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:

Question

The % increase in oxygen in steam volatile product with respect to phenol is ____ $10^{-1}\%$.

Answer 1

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

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.
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

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