Jones M., Fleming S.A. Organic Chemistry

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9.9 Rearrangements during HX Addition to Alkenes 389

AgNO

Neopentyl alcohol 2-Methyl-2-butanol

(97%)

Neopentyl iodide

FIGURE 9.48 The solvolysis of neopentyl iodide does not take a “normal” course.

AgI

path a

path b

AgNO

Neopentyl iodide

Primary carbocation

Tertiary carbocation

Direct formation of the tertiary

carbocation; the methyl group

migrates as iodide leaves

2-Methyl-2-butanol

AgI

FIGURE 9.49 Rearrangements are common in many reactions in which carbocations are involved. Path a shows the

stepwise process and path b is the concerted process.

Such rearrangements are so common that they have come to be diagnostic for reac-

tions involving carbocations. Molecules prone to such rearrangements are used to test

for carbocation involvement. Here is an example: Attempted S

1 solvolysis of neopentyl

iodide (1-iodo-2,2-dimethylpropane) in water leads not to neopentyl alcohol, as might

be anticipated, but instead to a rearranged alcohol, 2-methyl-2-butanol (Fig. 9.48).

Carbocation rearrangement E1

These results imply the scenario outlined in path a of Figure 9.49, in which an

unstable primary carbocation rearranges to a much more stable tertiary carbocation

through the shift of a methyl group. In fact, the presence of the primary ion is

so unsettling—they are most unstable and not likely to be formed—that a variant

of this mechanism has been proposed in which the methyl group migrates in a con-

certed fashion as the iodide departs (path b, Fig. 9.49). A concerted reaction is one

that has no intermediates.

At very low temperature in highly polar but nonnucleophilic solvents, some

carbocations can be observed spectroscopically. Under such conditions it is possi-

ble to determine that many carbocations are rapidly rearranging as we observe

them. For example, the two tertiary carbocations in Figure 9.50 rapidly intercon-

vert at 60 °C.

FSO

– 60 ⬚C

FIGURE 9.50 Rearranging tertiary

carbocations.

390 CHAPTER 9 Additions to Alkenes 1

9.10 Hydroboration

Now we pass on to another addition reaction, hydroboration, which is the addition

of hydrogen and boron across a π bond. The mechanism is not simple, and it will

take some effort for you to become comfortable with all the details. The mechanis-

tic discussion does make several worthwhile points,so it merits your attention for that

reason alone. But there is another, more compelling reason to master the hydrobora-

tion reaction—it is one of the most useful of synthetic reactions.Indeed H. C. Brown

(1912–2004) won a Nobel Prize in 1979 for the development of hydroboration.

Consider the following seemingly simple synthetic task outlined in Problem 9.13.

Borane

Boron trifluoride

FIGURE 9.51 Both borane and boron

trifluoride are Lewis acids and will

react like carbocations.

WORKED PROBLEM 9.13 Try to provide a synthesis of 2-methyl-1-propanol. You

may start from any alkene.

ANSWER Well, there isn’t one (yet). At this point in your study of organic chemistry,

you have no way to make this simple alcohol. If we start from the obvious alkene,

2-methylpropene, the hydration reaction we learned in this chapter (p. 380) can

only give the product of Markovnikov addition, tert-butyl alcohol.This nonanswer

to such a simple question points out how limited our synthetic skills are so far.

2-Methyl-1-propanol

WEB 3D

Alkene hydroboration

The hydroboration reaction solves both Problem 9.13 and the general diffi-

culty, so let’s see how it works.

In the previous sections, we saw both the protonation of alkenes and the addi-

tion of a carbocation to an alkene to generate a new carbocation.The familiar theme

of overlap between filled (alkene π) and empty (carbon 2p) orbitals was recapitulat-

ed: “Lewis bases react with Lewis acids.”Other reagents containing empty p orbitals,

all good Lewis acids, might also be expected to add to alkenes, and so they do.

We know another kind of molecule that has an empty 2p orbital, the trigonal

boranes, BF

and BH

(Fig. 9.51).There is no positive charge on these compounds,

but they should still be strong Lewis acids (electrophiles).

PROBLEM 9.14 Verify that boron in BF

and BH

is neutral and that each boron

atom has an empty 2p orbital.

Although BF

is known and even commercially available, free BH

is unavailable

because it spontaneously dimerizes to diborane (B

),an unpleasant-smelling,flam-

mable, and toxic gas. Diborane ( ) does not have a structure similarH

9.10 Hydroboration 391

WORKED PROBLEM 9.15 Propose a structure for the ether–borane complex, and

suggest a mechanism for its formation.

ANSWER Borane is a Lewis acid because the boron atom has an empty 2p orbital.

