Compound 'P ' undergoes the following sequence of reactions:
'P^' is:

Solution

Chemistry ma’am · 4 steps

Work backwards from the isocyanide: undo carbylamine, then Hofmann, then reach the amide's parent acid.

1. Undo carbylamine. The final product is cyclohexyl isocyanide, C₆H₁₁-NC. CHCl₃ with alcoholic KOH turns a primary amine into an isocyanide, so the amine before it was cyclohexylamine, C₆H₁₁NH₂.
2. Undo Hofmann. Br₂ with KOH is the Hofmann bromamide reaction. It turns an amide into an amine with one carbon fewer. So Q is the amide with one more carbon than cyclohexylamine: cyclohexanecarboxamide, C₆H₁₁CONH₂.

   C₆H₁₁CONH₂ + Br₂ + 4KOH → C₆H₁₁NH₂ + K₂CO₃ + 2KBr + 2H₂O

3. Undo ammonia and heat. A carboxylic acid reacts with NH₃ to give its ammonium salt, and heating that salt removes water to give the amide: C₆H₁₁COOH →(NH₃) C₆H₁₁COO⁻NH₄⁺ →(Δ) C₆H₁₁CONH₂.
4. So. P is cyclohexanecarboxylic acid, C₆H₁₁COOH, which is option (a).

The idea to keep. In a backwards chain, track the carbon count. Hofmann bromamide removes one carbon. Carbylamine keeps the R group of the amine. Acid plus NH₃, then heat, gives the amide.

Answer from the NTA final answer key. Solution by PrepWiser.