In the Reimer-Tiemann reaction, C6H5-O^- Na^+ + CHCl3 + NaOH -> salicylaldehyde (o-hydroxybenzaldehyde), the electrophile involved is:

Correct answer: (D) Dichlorocarbene, :CCl2. Answer: (d) Dichlorocarbene, :CCl2 Solution: This is the Reimer-Tiemann reaction. CHCl3 reacts with base (OH^-) to first form the trichloromethyl carbanion CCl3^-, which then loses Cl^- to generate the reactive electrophile dichlorocarbene, :CCl2. This carbene attacks the phenoxide ring to give, after hydrolysis, the ortho-hydroxybenzaldehyde. Hence (d).

NEET 2018 · ChemistryPrevious Year Question

In the Reimer-Tiemann reaction, $C X_{6} H X_{5} - O X^{-} Na X^{+} + CHCl X_{3} + NaOH$ salicylaldehyde (o-hydroxybenzaldehyde), the electrophile involved is:

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

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

Correct Answer

(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 base ( $OH X^{-}$ ) to first form the trichloromethyl carbanion $CCl X_{3} X^{-}$ , which then loses $Cl X^{-}$ to generate the reactive electrophile dichlorocarbene, $: CCl X_{2}$ . This carbene attacks the phenoxide ring to give, after hydrolysis, the ortho-hydroxybenzaldehyde. Hence (d).

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In the Reimer-Tiemann reaction, CX6​HX5​−OX− NaX++CHClX3​+NaOH​ salicylaldehyde (o-hydroxybenzaldehyde), the electrophile involved is:

Solution & Explanation

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In the Reimer-Tiemann reaction, $C X_{6} H X_{5} - O X^{-} Na X^{+} + CHCl X_{3} + NaOH$ salicylaldehyde (o-hydroxybenzaldehyde), the electrophile involved is: