocation. The first will be weighted

more heavily than the second, and most of the positive charge in the ion will

be at the secondary position. Perhaps it is this more positive position that is

attacked by the anion under kinetic control to give the 1,2-addition product. It

is this factor that is almost always cited as the reason 1,2-addition is preferred

kinetically.

There is a brilliant experiment that settles the issue, done by Eric Nordlander

(1934–1984) at Case Western Reserve University. He studied the addition of deu-

terium chloride ( ), not to 1,3-butadiene, but to 1,3-pentadiene.In this mol-

ecule, addition of D



yields a resonance-stabilized diene in which both resonance

Chloride is closer to

C(2) than to C(4)

and perhaps this is

the reason that 1,2-

addition is favored

–

1,2-Product

C(2) addition C(4) addition

Addition to C(4)Addition to C(2)

1,4-Product

C(2)

C(4)

C(2)

C(4)

FIGURE 12.44 After protonation, the chloride ion is closer to C(2) than to C(4).

Perhaps this proximity accounts for the preference for 1,2-addition.

540 CHAPTER 12 Dienes and the Allyl System: 2p Orbitals in Conjugation

forms are secondary cations. Indeed these resonance forms are identical except for

the remote isotopic deuterium label (Fig. 12.45).

PROBLEM 12.19 Predict the kinetic and thermodynamic products for HCl addi-

tion to 4-methyl-1,3-pentadiene. Be careful, thermodynamic products are not

always a result of 1,4-addition.

–

If the proximity of chloride

to C(2) is the reason for

the kinetic preference for

1,2-addition, it should

still be preferred in this

electronically symmetrical

cation, and it is!

1,4-Addition (longer

path for chloride)

(24%)

1,2-Addition (shorter

path for chloride)

(75.5%)

FIGURE 12.46 1,2-Addition is still

preferred in 1,3-pentadiene. The

kinetic preference for 1,2-addition

is a simple proximity effect.

Summary

Under thermodynamic (relatively high energy) conditions, the more stable prod-

uct is the major compound formed in addition reactions with conjugated dienes.

Under kinetic conditions, the major product comes from 1,2-addition—even if

it is the less stable product.Why? Proximity! In the formation of the allyl cation,

the nucleophile is “born” closer to the 2-position than to the 4-position.

–

These resonance forms are equivalent (except for the D)

and the two positions must share the positive charge equall

FIGURE 12.45 Protonation (D

this case) of 1,3-pentadiene gives

a resonance-stabilized allylic

carbocation in which both

contributing resonance forms are

secondary. Addition of chloride at

the two positions must take place

to give equal amounts of 1,2- and

1,4-addition.

If the preference for 1,2-addition depends on the difference between a secondary

and a primary carbocation, that preference should disappear in the reaction with

1,3-pentadiene. If proximity is at the root of the effect, 1,2-addition should still be

favored.For once simplicity wins out, and 1,2-addition is still preferred (Fig. 12.46).

12.11 The Allyl System: Three Overlapping 2p Orbitals 541

Allyl

Molecular orbitals for allyl

Resonance forms for allyl

FIGURE 12.47 The resonance and

molecular orbital treatment of the

allyl system.

We will now continue with an examination of a few more reactions in which

allylic stabilization is important.

12.11 The Allyl System: Three Overlapping

2p Orbitals

We have met the structure of the allyl system several times already (Problems 1.61,9.3),

but Figure 12.47 again summarizes both the molecular orbital treatment and the

resonance treatment of this system of three contiguous 2p orbitals.

PROBLEM 12.20 Add electrons to both the resonance and molecular orbital

descriptions in Figure 12.47 to form the allyl anion, radical, and cation.

12.11a The Allylic Cation: S

1 Solvolysis of Allylic Halides The reso-

nance stabilization of the allylic cation makes allylic halides relatively reactive under

1 conditions (Fig. 12.48).

WEB 3D

–

deprotonation

3-Iodopropene

(allyl iodide)

capture

by water

FIGURE 12.48 The S

1 solvolysis of allyl iodide in

water. This reaction proceeds much faster than the

solvolysis of propyl iodide.

542 CHAPTER 12 Dienes and the Allyl System: 2p Orbitals in Conjugation

As mentioned earlier, the allyl cation can be captured at either carbon sharing

the positive charge. Table 12.3 gives some average rates of S

1 solvolysis for a few

common structural types.Primary allylic halides react much faster than primary alkyl

halides.Secondary and tertiary allylic halides also ionize faster than their non-allylic

counterparts. Resonance stabilization does have an accelerating effect on the

ionizations.

