In the reaction C6H5OH + CHCl3 + NaOH -> salicylaldehyde the electrophile involved is

Correct answer: (D) (D) Dichlorocarbene :CCl2. Answer: (D) Dichlorocarbene :CCl2 Solution: This is the Reimer-Tiemann reaction. CHCl3 reacts with the base NaOH to generate the electrophile dichlorocarbene: CHCl3 + OH^- -> ^-CCl3 + H2O ^-CCl3 -> :CCl2 + Cl^- The electrophilic dichlorocarbene :CCl2 attacks the ortho position of the phenoxide ring; subsequent hydrolysis gives the -CHO group, producing salicylaldehyde (2-hydroxybenzaldehyde).

NEET 2018 · ChemistryPrevious Year Question

In the reaction $C X_{6} H X_{5} OH + CHCl X_{3} + NaOH salicylaldehyde$ the electrophile involved is

Answer: (D) (D) Dichlorocarbene $: CCl X_{2}$ . \textbf{Answer:} (D) Dichlorocarbene $: CCl X_{2}$ \textbf{Solution:} This is the Reimer-Tiemann reaction.

A.(A) Dichloromethyl anion $X_{-} X_{2}^{2 -} CHCl X_{2}$
B.(B) Formyl cation $X_{+} X_{2}^{2 +} CHO$
C.(C) Dichloromethyl cation $X_{+} X_{2}^{2 +} CHCl X_{2}$
D.(D) Dichlorocarbene $: CCl X_{2}$ ✓

Correct Answer

(D) (D) Dichlorocarbene $: CCl X_{2}$

Solution & Explanation

\textbf{Answer:} (D) Dichlorocarbene $: CCl X_{2}$ \textbf{Solution:} This is the Reimer-Tiemann reaction. $CHCl X_{3}$ reacts with the base $NaOH$ to generate the electrophile dichlorocarbene: $CHCl X_{3} + OH X^{-} X_{-} X_{2}^{2 -} CCl X_{3} + H X_{2} O$ $X_{-} X_{2}^{2 -} CCl X_{3} : CCl X_{2} + Cl X^{-}$ The electrophilic dichlorocarbene $: CCl X_{2}$ attacks the ortho position of the phenoxide ring; subsequent hydrolysis gives the $- CHO$ group, producing salicylaldehyde (2-hydroxybenzaldehyde).

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In the reaction CX6​HX5​OH+CHClX3​+NaOH​salicylaldehyde the electrophile involved is

Solution & Explanation

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In the reaction $C X_{6} H X_{5} OH + CHCl X_{3} + NaOH salicylaldehyde$ the electrophile involved is