Jones M., Fleming S.A. Organic Chemistry

10.2 Addition of H

2

and X

2

Reagents 419

Br

..

–

Br

..

+

H

Br

2

S

N

2

trans Br and OH

g

rou

p

s

H

2

O

..

H

H H H

OH

2

..

+

Br

..

O

+

..

H

Br

..

OH

H

Br

..

WEB 3D

FIGURE 10.16 The bromonium ion mechanism demands that the products be formed by anti addition, as they are.

This stereochemical experiment uses cyclohexene to show that the bromine and solvent molecule become attached

from different sides of the ring through the anti addition enforced by the presence of the bromonium ion.

Addition of X

2

in nucleophilic solvents presents us with another opportunity.

We can measure the regioselectivity of the reaction. For example, in the bromina-

tion of 2,3,3-trimethyl-1-butene in carbon tetrachloride, there is no way to deter-

mine the regiochemistry of the bromide attack of the bromonium ion. Bromination

in water allows us to see where the hydroxyl group goes as it opens the bromonium

ion. There are two possibilities (Fig. 10.17). The hydroxyl group can attach to the

more substituted or to the less substituted carbon of the original double bond.

AB

2,3,3-Trimethyl-1-butene

C

CH

3

C

H

3

C

CH

3

CH

3

CH

3

Br

2

CH

2

CH

3

Br

C

CH

2

H

3

C

CCl

4

C

CH

3

H

3

C

Br

2

Br

OH

C

CH

2

H

3

C

CH

3

H

3

C

OH

Br

or

C

CH

2

H

3

C

H

2

O

H

3

C

H

3

C

H

3

C

FIGURE 10.17 A test of the

regiochemistry of the addition of

Br

2

/H

2

O to alkenes. Halohydrin A

would result from water attacking

the more substituted carbon of the

bromonium ion and halohydrin B

would result from attack at the less

substituted carbon.

2. deprotonate

1.

C

CH

3

H

3

C

CH

3

Br

2

C

CH

2

H

3

C

CH

3

H

3

C

CH

3

OH

C

CH

2

H

3

C

H

2

O

Br

..

–

+

C

CH

2

CH

3

H

3

C

H

3

C

H

3

C

Br

..

Br

..

FIGURE 10.18 The bromine becomes

attached to the less substituted

carbon and the nucleophilic solvent

to the more substituted carbon of the

alkene. It is as if the reaction went

through a carbocationic intermediate.

In practice, the hydroxyl group becomes attached to the more substituted carbon. It

is as if the reaction were proceeding through a carbocation (Fig. 10.18).But we are quite

certain that it does not! The problem now is to reconcile the observed regioselectivity

with the intermediacy of a bromonium ion (recall trans addition and the lack of carbo-

cation rearrangements).

hydroxyl groups have a trans relationship. Such compounds are called halohydrins.

The bromonium ion mechanism fits all the data (Fig. 10.16).

The bromonium ion is symmetrical only when formed from an alkene in which

the two carbons of the double bond are the same. In all other cases, the two

carbon–bromine bonds must be of different strength. In resonance terms,we say that

the weighting factor (coefficient) for the more stable form is greater than that for the

420 CHAPTER 10 Additions to Alkenes 2 and Additions to Alkynes

This bond is longer and weaker…

Summary

structure

Tertiary

carbocation

(more important)

Primary

carbocation

(less important)

Br

..

Br

..

CH

2

CH

2

Br

+

CH

2

c

1

c

2

c

1

>>c

2

CH

2

…than this bond

..

Br

+

Br

+

..

Br

..

+

C

CH

2

CH

3

H

3

C

H

3

C

H

3

C

CH

3

H

3

C

H

3

C

H

3

C

CH

3

H

3

C

H

3

C

H

3

C

CH

3

H

3

C

H

3

C

H

3

C

CH

3

H

3

C

H

3

C

H

3

C

WEB 3D

FIGURE 10.19 A resonance

description of the bromonium ion.

HH

C

CH

3

CH

3

H

3

C

CH

2

H

3

C

CH

3

CH

3

H

3

C

HO

C

CH

2

H

3

C

Not much δ

+

on this primary

carbon

addition of water

breaks weaker

carbon–bromine bond

deprotonationaddition of

(60%)

(24%)

C

CH

3

CH

3

H

3

C

CH

2

H

3

C

Br

..

