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

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(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

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positively charged nitrogen at benzene ring.

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(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 series of reactions: $C_6H_5NH_2$ $\xrightarrow{NaNO_2/HCl}_{0 - 5\ ^\circ C} X$ $\xrightarrow{HNO_2}_{H_2O} Y + N_2 + HCl$. $X$ and $Y$ are respectively:

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