TABLE 12.3 Relative Rates of S

1 Solvolysis of RCl in 50% Ethyl Alcohol at 45 ⬚C

Molecule

The value for the primary substrate is suspect, and may include some contribution from S

2 reactions.

Molecule

Ion

Primary

Secondary

Allyl: both resonance

forms are primary

Allyl: both resonance

forms are primary

Allyl: one resonance

form is primary,

one secondary

Relative Rate

1.0

1.7

14.3

21.4

1300

Ion

Allyl: one resonance

form is primary,

one secondary

Tertiary

Allyl: one resonance

form is primary,

one tertiary

Allyl: one resonance

form is primary,

one tertiary

Relative Rate

1157

3 ⫻ 10

1.9 ⫻ 10

7.9 ⫻ 10

PROBLEM 12.21 Explain why the last two entries in Table 12.3 react at nearly the

same rate and are so much faster than the others. Can you devise a molecule that

would react even faster than the last two entries? Draw it.

Remember: We are doing a dangerous thing here. We are again equating ther-

modynamic stability (a secondary allylic cation is more stable than a non-resonance-

stabilized secondary cation) with kinetics (therefore secondary allylic halides form

cations faster than do typical secondary halides). Although justified here, we must

be clear on the reason why thermodynamics and kinetics are closely related in this

example. Work Problem 12.22.

PROBLEM 12.22 Explain carefully, with the use of an energy diagram, why in an

1 reaction of an alkyl halide the thermodynamic stability of the intermediate

cation is related to the ease of ionization of the corresponding alkyl halide.

Remember the Hammond postulate (p. 351), and consider the transition state

for ionization.

12.11b S

2 Reactions of Allylic Halides Not only do allylic halides react

faster than non-allylic halides in the S

1 reaction, but they are accelerated in S

12.11 The Allyl System: Three Overlapping 2p Orbitals 543

reactions as well. Table 12.4 gives some average rates of S

2 reactions for a variety

of halides.

Transition state for S

2 displacement

of an all

l halide

(

lower ener

gy)

Transition state for S

2 displacement

of a

rimar

halide

(

her ener

gy)

–

FIGURE 12.49 A comparison of the transition states for S

2 displacement of a primary iodide and allyl iodide.

TABLE 12.4 Relative Rates of S

2 Reaction with Ethoxide in Ethyl Alcohol at 45 ⬚C

Molecule

Relative Rate

1.0

37.0

33.0

Relative Rate

1.85

556

When we are considering a question involving reaction rates, the answer can

always be found through an analysis of the transition states. The rate is dependent

on the activation energy for the reaction, which is the height of the transition state

relative to the starting halide. Let’s compare the transition state for the reaction of

a 1-propyl halide with that for the reaction of an allyl halide (Fig. 12.49).

In the transition state for the S

2 reaction of the allyl halide, there is overlap

between the 2p orbital developed on the carbon at which displacement occurs and

the 2p orbitals of the double bond (Fig. 12.49). Therefore, this transition state con-

tains an allylic system. It will benefit energetically from the delocalization of elec-

trons, and will lie at lower energy than the transition state for S

2 displacement in

the 1-propyl system, in which there is no delocalization.

12.11c The Allyl Radical It is much easier to form a resonance-stabilized

allylic radical than an undelocalized radical. We saw the effects of delocalization in

reactions involving allylic radicals in Chapter 11 (p. 497). For example, the regiospe-

cific bromination of the allylic position of alkenes with NBS is possible because of

the relative ease of forming the delocalized allyl radical (Fig. 11.56).

12.11d The Allyl Anion Now let’s look at the acidity of allylic hydrogens to

see if the formation of a resonance-stabilized allylic anion has any effect on that

acidity.We would surely expect it to.If delocalization of electrons is important, then

544 CHAPTER 12 Dienes and the Allyl System: 2p Orbitals in Conjugation

removal of a proton at the allylic position should be easier than at a position not

leading to a delocalized anion. A comparison of the acidities of propane and

propene should test this idea (Fig. 12.50).

∼ 50–60

= 43

–

FIGURE 12.50 Propene is far more

acidic than propane.

