Aniline $C H_{3} C l - --- \to A l C l_{3}$
How many methyl groups will be substituted on aniline to form a final product?(Write your answer in integer number)

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Solution

Aniline reacts with aluminum chloride (the Lewis base used in Friedel crafts reaction) to form anilinium hydrochloride. This results in a positive charge on the nitrogen atom.
Hence, aniline does not undergo Friedel craft's reaction and no methyl group is substituted.
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Q. Number of methyl substitutions takes place in the ring when aniline is reacted with

C H_{3} C l

in the presence of

A l C l_{3}

Q. 1. N,N-dimethyl aniline
2. Triethylamine
3. N-methyl aniline
4. p-methyl aniline
How many of the given compound(s) will not give a positive carbylamine test?

Q. In case of substituted aniline the group which decreases the basic strength is :

Account for the following

(i) $p K_{b}$ of aniline is more than that of methyl amine.

(ii) Ethylamine is soluble in water whereas aniline is not

(iii) Methyl amine in water reacts with ferric chloride to precipitate hydrated ferric oxide

(iv) Although amino group is o- and p- directing in aromatic electrophilic substitution, aniline on nitration gives a substantial amount of m-nitroaniline

(v) Aniline does not undergo Friedel Crafts reaction

(vi) Diazonium salts of aromatic amines are more stable than those of aliphatic amines

(vii) Gabriel phthalimide synthesis is preferred for synthesising primary amines

Amino group is ortho, para directing for aromatic electrophilic substitution. On nitration of aniline with concentrated acids, a good amount of meta nitro aniline is obtained. This is due to

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Aniline CH3Cl−−−−→AlCl3How many methyl groups will be substituted on aniline to form a final product?(Write your answer in integer number)

Aniline reacts with aluminum chloride (the Lewis base used in Friedel crafts reaction) to form anilinium hydrochloride. This results in a positive charge on the nitrogen atom.Hence, aniline does not undergo Friedel craft's reaction and no methyl group is substituted.Ans-0

Aniline $C H_{3} C l - --- \to A l C l_{3}$
How many methyl groups will be substituted on aniline to form a final product?(Write your answer in integer number)

Aniline reacts with aluminum chloride (the Lewis base used in Friedel crafts reaction) to form anilinium hydrochloride. This results in a positive charge on the nitrogen atom.
Hence, aniline does not undergo Friedel craft's reaction and no methyl group is substituted.
Ans-0