Alcohols, Phenols and Ethers · Lesson 5 of 12
Acidity, Basicity and Ester Formation
“Use conjugate-base stability to explain proton transfer and distinguish it from acylation at oxygen.”
• Explain alcohol and phenol acidity using conjugate-base stability. • Interpret pKa values and calculate relative acid strengths. • Predict substituent effects on phenols and distinguish proton transfer from redox. • Write esterification and acetylation reactions with appropriate conditions and limitations.
The same oxygen can donate or accept a proton
An alcohol contains a polar O–H bond and oxygen lone pairs. A sufficiently strong base can remove the OH proton to give an alkoxide, RO⁻, so the alcohol acts as a Brønsted acid. In a strongly acidic solution, oxygen can donate a lone pair to H⁺, giving ROH₂⁺, so the alcohol acts as a Brønsted base. These are different reactions under different conditions. Protonated alcohol, neutral alcohol and alkoxide have different charges and reactivities; their formulas must not be interchanged.
For related small aliphatic alcohols in water, additional electron-donating alkyl groups and poorer solvation of a crowded alkoxide generally reduce acidity: methanol > many primary > secondary > tertiary alcohols. This is a qualified aqueous trend, not a solvent-independent law. Simple alcohols have acidities near that of water, and aqueous NaOH is not used to make them quantitatively into dry alkoxides. Use an appropriate stronger reagent or active metal when an alkoxide is required. Electron-withdrawing substituents can reverse simple hydrocarbon-based comparisons.
Active metals remove the hydroxyl hydrogen
Bubbles of hydrogen demonstrate removal of the hydroxyl hydrogen. They do not by themselves measure the equilibrium acidity, because reaction with a metal is also redox: Na is oxidised from 0 to +1 and hydrogen is reduced from +1 to 0. The balanced aluminium relation is 6ROH + 2Al → 2Al(OR)₃ + 3H₂ when the metal reacts. The oxygen remains bonded to its carbon group. Moisture can consume the active metal or protonate an alkoxide, so water is not an inert solvent for this preparation.
Why phenol reacts with hydroxide
Phenol is appreciably more acidic than ethanol because its conjugate base is more stabilised relative to its neutral acid. In ethoxide, negative charge is largely localised on oxygen. In phenoxide, oxygen can donate an electron pair into the ring, giving resonance contributors with negative charge at ortho and para carbons. The actual ion is a delocalised resonance hybrid. Oxygen-bearing contributors remain important because oxygen accommodates negative charge well; the charge is not equally divided among every atom.
Neutral phenol is also conjugated. Its charge-separated contributors have positive oxygen and negative ring positions, whereas phenoxide can delocalise an existing negative charge without creating an additional positive centre. Acidity compares the free-energy difference between an acid and its conjugate base; it is misleading to compare their isolated absolute “stabilities” and declare that one entire species is simply more stable than another with a different proton count. The relative stabilisation of phenoxide favours proton loss.
Phenol is still much weaker than a typical carboxylic acid. It does not normally liberate CO₂ from aqueous bicarbonate as a carboxylic acid does. Thus “acidic” is comparative: active sodium reacts with alcohols and phenols, hydroxide readily deprotonates phenol, and bicarbonate normally distinguishes phenol from stronger carboxylic acids. These observations are explained by different reaction driving forces rather than one universal yes/no acid label.
Use pKa as a logarithmic scale
A smaller pKa means a larger Ka and a stronger acid under comparable conditions. For two acids, divide their Ka values and use Ka=10^(−pKa). The ratio Ka(A)/Ka(B)=10^[pKa(B)−pKa(A)]. A difference of one pKa unit is a factor of ten in the dissociation constant, not merely one extra percent ionisation. Comparing Ka does not imply identical percentage dissociation at every concentration.
| Compound | Representative aqueous pKa |
|---|---|
| 2-Nitrophenol | 7.2 |
| 3-Nitrophenol | 8.3 |
| 4-Nitrophenol | 7.1 |
| Phenol | 10.0 |
| 2-Methylphenol | 10.2 |
| 3-Methylphenol | 10.1 |
| 4-Methylphenol | 10.2 |
| Ethanol | 15.9 |
A nitro group withdraws electrons by induction and can accept negative charge through resonance when it lies ortho or para to the phenoxide oxygen. The additional resonance pathway is absent in the meta arrangement, although induction still makes meta-nitrophenol stronger than phenol. A methyl group generally donates electron density, destabilising the phenoxide relative to phenol and reducing acidity. Methoxy can donate by resonance at ortho/para positions, with induction and solvation also contributing; it cannot simply be treated as another nitro group.
Problem
Using pKa 10.0 for phenol and 15.9 for ethanol, compare their acid dissociation constants.
- 1.Ka(phenol)/Ka(ethanol)=10^(15.9−10.0)=10^5.9.
- 2.10^5.9≈7.94×10⁵, so the phenol dissociation constant is roughly eight hundred thousand times larger. “About a million” is an order-of-magnitude description.
