Question

Given below are two statements:
Statement I: C–Cl bond is stronger in $\mathrm{CH_2 = CH{-}Cl}$ than in $\mathrm{CH_3{-}CH_2{-}Cl}$.
Statement II: The given optically active molecule, on hydrolysis, gives a solution that can rotate the plane polarized light.
In the light of the above statements, choose the correct answer from the options given below:

• A. Both Statement I and Statement II are false
• B. Statement I is true but Statement II is false
• C. Both Statement I and Statement II are true
• D. Statement I is false but Statement II is true
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Hint:

For Statement I, compare the percentage of s-character in the hybrid orbital used for the C-Cl bond in each molecule. For Statement II, think about what an SN2 backside attack does to a chiral centre during hydrolysis: the configuration flips, but the product does not stop being chiral.

Answer 1

The Correct Option is C

Step 1: Analysing 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, making it stronger than the C–Cl bond in ethyl chloride $\mathrm{(CH_3{-}CH_2{-}Cl)}$, where carbon is sp$^3$ hybridized.
Hence, Statement I is true.
Step 2: Analysing Statement II.
The given molecule is optically active. On hydrolysis, substitution occurs without destroying chirality, producing a chiral alcohol. Hence, the resulting solution remains optically active and can rotate plane polarized light.
Thus, Statement II is true.
Step 3: Final conclusion.
Both Statement I and Statement II are true, corresponding to option (3).

Answer 2

Concept:

• Hybrid orbitals with more s-character hold electrons closer to the nucleus, giving shorter and stronger bonds. Percentage s-character: $sp = 50\%$, $sp^2 = 33\%$, $sp^3 = 25\%$.
• In a nucleophilic substitution at a chiral carbon proceeding by the $S_N2$ pathway, the nucleophile attacks from the side opposite the leaving group, flipping the spatial arrangement of the other three groups (Walden inversion).
• Walden inversion changes which enantiomer is formed, but the product is still a single, chiral compound.

Step 1: Identify the hybridization of the carbon bonded to chlorine in each compound.
In $CH_2=CH-Cl$, the carbon attached to $Cl$ is part of a double bond, so it is $sp^2$ hybridized.
In $CH_3-CH_2-Cl$, the carbon attached to $Cl$ has only single bonds, so it is $sp^3$ hybridized.

Step 2: Compare bond strength using percentage s-character.
The $sp^2$ carbon has $33\%$ s-character, more than the $25\%$ s-character of the $sp^3$ carbon.
Higher s-character pulls the bonding electron pair closer to the carbon nucleus, giving a shorter and stronger $C-Cl$ bond.
So the $C-Cl$ bond in $CH_2=CH-Cl$ is stronger than in $CH_3-CH_2-Cl$, confirming Statement I.

Step 3: Analyze what happens to chirality during hydrolysis.
Hydrolysis of a chiral alkyl halide by the $S_N2$ pathway replaces the leaving group with $-OH$ through a single backside attack.
This attack inverts the spatial arrangement at that carbon, but all four groups attached to it remain different from one another.

Step 4: Conclude on the optical activity of the product.
Since the product carbon still carries four different groups, the resulting alcohol is still chiral, just with the configuration flipped relative to the starting molecule.
A single chiral compound in solution rotates plane polarized light, so the solution formed is optically active, confirming Statement II.

Final Answer: Both Statement I and Statement II are true.