Discover how the -OH group transforms a hydrocarbon into a molecule that hydrogen-bonds, dissolves in water, and behaves completely differently depending on where that hydroxyl sits on the carbon chain.
Four printable worksheets that build from the foundations up to exam-style questions — start at whatever level suits you.
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?
Classification rule: count the carbons directly bonded to the C-OH carbon — 1 = primary, 2 = secondary, 3 = tertiary. The -OH group is the same in all three; the difference is the carbon skeleton around it.
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 has only single bonds — 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°) |
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.
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 shape — which minimises surface area → weaker dispersion forces → lower BP.
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.
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 bonded to THREE carbons → Tertiary (3°). (b) Pentan-1-ol: C-OH at C1 bonded to ONE carbon → Primary (1°), straight chain. (c) Pentan-2-ol: C-OH at C2 bonded to C1 and C3 → Secondary (2°). (d) 2-methylbutan-1-ol: C-OH at C1 bonded to ONE carbon → 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: (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. 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. Practically insoluble.
(b) Hexane is non-polar. Hexan-1-ol in hexane: the 6-carbon non-polar chain is compatible with hexane via dispersion forces. 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 → miscible. Hexan-1-ol: 6C chain disrupts more water H-bonds than -OH compensates → insoluble. (b) Hexan-1-ol: long non-polar chain compatible with hexane (dispersion). Ethanol: polar -OH incompatible with non-polar hexane.
Problem: Three C₄H₁₀O compounds: A = CH₃CH₂CH₂CH₂OH (BP 118°C, partially miscible in water); B = CH₃CH(OH)CH₂CH₃ (BP 100°C, miscible in water); C = (CH₃)₃COH (BP 83°C, fully miscible in water). (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.
(a): 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 (straight chain): maximum surface area, strongest dispersion, highest BP. C (three methyl groups): most compact, minimum surface area, weakest dispersion, lowest BP.
(c) Water solubility C > A: Both have the same -OH. 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.
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.
All three can form hydrogen bonds (O–H···O), but chain length and branching determine dispersion force strength. Methanol (MW 32) has the fewest electrons and weakest London forces — lowest BP (65°C). Propan-2-ol (MW 60) has more electrons and greater surface area despite branching — BP 82°C. 2-methylpropan-2-ol (MW 74) has the greatest MW but a very compact spherical shape reduces surface contact, so its London forces are weaker than expected — BP 83°C. The tert-alcohol's compactness roughly cancels its mass advantage over propan-2-ol.
Complete the Learn phase to unlock Practice.
For each of the following alcohols: (i) classify as 1°, 2°, or 3°; (ii) identify the C-OH carbon and count its carbon neighbours.
Now rank all three in order of decreasing boiling point and justify your ranking.
A. Would you expect octan-1-ol (C8) to dissolve in water? Explain using IMF reasoning.
B. Compare the water solubility of 2-methylpropan-2-ol (tertiary, 4C) vs butan-1-ol (primary, 4C). Which is more water-soluble and why?
C. A student says "ethanol is soluble in water because it has an -OH group." Identify what this explanation is missing.
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?
1. (3 marks) Explain why butan-1-ol has a much higher boiling point than butane (C₄H₁₀), which has a similar molecular mass.
2. (4 marks) Describe and explain the trend in solubility as the carbon chain length increases from C1 to C6. Include reference to the relevant intermolecular forces.
3. (5 marks) Compare and contrast the boiling points and water solubility of butan-1-ol and 2-methylpropan-2-ol. Both have the molecular formula C₄H₁₀O. Explain how molecular shape determines both properties.
1. C — C-OH at central carbon bonded to ethyl, methyl, and ethyl groups = three carbons → tertiary.
2. C — Both have one -OH: equivalent H-bonding. Difference from dispersion forces. Pentan-1-ol straight chain → maximum surface area → highest BP.
3. C — Ethanol (2C, -OH dominant) → water-compatible. Hexan-1-ol (6C, chain dominant) → hexane-compatible.
4. B — The H-bond DONOR is the H atom bonded to oxygen. The oxygen is the H-bond ACCEPTOR (via its lone pairs).
5. A — Ethanol's -OH enables extensive H-bonding (~20–25 kJ/mol per H-bond), requiring much more energy to overcome than the weak London dispersion forces between ethane molecules.
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 [1].
Q2 (4 marks): As chain length increases from C1 to C6, water solubility decreases [1]. Short-chain alcohols (C1–C3) are fully miscible — the -OH group can form H-bonds with water, and the small non-polar chain creates minimal disruption to water's H-bond network [1]. As chain length increases (C4+), the non-polar chain grows while the -OH remains constant. 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 [1].
Q3 (5 marks): Both have one -OH group — both form hydrogen bonds of similar strength [1]. Butan-1-ol (1°, straight chain) has a higher boiling point (118°C vs 83°C) because its extended chain maximises surface area, enabling stronger London dispersion forces [1]. 2-methylpropan-2-ol (3°) has compact branched structure — minimum surface area — weaker dispersion — lower BP [1]. 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 [1].
Return to your hypothesis about methanol's specific toxicity. You should now know: methanol → methanal (formaldehyde) → methanoic acid (formic acid) — both highly toxic. Formaldehyde specifically destroys proteins in the optic nerve → permanent blindness. The toxicity is not about IMF or chain length — it is entirely about metabolic products.
What is a primary alcohol, and give one example?
Why do alcohols have much higher boiling points than alkanes of similar molecular mass?
Why does the boiling point order for C4H10O isomers follow: primary > secondary > tertiary?
Explain why ethanol is fully miscible in water but hexan-1-ol is practically insoluble.
In alcohol-alcohol hydrogen bonding, identify the donor and acceptor species.
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