In the Reimer-Tiemann reaction,phenol is conv | vedclass

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(D) The Reimer-Tiemann reaction involves the treatment of phenol with chloroform $(CHCl_3)$ in the presence of an aqueous base (like $NaOH$). $1$. The

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(D) The Reimer-Tiemann reaction involves the treatment of phenol with chloroform $(CHCl_3)$ in the presence of an aqueous base (like $NaOH$). $1$. The base deprotonates the phenol to form a phenoxide ion. $2$. The chloroform reacts with the base to generate a dichlorocarbene $(:CCl_2)$ intermediate,which acts as an electrophile. $3$. The phenoxide ion attacks the dichlorocarbene to form an ortho-dichloromethyl phenoxide intermediate,which is represented as a benzene ring with an $-O^-Na^+$ group and a $-CHCl_2$ group at the ortho position. $4$. This intermediate then undergoes hydrolysis with $NaOH$ followed by acidification to yield salicylaldehyde.

In the Reimer-Tiemann reaction,phenol is converted into salicylaldehyde through an intermediate. The structure of the intermediate is . . . . . . .

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