Question

Can $H_2/Pd/C$ reduce both double bond and carbonyl group on benzene or only carbonyl group?
The compound is:
 $C_6{H}_5-CH=CH-CO-OC{H}_3$

Answer 1

Hint: Chemical compounds are classified on the basis of functional groups. Alkenes are double bonded compounds that can be reduced by $H_2/Pd/C$ to alkanes. Carbonyl group means $C=O$ which cannot be reduced by $H_2/Pd/C$ , but can be reduced by $H_2/Pd/C$ when directly attached to any aromatic ring like benzene.

Complete answer:
Given compound is $C_6{H}_5-CH=CH-CO-OC{H}_3$ . It consists of an ester functional group along with the double bond between the carbon and carbon. Thus, alkenes and ester were the functional groups present in the given compound.
Alkenes can be reduced to alkanes by the action of hydrogenation in presence of palladium catalyst over the surface of charcoal. Thus, $H_2/Pd/C$ can reduce the double bond.
The carbonyl group means the carbon is bonded with an oxygen atom with a double bond. $H_2/Pd/C$ cannot reduce the carbonyl group in general. But it can reduce a carbonyl group when the carbonyl group is directly attached to any aromatic rings like benzene. In the given compound, the carbonyl group is not directly attached to the aromatic ring. It is present in the side chain.
Thus, $H_2/Pd/C$ cannot reduce both the double bond and carbonyl group. But reduce the double bond only.

Note:
Aromatic compounds mean the compounds which are cyclic, planar, conjugation of pi-electrons and obeying Huckel’s rule. Huckel’s rule is given by $\left(4n+2\right)\pi$ electrons. where n is any whole number. Benzene satisfies all the above conditions and can be considered as an aromatic compound.