JEE Main Chemistry Question (2026) — Solution

Step-by-step 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) NH_3, C_6H_ 11 CONH_2 (Q) C_6H_ 11 CONH_2 (Q) Br_2, KOH C_6H_ 11 NH_2 (Cyclohexylamine) C_6H_ 11 NH_2 CHCl_3, KOH C_6H_ 11 NC (Cyclohexyl isocyanide) Comparing with the options, 'P' is cyclohexanecarboxylic acid, which corresponds to option (4).