Question

\[C{{H}_{3}}C{{H}_{2}}Cl\xrightarrow{NaCN}X\xrightarrow{Ni/{{H}_{2}}}Y\xrightarrow{\text{acetic anhydride}}Z\]
Z in the above reaction sequence is:
a.) $C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}NHCOC{{H}_{3}}$
b.) $C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}N{{H}_{2}}$
c.) $C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}CONHC{{H}_{3}}$
d.) $C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}CONHCOC{{H}_{3}}$

Answer 1

Hint: To solve this type of question it is important to understand the reaction condition and the type of reaction. Different temperature and pressure lead to formation of different products.

Complete Solution :
Before solving this question, let's understand the reagents involved In the chemical reaction:
The first reagent used is NaCN, sodium cyanide is generally used to add to the aldehydes or ketones to form cyanohydrin. It is generally used in the presence of HCl.
The second reagent used is $Ni/{{H}_{2}}$ . This reagent is generally used to carry out the process of reduction. In the given reaction it will reduce the cyanide group to amine group.
The third reagent used is acetic anhydride. It is a colourless liquid which has a very strong smell similar to that of acetic anhydride.
- In the first step of the reaction $C{{H}_{3}}C{{H}_{2}}Cl$ reacts with NaCN and the chlorine group is replaced with a cyanide group. The reaction involved is mentioned below:
\[C{{H}_{3}}C{{H}_{2}}Cl\xrightarrow{NaCN}C{{H}_{3}}C{{H}_{2}}CN\]

- In the second step $C{{H}_{3}}C{{H}_{2}}CN$ reacts with $Ni/{{H}_{2}}$ which acts as a reducing agent and reduce $C{{H}_{3}}C{{H}_{2}}CN$ to \[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}N{{H}_{2}}\] . the reaction involved is mentioned below:
\[C{{H}_{3}}C{{H}_{2}}CN\xrightarrow{Ni/{{H}_{2}}}C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}N{{H}_{2}}\]

- In the third step \[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}N{{H}_{2}}\] dissolves in acetic anhydride and form $C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}NHCOC{{H}_{3}}$. The reaction involved is mentioned below:
\[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}N{{H}_{2}}\xrightarrow{\text{acetic anhydride}}C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}NHCOC{{H}_{3}}\]
Hence, the correct answer is option (A). i.e. z in the above sequence is $C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}NHCOC{{H}_{3}}$. So, the correct answer is “Option A”.

Note: We can also use potassium cyanide in place of sodium cyanide. They both perform the same reaction and generally dilute HCl is used along with NaCN or KCN in the reaction.