Question

The CORRECT order of carbonyl stretching frequencies for the following compounds is 

• A. II < I < III < IV
• B. I < III < II < IV
• C. IV < II < III < I
• D. III < IV < II < I

Hint:

The carbonyl stretching frequency increases with electron-donating groups and decreases with electron-withdrawing groups.

Answer 1

The Correct Option is D

Factors affecting carbonyl stretching frequency:

The C=O stretching frequency depends on:

1. Electron-withdrawing groups: Increase frequency (stronger C=O bond)
2. Electron-donating groups: Decrease frequency through resonance (weaker C=O bond)
3. Resonance delocalization: Decreases frequency (C=O has partial single bond character)

Analyzing each compound:

I - Benzoyl chloride (PhCOCl):

• Cl is electronegative and electron-withdrawing by induction
• Minimal resonance donation from Cl (poor overlap between Cl lone pairs and C=O π*)
• High carbonyl frequency (~1770-1790 cm⁻¹)

II - Phenyl acetate (PhOCOCH₃):

• Oxygen has two effects:
  • Electron-withdrawing by induction
  • Strong resonance donation from O lone pairs into C=O
• Resonance effect dominates, significantly reducing C=O bond order
• Lower frequency (~1760-1770 cm⁻¹)

III - N,N-Dimethylbenzamide (PhCON(CH₃)₂):

• Nitrogen has strong resonance donation into C=O
• Creates significant C-O single bond character
• Amides have the lowest carbonyl frequencies among these compounds
• Very low frequency (~1630-1680 cm⁻¹)

IV - Benzaldehyde (PhCHO):

• No heteroatom directly attached to carbonyl
• Conjugation with phenyl ring slightly lowers frequency
• Moderate frequency (~1690-1710 cm⁻¹)

Ordering by C=O stretching frequency (lowest to highest):

• III (amide): ~1650 cm⁻¹ - strongest resonance from N
• IV (aldehyde): ~1700 cm⁻¹ - no heteroatom effect
• II (ester): ~1765 cm⁻¹ - resonance from O
• I (acid chloride): ~1780 cm⁻¹ - weak resonance, inductive effect dominates

Order: III < IV < II < I

Answer: (D) III < IV < II < I