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In the reaction OH + CHCl3 + NaOH → O⁻Na⁺ + CHO, the electrophile involved is

1. Dichloromethyl cation (:CHCl2)
2. Formyl cation (:CHO)
3. Dichlorocarbene (:CCl2)
4. Dichloromethyl anion (:CHCl2⁻)

Correct answer: Dichlorocarbene (:CCl2)

Solution

In this reaction, the base (NaOH) deprotonates CHCl3, forming the highly reactive intermediate dichlorocarbene (:CCl2), which acts as the electrophile. This intermediate is key to the reaction mechanism.

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