Br

..

–

..

Br

..

Br

..

Br

O

+

..

CH

2

Most δ

+

is on this

tertiary carbon;

the bond between

it and Br is longer

and weaker

than the other

C

Br bond

..

Br

+

H

2

O

..

H

2

O

..

Nu = H

2

O

..

(Nu = Br

–

)

..

C

CH

2

CH

3

H

3

C

H

3

C

H

3

C

CH

3

H

3

C

H

3

C

H

3

C

FIGURE 10.20 Addition of

nucleophiles breaks the longer,

weaker bond.C

O

Br

less stable form (c

1

c

2

, Fig. 10.19).The two carbons share the positive charge, but

not equally.The charge is more stable on the more substituted carbon.Therefore,the

more substituted carbon–bromine bond is longer and weaker than the less substituted

bond in this positively charged intermediate. For this reason, the transition state for

7

breaking the bond to the more substituted carbon will be lower in energy than that

for breaking the bond to the less substituted carbon. Therefore nucleophiles such as

methyl alcohol, water, and halide add at the more substituted position (Fig. 10.20).

Summary

The critical difference between addition of halogen (X

2

) and the many HX addi-

tions we saw in Chapter 9 is the presence of the intermediate halonium (bromo-

nium or chloronium) ion. This intermediate is demanded by the observed

preference for anti addition. However, occasionally the cyclic halonium ion is less

stable than the corresponding open ion and the open ion will be favored. One

way to stabilize the open ion is through resonance. If the open ion is the inter-

mediate, the products of both cis and trans addition will be observed. When a

protic nucleophilic solvent is used in the reaction,the solvent as well as the halide

will add to the intermediate. In such cases halohydrins or halogenated ethers are

formed along with dihalides.

10.3 Hydration through Mercury Compounds: Oxymercuration 421

10.3 Hydration through Mercury Compounds:

Oxymercuration

For small-scale hydrations, it is convenient to use mercury salts, such as mercuric

acetate, Hg(OAc)

2

, to carry out the addition of across a double bond.

This process is called oxymercuration. In this two-step process, an alkene is first

treated with mercuric acetate, then the initial alkylmercury compound formed is

reduced with sodium borohydride (Na



BH

4

).

H

O

OH

1. Hg(OAc)

2

/H

2

O

2. Na

+

–

BH

4

H

C

OH

C

The abbreviation AcO



(or



OAc) represents the acetate ion (CH

3

COO



), the

conjugate base of acetic acid, CH

3

COOH (often abbreviated AcOH). Lewis struc-

tures are given in Figure 10.21, which shows the relationship between the acetate

ion and its conjugate acid (AcOH). Note the resonance stabilization of the acetate

ion—the two oxygen atoms are equivalent.

Acetic acid, the

conjugate acid of…

AcOH

…the acetate ion

H

3

C

CH

O

..

H

3

C

O

C

O

+

O

..

O

..

–

H

3

C

O

H

3

O

..

+

H

2

O

..

AcO

–

FIGURE 10.21 Note the resonance

stabilization of the acetate ion.

CONVENTION ALERT

PROBLEM 10.8 The pK

a

of acetic acid is 4.8. The corresponding two-carbon

alcohol, ethyl alcohol (CH

3

CH

2

OH), is a much weaker acid, as its pK

a

of

15.9 demonstrates. Explain why acetic acid is much more acidic than ethyl

alcohol.

422 CHAPTER 10 Additions to Alkenes 2 and Additions to Alkynes

The first step in the oxymercuration reaction is the formation of a cyclic mer-

curinium ion (Fig. 10.22). Open carbocations are unlikely intermediates, as they

would surely undergo carbocationic rearrangements, and these are not seen even in

systems normally prone to rearrangement.

CH

2

CH

2

H

Ring opening by breaking the

weaker carbon–mercury bond

A mercurinium ion

deprotonation

addition

Initial isolable

product

C

CH

2

OH

..

