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
- SN2. A iso-butyl ethyl ether; SN1, B = tert- butyl ethyl ether
- SN1, A tert-butyl ethyl ether; SN2, B = iso- butyl ethyl ether
- SN1, A = tert-butyl ethyl ether; SN1, B = 2- butyl ethyl ether
- SN2, A = 2-butyl ethyl ether; SN2, B = iso- butyl ethyl ether
The Correct Option is A
Solution and Explanation
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).
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