Skip to content
Question of 135

Q.Draw the resonating structure of phenol.

Rajasthan RbseRajasthan Board Senior Secondary Examination 2025Subjective· 2mImportance★★★★★
0% · 0/135 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Resonance in phenol arises from delocalisation of an oxygen lone pair into the aromatic ring, generating structures with negative charge at ortho/para carbons and positive charge on oxygen; this both weakens the O-H bond (raising acidity) and activates the ring toward electrophiles at o/p positions.

Phenol's oxygen has two lone pairs; one of them lies in a p-orbital that can overlap with (delocalise into) the pi-electron system of the benzene ring. This gives rise to five resonance (canonical) contributing structures, described here in place of a drawing:

Structure I (the standard Kekule/Lewis structure): C6H5-O-H, with both lone pairs localised on O and alternating double bonds in the ring, no charges.

Structures II & III: delocalisation of the O lone pair into the ring places a negative charge at the ORTHO carbons (two equivalent structures, one for each ortho position) and a corresponding positive charge on the oxygen (O+=). The C-O bond gains partial double-bond character.

Structure IV: delocalisation places the negative charge at the PARA carbon, again with positive charge on oxygen.

(No resonance structure places negative charge at the meta position, since that would require breaking rather than following the conjugated pi system.)

Consequences of this resonance:

  1. The C-O bond in phenol gets partial double-bond character (shorter/stronger than a typical alcohol C-O bond). …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.