Question

$C{H}_{3}CHO$ $\underset{\left(ii\right)H_{2}O}{\overset{\left(i\right)C{H}_{3}MgBr}{\to }}$ $\left(A\right)$ $\stackrel{H_{2}S{O}_4,\Delta }{\to }$ $\left(B\right)$ $\underset{Oxidation}{\overset{Hydroboration}{\to }}\left(C\right)$

Compound $A$ and $C$ in the given reaction are:

• A. Identical
• B. Positional Isomer
• C. Functional Isomers
• D. Optical Isomers

Answer 1

Answer: (B)

Positional Isomer

Acetaldehyde reacts with methyl magnesium bromide followed by hydrolysis to give propan-2-ol (compound A). This on heating with sulphuric acid undergoes dehydration to form 1-propene (compound B).

Hydroboration oxidation gives propan-1-ol (compound C).
Thus, compounds A and C are positional isomers.

Hence, the correct option is $\text{B}$