Question

(i)(ii)(iii)
The correct order of C = O bond length in ethyl propanoate(I), ethyl propenoate(II) and ethenyl propanoate(III) is

• A. I>II>III
• B. III>II>I
• C. I>III>II
• D. II>I>III

Answer 1

The Correct Option is D

Let’s understand the structure and resonance in each compound: 

1. Ethyl propanoate (I): CH₃CH₂COOCH₂CH₃ 
   → No conjugation with the carbonyl (C=O) group. So, it has a typical ester C=O bond length.
2. Ethyl propenoate (II): CH₂=CH–COOCH₂CH₃ 
   → The carbonyl group is conjugated with the adjacent C=C double bond (alkene), which causes resonance. 
   → This partial delocalization lengthens the C=O bond slightly, making it longer than in (I).
3. Ethenyl propanoate (III): CH₃CH₂COOCH=CH₂ 
   → The vinyl group is attached to the –OR part (not conjugated with the C=O group). 
   → But –OCH=CH₂ can still donate some electron density through resonance, increasing delocalization over the ester group.

Now consider the resonance effects:

• Conjugation with the C=O group increases delocalization and results in longer C=O bonds.
• More conjugation → more delocalization → longer C=O bond.

So, the bond length order is:

II (most conjugation, longest C=O) > I (normal ester) > III (least conjugation affecting C=O)

Correct order: II > I > III

Correct answer: Option (D)