[amp_mcq option1=”1-methoxycyclohexa-1, 4-diene” option2=”2-methoxycyclohexa-1, 3-diene” option3=”1-methoxycyclohexa-1, 3-diene” option4=”3-methoxycyclohexa-1, 4-diene” correct=”option1″]
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 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.