In Friedel-Crafts alkylation of aniline,one g | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Aniline $(C_6H_5NH_2)$ is a Lewis base due to the lone pair of electrons on the nitrogen atom.Friedel-Crafts catalysts like anhydrous $AlCl_3$ are

Vedclass · Accepted Answer

positively charged nitrogen at benzene ring.

Vedclass · Answer

(D) Aniline $(C_6H_5NH_2)$ is a Lewis base due to the lone pair of electrons on the nitrogen atom.Friedel-Crafts catalysts like anhydrous $AlCl_3$ are Lewis acids.When aniline reacts with $AlCl_3$,the lone pair on the nitrogen atom coordinates with the $AlCl_3$ to form a salt,resulting in a positively charged nitrogen atom $(C_6H_5NH_2^+-AlCl_3^-)$.Because of this positive charge,the nitrogen atom becomes strongly electron-withdrawing,which deactivates the benzene ring towards electrophilic substitution.Therefore,aniline does not undergo Friedel-Crafts alkylation or acylation.

In Friedel-Crafts alkylation of aniline,one gets ....

Similar Questions

In the following reactions,$P$,$Q$,$R$,and $S$ are the major products.
(image)
The correct statement about $P$,$Q$,$R$,and $S$ is

How many metamers are possible for the molecular formula $C_4H_{11}N$?

Which of the following statements about aniline is false?

Aniline reacts with excess $Br_2 / H_2O$ to give the major product.

Which one of the following compounds forms a quaternary salt on reacting with excess methyl iodide?

Vedclass Test Series

Exam Paper Generator

Online Exam Module