An ether, here tetrahydrofuran (THF), is a Lewis base.The reaction between the

Lewis base and Lewis acid produces the borane–THF complex.

Lewis acid

(electrophile)

Lewis base

(nucleophile)

Borane–THF complex

–

This structure cannot be B

because

there are only 12 electrons available

for bonding in this molecule, and this

formulation requires 14

AA'

FIGURE 9.52 The real structure of

diborane.

Borane–THF

complex

–

FIGURE 9.53 A schematic

hydroboration. The elements of

hydrogen and boron have been added

across the double bond.

to ethane, . An ethane-like structure requires 14 electrons to make

7 boron–hydrogen and boron–boron σ bonds. The 6 hydrogens in diborane con-

tribute 6 electrons and the 2 borons another 6 to give a total of 12, so we are 2 elec-

trons shy of the required number. The real structure is shown in Figure 9.52, along

with a resonance formulation. The molecule is a hybrid of the two resonance forms

A and A′. In this molecule, partial bonds (shown as dashed bonds in Fig. 9.52) are

used to connect the atoms.This kind of bonding is more common than once thought.

It abounds in inorganic chemistry, and in the organic chemistry of electron-

deficient species, such as carbocations.

When diborane dissolves in diethyl ether ( ) a

borane–ether complex is formed.This complex can be used as a source of BH

.Often

the cyclic ether tetrahydrofuran is used. Some borane–ether complexes are so stable

that they can even be distilled.

When the borane–ether complex is allowed to react with an alkene, there is a

rapid addition of the borane across the double bond. The initial product is an

alkylborane (Fig. 9.53). As in the addition of a carbocation to a double bond, the

initial interaction is between the empty 2p orbital (in this case on neutral boron,

392 CHAPTER 9 Additions to Alkenes 1

not positive carbon) and the filled π orbital of the alkene (Fig. 9.54a). If the mech-

anism were directly parallel to that of cation addition, we would form the charge-

separated molecule shown in Figure 9.54a.A hydride (H



) could then be delivered

from the negative boron to the positively charged carbon to complete the addition

(Fig. 9.54b). However, the mechanism is not exactly parallel to what we saw before.

In this case, the hydride is transferred as the carbon–boron bond is made (Fig. 9.55).

This reaction occurs in a single step.There is no intermediate.In the transition state,

partial bonds from carbon to boron and carbon to hydrogen are formed. Let’s see

how we know this mechanism satisfies the experimental data and, as we do this,

why the more conventional mechanism is inadequate.

(a)

(b)

C H

Nucleophile—

π bond

Electrophile (empty 2p orbital)

–

FIGURE 9.54 (a) A possible first step in

addition of BH

to an alkene.The π

bond of the alkene acts as nucleophile

and reacts with the strong Lewis acid,

. (b) A possible second step in the

hydroboration reaction is transfer

of a hydride (H



Transition state with

partial bonds (dashed)

FIGURE 9.55 A one-step or concerted mechanism for

hydroboration.

First of all,there are no rearrangements during hydroboration.The concerted mech-

anism outlined in Figure 9.55 cannot lead to rearrangements, but the intermediate car-

bocation of Figure 9.54 would surely do so. For example, the hydroboration of

3,3-dimethyl-1-butene leads exclusively to borane A, shown in path a of Figure 9.56.

Were a carbocation intermediate involved, rearrangement of the secondary carbocation

to the more stable tertiary carbocation would seem inevitable, and product B should be

observed (path b of Fig.9.56).But the reaction yields a regiospecific hydroboration.Only

one of two possible product boranes is formed.This phenomenon is general—in hydro-

boration the boron becomes attached to the less substituted carbon of the double bond.

Our one-step mechanism must be able to explain this regiochemical preference.

(CH

)

B not observed

Transition state with

partial bonds (dashed)

Tertiary carbocation

(more stable)

Secondary carbocation

(less stable)

rearrangement hydride shift

C CH

(CH

)

C CH

–

CCH

path b

path a

FIGURE 9.56 An intermediate carbocation of path b should lead to rearrangements in hydroboration, but

no evidence of such can be found, which suggests that the concerted path a is the correct mechanism.

9.10 Hydroboration 393

There are at least two factors, one simple, the other more complicated, that

explain the regiospecific addition in the hydroboration reaction. The simple analy-

sis is a steric argument. Consider the reaction of BH

with 2-methyl-1-butene.There

are steric differences in the two possible concerted additions. In the addition lead-

ing to the observed product (Fig. 9.57a), the larger end of the borane molecule, the

portion, is nearer to the smaller hydrogens than to the larger alkyl groups. In

the path leading to the product that is not observed (Fig. 9.57b), the larger end

opposes the relatively large methyl and ethyl groups.

