Cyclohexylamine when treated with nitrous acid yields (P). On treating (P) with PCC results in (Q). When (Q) is heated with dil. NaOH we get (R) The final product (R) is :

Solution

Chemistry ma’am · 5 steps

Amine to alcohol to ketone, then two cyclohexanones do an aldol and lose water on heating.

1. Amine to alcohol. A primary aliphatic amine with HNO₂ forms an unstable diazonium salt. It loses N₂, and water attacks to give an alcohol. So P is cyclohexanol.
2. Alcohol to ketone. PCC oxidises a secondary alcohol to a ketone and stops there. So Q is cyclohexanone, which has α-hydrogens on both sides of the C=O.
3. Aldol addition. Dilute NaOH removes an α-H from one cyclohexanone. This enolate attacks the C=O carbon of a second cyclohexanone. The product is a β-hydroxy ketone, 2-(1-hydroxycyclohexyl)cyclohexanone. The OH sits on the carbon of the second ring that was its carbonyl carbon.

   2 C₆H₁₀O →(dil. NaOH, Δ) C₁₂H₁₈O + H₂O

4. Heat dehydrates. On heating, the OH leaves together with the α-H. The new C=C forms between the two rings, conjugated with the C=O. This conjugated α,β-unsaturated ketone is the more stable product.
5. So. R is 2-cyclohexylidenecyclohexanone, with an exocyclic double bond joining the two rings next to the C=O. That is option (b).

The idea to keep. An aldol needs an α-H. With heat, the β-hydroxy carbonyl always loses water so the new C=C is conjugated with the C=O. Look for the α,β-unsaturated product.

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