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:

Resonance increases bond strength, and chirality is preserved if substitution does not cause racemization.

Answer 1

The Correct Option is C

To determine the correctness of the given statements, let's analyze each one separately:

1. Statement I: \(C–Cl\) bond is stronger in \(\mathrm{CH_2 = CH{-}Cl}\) than in \(\mathrm{CH_3{-}CH_2{-}Cl}\).
   • The molecule \(\mathrm{CH_2 = CH{-}Cl}\) has a sp² hybridized carbon (due to the double bond), whereas \(\mathrm{CH_3{-}CH_2{-}Cl}\) has a sp³ hybridized carbon.
   • In sp² hybridization, the s-character is higher compared to sp³ hybridization. This increased s-character leads to a stronger and shorter bond.
   • Therefore, the \(C–Cl\) bond in \(\mathrm{CH_2 = CH{-}Cl}\) is stronger than in \(\mathrm{CH_3{-}CH_2{-}Cl}\).
   • Conclusion: Statement I is true.
2. Statement II: The given optically active molecule, on hydrolysis, gives a solution that can rotate plane polarized light.
   • An optically active compound contains a chiral center and can rotate plane-polarized light.
   • Upon hydrolysis, the optical activity depends on the configuration of the products. In many cases, if a compound remains chiral after the reaction, the product would still be optically active.
   • Thus, if the molecule is initially optically active, it's likely that one or more hydrolysis products will also be optically active, thereby rotating plane-polarized light.
   • Conclusion: Statement II is true.

Based on our analysis, we can conclude that both Statement I and Statement II are true. Therefore, the correct answer is:

Both Statement I and Statement II are true