Question

Compound with molecular formula ${\mathrm{C}}_3{\mathrm{H}}_6\mathrm{O}$ can show-

• A. Both positional isomerism and metamerism
• B. Metamerism
• C. Positional isomerism
• D. Functional group isomerism

Answer 1

The correct option is D

Functional group isomerism

Explanation for the correct option:

(D) For a compound with the molecular formula ${\mathrm{C}}_{\mathrm{c}}{\mathrm{H}}_{\mathrm{h}}{\mathrm{X}}_{\mathrm{x}}{\mathrm{N}}_{\mathrm{n}}{\mathrm{O}}_{\mathrm{o}}$

Degree of unsaturation $=(\mathrm{C}+1)-\frac{\mathrm{H}}{2}-\frac{\mathrm{X}}{2}+\frac{\mathrm{N}}{2}$

Where, $\mathrm{X}=\mathrm{Halogen}$

$$\begin{gathered} \mathrm{C}=\mathrm{Number}\mathrm{of}\mathrm{carbon}\mathrm{atom} \\ \mathrm{H}=\mathrm{Number}\mathrm{of}\mathrm{hydrogen}\mathrm{atom} \\ \mathrm{N}=\mathrm{Number}\mathrm{of}\mathrm{nitrogen}\mathrm{atom} \\ \mathrm{O}=\mathrm{Number}\mathrm{of}\mathrm{of}\mathrm{oxygen}\mathrm{atom} \end{gathered}$$

In ${\mathrm{C}}_3{\mathrm{H}}_6\mathrm{O}$

Degree of saturation$=(3+1)-\frac{6}{2}=1$

Degree of saturation$=1$

Possible functional group isomers are ${\mathrm{CH}}_3-{\mathrm{CH}}_2-\mathrm{CH}=\mathrm{O}\left(\mathrm{propanone}\right)\&{\mathrm{CH}}_3-\mathrm{CO}-{\mathrm{CH}}_3\left(\mathrm{propanal}\right)$.

Therefore, functional group isomerism is the most appropriate.

Explanation for incorrect options:

(B) Metamerism has different alkyl chain groups around the functional group.

(C) Positional isomerism arises due to to a difference in the position of the substituent atom or a functional group on the carbon skeleton.

(A) From the above information, we can see that both metamerism and positional isomers are not possible.

Hence, the correct option is (D).