Alcohols are the gateway functional group of Module 7 — they connect back to the hydrocarbons you just studied, forward to aldehydes, ketones, carboxylic acids and esters, and sideways to the fermentation chemistry that has shaped human civilisation for ten thousand years.
Use the PDF for classwork, homework or revision. It includes key ideas, activities, questions, an extend task and success-criteria proof.
Methanol, ethanol, propan-1-ol, and butan-1-ol are all primary alcohols in the same homologous series. Methanol is a colourless liquid that is fatally toxic — a single tablespoon can cause permanent blindness, and a few tablespoons can kill. Ethanol is the same functional group, one carbon longer, and is consumed in billions of litres per year by humans without moderate acute toxicity. Propan-1-ol and butan-1-ol are used as industrial solvents.
How can four consecutive members of the same homologous series have such dramatically different biological effects? Write your hypothesis — what changes as the chain gets longer, and why might methanol be so specifically toxic?
📚 Core Content
The -OH group is the defining feature of alcohols — it is small, polar, and capable of hydrogen bonding, and those three characteristics together explain almost every physical property that separates alcohols from the hydrocarbons they are built on.
The carbon bearing -OH is sp³-hybridised — tetrahedral geometry, ~109.5° bond angles. The oxygen in -OH has two bonding pairs (one to C, one to H) and two lone pairs. The O-H bond is highly polar (O is much more electronegative than H — δ⁻ on O, δ⁺ on H).
This combination — a strong H-bond donor (O-H) and a strong H-bond acceptor (lone pairs on O) — means alcohol molecules form extensive hydrogen bond networks with each other and with water.
The classification of an alcohol as primary, secondary, or tertiary is not just a naming convention — it determines every reaction the alcohol can undergo, from what oxidation products it gives to whether it undergoes elimination or substitution, and how quickly.
The classification is determined by the carbon bearing the -OH group — specifically, how many OTHER carbon atoms that carbon is directly bonded to.
C-OH bonded to 1 other C
General: R-CH₂OH
Examples:
ethanol: CH₃CH₂OH
propan-1-ol: CH₃CH₂CH₂OH
C-OH bonded to 2 other C
General: R-CH(OH)-R'
Examples:
propan-2-ol: CH₃CHOHCH₃
butan-2-ol: CH₃CHOHCH₂CH₃
C-OH bonded to 3 other C
No H on C-OH
General: R-C(OH)(R')(R'')
Example:
2-methylpropan-2-ol: (CH₃)₃COH
| Alcohol | Formula | C-OH bonds to | Class |
|---|---|---|---|
| Methanol | CH3OH | 0 other C (3H) | Primary (1°) — by convention |
| Ethanol | CH3CH2OH | 1 other C | Primary (1°) |
| Propan-2-ol | CH3CHOHCH3 | 2 other C | Secondary (2°) |
| Butan-2-ol | CH3CHOHCH2CH3 | 2 other C | Secondary (2°) |
| 2-methylpropan-2-ol | (CH3)3COH | 3 other C | Tertiary (3°) |
| 2-methylbutan-2-ol | CH3C(OH)(CH3)CH2CH3 | 3 other C | Tertiary (3°) |
The boiling points of alcohols are dramatically higher than those of comparable hydrocarbons — and the reason is hydrogen bonding — but the variation in boiling point within the alcohol series, and between primary, secondary, and tertiary isomers, reveals additional layers of IMF subtlety.
Boiling point increases steadily with chain length: longer chains have greater surface area → stronger dispersion forces in addition to the H-bonding from -OH. The -OH contribution is roughly constant; the growing dispersion force from the chain drives BP upward.
For constitutional isomers (same molecular formula), the boiling point order is: Primary > Secondary > Tertiary.
All three classes have the same -OH group — the H-bonding contribution to BP is approximately equal for all three. The difference arises from dispersion forces:
Tertiary alcohols are the most branched — the most compact, spherical shape — which minimises surface area available for contact with neighbouring molecules. Less surface area → weaker dispersion forces → lower BP.
Primary alcohols (straight chain) have maximum surface area → strongest dispersion forces → highest BP.
Example — C4H10O isomers:
Short-chain alcohols dissolve in water because the -OH group H-bonds with water — but chain length is a limit, and understanding where solubility ends and why gives you the framework to predict solubility for every functional group in Module 7.
Branching and solubility: Branched-chain alcohols are generally MORE soluble in water than their straight-chain isomers because branching reduces the effective size of the non-polar region. 2-methylpropan-2-ol (tertiary, 4C) is fully miscible with water; butan-1-ol (primary, 4C, straight chain) is only partially miscible — the compact shape of the tertiary alcohol exposes less non-polar surface area to water.