C

H

R

+

HgOAc

Hg(OAc)

2

=C

CH

2

HgOAc

C

CH

2

HgOAc

+

C

H

C

CH

2

..

O

+

HgOAc

H

3

O

..

+

OH

2

..

OH

2

..

H

2

O

..

H

R

H

R

H

R

RR

+

FIGURE 10.22 Oxymercuration

begins by attack of the alkene on the

Lewis acid mercuric acetate to form a

cyclic mercurinium ion. The

nucleophilic solvent, water, then adds

in S

N

2 fashion to give the first

product, a mercury-containing

alcohol, where R is a simple alkyl

group.

CH

2

+

H

2

O

HgOAc

Hg

OAc

C

Secondary

carbocation

3-Methyl-1-butene

Rearranged alcohol

not formed!

H

2

O

+–

BH

4

Na

C

CH

2

CH

2

H

3

C

CH

3

H

Tertiary

carbocation

CH

3

HO

H

CH

2

C

CH

2

H

3

C

H

3

C

H

3

C

H

3

C

H

3

C

CH

3

HO

CH

2

CH

3

H

3

C

CC

C

+

With unsymmetrically substituted alkenes, the two bonds to mercury will not

be equally strong. (Recall our discussion of asymmetrical bromonium ions just a

few pages ago.) The bond between mercury and the more substituted carbon will be

longer and weaker than the bond to the less substituted carbon of the mercurinium

ion. It is at the more substituted carbon that water opens the ring. So, the first

product of this reaction is the mercury-containing alcohol shown in Figure 10.22.

WORKED PROBLEM 10.9 Design an experiment to test for carbocationic

rearrangements in the oxymercuration reaction.

ANSWER It’s simple. Just use any alkene in which migration of hydride (H



) or

migration of R



would give a more thermodynamically stable cation. 3-Methyl-

1-butene would work well. If addition gave an open carbocation rather than the

cyclic mercurinium ion, rearrangements would surely occur.That they are not observed

provides strong evidence that open cations are not involved in oxymercuration.

:

10.4 Other Addition Reactions Involving Three-Membered Rings 423

H

R

C

CH

2

OH

HgOAc

H

C

CH

2

OH

H

C

CH

2

OH

D

H

Na

+

–

BH

4

–

BD

4

Na

+

FIGURE 10.23 Reduction of the

mercury-containing products gives

alcohols.

Oxymercuration is synthetically useful because the reagent sodium borohydride

(Na



BH

4

) efficiently replaces the mercury with hydrogen. Figure 10.23 shows

the reaction with sodium borohydride and sodium borodeuteride (Na



BD

4

).

The use of the deuterated molecule allows us to see with certainty the position of

the entering deuterium.

So we now have a new hydration reaction, one that proceeds through a cyclic

mercurinium ion but ultimately gives the product of Markovnikov addition,the more

substituted alcohol.

Summary

Here are our three methods of hydration:

1. Direct hydration (p. 380) proceeds through an intermediate carbocation that is

captured by water to give the product of Markovnikov addition. The reaction is

limited in utility because rearrangements of the initially formed cation to more

stable species can lead to undesired, rearranged products.

2. Hydroboration/oxidation (p. 390) first generates an alkylborane that subse-

quently reacts with peroxide in base to give the product of anti-Markovnikov

addition.

3. Oxymercuration proceeds through a cyclic mercury-containing ion and also gives

the product of Markovnikov addition. No rearrangements take place, which is

sometimes a great advantage.

PROBLEM 10.10 Work out the consequences of applying the three procedures out-

lined above to 3-methyl-1-butene.

10.4 Other Addition Reactions Involving

Three-Membered Rings: Oxiranes and

Cyclopropanes

10.4a Oxiranes We have seen that many addition reactions begin with for-

mation of a three-membered ring. These intermediates have almost always been

rapidly converted into the final products through addition of a nucleophile. This

state of affairs will now change dramatically, as you learn about the addition reac-

tions that are the best routes to two types of stable three-membered rings: the

oxiranes (epoxides, p. 317) and cyclopropanes.

424 CHAPTER 10 Additions to Alkenes 2 and Additions to Alkynes

Hydrogen oxide

(water)

A carboxylic acid

A peracid

Hydrogen peroxide

CH

HH

R

O

..