(a)

(b)

Transition state

The sterically less

congested arrangement

leads to the observed

product

This sterically more

congested arrangement

leads to the product that

is not observed

CCH

Transition state

Observed product

Not observed

FIGURE 9.57 (a) Steric factors favor

addition of the larger BH

group to

the less congested end of the alkene.

(b) Steric interactions could hinder

formation of the nonobserved

product.

–

Tertiary δ

(more stable)

Observed product

This product is not formed

–

Primary δ

(less stable)

FIGURE 9.58 In the two possible

transition states for the concerted

addition, there will be partial positive

charge developed on carbon. This δ



will be more stable at the more

substituted position.

The more complicated factor has to do with the way in which the two new

bonds are made in the transition state for the hydroboration reaction. Two bonds

are formed in the same step, without a cationic intermediate, but there is no need

for the new bonds to develop to the same extent as the reaction proceeds. Indeed,

in a philosophical sense, they cannot, because they are different. In the transition

state, one bond will be further along in its formation than the other. This differ-

ential progress can be symbolized by bonds of different lengths, as in Figure 9.58.

394 CHAPTER 9 Additions to Alkenes 1

In such a mechanism, a full positive charge is not developed on carbon, but a partial

one is. All the factors that operate to stabilize a full positive charge also operate to sta-

bilize a partial positive charge. A more-substituted partial positive charge is more sta-

ble than a less-substituted partial positive charge,and this difference favors the observed

product. So, the initial product of hydroboration of an alkene is the monoalkylborane

formed by addition in which boron becomes attached to the less substituted end of

the double bond (Fig. 9.58). Does this result violate Markovnikov’s rule? No!

Here is an instance where rules can be confusing. When we say “Markovnikov

addition,” we are used to seeing the hydrogen attached to the less substituted car-

bon of the double bond and the X group to the other carbon (Fig. 9.59).

–

FIGURE 9.59 Typical Markovnikov

addition in which the more stable

cation is formed as an intermediate to

give the product in which H is

attached to the less substituted

carbon of the original double bond,

and X is attached to the more

substituted end.

When adds across a π bond, it is the boron that is the positive end of

the dipole, the electrophile in the addition. It is the boron, not the hydrogen, that

becomes attached to the less substituted end of the double bond. As Figure 9.60 tries

to show,it is the terminology that gets complicated,not the reaction mechanisms.There

is danger in learning rules; it is much safer to understand the reaction mechanism.

Transition state

CCH

–

FIGURE 9.60 This reaction is still

Markovnikov addition.The

electrophilic boron in this case adds

to the less substituted end of the

alkene. Positive charge develops on

the more substituted carbon.

There is another reason we know that the concerted mechanism for hydrobora-

tion is correct, and it involves the stereochemistry of the addition reaction.

Hydroboration can be shown to proceed in syn fashion. A syn addition describes

the result when two pieces (H and BH

in this case) are delivered to the same side of an

alkene. Figure 9.61 shows the hydroboration of cis-1,2-dideuterio-1-hexene and of

1-methylcyclopentene.These reactions exclusively produce the products of syn addition.

cis -1,2-Dideuterio-1-hexene

1-Methylcyclopentene

HBR

FIGURE 9.61 These two

stereochemical experiments show

that addition of BH

proceeds in a

syn fashion.The H and BH

groups

are delivered to the same side of the

alkene (R cyclohexyl).=

9.10 Hydroboration 395

Summary

The current mechanistic hypothesis for hydroboration, a concerted, one-step syn

addition to alkenes in which the two new bonds are partially formed to different

extents in the transition state, nicely rationalizes all the experimental data. A

mechanism involving formation of an open cation followed by hydride transfer

fails to account for the observed syn stereochemistry of addition, which is

demanded by the one-step mechanism.

Transition state

FIGURE 9.62 Syn addition is accommodated by a mechanism in which both new bonds are

forming in the transition state.There are no intermediates and therefore no chance for

rotation and scrambling of stereochemistry.

transfer

rapid

rotation

–

transfer

–

HBR

–

FIGURE 9.63 Rotation about carbon–carbon single bonds is very fast. A two-step

mechanism predicts that two products should be formed from the hydroboration of cis-1,2-

dideuterio-1-hexene. Both syn and anti addition products should be formed. However, only

the product of syn addition is found, indicating that there is no carbocation intermediate.

Of course,the one-step addition process must involve syn addition because both

new bonds, the one to boron and the one to hydrogen, are made at the same or nearly

the same time. Figure 9.62 shows the observed syn addition to deuterated 1-hexene.

