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

Identify an aromatic,mixed,$3^{\circ}$ amine among the following compounds.

The possible product$(s)$ obtained from the reaction of cyclobutyl amine with $HNO_{2}$ is/are:

Arrange the following in the increasing order of their basic strengths: $CH_{3}NH_{2}$,$(CH_{3})_{2}NH$,$(CH_{3})_{3}N$,$NH_{3}$.

Aniline reacts with acetaldehyde to form

In the given reaction,identify the reagent $A$:
$4-nitro-3-bromobenzenediazonium \ chloride \xrightarrow{A} 1-bromo-3-nitrobenzene$

Vedclass Test Series

Exam Paper Generator

Online Exam Module