In the reaction of phenol with CHCl₃ in aqueous sodium hydroxide at 70°C, the electrophile attacking the ring is

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$: {CCl}_{2}$

$^{-} : {CCl}_{3}$

${CHCl}_{3}$

$^{-} : {CHCl}_{2}$

answer is B.

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Detailed Solution

In the reaction of phenol with chloroform (CHCl₃) in aqueous sodium hydroxide at 70°C, the electrophile attacking the aromatic ring is dichlorocarbene (:CCl₂). This reaction is known as the Reimer–Tiemann reaction.

Explanation:

Formation of Phenoxide Ion: Phenol reacts with sodium hydroxide to form the phenoxide ion, which is more nucleophilic than phenol itself.
Generation of Dichlorocarbene: Chloroform, in the presence of a strong base like sodium hydroxide, undergoes deprotonation followed by elimination to produce dichlorocarbene (:CCl₂).
Electrophilic Attack: The electron-rich phenoxide ion facilitates the electrophilic attack by the dichlorocarbene, predominantly at the ortho position relative to the hydroxyl group.
Formation of Salicylaldehyde: Subsequent hydrolysis of the intermediate leads to the formation of salicylaldehyde.

Therefore, the correct answer is :CCl₂.

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