In the reaction given below:
The product 'X' is:

Solution

Chemistry ma’am · 3 steps

Go group by group: LiAlH4 reduces the ketone and the amide, but it leaves the C=C alone.

1. Spot the groups. The starting compound has three groups that could react: a ketone (C=O), a C=CH₂ double bond next to it, and a primary amide (-CONH₂).

   Group	With LiAlH₄, then H₃O⁺	In the product
   Ketone >C=O	Hydride adds to the carbonyl carbon and workup protonates the O	secondary alcohol >CH-OH
   Amide -CONH₂	The carbonyl oxygen is removed completely	amine -CH₂NH₂
   Alkene >C=CH₂	Hydride is a nucleophile and goes to the polar C=O, not to the C=C	stays as >C=CH₂

2. Amide to amine. After the first hydride adds, the oxygen leaves bound to aluminium and an iminium ion forms. A second hydride reduces it, so that carbon ends up as CH₂ next to NH₂.
3. So. X is CH₃CH₂CH(OH)-C(=CH₂)-CH(CH₃)-CH₂NH₂: an alcohol, an untouched C=CH₂, and an amine. That is option (a).

The idea to keep. LiAlH₄ turns ketones into alcohols and amides into amines, removing the amide oxygen fully. Its hydride attacks C=O groups, so a C=C usually survives.

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