Question

4. C2 JUN 09 Q7c

Propene reacts with hydrogen bromide to give 2-bromopropane as the major product.
(1) Using the reaction scheme below, show the mechanism of the reaction using curly
arrows and full negative and positive charges as appropriate.
(2)
H,C
H
c=c
H
H
H-Br
CH, H
H-C-C-H
Br H
CH, H
1 1
H-C-C-H
Br H
2-bromopropane
(ii) State briefly, why 2-bromopropane, rather than 1-bromopropane, is the main product
of this reaction.
[1]

Answer 1

Step-by-step explanation:

(1) The mechanism of the reaction between propene and hydrogen bromide to give 2-bromopropane is as follows:

Protonation of the alkene: A proton from hydrogen bromide (HBr) attacks the alkene, forming a carbocation intermediate.

Bromine addition: Bromine (Br) adds to the carbocation intermediate to form an intermediate bromonium ion.

Deprotonation: A proton from the bromonium ion is removed by a water molecule or another molecule, producing the final product 2-bromopropane.

The mechanism can be represented using curly arrows as follows:

H,C

H

c=c

H

H

H-Br

CH, H

H-C-C-H

Br H

CH, H

1 1

H-C-C-H

Br H

2-bromopropane

(2) 2-bromopropane is the main product of this reaction because of the stereochemistry of the reaction. When the carbocation intermediate forms, the bromine atom has a preference for adding to the face of the alkene that has the least number of hydrogen atoms. This leads to the formation of the 2-bromopropane, which has the bromine atom attached to the carbon atom with two hydrogen atoms. On the other hand, the formation of 1-bromopropane, which has the bromine atom attached to the carbon atom with three hydrogen atoms, is less favorable. This is why 2-bromopropane is the main product of this reaction.