O

..

O

..

O

..

O

..

O

..

O

..

O

..

HH

O

..

O

..

CH

O

..

O

..

H

Trifluoroacetic acid

Trifluoroperacetic acid

CH

O

..

C

F

3

C

F

3

C

WEB 3D

FIGURE 10.24 The prefix per- means

extra. So a “peroxide” and a “peracid”

each contain an extra oxygen atom,

one more than the oxide or acid.

An organic acid has the formula and is called a carboxylic acid.

Its structure is shown in Figure 10.24. We will study this functional group in detail

in Chapter 18. A peracid has the formula . Figure 10.24 shows the

difference in more detail.The “per”indicates an “extra”oxygen is present.Thus hydro-

gen peroxide is , whereas the normal oxide of hydrogen is water, HOH.HOO

H

R

O

COOOH

R

O

COOH

This open cation must lead to

rearrangements that are not observed

C

O

F

3

C

H

C

+

OH

..

H

3

O

..

O

..

C

+

C

O

OH

2

..

+

FIGURE 10.26 A stepwise (nonconcerted) mechanism for epoxidation does not happen.

(a)

(b)

epoxide

oxirane

oxacyclopropane

An

O

..

CF

3

Br

..

O

..

–

C

H

C

+

Br

..

Br

..

Br

+

F

3

C

O

C

+

O

..

O

C

FIGURE 10.25 (a) Bromination of

an alkene. (b) Epoxidation of an

alkene.There is an additional transfer

of hydrogen to give the isolable,

stable epoxide.

The epoxidation reaction resembles the formation of a bromonium ion. In both

reactions, a leaving group is displaced by the alkene acting as nucleophile in an S

N

2

reaction (Fig. 10.25). Simultaneous transfer of hydrogen to the carbonyl ( )

oxygen gives a stable three-membered ring, an epoxide.

C

P

O

It is important to be clear why we are so sure that epoxidation occurs in a single

step.Why not just form the carbocation and then close up to the oxirane (Fig.10.26)?

Alkene epoxidation

10.4 Other Addition Reactions Involving Three-Membered Rings 425

If a free cation were involved, carbocation rearrangements would be expected, and

they are not found. Moreover it is easy to monitor the stereochemistry of this reac-

tion. Concerted (no intermediates) formation of the oxirane explains the observed

syn addition very well. A reaction in which the two new bonds are formed at the same

time cannot change the stereochemical relationships of the groups in the original

alkene. On the other hand, a mechanism involving an open carbocation would very

likely lead to a mixture of stereoisomeric oxiranes (Fig. 10.27). If only one bond were

made, there would be a carbocationic intermediate in which rapid rotation about

carbon–carbon single bonds would take place. These rotations would scramble the

original stereochemical relationship present in the alkene, and this result is not

observed. Instead, oxirane formation occurs with retention of the stereochemical

relationships present in the starting alkene.

PROBLEM SOLVING

The mechanistic technique outlined in Figure 10.27 above is a general one,

often used to distinguish one-step from two-step mechanisms. Retention of

stereochemistry (cis cis and trans trans) implicates a one-step process,

whereas scrambling of stereochemistry (cis cis and trans, or trans cis and

trans) is indicative of a stepwise process proceding through an intermediate (or

intermediates) capable of rotation. We’ll see it again in a few pages. (Can you

spot it?) You should be alert for it, as it forms the basis for many mechanism-

based problems.

UU

WEB 3D

THE GENERAL CASE

A SPECIFIC EXAMPLE

rapid

rotation

cis Alkene

cis Epoxide

+

trans Epoxide

(Not observed)

cis Epoxide

H

HH

R

H

RR

R

a peracid

H

C

+

C

R

H

CC

concerted,

syn addition

concerted,

syn addition

CH

3

COOOH

20 C, 3 h

Oleic acid

(CH

2

)

7

COOH

(CH

2

)

7

COOH

CH

3

(CH

2

)

7

CH

3

(CH

2

)

7

HH

H

RR

R

H

CC

(82%)

H

HH

CC

H

CC

H

OH

..

C

+

C

OH

..

O

..

O

..

