Jones M., Fleming S.A. Organic Chemistry

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20.5 Sigmatropic Shift Reactions 1049

This kind of process, in which starting material rearranges into itself, is called a

degenerate reaction (p. 1031). Isotopically labeled molecules reveal the degenerate

reaction and allow a determination of the kinetic activation parameters. The reac-

tion requires an activation energy of 36–38 kcal/mol. Other labels are possible, of

course,and substituted molecules will reveal the reaction as well as the more sophis-

ticated isotopic labels. In most substituted cases the reaction is no longer degener-

ate and the equilibrium constant K for the reaction can no longer equal 1 (Fig.20.31).

PROBLEM 20.11 Which of the molecules in each of the two reactions of Figure 20.31

will be favored at equilibrium? Explain your choice. Draw the product of heating

(Z)-3-methyl-1,3-pentadiene. Is this reaction degenerate or not?

WORKED PROBLEM 20.12 Estimate the bond dissociation energy (BDE) for the

migrating carbon–hydrogen bond in 1,3-pentadiene. Why is the observed activa-

tion energy of 36–38 kcal/mol for the [1,5] shift of hydrogen shown in Figure

20.30a so much lower than your answer?

ANSWER The carbon–hydrogen bond in ethane has a BDE of 101 kcal/mol

(Table 8.2). But the pentadienyl radical is resonance stabilized and that stabiliza-

tion will begin to be felt as the bond breaks, further decreasing the BDE. If allylic

resonance is worth about 13 kcal/mol (p. 475), we might expect the BDE to be

about 101  13  88 kcal/mol.

FIGURE 20.31 Substituted molecules also reveal the reaction, but the rearrangement is no longer degenerate,

because starting material and product are different.

BDE = 101 kcal/mol

BDE (first guess) = ~101 kcal/mol

(continued)

1050 CHAPTER 20 Reactions Controlled by Orbital Symmetry

Yet the real BDE is much, much lower than this, about 37 kcal/mol. The reason

is that as the carbon–hydrogen bond begins to break, the new carbon–hydrogen

bond at C(1) is forming. It is not necessary to break the carbon–hydrogen bond

fully to make the very unstable hydrogen radical.

Instead, formation of the new bond is initiated prior to breaking of the

original bond. This bond formation decreases the energy required to

break the bond.

The deuterium labeling experiment of Figure 20.30b reveals that much is hap-

pening before the onset of high-temperature radical reactions. A hydrogen (deuterium,

in this case) atom is moving from one position in the molecule to another.This kind

of reaction has come to be known as a sigmatropic shift.

A sigmatropic shift is the

movement of a sigma bond from one atom to another. These reactions can involve

hydrogens, carbons, or heteroatoms.

How are sigmatropic shifts formally described? The first step is to identify the

bond that is broken in the reaction and the bond that is made. In other words, write

an arrow formalism. The bond that breaks in 1,3-pentadiene could be designated

the “1–1”bond. Remember that these numbers have nothing to do with the IUPAC

name. Next, note where the new bond is formed. In Figure 20.32a, the new bond is

[1,5],in (b) it is [2,3], and in (c) it is [3,3].The numbers denoting the shift are always

enclosed in brackets and separated by a comma.

C(1)

C(5)

It must be admitted that the Woodward–Hoffmann theory is filled with jargon. There is nothing to do but

learn it.

[1,5]

Starting point

Other starting point

Terminus

Bond

broken

New bond

(a)

[2,3]

New bond

(b)

[3,3]

(c)

–

FIGURE 20.32 (a) In a [1,5]

sigmatropic shift of hydrogen, the

migrating hydrogen atom (atom 1)

moves to atom 5.The new sigma

bond is between atom 1 and atom 5,

hence the shift is [1,5]. (b) In the

ammonium ion shown the reaction

is a [2,3] sigmatropic shift, and in

CONVENTION ALERT

20.5 Sigmatropic Shift Reactions 1051

PROBLEM 20.13 Classify the following reactions as [x,y] sigmatropic shifts.

Which, if any, of the reactions are degenerate?

Parts (c) and (d)

are not [1, x ] shifts!

