Question

The product predominates in?

a)

b)
     

c)
  

d)

Answer 1

Hint: The given question has a benzene group with a functional group which is a carbonyl group. The reagent used is $ LiAl{H_4}\ $ which is a strong reducing agent. It is also abbreviated as $ LAH\ $ in modern organic synthesis.

Complete answer:
In this reaction, the $ C = O $ gets reduced in presence of $ LiAl{H_4}\ $ and the bond breaks. The electron of that bond moves to oxygen where an attack of hydrogen occurs. This way the Oh bond is established. The other hydrogen forms a bond with the carbocation and hence all the electrons are fully filled. we also observe that there is another double bond present, actually this reducing agent can not reduce an isolated non-polar multiple bond like $ c = c $ . Therefore, the other bond will not reduce.
  $ {C_6}{H_5} - CH = CH - CHO\xrightarrow{{LiAl{H_4}}}{C_6}{H_5}C{H_2}C{H_2}C{H_2}OH\ $
Reduction of alpha and beta unsaturated carbonyl compounds generally gives 1,2-reduction product (i.e; double bond or triple bond is not reduced) when a phenyl group is attached to the beta carbon atom of a, b unsaturated carbonyl group, the double bond is also reduced.
 $ {C_6}{H_5} - CH = CH - CHO\xrightarrow{{LiAl{H_4}}}{C_6}{H_5}C{H_2}C{H_2}C{H_2}OH\ $
In such cases reduction of the double bond of allylic alcohol is thought to proceed through a cyclic organoaluminum compound. The cyclic intermediate reacts with allyl alcohol to give the product.
 $ 2{C_6}{H_5} - CH = CH - CHO\xrightarrow{{LiAl{H_4}}}{({C_6}{H_5}C{H_2}C{H_2}C{H_2}O)_2}Al{H_2}\ $
So option, C is correct.

Note:
 $ LiAl{H_4}\ $ is a nucleophilic reducing agent, best used to reduce polar multiple bonds like present in the carbonyl group. This group reduces the aldehydic group into primary alcohol, amides and nitriles into amines, carboxylic acids and esters to primary alcohols, and ketones into secondary alcohols.