O

..

O

..

+

OH

..

CC

+

OH

..

Formation

of an open

carbocation

1

Closure

2

Deprotonation

3

C

FIGURE 10.27 A test of the mechanism

of epoxidation. A stepwise mechanism

predicts scrambling of stereochemistry,

whereas a concerted mechanism must

retain the stereochemistry of the

starting alkene. Because cis alkenes

give only cis epoxides, the reaction

must be concerted.

H

OH

HO

Ti[OCH(CH

3

)

2

]

4

(CH

3

)

3

COOH

CH

2

Cl

2

H

O

OH

The allylic alcohol

(S,S) Enantiomer

of diethyl tartrate

(78%)

+

COOEt

EtOOC

H

O

OH

FIGURE 10.28 Asymmetric (Sharpless) epoxidation of alkenes (Et CH

3

CH

2

).=

10.4b Asymmetric (Sharpless) Epoxidation In recent years, an enormously

useful synthetic technique for producing enantiomeric oxiranes (epoxides) has appeared.

Discovered by Barry Sharpless (b. 1941), now at the Scripps Research Institute, it uses

a witches’ brew of titanium isopropoxide, tert-butyl peroxide, and one enantiomer of a

tartaric ester to react with an allylic alcohol ( ). Part of the

great utility of this procedure comes from the observation that the two enantiomers of

tartaric ester lead to two different stereochemistries of product oxirane (Fig. 10.28).

Professor Sharpless shared the Nobel prize in chemistry in 2001 for his discovery of

chiral catalysts.

R

O

CH

P

CH

O

CH

2

OH

426 CHAPTER 10 Additions to Alkenes 2 and Additions to Alkynes

WEB 3D

OH

A 1,2-diol

A

n epoxide

..

H

2

O / H

3

O

..

–

..

H

2

O /

..

O

HO

..

+

FIGURE 10.29 Epoxides (oxiranes)

open in either acid or base.

The mechanism of this complex reaction involves the titanium compound act-

ing as a clamp,holding the alkene,peroxide,and (S,S)-tartaric ester together.Because

the ester is asymmetric, the clamped combination of molecules is also asymmetric.

In one cluster, the oxirane oxygen is delivered from one side, whereas in the enan-

tiomeric cluster formed from (R,R)-tartaric ester, the oxygen comes from the other

side. These and other epoxides can then be transformed into all manner of com-

pounds, as we will now see.

PROBLEM 10.11 The (R,R) enantiomer of diethyl tartrate gives a diastereomer of

the product shown in Figure 10.28. Draw this diastereomer.

10.4c Further Reactions of Oxiranes Unlike the closely related bromonium

ions, oxiranes can be isolated under many reaction conditions.The bromonium ion

is doomed to bear a positive charge and is therefore a more powerful Lewis acid

than the neutral oxirane. Oxiranes will react, however, when treated in a second

step with either acids or bases. For example, reaction with either H

3

O



/H

2

O or

HO



/H

2

O leads to opening of the three-membered ring and formation of a

1,2-diol (Fig. 10.29).

The mechanisms of these ring openings are straightforward extensions of

reactions you already know. In base, the strongly nucleophilic hydroxide ion

attacks the oxirane and displaces the oxygen atom from one carbon in an S

N

2

Basic epoxide ring opening

10.4 Other Addition Reactions Involving Three-Membered Rings 427

OH

..

HO

..

+

OH

..

S

N

2

protonation

The mechanism of epoxide opening in base

protonation

H

HH

H

OH

..

HO

..

OH

2

..

deprotonation

The mechanism of epoxide opening in acid

O

+

..

H

2

O

OH

..

HO

..

O

+

–

OH

..

–

O

HO

..

H

2

O

..

H

2

O

..

H

3

O

..

+

..

S

N

2

FIGURE 10.30 Mechanisms for

epoxide opening in base and acid.

reaction (Fig. 10.30). In acid, there is no strong nucleophile, but protonation of

the oxygen atom of the oxirane creates a good Lewis acid. Water is a strong

enough nucleophile to open the protonated oxirane to give, after deprotonation,

the 1,2-diol (Fig. 10.30).

