JEE Chemistry 2026 Question with Solution

Given: Two statements are to be checked for correctness.

Find: Which option correctly identifies the truth values of Statement I and Statement II.

Statement I: In vinyl chloride $\mathrm{(CH_2=CH{-}Cl)}$, the carbon bonded to chlorine is $sp^2$ hybridized. The C–Cl bond has partial double bond character due to resonance, so this bond is stronger than the C–Cl bond in ethyl chloride $\mathrm{(CH_3{-}CH_2{-}Cl)}$, where the carbon is $sp^3$ hybridized.

Therefore, Statement I is true.

Statement II: The given molecule is optically active. On hydrolysis, substitution occurs without destroying chirality, so a chiral alcohol is formed. Hence, the resulting solution remains optically active and can rotate plane polarized light.

Therefore, Statement II is true.

So, both Statement I and Statement II are true. The correct option is C.

Given:

• Statement I compares the strength of the C–Cl bond in $\mathrm{CH_2=CH{-}Cl}$ and $\mathrm{CH_3{-}CH_2{-}Cl}$.
• Statement II asks whether hydrolysis of the given optically active molecule gives a solution that can rotate plane polarized light.

Find: The correct option.

1. Analyse Statement I

In vinyl chloride $\mathrm{CH_2=CH{-}Cl}$, the chlorine atom is attached to an $sp^2$ hybridized carbon.

Because of resonance, the C–Cl bond acquires partial double bond character.

This makes the bond shorter and stronger than the C–Cl bond in ethyl chloride $\mathrm{CH_3{-}CH_2{-}Cl}$, where chlorine is attached to an $sp^3$ hybridized carbon.

So, Statement I is true.

2. Analyse Statement II

The molecule is stated to be optically active.

On hydrolysis, substitution gives a chiral alcohol without loss of chirality.

Therefore, the product solution remains optically active and can rotate plane polarized light.

So, Statement II is true.

3. Conclude

Both statements are true.

Hence, the correct option is C, that is, option (3).