Question

Z in the reaction is:
$\text{C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{Cl}\xrightarrow{\text{NaCN}}\text{X}\xrightarrow{\text{Ni/}{{\text{H}}_{2}}}\text{Y}\xrightarrow{\text{Acetic Anhydride}}\text{Z}$,
A.$\text{C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{C}{{\text{H}}_{\text{2}}}\text{NHCOC}{{\text{H}}_{\text{3}}}$
B.$\text{C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{C}{{\text{H}}_{\text{2}}}\text{N}{{\text{H}}_{\text{2}}}$
C.$\text{C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{C}{{\text{H}}_{\text{2}}}\text{CONHC}{{\text{H}}_{\text{3}}}$
D.$\text{C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{C}{{\text{H}}_{\text{2}}}\text{CONHCOC}{{\text{H}}_{\text{3}}}$

Answer 1

Sodium cyanide is used in Organic chemistry to add the cyanide group to any organic molecule while nickel catalyst along with hydrogen gas is used to hydrogenate any compound. Acetic anhydride is used for the acetylation of any compound.

Complete answer:
The given compound in the question is ethyl chloride where the chloride group acts as the leaving group on reaction with sodium cyanide as the ethyl chloride undergoes nucleophilic bimolecular substitution reaction with sodium cyanide to form ethyl cyanide. In the next step, the hydrogen molecules that are absorbed on the surface of the nickel metal undergo bond dissociation to form hydrogen atoms which attach to the cyanide group to form the amine group, due to the reduction of the cyanide group. This leads to the formation of propyl amine and hence one carbon atom in the chain is increased. So cyanide groups are used to increase the number of carbon atoms in the chain of the organic compound. Lastly, the acetylation of the propyl amine by the acetic anhydride leads to the formation of the N-propylacetamide.
The formula for the compound is $\text{C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{C}{{\text{H}}_{\text{2}}}\text{NHCOC}{{\text{H}}_{\text{3}}}$.
the mechanism of the reaction is as follows:
$\text{C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{Cl}\xrightarrow{\text{NaCN}}\text{C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{CN}\xrightarrow{\text{Ni/}{{\text{H}}_{2}}}\text{C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{C}{{\text{H}}_{\text{2}}}\text{N}{{\text{H}}_{\text{2}}}\xrightarrow{\text{Acetic Anhydride}}\text{C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{C}{{\text{H}}_{\text{2}}}\text{NHCOC}{{\text{H}}_{\text{3}}}$

Hence, the correct answer is option A.

Note:
The use of Grignard’s reagent which is alkyl magnesium halide in dry ether is also used to increase the number of carbon atoms in the molecule. In the Grignard’s reagent, the organic group can also be any cyclic compound as well as a compound containing double bond.