Question:

Given below are two statements.

In the light of the above statements, choose the correct answer from the options given below:

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Key concepts: \begin{itemize} \item Dipole moment depends on charge magnitude and separation distance \item Conjugation affects both electronic distribution and bond lengths \item Partial double bond character decreases bond length \end{itemize}
Updated On: Nov 20, 2025
  • Statement I is false but Statement II is true
  • Both Statement I and Statement II are false
  • Statement I is true but Statement II is false
  • Both Statement I and Statement II are true
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The Correct Option is C

Approach Solution - 1

1. Analysis of Statement I:

  • For \({CH_3-CH=CH-CH=O}\): $\mu = q \times d$
  • Conjugated system creates greater charge separation
  • More distance between charges than in saturated compound

Therefore, greater dipole moment

2. Analysis of Statement II:

  • In \({CH_3-CH=CH-CH=O}\), $C_1-C_2$ has partial double bond character
  • Double bond character means shorter bond length
  • Compared to pure single bond in \chemfig{CH_3-CH_2-CH_2-CH=O}
    Statement II is FALSE
    (actual bond length is shorter)
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Approach Solution -2

We compare two molecules:
(1) \( \mathrm{CH_3-CH=CH-CHO} \) (crotonaldehyde; conjugated enal)
(2) \( \mathrm{CH_3-CH_2-CH_2-CHO} \) (butyraldehyde; saturated aldehyde)

Concept Used:

Conjugation between \( \mathrm{C=O} \) and \( \mathrm{C=C} \) in an enal produces resonance forms that increase charge separation and give partial double-bond character to the \( \mathrm{C_1{-}C_2} \) bond (between the carbonyl carbon \( \mathrm{C_1} \) and the adjacent carbon \( \mathrm{C_2} \)). Thus the \( \mathrm{C_1{-}C_2} \) bond becomes shorter than a normal single bond. The important resonance pair is:

\[ \mathrm{O{=}C{-}C{=}C \ \rightleftharpoons \ O^{-}\!-\!C^{+}{-}C{=}C} \]

The zwitterionic contributor (right) shows larger charge separation, which generally enhances the molecular dipole moment of the conjugated enal relative to the saturated aldehyde.

Step-by-Step Solution:

Step 1 (Dipole comparison): In \( \mathrm{CH_3-CH=CH-CHO} \), the conjugation allows the zwitterionic form \( \mathrm{O^{-}\!-\!C^{+}{-}CH{-}CH_3} \), increasing charge separation. Hence the net dipole of the conjugated aldehyde is larger than that of the non-conjugated \( \mathrm{CH_3-CH_2-CH_2-CHO} \).

\[ \mu\!\left(\mathrm{CH_3-CH=CH-CHO}\right) \;>\; \mu\!\left(\mathrm{CH_3-CH_2-CH_2-CHO}\right) \]

Statement I is true.

Step 2 (Bond-length comparison): Because of the resonance shown above, the \( \mathrm{C_1{-}C_2} \) bond in the enal has partial double-bond character and is therefore shorter than a normal \( \mathrm{C{-}C} \) single bond. In the saturated aldehyde there is no such conjugation, so \( \mathrm{C_1{-}C_2} \) is a normal single bond.

\[ \ell\!\left(\mathrm{C_1{-}C_2\ in\ CH_3-CH=CH-CHO}\right) \;<\; \ell\!\left(\mathrm{C_1{-}C_2\ in\ CH_3-CH_2-CH_2-CHO}\right) \]

Statement II (which claims it is greater) is false.

Final Computation & Result

Correct conclusion: Statement I is true; Statement II is false. Statement II is not a correct explanation of Statement I.

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