"Tertiary alcohols have higher boiling points than primary because they have more alkyl groups." Wrong. More alkyl groups = more branching = more compact shape = less surface area = weaker dispersion forces = LOWER boiling point. Primary > Secondary > Tertiary.
"Tertiary alcohols cannot form hydrogen bonds." Wrong. All alcohols have the -OH group and form hydrogen bonds via the same O-H donor / lone-pair acceptor mechanism. Classification (1°/2°/3°) does not affect hydrogen bond strength.
"Methanol is classified as a 0° (zeroth) alcohol because C-OH has no other carbon neighbours." By convention, methanol is PRIMARY. Zero other carbons satisfies the ≤1 condition for primary classification.
"The oxygen donates the hydrogen bond in alcohols." The HYDROGEN (on the O-H bond, δ⁺) is the donor; the oxygen's lone pairs are the acceptors. H-bond donor = the δ⁺ H bonded to a highly electronegative atom.
🧠 Worked Examples
Problem: Classify each alcohol and arrange in order of decreasing boiling point: (a) 2-methylbutan-2-ol, (b) pentan-1-ol, (c) pentan-2-ol, (d) 2-methylbutan-1-ol.
Classify each:
(a) 2-methylbutan-2-ol: C-OH at C2 bonded to CH3, CH3 (branch), and CH2CH3 — THREE carbons → Tertiary (3°).
(b) Pentan-1-ol: C-OH at C1 bonded to ONE carbon (C2) → Primary (1°), straight chain.
(c) Pentan-2-ol: C-OH at C2 bonded to C1 and C3 — TWO carbons → Secondary (2°).
(d) 2-methylbutan-1-ol: C-OH at C1 bonded to ONE carbon (C2) → Primary (1°), branched.
All four are C5H12O — constitutional isomers. All have one -OH group — equivalent H-bonding. BP differences come from dispersion forces → determined by branching/shape.
Rank by surface area (most extended → highest BP): Pentan-1-ol (straight, 1°) > 2-methylbutan-1-ol (1°, one branch) > pentan-2-ol (2°, compact) > 2-methylbutan-2-ol (3°, most compact).
Answer: Classifications: (a) 3°, (b) 1°, (c) 2°, (d) 1°. Decreasing BP: pentan-1-ol (138°C) > 2-methylbutan-1-ol (129°C) > pentan-2-ol (119°C) > 2-methylbutan-2-ol (102°C). All have equivalent -OH H-bonding; BP differences driven by branching → surface area → dispersion forces.
Problem: (a) Explain why ethanol is fully miscible with water but hexan-1-ol is practically insoluble. (b) Explain why hexan-1-ol dissolves readily in hexane but ethanol does not.
(a) Both have -OH: Both can form H-bonds with water (O-H donates to water; lone pair accepts from water). The difference is the non-polar chain.
Ethanol (C2): The 2-carbon chain creates a small non-polar region. The -OH interaction with water is energetically sufficient to compensate for disrupting water's H-bond network around the small ethyl group. Fully miscible.
Hexan-1-ol (C6): The 6-carbon chain must be accommodated within water's H-bond network. Disrupting water-water H-bonds to accommodate six carbons requires more energy than the single -OH can compensate for. The non-polar chain overwhelms the -OH. Practically insoluble.
(b) Hexane is non-polar (dispersion forces only). Hexan-1-ol in hexane: the 6-carbon non-polar chain is compatible with hexane via dispersion forces; the -OH is a minor polar feature. Dissolves readily. Ethanol in hexane: ethanol's dominant -OH makes it polar — hexane cannot compensate for breaking ethanol-ethanol H-bonds. Poorly miscible.
Answer: (a) Ethanol: short 2C chain — -OH H-bonding compensates for water H-bond disruption → miscible. Hexan-1-ol: 6C chain disrupts more water H-bonds than the single -OH can compensate → insoluble. (b) Hexan-1-ol: long non-polar chain dominates → compatible with hexane (dispersion). Ethanol: polar -OH dominates → incompatible with non-polar hexane.
Problem (7 marks): Three C4H10O compounds: A = CH3CH2CH2CH2OH (BP 118°C, partially miscible in water, miscible in hexane); B = CH3CH(OH)CH2CH3 (BP 100°C, miscible in water, partially miscible in hexane); C = (CH3)3COH (BP 83°C, fully miscible in water, poorly miscible in hexane). (a) Name and classify each. (b) Explain BP trend A > B > C. (c) Explain why C is fully water-miscible while A is only partially miscible. (d) Explain hexane solubility trend A > B > C.
