Question

(i) Predict the products A and B in the hydroboration-oxidation reaction. 
(ii) Explain the preparation of phenol from cumene.

• A.
• B.
• C.
• D.

Hint:

Important Points: Hydroboration–oxidation gives alcohols by anti-Markovnikov addition. Cumene process is an important industrial method for preparing phenol. Products of cumene process: Phenol + Acetone.

Answer 1

The Correct Option is A

Concept:
Hydroboration–oxidation is an important reaction of alkenes that produces alcohols through anti-Markovnikov addition. Phenol can also be prepared industrially from cumene (isopropyl benzene) through oxidation followed by acid cleavage.

Step 1: Hydroboration–Oxidation Reaction.
In the first step, propene reacts with diborane (BH3)2 to form tripropylborane (A).

3 CH3CH=CH2 → (CH3CH2CH2)3B   (in presence of diborane (BH3)2)

In the second step, oxidation with alkaline hydrogen peroxide (H2O2/OH⁻) converts it into propan-1-ol (B).

(CH3CH2CH2)3B → 3 CH3CH2CH2OH   (H2O2 / OH⁻)

This reaction follows the anti-Markovnikov rule, where the OH group attaches to the less substituted carbon atom.

Step 2: Preparation of Phenol from Cumene.
Cumene (isopropyl benzene) is oxidized with oxygen from air to form cumene hydroperoxide.

Cumene → Cumene Hydroperoxide   (O2 from air)

The hydroperoxide undergoes acid-catalysed cleavage to give phenol and acetone.

Cumene Hydroperoxide → Phenol + Acetone   (H⁺ / H2O)

This method is widely used industrially for the large-scale production of phenol.