The major product formed in the reaction of anisole with lithium, liquid ammonia and t-butanol is A. 1-methoxycyclohexa-1, 4-diene B. 2-methoxycyclohexa-1, 3-diene C. 1-methoxycyclohexa-1, 3-diene D. 3-methoxycyclohexa-1, 4-diene

1-methoxycyclohexa-1, 4-diene
2-methoxycyclohexa-1, 3-diene
1-methoxycyclohexa-1, 3-diene
3-methoxycyclohexa-1, 4-diene

The correct answer is A. 1-methoxycyclohexa-1, 4-diene.

The reaction of anisole with lithium, liquid ammonia and t-butanol is a Friedel-Crafts alkylation reaction. In this reaction, the alkyl group from t-butanol is transferred to the aromatic ring of anisole. The major product formed in this reaction is 1-methoxycyclohexa-1, 4-diene.

The mechanism of the reaction is as

132.3c6.3 23.7 24.8 41.5 48.3 47.8C117.2 448 288 448 288 448s170.8 0 213.4-11.5c23.5-6.3 42-24.2 48.3-47.8 11.4-42.9 11.4-132.3 11.4-132.3s0-89.4-11.4-132.3zm-317.5 213.5V175.2l142.7 81.2-142.7 81.2z"/> Subscribe on YouTube
follows:
  1. The lithium metal reacts with t-butanol to form lithium t-butoxide.
  2. The lithium t-butoxide reacts with anisole to form a lithium methoxycyclohexadienolate intermediate.
  3. The lithium methoxycyclohexadienolate intermediate loses lithium ion to form 1-methoxycyclohexa-1, 4-diene.

The other options are incorrect because they do not represent the major product of the reaction. Option B, 2-methoxycyclohexa-1, 3-diene, is a minor product of the reaction. Option C, 1-methoxycyclohexa-1, 3-diene, is not a product of the reaction. Option D, 3-methoxycyclohexa-1, 4-diene, is not a product of the reaction.

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