Acidity of alcohols and phenols; esterification
Alcohols and phenols lose H⁺ from O-H. Phenol is more acidic because the phenoxide ion is resonance stabilised. Electron-withdrawing groups raise acidity; electron-releasing groups lower it.
Reactions with active metals and with bases
- 12CH₃CH₂OH + 2Na → 2CH₃CH₂ONa + H₂Sodium ethoxide forms; hydrogen gas bubbles off
- 22C₆H₅OH + 2Na → 2C₆H₅ONa + H₂Sodium phenoxide forms
- 3★C₆H₅OH + NaOH → C₆H₅ONa + H₂OPhenol reacts with aqueous NaOH; alcohols do not. This shows phenol is the stronger acid
- 4R-O⁻ + H-OH → R-OH + OH⁻An alkoxide takes a proton from water, so alcohols are weaker acids than water, and alkoxides are stronger bases than OH⁻
★ key step
Why phenol is more acidic than ethanol
In phenol, the -OH is on an sp² carbon, which pulls electrons more than an sp³ carbon, so the O-H bond is more polar. More importantly, look at the ions left after H⁺ leaves. In the phenoxide ion, the negative charge is delocalised (spread) over the oxygen and the ring by resonance, which makes the ion stable. In the ethoxide ion, the charge stays on oxygen alone, and the ethyl group pushes in more electron density (+I effect). A more stable anion means a stronger acid, so phenol (pKa 10.0) is far more acidic than ethanol (pKa 15.9).
How to draw the resonance structures of the phenoxide ion
- Draw the phenoxide ion with -O⁻ on the ring, in both Kekulé forms (two structures).
- Draw a curved arrow from a lone pair on O⁻ to make C=O. Move a ring π bond so the negative charge lands on an ortho carbon.
- Repeat to put the charge on the para carbon, and then on the other ortho carbon.
- Label the charge positions: oxygen, two ortho carbons and the para carbon. Write 'charge delocalised, ion stabilised'.
Effect of substituents on acidity
| Compound | pKa | Reason |
|---|---|---|
| Ethanol | 15.9 | Charge on ethoxide stays on O; the alkyl group's +I effect makes it worse |
| o-, m-, p-Cresol (methylphenols) | 10.2, 10.1, 10.2 | Electron-releasing -CH₃ destabilises the phenoxide ion: weaker than phenol |
| Phenol | 10.0 | Phenoxide ion stabilised by resonance |
| m-Nitrophenol | 8.3 | -NO₂ withdraws electrons by −I effect only; from meta it cannot take the charge by resonance |
| o-Nitrophenol | 7.2 | -NO₂ withdraws by −I and −R (resonance); the charge spreads onto NO₂ |
| p-Nitrophenol | 7.1 | Same as ortho: strong stabilisation of the ion |
| 2,4,6-Trinitrophenol (picric acid) | far lower | Three -NO₂ groups; a strong acid |
Esterification and acetylation
- 1R-OH + R'-COOH ⇌ R'-COOR + H₂OA little conc H₂SO₄ as catalyst. The reaction is reversible, so water is removed as it forms
- 2R-OH + (R'CO)₂O ⇌ R'-COOR + R'COOHAcid anhydride, with H⁺ as catalyst
- 3R-OH or Ar-OH + R'COCl → R'-COOR (or R'-COOAr) + HClWith pyridine, a base that removes the HCl and pushes the reaction forward
- 4★HOC₆H₄COOH + (CH₃CO)₂O → CH₃COOC₆H₄COOH + CH₃COOHAcetylation (bringing in a CH₃CO group) of salicylic acid gives aspirin (acetylsalicylic acid); H⁺ catalyst
★ key step
Aspirin, a common tablet for pain and for thinning blood, is made by acetylating salicylic acid: an esterification of a phenolic -OH group.
3 marks: “Explain why phenol is more acidic than ethanol. Draw the resonance structures of the phenoxide ion.” Phenoxide's negative charge is delocalised over O and the ortho and para carbons (five resonance structures), so it is stable. Ethoxide's charge stays on O and the +I effect of ethyl destabilises it.
2 marks: “Arrange in increasing order of acidity, with reasons: ethanol, p-nitrophenol, phenol, p-cresol.” Ethanol < p-cresol < phenol < p-nitrophenol. -CH₃ releases electrons and destabilises the ion; -NO₂ withdraws electrons and stabilises it.
2 marks: “What is acetylation? Write the equation for the preparation of aspirin.” Bringing a CH₃CO group into an alcohol or phenol. HOC₆H₄COOH + (CH₃CO)₂O (H⁺) → CH₃COOC₆H₄COOH + CH₃COOH.
Spread the charge, strengthen the acid. NO₂ at ortho or para pulls hardest; CH₃ pushes back.