Question

Ethyl isocyanide on hydrolysis with dilute sulphuric acid gives:
This question has multiple correct options
A) ethylamine
B) propanoic acid
C) formic acid
D) methylamine

Answer 1

Hint: Isocyanide group is hydrolysed. Two molecules of water add nitrogen and carbon atoms to the isocyanide group.

 Complete step by step answer:
Alkyl isocyanide is represented by the general formula $$\text{R}-\text{NC}$$ whereas an alkyl amine is represented by the general formula $$\text{R}-\text{N}{\text{H}}_2$$.
Alkyl isocyanide contains carbon-nitrogen triple bond. An alkyl group is attached to the nitrogen atom. Thus, in alkyl isocyanide, nitrogen atom forms bonds with two carbon atoms. In alkyl amine, carbon-nitrogen single bond is present between the carbon atom of the alkyl group and nitrogen atom of the amino group.
The hydrolysis of alkyl isocyanides in presence of dilute sulphuric acid gives alkyl amine and formic acid.

$$\text{R}-\text{NC + }{\text{H}}_2\text{O }\stackrel{\text{dil }{\text{H}}_{\text{2}}\text{S}{\text{O}}_4}{\to }\text{ R}-\text{N}{\text{H}}_2\text{ + HCOOH}$$

Similarly, the hydrolysis of ethyl isocyanide $$\left({\text{C}}_2{\text{H}}_5-\text{NC}\right)$$ in presence of dilute sulphuric acid gives ethyl amine $$\left({\text{C}}_2{\text{H}}_5-\text{N}{\text{H}}_2\right)$$ and formic acid $$\left(\text{ HCOOH}\right)$$.

$${\text{C}}_2{\text{H}}_5-\text{NC + }{\text{H}}_2\text{O }\stackrel{\text{dil }{\text{H}}_{\text{2}}\text{S}{\text{O}}_4}{\to }{\text{C}}_2{\text{H}}_5-\text{N}{\text{H}}_2\text{ + HCOOH}$$

In the above reactions, the isocyanide group is transformed into an amino group.

Thus, option A and C are the correct answers.
Additional information: Formation of propanoic acid $$\left({\text{C}}_2{\text{H}}_5-\text{COOH}\right)$$ is not possible in this reaction. Methyl amine $$\left(\text{C}{\text{H}}_3-\text{N}{\text{H}}_2\right)$$ is obtained on the hydrolysis of methyl isocyanide in presence of dilute sulphuric acid.

Note:
As we know that, Ethyl isocyanide contains three carbon atoms. During hydrolysis, one carbon atom is lost in the form of formic acid. Thus, the product ethylamine has only two carbon atoms. Hence, you can simply solve the question by finding the option, in which the number of carbon atoms is one less than that in ethyl isocyanide.