Acidic epoxide ring opening

–

+

(Not observed)

H

HH

OH

2

..

protonation deprotonation

O

+

..

S

N

2

OH

..

O

+

O

HO

..

O

..

H

2

O

..

H

2

O

..

H

2

O

..

H

3

O

..

+

H

3

O

..

+

HO

..

FIGURE 10.31 Water is not a strong

enough nucleophile to open the

unprotonated oxirane, because RO



is too poor a leaving group. Acid

catalysis changes the leaving group to

ROH, which is much easier to displace.

There is a strong parallel here to reactions we have already seen. Hydroxide is a

poor leaving group. Treatment with acid or formation of sulfonate esters converts

hydroxide into water or another good leaving group, and many reactions require this

transformation (see p. 281 for examples).The opening of the oxirane in acid involves

exactly the same kind of transformation; protonation facilitates the opening by mak-

ing the leaving group ROH rather than RO



(Fig. 10.31).

Unsymmetrical oxiranes can open in two ways and the regiochemical result is gen-

erally different in acid and base (Fig. 10.32). How does the difference in regiochem-

istry arise? Why should the two openings be different? In base,the ring opening is an

In acid, the nucleophile

generally becomes

attached to the more

substituted carbon

In base, the nucleophile

generally becomes

attached to the less

substituted carbon

H

2

Nu

CH

2

OH

O

C

R

C

CH

2

OH

C

CH

2

HNu

HNu / Nu

Nu

..

+

–

FIGURE 10.32 The regiochemistry

of epoxide opening is different in acid

and base.

428 CHAPTER 10 Additions to Alkenes 2 and Additions to Alkynes

THE GENERAL CASE

A SPECIFIC EXAMPLE

Na

+

N

3

–

H

2

O

..

H

3

C

H

3

C

H

3

C

H

3

C

–

O

..

H

3

C

HNu

Nu

N

3

CH

2

(41%)

C

Nu

CH

2

C

H

3

C

OH

..

CH

2

+

H

3

C

H

3

C

H

3

CH

3

C

H

2

O/dioxane

80 ⬚C

–

O

..

C

CH

2

C

N

3

H

3

C

H

3

C

OH

..

CH

2

C

CH

2

+

..

HO

O

..

O

..

Nu

..

–

Nu

..

–

FIGURE 10.33 In base, the nucleophile adds at the sterically less encumbered carbon.

THE GENERAL CASE

A SPECIFIC EXAMPLE

HX

+

Cl

..

Cl

..

H

3

C

H

3

C

HCl

..

(55%) (45%)

..

H

3

C

CH

3

CH

2

C

H

CH

2

C

H

CH

2

C

H

3

C

H

3

C

OH

..

OH

..

CH

2

C

+

This weaker carbon–oxygen

bond will be the major bond

broken by an adding nucleophile

O

C

CH

2

H

3

C

O

C

CH

2

H

3

CH

3

C

H

3

C

H

3

C

O

C

CH

2

..

O

..

O

..

+

H

3

C

H

3

C

H

3

C

FIGURE 10.34 Both carbons help bear the positive charge in the protonated oxirane. Most of the positive

charge is on the more substituted carbon, and that is where the nucleophile adds.This reaction is much like

the opening of an unsymmetrical bromonium ion by water or alcohol.

S

N

2 displacement by hydroxide on the unprotonated oxirane. As in any S

N

2 reaction,

steric matters are important, and the ring opens at the sterically more accessible, less

substituted carbon (Fig. 10.33). In acid, the first step is the protonation of the oxygen

of the three-membered ring (Fig. 10.34). Recall the discussion of the regiochemistry

of the opening of unsymmetrical bromonium ions (p. 419) and mercurinium ions

(p. 422).The positive charge is borne not only by the oxygen but also by the two car-

bons of the oxirane. Most of the positive charge on carbon will reside at the more sub-

stituted position.The more substituted carbon–oxygen bond will be longer and weaker

than the other, less substituted carbon–oxygen bond (Fig. 10.34). As in unsymmetri-

cal bromonium ions, the transition state for addition at the more substituted position

will be lower in energy than that for addition at the less substituted position.

Jones M., Fleming S.A. Organic Chemistry

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