Question

Predict the alkanes obtained by
-The action of sodium metal on n-propyl chloride
-catalytic hydrogenation of isobutylene.

Answer 1

In the first case, wurtz reaction takes place in which two alkyl chloride combines to form a single alkane, hexene.
Double bond breaks to form a single bond due to addition of hydrogen.

Complete step by step answer:
When sodium metal reacts with n-propyl chloride, it results in a formation of n-hexane and the reaction is known as wurtz reaction.
$2{C_3}{H_7}Cl\xrightarrow{{Na}}{C_6}{H_{14}}$
Here in the given reaction two alkyl chloride combines and forms a single alkane ( hexane ) as a major product.
Thus the above reaction leads to the formation of hexane.
Catalytic hydrogenation of isobutylene leads to the formation of isobutene as a major product.
$CH = {C_3}{H_5}\xrightarrow[{Pt}]{{{H^{^ + }}/Pd}}{C_4}{H_8}$
In the given reaction there is addition of hydrogen in place of alkene bond.
Thus the above reaction leads to the formation of isobutene.

Additional Information:
Wurtz reaction: It is a coupling reaction of two compounds in organic chemistry, in which two alkyl halides in reaction with sodium metal in solution of dry ether forms a higher alkane. In this reaction sodium metal is treated in a solution of dry ether to produce higher alkanes.
Catalytic hydrogenation: Catalytic hydrogenation is addition of hydrogen to a compound with a double or triple bond with the help of catalyst. The given compound gets reduced and leaves lesser bonds between the carbons. The process of catalytic hydrogenation uses a metal catalyst such as palladium, nickel or platinum.

Note:
It is a simple wurtz reaction. You just need to have clear knowledge of organic chemistry.
Hydrogen, platinum and palladium just act as catalysts. Don’t include them in the main reaction.