Question

The reaction involves indene, fluorochlorobromomethane (CH(F)(Cl)(Br)), and potassium methoxide (MeOK). What is the major product of this reaction?

• A. 2-Fluoronaphthalene.
• B. A cyclopropane fused indene derivative with F and Cl on the cyclopropane carbon.
• C. A cyclopropane fused indene derivative with Cl and Br on the cyclopropane carbon.
• D. 2-Bromonaphthalene.

Answer 1

Answer: (B)

A cyclopropane fused indene derivative with F and Cl on the cyclopropane carbon.

Answer 2

The reaction proceeds via carbene formation and subsequent cyclopropanation.

1. Carbene Generation: Potassium methoxide (MeOK) deprotonates CH(F)(Cl)(Br) to form a carbanion. This carbanion then undergoes alpha-elimination, losing the best leaving group (Br$^-$) to form fluorochlorocarbene (:C(F)(Cl)). The order of leaving group ability is I > Br > Cl > F. $CH(F)(Cl)(Br) + MeO^- \rightarrow ^-C(F)(Cl)(Br) + MeOH$ $^-C(F)(Cl)(Br) \rightarrow :C(F)(Cl) + Br^-$

2. Cyclopropanation: The generated fluorochlorocarbene adds across the double bond of indene in a [1+2] cycloaddition reaction. This forms a new three-membered cyclopropane ring fused to the five-membered ring of indene, with the cyclopropane carbon bearing the fluorine and chlorine atoms.

Therefore, the major product is a cyclopropane fused indene derivative with F and Cl on the cyclopropane carbon.