The Reimer-Tiemann reaction involves a

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(B) The Reimer-Tiemann reaction involves the reaction of phenol with chloroform $(CHCl_3)$ in the presence of an aqueous base ($NaOH$ or $KOH$).In the

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Carbene intermediate

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(B) The Reimer-Tiemann reaction involves the reaction of phenol with chloroform $(CHCl_3)$ in the presence of an aqueous base ($NaOH$ or $KOH$).In the first step,the base abstracts a proton from chloroform to form a trichloromethyl carbanion $(:CCl_3^-)$,which subsequently loses a chloride ion to generate dichlorocarbene $(:CCl_2)$ as an electrophilic intermediate.This dichlorocarbene then attacks the phenoxide ion to form the product.Therefore,the correct answer is $(B)$.

Acid	$pKa$
$(a)$ Phenol	$(i)$ $16$
$(b)$ $p$-Nitrophenol	$(ii)$ $0.78$
$(c)$ Ethanol	$(iii)$ $10$
$(d)$ Picric acid	$(iv)$ $7.1$

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