A mechanism involving an intermediate carbocation has great problems account-

ing for syn addition. If a planar cation were formed on addition to the deuterat-

ed 1-hexene, we would expect the rapid ( 10

1

) rotation about carbon–

carbon single bond to produce the two diastereomeric products shown in

Figure 9.63.

396 CHAPTER 9 Additions to Alkenes 1

WORKED PROBLEM 9.16 Draw Energy versus Reaction progress diagrams for the

concerted and two-step mechanisms for the addition of BH

to alkenes.

ANSWER The crucial difference is the presence of an intermediate in the two-step

process.The one-step (concerted) reaction simply passes from starting material to

product over a single transition state.

Reaction progress

Energy

One-step

(concerted)

Two-step

Reaction progress

Intermediate

–

Transition state

–

This example is typical of the use of stereochemistry in determining reaction mech-

anism. One reason we spent so much time on the stereochemical relationships of mol-

ecules of various shapes (diastereomers, enantiomers, and meso forms) was that we will

use these relationships over and over again as we dig into questions of reaction mech-

anism. If it is not obvious what we are talking about in the last few paragraphs, and if

Figures 9.61–9.63 are at all unclear, be sure to go back to Chapter 4 and work over the

stereochemical arguments again.Otherwise,you run the risk of being lost in further dis-

cussion.The key points to see are the difference between the two hypothetical products

shown in Figure 9.63 and their required formation from the rotating cation.A good test

is to see if you can do Problems 9.17 and 9.18 easily. If you can, you are in good shape.

PROBLEM 9.17 Bromine adds to alkenes to give dibromo compounds. Based on

the products from the reaction of bromine with the 2-butenes, determine if the

addition is syn or anti. This problem does not pose a mechanistic question—you

do not have to draw arrow formalisms. You only need to determine from the

structure of the products the direction from which the two bromines have added.

+ Br

H H

+ Br

9.10 Hydroboration 397

PROBLEM 9.18 Devise a way to determine the stereochemistry of the addition of hydro-

gen bromide to alkenes. Assume you have access to any necessary starting materials.

Still a Lewis acid!

repeat the

hydroboration

A dialkylborane

CCH

repeat the

hydroboration

RBH

A trialkylborane

FIGURE 9.64 When BH

is used, the

initially formed alkylboranes are

Lewis acids and can undergo further

hydroborations.

A single addition isn’t the end of the hydroboration story.The initially produced

monoalkylborane still has a trigonal boron; therefore, it has an empty 2p orbital

(Fig. 9.64). It’s still a Lewis acid, and it can participate in a second hydroboration

reaction to give the product of two hydroborations of the alkene, a dialkylborane.

Nor must the reaction stop here. The dialkylborane is also a Lewis acid, and it can

do one more hydroboration to give a trialkylborane. Now, however, all the original

hydrogens that were on the boron are used up and there can be no further hydro-

borations.In practice,the number of hydroborations depends on the size of the alkyl

groups in the alkene. When the groups are rather large, further reaction is retarded

by steric effects—the alkyl groups just get in the way. For example, 2-methyl-

2-butene hydroborates only twice, and 2,3-dimethyl-2-butene only once (Fig. 9.65).

2-Methyl-2-butene

2,3-Dimethyl-2-butene

2-Methyl-2-butene

repeat

A monoalk

lborane

A dialkylborane

FIGURE 9.65 The number of

hydroborations depends on steric

effects.

398 CHAPTER 9 Additions to Alkenes 1

WORKED PROBLEM 9.19 Provide a mechanism for the following reaction:

ANSWER The first hydroboration goes in normal fashion.

Now a second,intramolecular hydroboration takes place as the remaining dou-

ble bond in the ring attacks the boron.

It takes some unraveling to see the structure of the final product. It is

important to begin to hone your skills at this kind of thing, and this problem

provides practice.

Alcohol Boric acid

/ HO

B(OH)

–

3 R

FIGURE 9.66 Formation of an alcohol

from a borane.

9.11 Hydroboration in Synthesis:

Alcohol Formation

It would be presumptuous to attempt to summarize fully the utility of the hydro-

boration reaction here.There are many variations of this reaction, each designed

to accomplish a specific synthetic transformation. A book could be written on

the subject. In fact, Professor H. C. Brown has written just such a book. We’ll

only mention one especially important reaction here, one that leads to a new

process for the synthesis of alcohols (recall Problem 9.13, p. 390). When one

molecule of trialkylborane is treated with hydrogen peroxide (H

) and hydrox-

ide ion (HO



), three molecules of an alcohol are formed along with boric acid

(Fig. 9.66).