(a)

(b)

(c)

(d)

(e)

Sigmatropic shifts respond to attempted mechanistic experiments in true no-

mechanism fashion.Neither acids nor bases strongly affect the reaction, and the sol-

vent polarity, or even the presence of solvent, is usually unimportant. The reaction

proceeds quite nicely in the gas phase. It is simple to construct an arrow formalism

picture of the reaction, but this device does little more than point out the overall

change produced by the migration. In these figures,the arrow formalism is even more

of a formalism than usual.For example, the arrows of Figure 20.33 could run in either

direction,clockwise or counterclockwise.That is not true for a polar reaction in which

the convention is to run the arrows from nucleophile (the Lewis base) toward the

electrophile (Lewis acid).

It is already possible to identify one curious facet of this reaction, and a little

experimentation reveals others. The product of a [1,3] shift is absent. Why, if the

hydrogen is willing to travel to the 5-position, does it never stop off at the 3-posi-

tion? An arrow formalism can easily be written for a [1,3] shift (Fig. 20.34), and it

might reasonably be argued that the [1,3] shift, requiring a shorter path than the

[1,5] shift, should be easier. Why are [1,3] shifts not observed?

[1,5]

FIGURE 20.33 Two arrow formalism

descriptions of the [1,5] shift of

hydrogen in (Z)-1,3-pentadiene.

Not observed

[1,3]

FIGURE 20.34 An arrow formalism

for the [1,3] shift of hydrogen that

does not occur.

A second strange aspect of sigmatopic shifts appears in photochemical experi-

ments. As you already know, it is possible to deliver energy to a molecule in more

than one way. One way is to heat up the starting material.However, conjugated mol-

ecules absorb light (p. 529), light quanta contain energy, and this gives us another

1052 CHAPTER 20 Reactions Controlled by Orbital Symmetry

Product of

a [1,3] shift

Product of

a [1,5] shift

[1,3] H

shift

not

hν

THE GENERAL CASE

A SPECIFIC EXAMPLE

FIGURE 20.35 Photolysis of

1,3-dienes induces [1,3] shifts.

way to transmit energy to the molecule. Curiously, when 1,3-pentadienes are irra-

diated, the products of the reaction include the molecules formed through [1,3]

shifts, but not those of [1,5] shifts (Fig. 20.35). The reactions are sometimes com-

plicated, because other thermal or photochemical reactions often take place, but if

we just concentrate on the products of shifts, this strange dependence on the method

of energy delivery appears.

WORKED PROBLEM 20.14 Perhaps it could be argued that the loss of conjugation

in the product of the [1,3] shift results in an energetic favoring of the [1,5]

process. Design a substituted 1,3-pentadiene that would test this (incorrect)

surmise.

ANSWER This problem may be hard because you are not used to thinking about

designing reactions.The hypothesis is that the [1,5] shift is preferred to the [1,3]

shift because it preserves the conjugation present in a 1,3-diene. In order to test

this surmise,we need to design a molecule in which both the [1,5] and [1,3] shifts

preserve the same amount of conjugation.The deuterium label is needed to reveal

the [1,5] or [1,3] nature of the reaction. Of course a [1,7] shift might also occur

in this reaction.

These are the same molecule

except for the isotopic label

[1,5]

[1,3]

20.5 Sigmatropic Shift Reactions 1053

In summary, thermal energy induces [1,5] shifts and photochemistry favors [1,3]

shifts. So, any mechanism must provide an explanation for these observations.

Let’s start our analysis by examining the early stages of the [1,5] sigmatropic

shift. As the reaction begins, a carbon–hydrogen bond of the methyl group begins

to stretch (Fig. 20.36a). If this stretching were continued to its limit, it would pro-

duce two radicals, a hydrogen atom and the pentadienyl radical (Fig. 20.36a). We

know this cleavage to give radical intermediates does not happen because the prod-

ucts of recombination of these radicals are not found (Fig. 20.36b).

break the

bond

Radical dimer, not formed

A pair of

radicals

stretch the

bond

(a)

(b)

FIGURE 20.36 (a) In the

earliest stages of the

reaction, a carbon–hydrogen

bond begins to stretch.

Ultimately a pair of radicals

would result. (b) We know

this bond breaking does not

take place because the

radical dimer is not found.

We should stop right here to question the assumption that the carbon–hydrogen

bond is breaking in homolytic (radical) fashion. Why not write a heterolytic cleav-

age to give a pair of ions rather than radicals (Fig. 20.37)?