- 3.This comparison uses the same solvent and comparable definitions. It is a ratio of equilibrium constants, not a claim that every solution of phenol contains eight hundred thousand times as much H₃O⁺.
Problem
Arrange propan-1-ol, 4-methylphenol, phenol, 3-nitrophenol, 3,5-dinitrophenol and 2,4,6-trinitrophenol in increasing acidity.
- 1.Propan-1-ol lacks phenoxide resonance stabilisation and is weakest. A methyl group makes 4-methylphenol weaker than phenol.
- 2.An electron-withdrawing nitro group increases acidity even at meta through induction; two meta nitro groups increase it further. Three nitro groups including ortho/para resonance pathways strongly stabilise the conjugate base.
- 3.The order is propan-1-ol < 4-methylphenol < phenol < 3-nitrophenol < 3,5-dinitrophenol < 2,4,6-trinitrophenol. Each comparison concerns the conjugate-base/acid relationship, not the number of OH groups.
Ester formation changes an O–H bond into an O–acyl bond
An ester contains R′C(=O)–OR, with an oxygen bonded to a carbonyl carbon. It is not an ether R–O–R′. In acylation of an alcohol, the alcohol oxygen attacks an activated acyl carbon and ultimately becomes the single-bond oxygen of the ester. The alcohol O–H bond is lost, while its original R–O connection remains. Acetylation specifically introduces CH₃C(=O)–, called an acetyl group.
Acid activates the carboxylic carbonyl toward attack by alcohol oxygen. Addition, proton transfers and elimination of water restore the carbonyl and regenerate acid. Because the sequence is reversible, an equilibrium shift matters. Phenols are weaker oxygen nucleophiles than ordinary alcohols and do not generally give useful direct Fischer esterification with carboxylic acids; use an activated acyl reagent for a phenolic ester. Do not extend a general alcohol equation uncritically to every ArOH.
Acid chloride reactions are driven by an activated acyl group with a good leaving group. The added base traps HCl and prevents unwanted protonation; this is not the same equilibrium as acid-catalysed carboxylic acid esterification. An anhydride produces carboxylic acid, not water as its organic acyl-transfer coproduct. The reagent class determines which by-product belongs in the equation.
Problem
Explain formation of aspirin from salicylic acid and ethanoic anhydride.
- 1.Salicylic acid is 2-hydroxybenzoic acid, containing both a ring OH and COOH. Acetylation uses ethanoic anhydride and an acid catalyst.
- 2.The phenolic oxygen becomes –OC(=O)CH₃ while the carboxylic acid remains: C₆H₄(OH)COOH + (CH₃CO)₂O → C₆H₄(OC(=O)CH₃)COOH + CH₃COOH.
- 3.The product is 2-(acetyloxy)benzoic acid, commonly acetylsalicylic acid. Its chemical identity follows the acylation site; the equation is not a recommendation to prepare or use a medicine.
Quick Quiz
Quiz
Why is phenol more acidic than ethanol?
Explanation: Phenoxide delocalisation favours deprotonation without requiring C–O cleavage.
Which has the smallest listed pKa?
Explanation: The listed value 7.1 is lower than 10.0, 15.9 and 10.2.
Two acids differ by 3 pKa units. How much larger is Ka for the stronger one?
Explanation: The logarithmic relationship gives 10³.
What is the coproduct when ethanoic anhydride acetylates an alcohol?
Explanation: Acyl transfer from the anhydride leaves the other acyl fragment as a carboxylic acid.
Which is a reliable route to a phenolic ester?
Explanation: Activated acyl reagents overcome the low oxygen nucleophilicity of phenol.
Practice Problems
- Write reactions demonstrating phenol acidity with sodium and with hydroxide; explain their different driving forces.
- Compare Ka for 4-nitrophenol and 3-nitrophenol using pKa 7.1 and 8.3.
- Why is 2-nitrophenol more acidic than 2-methoxyphenol in the usual comparison?
- Prepare ethyl ethanoate from ethanol and explain how to favour the product.
- Write phenol acetylation with ethanoyl chloride and explain the role of pyridine.
- Does hydrogen evolution with sodium prove an alcohol is stronger than phenol?
2C₆H₅OH + 2Na → 2C₆H₅ONa + H₂ is coupled to metal oxidation and hydrogen reduction. C₆H₅OH + NaOH ⇌ C₆H₅ONa + H₂O is proton transfer, favoured because phenol is more acidic than water under these conditions. Both remove the OH proton while retaining C–O, but they are not identical kinds of evidence.
Key Takeaways
• An alcohol can act as a weak acid or as a base at oxygen, depending on its reaction partner. • Metal reactions form alkoxides or phenoxides with hydrogen evolution and include redox. • Phenoxide is resonance-stabilised, making phenol more acidic than ordinary ethanol. • Acidity compares an acid with its conjugate base under specified conditions. • Lower pKa means stronger acid; each unit represents a tenfold Ka difference. • Nitro groups generally strengthen phenol acidity; methyl groups generally weaken it. • Esterification transfers an acyl group to oxygen and retains the alcohol carbon–oxygen bond. • Phenolic esters are reliably prepared with activated acyl reagents.