Difficulty: Intermediate | Prerequisites: IUPAC nomenclature basics, SN1/SN2 mechanisms, acid-base fundamentals (Chapters 5–7)
Alcohols are among the most common functional groups in organic chemistry and serve as a gateway to dozens of reactions you will use for the rest of the course. Chapter 8 introduces how to name, classify, and synthesise alcohols, and how their dual acid-base character shapes their reactivity. If you can handle the material here, the oxidation, reduction, and organometallic content in Parts 2 and 3 will fall into place much more readily. You should already be comfortable with substitution and elimination mechanisms before starting.
Alcohols (R-OH) are classified as primary, secondary, or tertiary depending on the carbon bearing the hydroxyl group. They are amphoteric (both weakly acidic and weakly basic), act as weak nucleophiles, and can be synthesised from alkyl halides via SN1/SN2 or from esters via hydrolysis. Their physical properties, especially high boiling points, stem from hydrogen bonding.
Alcohol
A compound containing a hydroxyl group (-OH) bonded to an sp3-hybridised carbon. In simple terms, any organic molecule with an -OH on a saturated carbon.
Primary alcohol (1°)
An alcohol where the -OH-bearing carbon is attached to one other carbon. Think of it as the -OH sitting at or near the end of a chain.
Secondary alcohol (2°)
An alcohol where the -OH-bearing carbon is attached to two other carbons. In simple terms, the -OH is in the middle of a chain or at a branch point with two carbon neighbours.
Tertiary alcohol (3°)
An alcohol where the -OH-bearing carbon is attached to three other carbons. Think of it as the most substituted, most sterically crowded version.
Tautomerisation (keto-enol)
The equilibrium interconversion between an enol form (C=C with -OH) and a keto form (C=O). In simple terms, -OH on a double-bond carbon is unstable and rearranges to a carbonyl, because C=O (128 kcal/mol) is much stronger than C-O single bond (85.5 kcal/mol).
Phenol
The exception to the "no -OH on sp2 carbon" rule. A hydroxyl group bonded directly to a benzene ring. Stable because aromatic rings are very stable (covered further in Ch. 15).
Amphoteric
A substance that can act as both an acid and a base. Alcohols are amphoteric: they donate a proton (acid) or use their lone pair to accept a proton (base).
Alkoxide
The conjugate base of an alcohol, formed by removing the -OH proton (R-O⁻). Think of it as the deprotonated alcohol.
Polar protic solvent
A solvent that has O-H or N-H bonds and can hydrogen-bond. Alcohols themselves are polar protic solvents and are good solvents for SN reactions.
Ester hydrolysis
Breaking an ester (R-CO-OR') with water (or aqueous base) to produce an alcohol and a carboxylic acid (or carboxylate). This is a way to make alcohols while avoiding E2 competition.
IUPAC naming follows the standard nomenclature guide; the suffix is "-ol."
Classification is based on the degree of substitution at the carbon bearing the -OH:
1° : one carbon neighbour
2° : two carbon neighbours
3° : three carbon neighbours
Avoid placing -OH on an sp2 carbon. An enol (C=C-OH) will tautomerise to the keto form because the C=O bond (128 kcal/mol) is much stronger than C-O (85.5 kcal/mol).
The one major exception is phenol, where aromatic stabilisation keeps the -OH on the ring.
Alcohols are weak nucleophiles.
They are good solvents for SN reactions.
They are polar protic solvents (can donate H-bonds).
Hydrogen bonding raises their boiling points compared to analogous ethers or alkanes of similar molecular weight.
The -OH group is hydrophilic (water-loving); the hydrocarbon chain is hydrophobic (water-avoiding). This dual character governs solubility.
As weak acids:
Acidity order: methanol > 1° > 2° > 3° (methanol is the most acidic simple alcohol).
Two factors explain why more substituted alcohols are weaker acids:
Induction: C-H and C-C bonds are electron-donating, which pushes electron density toward the oxygen and destabilises the resulting alkoxide.
Solvation: bulky alkyl groups around the oxygen prevent solvent molecules from stabilising the alkoxide via dipole interactions or H-bonding.
As weak bases:
The lone pair on oxygen can accept a proton from a strong acid (e.g. HCl).
The protonated alcohol (R-OH2⁺) is analogous to H3O⁺.
Method 1: SN2 from primary alkyl halides
A primary alkyl halide reacts with water (or hydroxide) to give a primary alcohol.
Works well because primary substrates favour SN2.
Method 2: SN1 from secondary or tertiary alkyl halides
Tertiary or secondary halides with NaOH/H2O can give alcohols via SN1.