In fact,the pK

of propene is 43,some 7–17 pK

units lower than that for propane

(pK

50–60). The uncertainty arises from the difficulty of measuring the pK

an acid as weak as a saturated hydrocarbon. In any event, propene is a much stronger

acid than propane, and it is reasonable to suppose that the formation of the delo-

calized allyl anion is responsible for the increased acidity of the alkene.

12.12 The Diels–Alder Reaction of Conjugated Dienes

In 1950, Otto Diels (1876–1954) and his student Kurt Alder (1902–1958) received

the Nobel prize for their work on what has appropriately come to be known as the

Diels–Alder reaction, which they discovered in 1928. (So much for timeliness; at

least they lived long enough to outlast the Nobel committee’s renowned conser-

vatism.) We will have more to say about its mechanism in Chapter 20, but the

Diels–Alder reaction is characteristic of conjugated dienes and is of enormous syn-

thetic utility. Therefore we must outline it here.

When energy is supplied to a mixture of a conjugated diene and an alkene, a

ring-forming reaction takes place to produce a cyclohexene.The alkene in this reac-

tion is called a dienophile, because it has demonstrated an affinity for the diene.

The product of the reaction is often called the adduct, which is another word for

product in a reaction between two molecules. Our discussion of the Diels–Alder

mechanism begins with the arrow formalism,which points out which bonds are bro-

ken and shows where the new bonds are made in the forward reaction (Fig. 12.51).

Which orbitals are involved? The Diels–Alder reaction begins with the overlap

of the π systems of the dienophile and the diene.The two reaction partners approach

in parallel planes and as the bonds form,the end carbons of both the diene and alkene

rehybridize from sp

to sp

(Fig. 12.52).

Now sp

Still sp

⌬

FIGURE 12.52 In the Diels–Alder

reaction, the two participants

approach each other in parallel planes.

As the reaction occurs, the end atoms

of the diene and the dienophile

rehybridize from sp

to sp

⌬

FIGURE 12.51 The arrow formalism

for the Diels–Alder reaction.

Diels–Alder reaction

12.12 The Diels–Alder Reaction of Conjugated Dienes 545

PROBLEM 12.23 Curiously, it matters just how energy is supplied to the molecules

in the Diels–Alder reaction. If we heat the reactants up, the reaction occurs, but

if we attempt to supply energy photochemically, it generally fails. We will exam-

ine such matters in detail in Chapter 20, but we might start to consider them now.

When we think about reactions between atoms, we do not consider every electron

in the atom, just those most loosely held, the valence electrons. Similarly, in

molecular reactions it is the most weakly held electrons, those in the HOMO,

that control reactivity. The electrons in the HOMO might well be thought of as

the valence electrons of molecules. Remember what Figure 12.25 tells us about

the interaction of light and molecules. First, consider the phase relationships in

the possible HOMO–LUMO interactions in the transition state for a simple

Diels–Alder reaction.Then think about how the absorption of a single light pho-

ton might change those interactions and offer an explanation for why light

energy does not promote the Diels–Alder reaction.

PROBLEM 12.24 Write an arrow formalism for the reverse Diels–Alder reaction.

It is possible to estimate the thermochemistry in the Diels–Alder reaction by

comparing the bonds made and broken in the reaction.Three π bonds in the start-

ing material are converted into one π bond and two σ bonds in the product.

Accordingly, the reaction is calculated to be exothermic by approximately 28 kcal/mol

(Fig. 12.53).

FIGURE 12.53 The simplest Diels–Alder reaction is exothermic by about

28 kcal/mol.

PROBLEM 12.25 Why is the bond strength of the new σ bonds in cyclo-

hexene (80 kcal/mol each) used in Figure 12.53 lower than the bond in

ethane (90 kcal/mol)?

As shown in the previous figures, the Diels–Alder reaction requires the s-cis

form of the diene. However, it is the s-trans arrangement that is the more sta-

ble one, by approximately 3 kcal/mol (p. 523). If equilibrium favors the s-trans

form, why is the Diels–Alder reaction observed at all? Even though there is

little s-cis form at equilibrium, reaction with the dienophile disturbs the equi-

librium between the s-trans and s-cis form by depleting the s-cis partner.

