1-methoxycyclohexa-1, 4-diene
2-methoxycyclohexa-1, 3-diene
1-methoxycyclohexa-1, 3-diene
3-methoxycyclohexa-1, 4-diene
Answer is Wrong!
Answer is Right!
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
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follows:
- The lithium metal reacts with t-butanol to form lithium t-butoxide.
- The lithium t-butoxide reacts with anisole to form a lithium methoxycyclohexadienolate intermediate.
- 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.