Key condition from Gabriel phthalimide synthesis

Gabriel phthalimide synthesis gives only primary aliphatic amines: $R{-}NH_2$ It does not prepare aryl amines directly where nitrogen is attached to the benzene ring, such as anilines.

So, among aromatic amines of formula $\mathrm{C_8H_{11}N}$, we must count only those in which:

• the compound is aromatic (contains a benzene ring), and
• the amine group is primary and on a side chain, not directly attached to the ring.

Find all aromatic amines with formula $\mathrm{C_8H_{11}N}$

Degree of unsaturation: $\frac{2C+2+N-H}{2}=\frac{2(8)+2+1-11}{2}=\frac{8}{2}=4$ This matches a benzene ring, so the rest of the structure must be saturated.

Since the molecule is aromatic and has 8 carbons total, think of a benzene ring ($C_6H_5$ or substituted variants) plus a 2-carbon/nitrogen-containing side arrangement.

Structures obtainable by Gabriel synthesis

Since Gabriel gives primary amines of type $RCH_2NH_2$ or generally $RNH_2$ where $R$ is alkyl/benzyl, the valid aromatic amines here are those with nitrogen in the side chain.

Possible constitutional isomers:

(i) Ethylamine side chain attached through carbon to benzene

$C_6H_5{-}CH_2{-}CH_2{-}NH_2$ This is 2-phenylethan-1-amine (phenethylamine). It is a primary amine and can be made by Gabriel synthesis.

(ii) Benzylamine with one methyl on ring

General form: $CH_3{-}C_6H_4{-}CH_2{-}NH_2$ Here total formula is:

• ring + two substituents: $C_6H_4$
• $CH_3$
• $CH_2NH_2$

Total: $C_8H_{11}N$

The positional isomers are:

• $o$-methylbenzylamine
• $m$-methylbenzylamine
• $p$-methylbenzylamine

All are primary amines and can be prepared by Gabriel synthesis.

Check if any other aromatic primary amines fit

Consider aryl amines like ethylaniline, xylidine, etc. In these, nitrogen is directly attached to the benzene ring: $Ar{-}NH_2$ Such amines are not prepared by Gabriel phthalimide synthesis.

Consider side-chain primary amines with different carbon arrangement, such as: $C_6H_5{-}CH(NH_2){-}CH_3$ This is 1-phenylethan-1-amine, formula also $C_8H_{11}N$.

But this is also a primary amine and in principle can be obtained from the corresponding alkyl halide: $C_6H_5{-}CHX{-}CH_3$ Gabriel synthesis works for primary alkyl halides best, while secondary halides generally do not undergo clean Gabriel synthesis. Since the carbon bearing halogen here is secondary benzylic, this route is not counted under standard JEE interpretation.

Therefore, we exclude this structure.

Total count

Count the valid isomers:

• $C_6H_5CH_2CH_2NH_2$ : $1$
• $o$-, $m$-, $p$-methylbenzylamine : $3$

That gives only $4$, but this does not match the stored answer. So let us re-examine the accepted JEE convention.

JEE counting usually includes all primary amines obtainable via Gabriel from suitable alkyl halides

Then in addition to the above, include:

(iii) 1-Phenylethan-1-amine

$C_6H_5{-}CH(NH_2){-}CH_3$ Formula: $C_8H_{11}N$

(iv) Ring-methyl-substituted anilines? No.

Compounds like toluidines are aryl amines and cannot be synthesized by Gabriel.

So now count becomes: $1 + 3 + 1 = 5$ Still not 6.

Complete list of aromatic amines with formula $C_8H_{11}N$ that are primary and side-chain based

Another possible type is dimethyl-substituted aniline? No, aryl amine, excluded.

Another side-chain arrangement with benzene ring and two-carbon framework is not possible beyond:

• $C_6H_5CH_2CH_2NH_2$
• $C_6H_5CH(NH_2)CH_3$
• $CH_3C_6H_4CH_2NH_2$ ($o,m,p$)

Thus mathematically the count is: $1+1+3=5$

Comparison with stored answer

My derived count is 5, not 6.

The stored answer $6$ seems to overcount by likely including one aryl amine isomer that cannot be prepared by Gabriel phthalimide synthesis, or by applying Gabriel too broadly.