(a) Names and classifications: A = butan-1-ol (1° — C-OH at C1, one carbon neighbour). B = butan-2-ol (2° — C-OH at C2, two carbon neighbours). C = 2-methylpropan-2-ol (3° — C-OH bonded to three CH3 groups).
(b) BP trend: All three are C4H10O with one -OH — equivalent H-bonding. Differences from dispersion forces. A (butan-1-ol): straight chain, maximum surface area, strongest dispersion forces, highest BP. B (butan-2-ol): -OH at C2, slightly more compact. C (2-methylpropan-2-ol): three methyl groups around central C-OH — most compact spherical shape, minimum surface area, weakest dispersion forces, lowest BP.
(c) Water solubility C > A: Both have the same -OH (identical H-bonding with water). C's compact branched structure exposes less non-polar surface to water than A's extended straight chain — less disruption to water's H-bond network. The -OH compensates more effectively in C than in A despite the same carbon count.
(d) Hexane solubility A > B > C: Solubility in non-polar hexane increases with non-polar character. A (butan-1-ol): long straight chain creates significant non-polar surface compatible with hexane. C (2-methylpropan-2-ol): compact structure centralises the polar -OH, maximising its influence and making it the least compatible with hexane of the three.
Answer: (a) A = butan-1-ol (1°); B = butan-2-ol (2°); C = 2-methylpropan-2-ol (3°). (b) Equal -OH H-bonding; BP decreases A→B→C due to increasing branching → decreasing surface area → decreasing dispersion forces. (c) C's compact shape disrupts fewer water H-bonds than A's extended chain — same -OH compensates more effectively despite same carbon count. (d) A is most hexane-soluble (straight non-polar chain compatible with hexane); C is least (polar -OH centralised, highest effective polarity).
✏️ Activities
Now rank all three alcohols above in order of decreasing boiling point and justify your ranking.
🧠 Check Your Understanding
1. Which of the following correctly classifies the alcohol in CH3CH2C(OH)(CH3)CH2CH3?
2. Pentan-1-ol (BP 138°C) and 2-methylbutan-2-ol (BP 102°C) have the same molecular formula C5H12O. Which explanation best accounts for the 36°C difference?
3. Which result is most consistent with IMF principles for alcohols?
4. In a hydrogen bond between two alcohol molecules, which species is the hydrogen bond DONOR?
5. A student is comparing ethanol (BP 78°C) with ethane (BP −89°C). Which statement correctly explains the 167°C difference?
📝 Short Answer
1. Explain why butan-1-ol has a much higher boiling point than butane (C4H10), which has a similar molecular mass. 3 MARKS
2. A student dissolves a series of alcohols in water. Describe and explain the trend in solubility as the carbon chain length increases from C1 to C6. Include reference to the relevant intermolecular forces. 4 MARKS
3. Compare and contrast the boiling points and water solubility of butan-1-ol and 2-methylpropan-2-ol. Both have the molecular formula C4H10O. In your response, explain how molecular shape determines both properties, with reference to relevant intermolecular forces. 5 MARKS
CH3CH2C(OH)(CH3)CH2CH3: C-OH is the central carbon, bonded to CH3CH2- (one side), CH3 (branch), and -CH2CH3 (other side) = THREE carbons → Tertiary (3°).
(CH3)2CHCH2OH: C-OH is the CH2OH carbon (terminal), bonded to ONE other carbon (the CH(CH3)2 group) → Primary (1°).
CH3CH(OH)CH(CH3)CH3: C-OH is C2, bonded to C1 (CH3) and C3 (CH(CH3)CH3) = TWO carbons → Secondary (2°).
BP ranking (highest to lowest): (CH3)2CHCH2OH (1°, branched) > CH3CH(OH)CH(CH3)CH3 (2°) > CH3CH2C(OH)(CH3)CH2CH3 (3°, most compact). All have equivalent -OH H-bonding; BP differences arise from dispersion forces — the 1° branched alcohol has more extended shape and greater surface area than the 3° tertiary alcohol.
A. Octan-1-ol (C8) would NOT dissolve in water (practically insoluble). The 8-carbon non-polar chain disrupts a large portion of water's H-bond network when attempting to dissolve — far more than can be compensated by the single -OH forming H-bonds with water. The non-polar character of the molecule dominates over the polar -OH.
B. The tertiary alcohol (2-methylpropan-2-ol) is more water-soluble because its compact, branched molecular shape reduces the effective non-polar surface area that must be accommodated within water's H-bond network. Despite having the same four carbon atoms as butan-1-ol, the spherical shape of the tertiary alcohol disrupts fewer water-water H-bonds per molecule, allowing the -OH group to compensate more effectively → full miscibility. Butan-1-ol's extended straight chain disrupts more water H-bonds → only partial miscibility.
