The final product[B] is:

Solution

Chemistry ma’am · 4 steps

Benzoyl chloride makes an amide, then LiAlH4 turns that C=O into CH2, not into CH–OH.

1. Benzoylation. The starting amine is C₆H₁₁CH₂NH₂. With C₆H₅COCl and aqueous NaOH (Schotten–Baumann reaction), the NH₂ is acylated and NaOH mops up the HCl formed.
2. Product A. [A] is the amide C₆H₅CO-NH-CH₂C₆H₁₁, that is N-(cyclohexylmethyl)benzamide. The C=O sits next to the benzene ring because the benzoyl group came from the acid chloride.

   C₆H₅CONHCH₂C₆H₁₁ →(LiAlH₄, H₂O) C₆H₅CH₂NHCH₂C₆H₁₁

3. Amide reduction. LiAlH₄ reduces an amide all the way to an amine. The whole C=O becomes CH₂ and the oxygen leaves, so no OH group is left behind.
4. So. [B] is the secondary amine N-benzyl-1-cyclohexylmethanamine, C₆H₅CH₂NHCH₂C₆H₁₁, which is option (a).

The idea to keep. LiAlH₄ on an amide always gives an amine: the C=O becomes CH₂ and the nitrogen stays attached. Acylate first, then reduce, and you have added an alkyl group to nitrogen.

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