The stability of ions depends greatly on solvation. It is inconceivable that a het-

erolytic bond breaking would not be highly influenced by solvent polarity. As we

have just seen, a change in solvent polarity affects the rate of this reaction only very

slightly. Only homolytic bond breaking is consistent with the lack of a solvent effect.

As the methyl carbon–hydrogen bond begins to stretch (Fig. 20.36a), some-

thing must happen that prevents the formation of the pair of radicals. As we have

already concluded in Problem 20.12, the developing hydrogen atom is intercept-

ed by reattachment at atom 5.In orbital terms, we would say that as the

bond breaks, the developing 1s orbital on hydrogen begins to overlap with the p

orbital on the carbon at position 5. Now we have a first crude model for the tran-

sition state for this sigmatropic shift (Fig. 20.38). Notice that the transition state

is cyclic.

C(1)

–

FIGURE 20.37 There are two possible

ways to break the carbon–hydrogen

bond heterolytically to give a pair of

ions. Either way depends strongly on

solvent polarity. Because there is

essentially no dependence of the rate

of this reaction on solvent polarity,

this mechanistic possibility is

excluded.

Transition

state

Hydrogen

1s orbital

FIGURE 20.38 Radical formation can be thwarted by partial bonding of the migrating

hydrogen to the orbitals on carbon atoms 1 and 5 in the transition state. As the

bond stretches, the hydrogen 1s orbital begins to overlap with an orbital on C(5).

C(1)

1054 CHAPTER 20 Reactions Controlled by Orbital Symmetry

This model leaves unanswered the serious questions of mechanism: Why does

the migrating hydrogen reattach at the 5-position in thermal reactions and at the

3-position in photochemical reactions? We have answered neither the questions of

regiochemistry ([1,5] or [1,3]) nor of dependence on the mode of energy input (ther-

mal or photochemical).

Let’s look more closely at our transition state model. In essence, it describes a

hydrogen atom moving across the π system of a pentadienyl radical (Fig. 20.39).

Pentadienyl

radical

(HOMO)

Energy

π Molecular orbitals of the pentadienyl system

(for the radical, the HOMO is Φ

)

Transition

state

Hydrogen

1s orbital

FIGURE 20.39 A model for the

transition state shows a hydrogen 1s

orbital migrating across a pentadienyl

radical, for which the π molecular

orbitals are shown.

The hydrogen atom is simple to describe—it merely consists of a proton and a sin-

gle electron in a 1s atomic orbital.The pentadienyl molecular orbital system isn’t dif-

ficult either.There are five π molecular orbitals and these can be constructed by taking

combinations of five carbon 2p orbitals. We assume again that it is the highest energy

electrons, those most loosely held, that are the most important in the reaction.These

are the electrons in the HOMO. If we accept this assumption, we can elaborate our

picture of the transition state for the reaction to give Figure 20.40. Notice that the

transition state for a [1,5] shift involves six electrons (4n  2). They come from the

sigma bond that is broken and the two double bonds in the π system.

HOMO, Φ

pentadienyl

Transition

state

FIGURE 20.40 An elaboration of our model for the transition state, showing a crude outline of the interaction of the

hydrogen 1s orbital with the lobes of the HOMO for pentadienyl,

.£

20.5 Sigmatropic Shift Reactions 1055

of Pentadienyl (HOMO)

Positive overlap

is possible at

C(5)—the orbital

phases are correct!

FIGURE 20.41 Now we can elaborate

our transition state model by filling

in the lobes of the HOMO of

pentadienyl,

.£

Notice a very important thing about the transition state outlined in Figure 20.40.

In the starting pentadiene, the bond made between the 1s orbital of hydrogen and

the hybrid orbital on carbon atom 1 must of course be made between lobes of the

same phase, which is represented by color coding (Fig. 20.41). As this bond stretches,

the phases of the lobes do not change: The 1s orbital is always fully or partially bonded

to a lobe of the same phase (Fig. 20.41).

In this way, we know the relative symmetries of some of the lobes in the transi-

tion state for the reaction. We also know the symmetry (from Fig. 20.39, or by fig-

uring it out) of Φ

, the HOMO of pentadienyl. So, we can elaborate the picture of

the transition state to include the symmetries of all the orbital lobes. The symme-

try of the lobes allows the [1,5] shift to take place as shown.

Notice that the process through which the hydrogen detaches from carbon 1 and

reattaches to carbon 5 from the same side allows the maintenance of bonding

(same sign) lobal interactions at all times.This reaction is “orbital symmetry allowed.”

