JEE Chemistry 2026 Question with Solution

Given: Compound $X$ has molecular formula $\mathrm{C_3H_6Cl_2}$. On treatment with excess $\mathrm{NaNH_2}$ it gives compound $Y$. Compound $Y$ on hydration with dilute $\mathrm{H_2SO_4}$ and $\mathrm{Hg^{2+}}$ gives compound $Z$, and on passing through a red hot iron tube gives compound $Q$.

Find: Whether Statement I and Statement II are true.

From the given reaction sequence, $X = \mathrm{C_3H_6Cl_2}$ is a vicinal or geminal dihalide. Excess $\mathrm{NaNH_2}$ causes double dehydrohalogenation, so compound $Y$ is propyne.

Propyne undergoes acid-catalysed hydration in the presence of $\mathrm{Hg^{2+}}$. Hydration of a terminal alkyne gives a methyl ketone. Therefore compound $Z$ is acetone.

Acetone gives a yellow precipitate of iodoform with NaOI. Hence Statement I is true.

When propyne is passed through a red hot iron tube, it undergoes cyclotrimerisation to form mesitylene, that is $1,3,5$-trimethylbenzene.

In mesitylene, the number of aromatic hydrogens is $3$ and the number of aliphatic hydrogens from three $\mathrm{CH_3}$ groups is $9$.

Hence Statement II is true.

Therefore, both Statement I and Statement II are true. The correct option is A.

Given: The sequence starts from a dihalide $\mathrm{C_3H_6Cl_2}$ and forms $Y$ with excess $\mathrm{NaNH_2}$.

Find: The truth values of the two statements.

Use standard reaction patterns.

1. Excess $\mathrm{NaNH_2}$ on a dihalide gives an alkyne, so $Y$ is propyne.
2. Terminal alkyne hydration with dilute $\mathrm{H_2SO_4}$ and $\mathrm{Hg^{2+}}$ gives a methyl ketone, so $Z$ is acetone, which gives the iodoform test.
3. Cyclotrimerisation of propyne in a red hot iron tube gives mesitylene, which has $3$ aromatic and $9$ aliphatic hydrogens.

Thus both statements are true, so the correct option is A.