The major product of the above reactions is :

Solution

Chemistry ma’am · 5 steps

Here OH⁻ is only a base: it removes the benzylic H, and that carbanion adds to cyclohexanone.

1. Cyanide substitution. CN⁻ replaces the benzylic bromide by S_N2, giving ArCH₂CN with Ar = 4-methoxyphenyl. The chain is now one carbon longer.
2. Make the carbanion. The CH₂ between the ring and C≡N is acidic, because both groups stabilise the negative charge. OH⁻ removes one H to give the carbanion ArC(-)H-CN.
3. Nucleophilic addition. This carbanion attacks the carbonyl carbon of cyclohexanone. The alkoxide picks up a proton, so the product is a tertiary alcohol: the ring carbon now carries OH and the CH(Ar)CN group.

   ArCH₂CN →(OH⁻) C₆H₁₀(OH)-CH(Ar)-CN →(H₂, Ni) C₆H₁₀(OH)-CH(Ar)-CH₂NH₂

4. Reduce the nitrile. H₂/Ni reduces -C≡N to -CH₂NH₂. It leaves the tertiary OH and the aromatic ring untouched.
5. So. The major product is 1-[2-amino-1-(4-methoxyphenyl)ethyl]cyclohexanol, with OH on the ring carbon and NH₂ on a terminal CH₂. That is option (d).

The idea to keep. A CH₂ sitting between an aryl ring and a CN group is acidic. With base it becomes a carbanion that adds to C=O, and H₂/Ni later turns the CN into CH₂NH₂.

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