Question

2-Methyl propyl bromide reacts with C₂H₅O^- and gives 'A' whereas on reaction with C₂H₅OH it gives 'B'. The mechanism followed in these reactions and the products 'A' and 'B' respectively are

• A. S_N2. A iso-butyl ethyl ether; S_N1, B = tert- butyl ethyl ether
• B. S_N1, A tert-butyl ethyl ether; S_N2, B = iso- butyl ethyl ether
• C. S_N1, A = tert-butyl ethyl ether; S_N1, B = 2- butyl ethyl ether
• D. S_N2, A = 2-butyl ethyl ether; S_N2, B = iso- butyl ethyl ether

Answer 1

The Correct Option is A

2-Methyl propyl bromide, also known as isobutyl bromide (\(\text{CH}_3\text{CH}_2\text{CH}(\text{CH}_3)\text{Br}\)), reacts differently with \( \text{C}_2\text{H}_5\text{O}^- \) and \( \text{C}_2\text{H}_5\text{OH} \) due to their differing nucleophilicity and basicity.

Reaction with \( \text{C}_2\text{H}_5\text{O}^- \) (Ethoxide Ion):

Ethoxide ion is a strong, bulky nucleophile.

The reaction proceeds via an SN₂ mechanism due to the strong nucleophilic nature of \( \text{C}_2\text{H}_5\text{O}^- \).

The substitution occurs at the less hindered primary carbon, leading to the formation of iso-butyl ethyl ether (\(\text{CH}_3\text{CH}_2\text{CH}(\text{CH}_3)\text{O}\text{C}_2\text{H}_5\)).

Reaction with \( \text{C}_2\text{H}_5\text{OH} \) (Ethanol):

Ethanol is a weak nucleophile and cannot effectively participate in SN₂ reactions.

The reaction proceeds via an SN₁ mechanism, which involves the formation of a carbocation intermediate.

Upon ionization, the secondary carbocation formed can rearrange to a more stable tertiary carbocation.

This leads to the formation of tert-butyl ethyl ether (\(\text{(CH}_3)_3\text{C}\text{O}\text{C}_2\text{H}_5\)).

Therefore, the correct mechanism and products are:

A = iso-butyl ethyl ether via SN₂ mechanism.

B = tert-butyl ethyl ether via SN₁ mechanism.

This corresponds to option (3).