Question

An alkene CH2CH=CH, is treated with B H in
presence of H2O2. The final product formed is
(a) CH3CH2CHO by CH3CH(OH)CH3
(c) CH3CH2CH2OH (d) (CH3CH2CH2)3B​

Answer 1

Explanation:

Hydroboration-oxidation reaction follows anti-Markovnikov's addition of H−OH across C=C to give alcohol.

Thus an alkene CH

3

CH=CH

2

when treated with B

2

H

6

in presence of H

2

O

2

will yield the final product as CH

3

CH

2

CH

2

OH

Answer 2

(d) 1-Bromopropane will be the product obtained in the reaction including peroxide.

Explanation:

• When any alkene reacts with Hydrogen bromide (HBr), then it yields 2-Bromoalkane. It would be a 2° compound.
• If we repeat the same process in the presence of peroxide (H2O2), then we will get 1-Bromoalkane. This compound would be a 1° compound.
• Reaction to this process is:

        $CH_{2}$−CH=$CH_{2}$ + HBr ⇒ $CH_{3}$−$CH_{2}$−$CH_{2}$Br

• The main reason behind this is that It follows the anti-Markownikoffs addition reaction or the peroxide effect.
• This rule follows when any peroxide is added to any organic compound reaction, and the halogen (Br) is attached to the 1° carbon atom instead of the 2° carbon atom.