JEE Main 2026 — Amines Question with Solution

The final product is cyclohexyl isocyanide ($C_6{H}_{11}NC$).
Working backwards from the final product:
The last step involves $CHC{l}_3$ and $KOH$ (alc), which is the Carbylamine reaction.
This reaction converts a primary amine ($R-N{H}_2$) into an isocyanide ($R-NC$). Thus, the intermediate before this step must be cyclohexylamine ($C_6{H}_{11}N{H}_2$).
The step before cyclohexylamine involves $B{r}_2$ and $KOH$, which is the Hoffmann Bromamide Degradation reaction.
This reaction converts an amide ($R-CON{H}_2$) into a primary amine ($R-N{H}_2$) with one less carbon atom.
Therefore, the amide 'Q' must be cyclohexanecarboxamide ($C_6{H}_{11}CON{H}_2$).
The first step converts 'P' into 'Q' using $N{H}_3$ and heat.
Carboxylic acids react with $N{H}_3$ followed by heating to form amides.
Thus, 'P' must be cyclohexanecarboxylic acid ($C_6{H}_{11}COOH$).
Reaction sequence:
$C_6{H}_{11}COOH$ (P) $\stackrel{N{H}_3,\Delta }{\to }{C}_6{H}_{11}CON{H}_2$ (Q)
$C_6{H}_{11}CON{H}_2$ (Q) $\stackrel{B{r}_2,KOH}{\to }{C}_6{H}_{11}N{H}_2$ (Cyclohexylamine)
$C_6{H}_{11}N{H}_2\stackrel{CHC{l}_3,KOH}{\to }{C}_6{H}_{11}NC$ (Cyclohexyl isocyanide)
Comparing with the options, 'P' is cyclohexanecarboxylic acid, which corresponds to option (4).