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Electrophilic Additions of Alkenes: Markovnikov and Anti-Markovnikov Reactions, Lecture notes of Stereochemistry

An introduction to electrophilic additions of alkenes, focusing on the formation of Markovnikov and anti-Markovnikov products. It explains the concept of electrophilic additions, the role of strong electrophiles, and the preference for addition at the most substituted carbon (Markovnikov's rule). The document also discusses the possibility of anti-Markovnikov products and the conditions under which they can form. It is a valuable resource for students studying organic chemistry, particularly those focusing on alkene reactions.

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ELECTROPHILIC ADDITIONS OF ALKENES AS THE
COUNTERPART OF ELIMINATIONS
INTRODUCTION
- Chapter 8 is mostly about alkene reactions. That is, how one can transform alkenes into other functional groups. Most
of these reactons are electrophilic additions, or the addition of electrophiles across the double bond. Several of these reactions amount to
addition of water to the p-bond. The result is the transformation of alkenes into alcohols. We now look at the generalities of this type of reaction
(electrophilic addition) and then we look at each reaction individually.
The C=C p-bond of alkenes is a source of electrons. It is considered a weak base or nucleophile. As such, it can react with strong electrophiles.
We have already learned how to identify electrophiles, but the p-bond requires strong electrophiles to react with. These can be strong proton
acids, or species containing atoms with incomplete octets (Lewis acids). Examples of strong proton acids are HBr and H2SO4. Examples of
Lewis acids are BH3, transition metal salts such as HgSO4, and carbocations.
The addition of strong electrophiles to the C=C p-bond can be viewed as the opposite of the elimination reaction, as illustrated below.
H3C C CH3
CH3
Br
E2
H3C
C
CH2
CH3
+HBr
Elimination results in net loss of HBr to form a new C=C bond.
Addition of HBr (a strong electrophile) across the p-bond
forms a new functional group, in this case an alkyl halide.
H3C
C
CH2
CH3
+HBr Addition
H3C C CH3
CH3
Br
pf3
pf4
pf5
pf8
pf9

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ELECTROPHILIC ADDITIONS OF ALKENES AS THE

COUNTERPART OF ELIMINATIONS

INTRODUCTION - Chapter 8 is mostly about alkene reactions. That is, how one can transform alkenes into other functional groups. Most

of these reactons are electrophilic additions , or the addition of electrophiles across the double bond. Several of these reactions amount to

addition of water to the p-bond. The result is the transformation of alkenes into alcohols. We now look at the generalities of this type of reaction

(electrophilic addition) and then we look at each reaction individually.

The C=C p-bond of alkenes is a source of electrons. It is considered a weak base or nucleophile. As such, it can react with strong electrophiles.

We have already learned how to identify electrophiles, but the p-bond requires strong electrophiles to react with. These can be strong proton

acids, or species containing atoms with incomplete octets (Lewis acids). Examples of strong proton acids are HBr and H 2 SO 4. Examples of

Lewis acids are BH 3 , transition metal salts such as HgSO 4 , and carbocations.

The addition of strong electrophiles to the C=C p-bond can be viewed as the opposite of the elimination reaction, as illustrated below.

H 3 C C CH 3

CH 3

Br

E

H 3 C

C

CH 2

CH 3

+ HBr

Elimination results in net loss of HBr to form a new C=C bond.

Addition of HBr (a strong electrophile) across the p-bond

forms a new functional group, in this case an alkyl halide.

H 3 C

C

CH 2

CH 3

+ HBr

Addition

H 3 C C CH 3

CH 3

Br

According to Saytzeff’s rule , in elimination reactions where formation of several alkenes is possible, the most highly substituted alkene

predominates as a product.

Conversely, Markovnikov’s rule says that in addition reactions of proton acids to alkenes, the proton of the strong acid preferentially

bonds to the carbon in the p -bond that already holds the greater number of hydrogens on it. In the alkene shown below, C-2 has more

protons attached to it (one) than C-1 (none). Accordingly, the hydrogen from HBr bonds to C-2 and the bromine bonds to C-1.

This preferred orientation is referred to as Markovnikov orientation. This suggests the possibility that another product with opposite orientation,

called the anti-Markovnikov product , might form. Given that this product does not normally form under ordinary conditions, the question then

is, are there special conditions under which it could form? The answer is yes, but with a very limited scope. We’ll address that point later.

Br

Markovnikov's product,

1 2 a tertiary bromide

HBr 1

Another way to state Markovnikov’s rule, in this case, is to say the the tertiary bromide will form preferentially over the secondary bromide. If

the electrophilic part of the reactant (H+) adds to the least substituted carbon (C-2), then the nucleophilic part (Br-) adds to the most substituted

carbon (C-1). The degree of substitution in this case refers to the number of alkyl groups originally attached to the p-bond. This is another

way by which Markovnikov’s rule becomes the counterpart of Saytzeff’s rule. The two statements are complementary.

