JEE Chemistry 2026 Question with Solution

Given: A hydrocarbon P with formula $\text{C}_4\text{H}_8$ reacts with HCl to give optically active Q with formula $\text{C}_4\text{H}_9\text{Cl}$. Then Q reacts with one mole of ammonia to give R with formula $\text{C}_4\text{H}_{11}\text{N}$. Finally, R on diazotization followed by hydrolysis gives S.

Find: Identify P, Q, R and S.

The key clue is that Q is optically active, so it must contain a chiral carbon.

From $\text{C}_4\text{H}_8$, possible alkene isomers include but-1-ene, but-2-ene, and 2-methylpropene.

• But-1-ene + HCl can give $\text{CH}_3\text{CH}_2\text{CH(Cl)CH}_3$, which is 2-chlorobutane and is chiral.
• But-2-ene + HCl also gives $\text{CH}_3\text{CH}_2\text{CH(Cl)CH}_3$, which is 2-chlorobutane and is chiral.
• 2-Methylpropene + HCl gives tert-butyl chloride, which is not chiral.

So Q must be 2-chlorobutane, with structure $\text{CH}_3\text{CH}_2\text{CH(Cl)CH}_3$.

Now ammonia replaces the $\text{-Cl}$ group by $\text{-NH}_2$, so R is butan-2-amine:

On diazotization followed by hydrolysis, a primary aliphatic amine converts into the corresponding alcohol. Therefore S is butan-2-ol:

Matching these with the options, option C is the only complete match.

Therefore, the correct option is C.

Given: The product Q formed after addition of HCl is optically active.

Find: Which sequence matches all four compounds.

Use the stereochemical clue first.

1. If P were 2-methylpropene, addition of HCl would give tert-butyl chloride, which has no chiral center. So this possibility is rejected.
2. For an optically active chloride, Q must be 2-chlorobutane, $\text{CH}_3\text{CH}_2\text{CH(Cl)CH}_3$.
3. Reaction of Q with one mole of ammonia gives butan-2-amine, $\text{CH}_3\text{CH}_2\text{CH(NH}_2\text{)CH}_3$.
4. Diazotization followed by hydrolysis converts that amine into butan-2-ol, $\text{CH}_3\text{CH}_2\text{CH(OH)CH}_3$.
5. Among the options, only C lists Q, R, and S correctly, with P as but-2-ene.

Therefore, the correct option is C.