Given below are two statements:
Statement I: The dipole moment of $$\overset{4}{\mathrm{CH_3}} - \overset{3}{\mathrm{CH}} = \overset{2}{\mathrm{CH}} - \overset{1}{\mathrm{CH}}=O$$ is greater than $$\overset{4}{\mathrm{CH_3}} - \overset{3}{\mathrm{CH_2}} - \overset{2}{\mathrm{CH_2}} - \overset{1}{\mathrm{CH}}=O$$.
Statement II: $$C_1-C_2$$ bond length of $$\underset{4}{\mathrm{CH_3}} - \underset{3}{\mathrm{CH}} = \underset{2}{\mathrm{CH}} - \underset{1}{\mathrm{CH}}=O$$ is greater than $$C_1-C_2$$ bond length of $$\underset{4}{\mathrm{CH_3}} - \underset{3}{\mathrm{CH_2}} - \underset{2}{\mathrm{CH_2}} - \underset{1}{\mathrm{CH}}=O$$.
In the light of the above statements, choose the correct answer from the options given below

Statement I compares the dipole moments of two aldehydes.
  • $$CH_3-CH=CH-CH=O$$ is an α,β-unsaturated (conjugated) aldehyde.
  • $$CH_3-CH_2-CH_2-CH=O$$ is a saturated aldehyde.

Dipole moment ( $$\mu$$ ) depends on the magnitude of charge separation and on the distance between the centres of positive and negative charge.
  • In the conjugated aldehyde the $$C=O$$ group is conjugated with the $$C=C$$ bond. By resonance the electron pair of the $$C=C$$ bond can shift towards the oxygen, giving forms such as
    $$CH_3-CH=CH-CH^{+}-O^{-}$$.
  • This resonance increases the positive character on the carbonyl carbon and the negative character on the oxygen, thereby increasing the $$C=O$$ dipole moment.
  • Any small dipole associated with the $$C=C$$ bond is approximately parallel to the carbonyl dipole, so the vectors add rather than cancel.

Therefore $$\mu\bigl(CH_3-CH=CH-CH=O\bigr) \gt \mu\bigl(CH_3-CH_2-CH_2-CH=O\bigr)$$ and Statement I is true.

Statement II compares the $$C_1-C_2$$ bond lengths.

• In $$CH_3-CH_2-CH_2-CH=O$$ both $$C_1$$ and $$C_2$$ are $$sp^3$$ hybridised. An $$sp^3-sp^3$$ single bond length is about $$1.54\ \text{Å}$$.
• In $$CH_3-CH=CH-CH=O$$, $$C_1$$ is $$sp^3$$ but $$C_2$$ is $$sp^2$$ (because it is part of the $$C=C$$ bond). An $$sp^3-sp^2$$ single bond has more $$s$$-character on one side and is therefore shorter (≈ $$1.50\ \text{Å}$$) than an $$sp^3-sp^3$$ bond.

Thus $$\text{bond length}(C_1-C_2)\bigl(CH_3-CH=CH-CH=O\bigr)$$ is smaller, not greater, than that in $$CH_3-CH_2-CH_2-CH=O$$. Statement II is false.

So, Statement I is true while Statement II is false. The correct choice is Option C.