C. The explanation is incomplete. It correctly identifies that the -OH group can form H-bonds with water, which contributes to solubility [1 mark]. However, it omits the mechanism of H-bonding (O-H donates H-bond to water's lone pair; water O-H donates H-bond to alcohol's lone pair), the chain length effect (the small 2-carbon non-polar chain of ethanol creates minimal disruption to water's H-bond network — this is the key reason ethanol is fully miscible rather than only partially miscible), and the "like dissolves like" principle that explains why long-chain alcohols are NOT fully water-soluble despite having the same -OH group.
1. C — CH3CH2C(OH)(CH3)CH2CH3: C-OH bonded to three other carbons (ethyl, methyl branch, ethyl) → tertiary. Option B misses the methyl branch.
2. C — Both have one -OH: equivalent H-bonding. Difference comes from dispersion forces. Pentan-1-ol's straight chain → maximum surface area → strongest dispersion forces → highest BP. 2-methylbutan-2-ol's branched structure → compact shape → minimum surface area → weakest dispersion → lowest BP. Option A is wrong — alcohol class (1°/2°/3°) does not change H-bond strength. Option D is wrong — tertiary alcohols DO form H-bonds via the same -OH group.
3. C — Ethanol (2C, -OH dominant) → polar, water-compatible, hexane-incompatible. Hexan-1-ol (6C, chain dominant) → non-polar character dominates, hexane-compatible, water-incompatible. Option B is wrong — a C10 alcohol is not water-miscible despite having -OH (chain length overcomes -OH contribution). Option D ignores chain length entirely.
4. B — The H-bond DONOR is the H atom bonded to oxygen. The δ⁺ H on the O-H bond forms the electrostatic interaction with a lone pair on a neighbouring atom. The oxygen is the H-bond ACCEPTOR (via its lone pairs). Lone pairs don't donate H-bonds — they receive them.
5. A — Ethanol's -OH enables extensive H-bonding between molecules (~20–25 kJ/mol per H-bond), requiring much more energy to overcome than the weak London dispersion forces between ethane molecules. The 167°C difference reflects this dramatic increase in IMF strength from adding one -OH group.
Q1 (3 marks): Butan-1-ol contains a hydroxyl group (-OH) [1]. The O-H bond is highly polar, making the H a hydrogen bond donor (δ⁺) and the oxygen's lone pairs hydrogen bond acceptors (δ⁻). Butan-1-ol molecules form hydrogen bonds with each other (~20–25 kJ/mol each) [1]. Butane has only C-H and C-C bonds — only weak London dispersion forces act between its molecules. Much more energy is required to overcome the H-bond network in butan-1-ol than to overcome dispersion forces in butane, resulting in a dramatically higher boiling point [1].
Q2 (4 marks): As chain length increases from C1 to C6, water solubility decreases [1]. Short-chain alcohols (C1–C3: methanol, ethanol, propan-1-ol) are fully miscible with water — the -OH group can form H-bonds with water (O-H donating and lone pairs accepting), and the small non-polar chain creates minimal disruption to water's H-bond network [1]. As chain length increases (C4+), the non-polar hydrocarbon chain grows while the -OH group remains constant at one per molecule. Accommodating the longer chain within water's H-bond network requires breaking more water-water H-bonds than the single -OH can compensate for [1]. Beyond ~C5, the non-polar chain dominates molecular character — water solubility becomes negligible because the energy cost of disrupting water's H-bond network outweighs the energy gained from -OH/water H-bond formation [1].
Q3 (5 marks): Both butan-1-ol and 2-methylpropan-2-ol are C4H10O with one -OH group — both form hydrogen bonds of similar strength with other molecules (O-H donor, lone pairs acceptor) [1]. Butan-1-ol (1°, straight chain) has a higher boiling point (118°C vs 83°C) because its extended straight chain maximises molecular surface area, enabling stronger London dispersion forces with neighbouring molecules in addition to H-bonding [1]. 2-methylpropan-2-ol (3°) has a compact, branched, spherical structure — minimum surface area — weaker dispersion forces — lower boiling point [1]. For water solubility, 2-methylpropan-2-ol is fully water-miscible while butan-1-ol is only partially miscible, despite the same carbon count [1]. The compact spherical shape of 2-methylpropan-2-ol exposes less non-polar surface area to water, causing less disruption to water's H-bond network — the -OH compensates more effectively. Butan-1-ol's extended chain disrupts more water H-bonds per molecule, making complete dissolution energetically less favourable [1].
Return to your hypothesis about methanol's specific toxicity. You should now be able to give a precise biochemical explanation:
Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?
Tick when you've finished the activities and checked the model answers.