JEE Chemistry 2023 Question with Solution

Given: Compound A has molecular formula $C_{10}H_{12}O_2$. It gives neutral FeCl$_3$ test positive, reacts with NaOH and CH$_3$Br to give $C_{11}H_{14}O_2$, gives methyl iodide with hydroiodic acid, and decolourises alkaline KMnO$_4$.

Find: The number of $\pi$ bonds present in compound A.

A positive neutral FeCl$_3$ test indicates the presence of a phenolic group. Decolourisation of alkaline KMnO$_4$ indicates the presence of a double bond.

Treatment with NaOH and CH$_3$Br gives $C_{11}H_{14}O_2$, which supports methylation. Formation of methyl iodide with hydroiodic acid confirms the presence of a methoxy ether group.

Thus, the structure contains one aromatic benzene ring and one aliphatic double bond.

Therefore, the number of $\pi$ bonds present in compound A is $4$.

Given: Compound A is a trisubstituted compound with formula $C_{10}H_{12}O_2$.

Find: Total number of $\pi$ bonds in A.

From the observations in the solution:

1. Neutral FeCl$_3$ test positive $\Rightarrow$ phenolic $-OH$ group is present.
2. Reaction with NaOH and CH$_3$Br gives $C_{11}H_{14}O_2$ $\Rightarrow$ methylation occurs.
3. Hydroiodic acid gives methyl iodide $\Rightarrow$ a methoxy $(-OCH_3)$ group is present.
4. Alkaline KMnO$_4$ is decolourised $\Rightarrow$ an additional double bond is present.

Hence, compound A contains:

• one benzene ring, contributing $3$ $\pi$ bonds,
• one extra aliphatic double bond, contributing $1$ $\pi$ bond.

Therefore, the required number of $\pi$ bonds is $4$.