Now look at the [1,3] shift (Fig. 20.42). Notice that it involves only four electrons

(4n) during the reaction: two electrons in the sigma bond that breaks and two in

the π system. To do the analogous [1,3] shift, keeping the shifting hydrogen on the

same side of the molecule requires the overlap of lobes of different symmetry—the

formation of an antibond. This reaction is “orbital symmetry forbidden”!

More important jargon enters here: The motions we have been describing in

which the migrating atom or group of atoms leaves and reattaches from the same

side of the π system are called suprafacial. If the migrating atom or group of atoms

leaves from one side of the π system and arrives at the other, the process is called

antarafacial (Fig. 20.43). In the [1,5] hydrogen shift, the typical sigmatropic shift

involving 4n  2 electrons, suprafacial motion is allowed by the symmetries of the

orbitals; in a [1,3] shift of a hydrogen, a typical 4n electron process,it is not (Figs.20.41

and 20.42).

Overlap is

antibonding

here

Overlap is

bonding here

FIGURE 20.42 By contrast, the [1,3]

shift, which is a four-electron

process, involves the formation of an

antibond if the hydrogen reattaches

to carbon 3 from the same side.

Bottom to bottom

Top to top

Bottom to top

Top to bottom

Suprafacial Antarafacial

FIGURE 20.43 Suprafacial motion

involves bond breaking and bond

making on the same side of the π

system. Antarafacial motion involves

bond breaking on one side and bond

making on the other.

1056 CHAPTER 20 Reactions Controlled by Orbital Symmetry

Why not migrate in a different fashion? Why can’t the hydrogen depart from

one side of the π system and reattach at the other? This antarafacial migration

would result in an orbital symmetry–allowed [1,3] migration (Fig. 20.44). The

problem is that the 1s orbital is small and cannot effectively span the distance

required for an antarafacial migration.There is nothing electronically disallowed

or forbidden about the antarafacial [1,3] migration, but there is an insuperable

steric problem.

The [1,5] sigmatropic shift is intramolecular and the maintenance of bonding

interactions explains why [1,5] hydrogen shifts are possible and [1,3] hydrogen shifts

are exceedingly rare. But we have not yet been able to deal with the dependence of

the course of this reaction on the mode of energy input—thermal or photochemical.

PROBLEM 20.15 Do you expect the [1,3] shift shown below to occur when

propene is heated? Explain carefully, using a molecular orbital argument.

of Pentadienyl

(HOMO)

Bonding

FIGURE 20.44 The antarafacial [1,3]

shift is allowed by orbital symmetry,

but the small size of the 1s orbital

makes the stretch impossible.

[1,3]

Now comes a most important point. Our model not only rationalizes known

experimental data, but includes the roots of a critical prediction. The Woodward–

Hoffmann explanation predicts that the [1,5] shift occur in a suprafacial fashion.

That is the only way in which bonding interactions can be maintained throughout

the migration of a hydrogen atom from carbon 1 to carbon 5. There is nothing so

far in the available experimental data that allows us to tell whether the observed shifts

are suprafacial, antarafacial, or a mixture of both modes.

Let’s set about finding an experiment to test the requirement of the theory for

suprafacial [1,5] motion. What will we know at the end of the experiment? If we

find that the [1,5] shift is indeed strictly suprafacial, the theory will be supported

(not proved!), and we will certainly feel better about the mechanistic hypothesis. If

we find antarafacial motion, or both suprafacial and antarafacial motions, the the-

ory will be proved (yes, proved) wrong.There is no way our hypothesis can accom-

modate antarafacial motion; there is an absolute demand for suprafaciality,

which nicely illustrates the precarious life of a theory. It can always be dis-

proved by the next experiment, and it can never become free of this state

of affairs.

The crucial test was designed by a German chemist, W. R Roth

(1930–1997).

It is the first of a series of beautiful experiments you will

encounter in this chapter. Roth and his co-workers spent some years in

developing a synthesis of the labeled molecule shown in Figure 20.45.

Notice that the molecule has the (S) configuration at carbon 1,and the (E)

stereochemistry at the double bond.

Now let’s work out the possible products from suprafacial and antarafacial

[1,5] shifts in this molecule. The situation is a little complicated because there

C(4)

C(5)

Roth’s diene

(

)

(

)

WEB 3D

FIGURE 20.45 The 1,3-diene used

by Roth and his co-workers to see

whether the [1,5] shift proceeded

in a strictly suprafacial manner.

