Compound ' P ' undergoes the following sequence of reactions : ![JEE Main 2026 Chemistry, Amines — question figure](https://prepforbharat.s3.ap-south-1.amazonaws.com/exam/615f0e999476412f48314daf/Chemistry/images/Amines/6983c1968bfdbb243563969b/question_1__q_6983c1968bfdbb243563969b__cdn-question-pool.getmarks.app__23S1_q_64_1__f9944ecc28_final_ppt_sync.png) ' $\mathrm{P}^{\prime}$ is :

JEE Main 2026 — Chemistry Organic Chemistry

2026mcqmedium

Compound ' P ' undergoes the following sequence of reactions :

JEE Main 2026 Chemistry, Amines — question figure

' $\mathrm{P}^{\prime}$ is :

Official previous-year question

Held on 23 Jan 2026 · Verified 6 Jul 2026.

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Solution

The final product is cyclohexyl isocyanide ( $C_6H_{11}NC$ ).

Working backwards from the final product:

The last step involves $CHCl_3$ and $KOH$ (alc), which is the Carbylamine reaction.

This reaction converts a primary amine ( $R-NH_2$ ) into an isocyanide ( $R-NC$ ). Thus, the intermediate before this step must be cyclohexylamine ( $C_6H_{11}NH_2$ ).

The step before cyclohexylamine involves $Br_2$ and $KOH$ , which is the Hoffmann Bromamide Degradation reaction.

This reaction converts an amide ( $R-CONH_2$ ) into a primary amine ( $R-NH_2$ ) with one less carbon atom.

Therefore, the amide 'Q' must be cyclohexanecarboxamide ( $C_6H_{11}CONH_2$ ).

The first step converts 'P' into 'Q' using $NH_3$ and heat.

Carboxylic acids react with $NH_3$ followed by heating to form amides.

Thus, 'P' must be cyclohexanecarboxylic acid ( $C_6H_{11}COOH$ ).

Reaction sequence:

$C_6H_{11}COOH$ (P) $\xrightarrow{NH_3, \Delta} C_6H_{11}CONH_2$ (Q)

$C_6H_{11}CONH_2$ (Q) $\xrightarrow{Br_2, KOH} C_6H_{11}NH_2$ (Cyclohexylamine)

$C_6H_{11}NH_2 \xrightarrow{CHCl_3, KOH} C_6H_{11}NC$ (Cyclohexyl isocyanide)

Comparing with the options, 'P' is cyclohexanecarboxylic acid, which corresponds to option (4).

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Solution

The final product is cyclohexyl isocyanide ( $C_6H_{11}NC$ ).

Working backwards from the final product:

The last step involves $CHCl_3$ and $KOH$ (alc), which is the Carbylamine reaction.

This reaction converts a primary amine ( $R-NH_2$ ) into an isocyanide ( $R-NC$ ). Thus, the intermediate before this step must be cyclohexylamine ( $C_6H_{11}NH_2$ ).

The step before cyclohexylamine involves $Br_2$ and $KOH$ , which is the Hoffmann Bromamide Degradation reaction.

This reaction converts an amide ( $R-CONH_2$ ) into a primary amine ( $R-NH_2$ ) with one less carbon atom.

Therefore, the amide 'Q' must be cyclohexanecarboxamide ( $C_6H_{11}CONH_2$ ).

The first step converts 'P' into 'Q' using $NH_3$ and heat.

Carboxylic acids react with $NH_3$ followed by heating to form amides.

Thus, 'P' must be cyclohexanecarboxylic acid ( $C_6H_{11}COOH$ ).

Reaction sequence:

$C_6H_{11}COOH$ (P) $\xrightarrow{NH_3, \Delta} C_6H_{11}CONH_2$ (Q)

$C_6H_{11}CONH_2$ (Q) $\xrightarrow{Br_2, KOH} C_6H_{11}NH_2$ (Cyclohexylamine)

$C_6H_{11}NH_2 \xrightarrow{CHCl_3, KOH} C_6H_{11}NC$ (Cyclohexyl isocyanide)

Comparing with the options, 'P' is cyclohexanecarboxylic acid, which corresponds to option (4).

JEE Main 2026 — Chemistry Organic Chemistry

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JEE Main 2026 — Chemistry Organic Chemistry

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