Question

What is the product of the following reaction?

A.

B.

C.

D.

Answer 1

Hint: In the reaction, the reactants provided to us is an unsaturated ketone. The reagent given is lithium aluminum hydride \[\left( LiAl{{H}_{4}} \right)\]. \[\left( LiAl{{H}_{4}} \right)\]is a very strong reducing reagent. It is generally a grey solid and is widely used to reduce different carboxyl groups from one form to the other.

Complete step-by-step answer: Lithium Aluminum hydride is majorly used in organic chemistry as a strong reducing agent. It is more strong and powerful as compared with the sodium borohydride. This is because the Al-H bond is generally weak as compared with the B-H bond.
In diether solution it reduces esters, carboxylic acids, acyl chlorides, aldehydes and ketones to the corresponding alcohols. Similarly, it converts amide, nitro, nitrile, oxime and imine into the amines. It also reduces the quaternary ammonium cation to the corresponding tertiary amines.
In the question, it is given that the ketone with a double bond is treated with the lithium aluminum hydride. In this reaction the lithium aluminum hydride will act as a strong reducing agent and thus converts the ketones into the alcohol (C=O group changes to C-OH).
The reaction will occur as follows:

Lithium Aluminum hydride also reduces alkyl halides to the alkanes. This should be noted that here the reagent will not reduce the double bonds. The double bonds will remain as it is. Therefore, these reagents are reduced in nature but they do not reduce the double bonds.

Hence, the correct option is option B.

Note: It should be noted that $LiAl{{H}_{4}}$is more reactive than $NaB{{H}_{4}}$. $NaB{{H}_{4}}$reduces only aldehydes, ketones, acid chlorides, alkyl halides. On the other hand, $LiAl{{H}_{4}}$reduces aldehydes, ketones, acid chlorides, alkyl halides, esters, amides, nitriles, nitro compounds etc. thus it is more reactive and more expensive than the sodium borohydride.