Anti-Markovnikov's product,

1 2 a secondary bromide

HBr 1

Br

E1 or E

Saytzeff's product

Br

Refer to section 8-3B of the Wade textbook (p. 3 19 of the 5 th ed.) and remember the following points.

a) The anti-Markovnikov addition of proton acids to alkenes works only with HBr.

b) It requires the presence of peroxides, which are free radical initiators.

c) The reaction follows a free radical mechanism, where the bromine is first to add to the alkene with formation of the most stable

free radical , which eventually leads to the anti-Markovnikov product.

Other proton acids that follow Markovnikov’s rule include HCl. HI, and H 2 O (acid-catalyzed) to form alcohols. We now summarize all the

important addition reactions of alkenes, including the markovnikov addition of proton acids and water. Then we address some concepts and

mechanistic considerations relevant to these reactions.

CH 3

HBr

ROOR'

CH 3

Br

Another example:

SUMMARY OF ALKENE REACTIONS

Alkenes are primarily prepared by elimination reactions of molecules that contain good leaving groups attached to sp 3 carbons.

Examples of such reactions are dehydrohalogenations with strong base, and acid-catalyzed dehydrations of alcohols. The opposite of an

elimination is an addition reaction. In an addition reaction an alkene adds elements to each of the carbons involved in the π -bond,

resulting in formation of sp 3 carbons from sp^2 carbons. This is one of the most important types of reactions that alkenes undergo.

Another important type of reaction involving alkenes is oxidative cleavage. In such reactions the carbon-carbon π-bond is completely broken

by the action of an oxidizing agent, resulting in more oxidized forms of carbon, such as aldehydes, ketones, and carboxylic acids. All these

reactions are covered in chapter 8 of the Wade text, 4th ed.

The following is a summary of the most important representative types of these reactions. For full details, please refer to the textbook.

1. ADDITION OF HBr, HCl, and HI. The addition of these substances to an alkene proceeds by an ionic mechanism, with formation of the

most stable carbocation. Therefore, it follows Markovnikov’s Rule.

CH 3

HBr

H 3 C Br

An Anti-Markovnikov variation requires the presence of peroxides as free radical initiators, and can only be performed with HBr. The

mechanism proceeds with formation of the most stable free radical, which results in formation of the Anti-Markovnikov product.

peroxides (ROOR')

CH 3

HBr

CH 3

Br

6. HALOGEN ADDITION IN THE PRESENCE OF WATER (halohydrin formation). Produces Markovnikov alcohols with a neighboring

halogen. The OH and the halogen add with anti -stereochemistry. If chiral centers result from this reaction, the product is obtained as an

enantiomeric mixture.

CH 3

Br (^2)

H 2 O

H 3 C OH

Br

THE NEXT TWO REACTIONS ADD TWO HYDROXYL GROUPS (OH) ACROSS THE DOUBLE BOND

7. SYN HYDROXYLATION. Adds two hydroxyl groups accross the π-bond with syn -stereochemistry. It can be performed with two reagent

mixtures: pot assium permanganate in basic medium, or osmium tetroxide in the presence of hydrogen peroxide. If chiral centers result

from this reaction, the product is obtained as an enantiomeric mixture.

CH 3 H 3 C OH

KMnO 4

H 2 O, OH

OH

( cis -diol)

CH 3 H 3 C OH

OsO (^4)

H 2 O (^2)

OH

( cis -diol)

8. ANTI-HYDROXYLATION SEQUENCE. Adds two hydroxyl groups accross the π-bond with anti -stereochemistry. It proceeds in two steps

via an epoxide (three-membered ring ether). If chiral centers result from this reaction, the product is obtained as an enantiomeric mixture.

CH 3 H

3 C^ OH

RCO 3 H

(peroxyacid)

( trans -diol)

H 3 C

O

epoxide

H 3 O

OH

9. CATALYTIC HYDROGENATION. Adds hydrogen accross the π-bond with syn -stereochemistry. It basically transforms an akene into an

alkane.

CH 3

H 2

Pt, Pd, or Ni

CH 3

MAIN OXIDATIVE CLEAVAGE REACTIONS

1. CLEAVAGE WITH HOT, CONCENTRATED POTASSIUM PERMANGANATE yields ketones and/or carboxylic acids, depending on

whether the carbon in question is in the middle of a chain or at the end of a chain in the product.

CH 3

KMnO 4

hot, conc. H^3 C^ OH

O

KMnO 4

hot, conc. (^) H 3 C OH

O

H 3 C CH 3

O

KMnO 4

hot, conc. H^3 C^ COOH

O