No typo here. There is no period after the R. Roth claimed that he was advised by his postdoctoral advisor

at Yale, William Doering, to take a middle initial in order not to be lost among the myriad “W. Roth’s” in

German chemical indexes. Others with common last names, and who were at Yale at the time, rejected sim-

ilar advice.

20.5 Sigmatropic Shift Reactions 1057

suprafacial

[1,5]

(

)

suprafacial

[1,5]

(

)

(

)

(

)

(

)

(

)

(

)

(

)

FIGURE 20.46 The predicted products formed by suprafacial migration in the two

rotational isomers of Roth’s diene.

are two conformational isomers of this molecule that can undergo the reaction,

shown in Figure 20.46. In one, the methyl group is aimed at the

double bond. In the other, it is the ethyl group that is directed toward the

double bond. In suprafacial migration of hydrogen in one confor-

mation, the product has the (R) configuration at C(5) and the (E) stereochem-

istry at the new double bond located between C(1) and C(2). Another possible

suprafacial migration starts from the second conformation. Here it is the oppo-

site stereochemistries that result. The configuration of C(5) is (S) and the new

double bond is (Z).

C(4)

C(5)

C(4)

C(5)

In antarafacial migrations, the hydrogen leaves from one side and is delivered

from the other. Figure 20.47 shows the results of the two possible antarafacial migra-

tions. Note that the products are different from each other and also different from

those predicted from suprafacial migration.

antarafacial

[1,5]

antarafacial

[1,5]

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

FIGURE 20.47 The possible products of antarafacial migration in Roth’s diene.

Roth and his co-workers had the experimental skill necessary to sort out the

products and found that only the products of suprafacial migration were formed.

The theory is beautifully confirmed by this experiment.

At last we come to the remaining question. Why should the source of ener-

gy make a difference in the mode of migration? Remember, the common, ther-

mally induced shifts are [1,5], whereas the photochemical shifts are [1,3]. A

molecule absorbs a photon of light and an electron is promoted from the

HOMO to the LUMO, producing a new HOMO and changing the symmetry

of the lobes at the ends of the π system. What happens when the symmetry of

the HOMO changes?

1058 CHAPTER 20 Reactions Controlled by Orbital Symmetry

We can now put together Table 20.3 that compares the types of sigmatropic shifts

that are possible and links the stereochemical outcome to the number of electrons

involved in the process and the type of energy used to drive the reaction.

Photochemical

HOMO

Antarafacial motion

is required for

thermal reaction

Suprafacial motion

is required for

photochemical reaction

Energy

hν

FIGURE 20.48 Suprafacial motion of a hydrogen is possible in

, the photochemical HOMO for this reaction.£

Figure 20.48 shows the situation for the simplest [1,3] shift in propene.

Absorption of a photon by propene promotes an electron,shown below in

of allyl.

Migration can now occur in a suprafacial fashion, and the steric barrier to thermal,

necessarily antarafacial [1,3] shifts is gone. The hydrogen can migrate in an easy

suprafacial manner.

Summary

We now understand three seemingly magical concerted reactions. Each is con-

trolled by orbital symmetry. Electrocyclic reactions occur in a conrotatory or

disrotatory fashion depending on the number of electrons in the system and

whether heat or light is used to drive them.The cycloaddition reactions occur

with orbital overlap between the reacting partners and the symmetry of the

HOMO and the LUMO determines the outcome of the reaction. Sigmatropic

shifts are perhaps the most challenging to understand. As long as you keep

in mind the necessity to form bonds and not antibonds, analysis of the

HOMO gives you the positions available for reattachment of a migrating

hydrogen or group. One can determine whether the reaction is allowed in a

suprafacial or antarafacial fashion. Add to this analysis the idea that hydro-

gen has only a small 1s orbital, and therefore any short antarafacial shifts are

generally impossible.

TABLE 20.3 Rules for Allowed Sigmatropic Reactions

Reaction Number of Electrons Thermal Photochemical

Typical 4n reactions 4n Antarafacial Suprafacial

Typical 4n  2 reactions 4n  2 Suprafacial Antarafacial

(continued)

PROBLEM 20.16 Provide mechanisms for the reactions shown on the next page.

Hint: Reactions (a) and (b) each require more than one sigmatropic shift.