The major product 'P' for the following sequence of reactions is :

Solution

Chemistry ma’am · 4 steps

Clemmensen strips the ketone to CH₂ first, then LiAlH₄ turns the amide into CH₂NH₂.

1. Two carbonyls. The starting compound has a ketone next to the benzene ring and an amide, CONH₂, at the other end of the chain. Each reagent attacks a different one.
2. Clemmensen step. Zn/Hg with HCl reduces an aldehyde or ketone C=O all the way to CH₂. It leaves the amide alone, which gives Ph-CH₂CH₂CH₂CH₂-CONH₂.
3. LiAlH4 step. LiAlH₄ reduces an amide by removing its oxygen. The C=O becomes CH₂ and the nitrogen stays, so -CONH₂ becomes -CH₂NH₂. H₃O⁺ only works up the reaction.

   Ph-CO-(CH₂)₃-CONH₂ →(Zn/Hg, HCl) Ph-(CH₂)₄-CONH₂ →(1) LiAlH₄, 2) H₃O⁺) Ph-(CH₂)₅-NH₂

4. So. P is 5-phenylpentan-1-amine. It has no OH group anywhere and an NH₂ at the end of a five-carbon chain, which is option (b).

The idea to keep. Clemmensen turns a ketone C=O into CH₂ and leaves amides alone. LiAlH₄ turns an amide into an amine with one more CH₂: RCONH₂ → RCH₂NH₂. The oxygen goes, the nitrogen stays.

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