C6H5OH+CHCl3+NaOH arrow Salicylaldehyde the electrophile involved in the above reaction is.

Question

C6H5OH+CHCl3+NaOH arrow Salicylaldehyde the electrophile involved in the above reaction is. (A) Dichloromethyl cation (C H C l2 ) (B) Dichlorocarbene (: C C l2) (C) Trichloromethyl anion barCCl3 (D) Formyl cation (C H O)

Tardigrade Admin · Accepted Answer

The alpha elimination reaction takes place between chloroform and sodium hydroxide.

C H C l_{3} + N a O H \to: CC l_{2} + N a Cl + H_{2} O

The above reaction of formation of dichlorocarbene is the first step of the Riemer Tiemann reaction.

Q. $C_{6} H_{5} O H + C H C l_{3} + N a O H \to$ Salicylaldehyde the electrophile involved in the above reaction is.

Dichloromethyl cation $(C H C l_{2})$

Dichlorocarbene $(: CC l_{2})$

Trichloromethyl anion $\overset{ˉ}{C} C l_{3}$

Formyl cation $(C H O)$

The alpha elimination reaction takes place between chloroform and sodium hydroxide.
$C H C l_{3} + N a O H \to: CC l_{2} + N a Cl + H_{2} O$
The above reaction of formation of dichlorocarbene is the first step of the Riemer Tiemann reaction.

Q. C6​H5​OH+CHCl3​+NaOH→ Salicylaldehyde the electrophile involved in the above reaction is.

Dichloromethyl cation (CHCl2​)

Dichlorocarbene (:CCl2​)

Trichloromethyl anion CˉCl3​

Formyl cation (CHO)