Reestablishment of equilibrium generates more s-cis molecules, which can react

in turn. Eventually, all the diene can be converted into Diels–Alder product,

546 CHAPTER 12 Dienes and the Allyl System: 2p Orbitals in Conjugation

no reaction

with dienophile

Reaction

ress

Energy

Transition state for the

Diels–Alder reaction

s-trans

Conformation

s-cis

Conformation

⌬

FIGURE 12.54 Even though

the unreactive s-trans form of

1,3-butadiene is favored at

equilibrium, the small amount of

the s-cis form present can lead to

product.The vertical axis of the

figure is not to scale.The squiggly

lines indicate gaps.

⌬

FIGURE 12.55 When there can be no

s-cis form of the diene, the

Diels–Alder reaction fails.

WORKED PROBLEM 12.26 Identify the source of strain in the putative product

shown for Diels–Alder addition to 3-methylenecyclohexene in Figure 12.55. Use

your models.

ANSWER Although the paper puts up with the double bond drawn in Figure

12.55, Nature won’t.The orbitals making up this bridgehead double bond do not,

in reality, overlap.

Poor

overlap

even though at any moment there is very little of the active s-cis form present

(Fig. 12.54).

When there is no possibility of any s-cis form, there can be no Diels–Alder reac-

tion. For example, although 1,2-dimethylenecyclohexane reacts normally with

dienophiles, 3-methylenecyclohexene does not. The product of this reaction would

contain a double bond far too strained to be formed (Fig. 12.55).

12.12 The Diels–Alder Reaction of Conjugated Dienes 547

Another serious mechanistic question involves the timing of the formation of the

two new σ bonds in the cyclohexene. Are they made simultaneously in a concerted

fashion, or are they formed in two separate steps? In principle, the mechanism of the

Diels–Alder reaction could either be concerted or involve two steps (Fig. 12.56).

step two

step one

In this nonconcerted, two-step

mechanism, the two

σ bonds

are formed in two separate steps

A concerted mechanism;

both new

σ bonds are

formed simultaneously

⌬

FIGURE 12.56 Arrow formalisms for

concerted and two-step mechanisms

for the Diels–Alder reaction.

This question asks if there

is an intermediate in the

reaction or not. Figure

12.56 shows the two arrow

formalisms and Figure

12.57 gives Energy versus

Reaction progress diagrams

for both the concerted and

two-step processes.

We might immediately

be suspicious of the two-

step mechanism. As it

involves an intermediate

allylic radical (Figs. 12.56

and 12.57), shouldn’t we

observe closure of the

diradical at both possible

positions? Should there not

be a vinylcyclobutane pro-

duced as well as the cyclo-

hexene (Fig. 12.58)?

One step

Reaction progress

Energy

Transition state

(no intermediate)

Reaction progress

Two step

Energy

Allylic radical and a

primary radical within

the same molecule (a

diradical intermediate)

(b)

path b

(a)

path a

step two

step one

Vinylcyclobutane

(not observed)

Normal

Diels–Alder

product

FIGURE 12.58 The two-step reaction

could lead to vinylcyclobutane.This

product is not observed.

FIGURE 12.57 Energy versus Reaction progress diagrams for the concerted and two-step mechanisms.

548 CHAPTER 12 Dienes and the Allyl System: 2p Orbitals in Conjugation

PROBLEM 12.27 Write an Energy versus Reaction progress diagram for the two-

step Diels–Alder reaction that includes formation of the vinylcyclobutane.

H R

cis

Diene

Dienophile

A one-step, concerted mechanism must

retain stereochemistry

A cyclohexene

trans

Still trans

Still cis

FIGURE 12.59 A concerted mechanism must preserve the

stereochemical relationships of the groups on the dienophile.

In fact, Diels–Alder reactions do not lead to vinylcyclobutanes, but give

cyclohexenes exclusively. Although this argument would seem at first to favor

the concerted mechanism, which can only give cyclohexenes, it could be

countered that the transition state for closing the four-membered ring will

contain much of the strain present in all cyclobutanes, and will therefore be high

in energy.

As is so often the case, the mechanistic question of the timing of new bond

formation is finally answered with a stereochemical experiment. A one-step reac-

tion must preserve the stereochemical relationships present in the original alkene.

For example, in a concerted reaction, a cis alkene must give a cis disubstituted

cyclohexene. If the reaction is concerted,which means the two new σ bonds form

simultaneously, there can be no change in the stereochemical relationship of

groups on the alkene (Fig. 12.59). If the bonds are formed in two steps, there

should be time for rotation about the carbon–carbon σ bond in the dienophile.