Problem: E2 elimination competes, especially with strong bases and 2° or 3° leaving groups.
Method 3: Ester hydrolysis
Advantage: avoids E2 competition because it uses a weak base/strong nucleophile approach.
Mechanism (covered fully in Ch. 20):
Resonance in the ester makes the carbonyl carbon a good electrophile.
Nucleophilic attack (addition) by hydroxide forms a tetrahedral intermediate.
Elimination of the alkoxide leaving group follows.
The alkoxide can act as a leaving group here because the solution is already basic and the driving force is re-formation of the strong C=O bond (ABC driving force).
The condensed "PTP" (Protect-Transform-Produce) form groups the ester hydrolysis and workup into a clean sequence.
Stereochemistry note: if you need retention of configuration at a stereogenic centre, two sequential SN2 reactions (double inversion) can achieve this.
Students often think any -OH on a carbon makes it an alcohol. It does not: if the carbon is sp2-hybridised (as in an enol), the compound tautomerises to a ketone unless aromatic stabilisation is present (phenol).
Students frequently confuse the acidity trend. More substituted alcohols are weaker acids, not stronger. The alkyl groups destabilise the alkoxide, not stabilise it.
Students sometimes forget that SN1 from secondary/tertiary halides competes with elimination. Ester hydrolysis exists partly to sidestep this problem.
The alkoxide leaving group in ester hydrolysis seems wrong because alkoxides are basic ("bad leaving groups"). The trick is that the solution is already basic and the ABC driving force (re-forming C=O) pushes the reaction forward.
⚠️ Classification (1°, 2°, 3°) determines which reactions and mechanisms apply. Know how to classify instantly.
⚠️ The keto-enol tautomerisation and why enols are disfavoured (bond-energy argument) is a recurring concept that reappears in carbonyl chemistry (Ch. 20+).
⚠️ The acidity order of alcohols, including the two reasons (induction and solvation), is a classic exam question.
⚠️ Ester hydrolysis mechanism, including why alkoxide can be a leaving group, is tested in both Ch. 8 and Ch. 20.
True or false: A tertiary alcohol is more acidic than methanol.
Fill in the blank: Enols tautomerise to ketones because the C=___ bond is stronger than the C-___ bond.
True or false: Ester hydrolysis avoids E2 competition.
Fill in the blank: Alcohols are ___________ , meaning they can act as both acids and bases.
True or false: Phenol tautomerises to a keto form.
Answers: 1. False (methanol is most acidic). 2. C=O, C-O. 3. True. 4. Amphoteric. 5. False (aromatic stability prevents this).
Q: Rank the following in order of increasing acidity: tert-butanol, ethanol, methanol, isopropanol.
A: tert-butanol < isopropanol < ethanol < methanol. More alkyl substitution destabilises the conjugate base (alkoxide) through induction and steric hindrance to solvation.
Q: Why does an enol tautomerise to a ketone, and what is the one major exception?
A: The C=O bond (128 kcal/mol) is much stronger than the C-O single bond (85.5 kcal/mol), so the keto form is thermodynamically favoured. The exception is phenol, where the aromatic ring provides enough stabilisation to keep the -OH on the sp2 carbon.
Q: In ester hydrolysis, why can alkoxide act as a leaving group even though it is a strong base?
A: The solution is already basic, and the driving force is reformation of the strong carbonyl (C=O) bond. The alkoxide departing is thermodynamically acceptable because the overall process is downhill in energy.
Q: You need to synthesise a primary alcohol from a primary alkyl halide. Which mechanism operates, and what reagent would you use?
A: SN2, using NaOH/H2O or simply H2O. The primary substrate and strong/moderate nucleophile favour the SN2 pathway.
Acid-base chemistry of alcohols connects directly to the pKa and conjugate base stability concepts from Chapter 5.
SN1/SN2 and E2 competition (Chapters 6–7) determines which alcohol synthesis route is practical.
Ester hydrolysis reappears in full mechanistic detail in Chapter 20 (carboxylic acid derivatives).
alcohol, hydroxyl group, -OH, primary alcohol, secondary alcohol, tertiary alcohol, 1° 2° 3° alcohol, tautomerisation, keto-enol, enol form, keto form, phenol, amphoteric, alkoxide, conjugate base, acidity of alcohols, induction effect, solvation, polar protic solvent, hydrogen bonding, boiling point, SN1, SN2, ester hydrolysis, tetrahedral intermediate, leaving group, CHEM 2510, Chapter 8, organic chemistry, Baldwin, Ohio State