8 exam-style question sets on Organic chemistry (Paper 2), each with a hint and a fully worked answer. The app holds 40 questions on this section in total, including variants of every set below, and lets you mark yourself part by part.
Practise this section in the app →
Question 1
The letters A to H in the table below represent eight organic compounds.
| Letter | Compound |
|---|---|
| A | CH₃CH₂CH₂CH₂CHO |
| B | CH₃CHClCH(CH₃)CH₂Cl |
| C | Pentan-3-one |
| D | C₃H₈O |
| E | CH₃C≡CCH(CH₃)CH₃ |
| F | CH₃CH₂COOCH₃ |
| G | CH₃CH(CH₃)CH₂CH₃ |
| H | Shown in the diagram |
Use the table to answer the questions that follow.
- aWrite down the LETTER of a compound in the table that: 1. is an alcohol 2. contains a formyl group 3. is an unsaturated compound (3)
- bWrite down the IUPAC name of compound B. (3)
- cWrite down the IUPAC name of compound E. (3)
- dDefine the term functional isomers. (2)
- eTwo compounds in the table are functional isomers of each other. Write down the TWO letters that represent these compounds. (1)
- fCompound F is prepared when a carboxylic acid reacts with compound X in the presence of an inorganic catalyst. Write down the NAME or FORMULA of the catalyst AND the type of reaction that takes place. (2)
- gFor the reaction in QUESTION (f), write down the STRUCTURAL FORMULA of the carboxylic acid used AND the IUPAC name of compound X. (4)
This question has a diagram, shown in the app.
Hint
Match each entry to its homologous series first; for naming, pick the longest chain containing the functional group and number for the lowest locants.
Worked answer
a) 1. D ✓ (C₃H₈O is an alcohol in the CAPS families: propan-1-ol or propan-2-ol) 2. A ✓ (ends in –CHO, an aldehyde) 3. E ✓ (contains a C≡C triple bond)
b) 1,3-dichloro-2-methylbutane ✓✓✓ (4-carbon chain ✓; number from the end giving the lowest locant set: Cl on C1 and C3 ✓; methyl on C2 ✓)
c) 4-methylpent-2-yne ✓✓✓ (5-carbon chain containing C≡C ✓; triple bond locant 2 ✓; methyl on C4 ✓)
d) Functional isomers are compounds with the same molecular formula but different functional groups. ✓✓
e) A and C ✓ — both are C₅H₁₀O: A is the aldehyde (pentanal) and C is the ketone (pentan-3-one).
f) Catalyst: concentrated H₂SO₄ (concentrated sulphuric acid) ✓. Type of reaction: esterification (condensation). ✓
g) F is methyl propanoate, so the acid is propanoic acid. Structural formula: a 3-carbon chain with all bonds shown; C1 carries a C=O and an O–H (condensed: CH₃CH₂COOH). ✓✓ Compound X (the alcohol): methanol ✓✓
Modelled on Nov 2024 Paper 2, Question 2 — same skills, new scenario.
Question 2
The vapour pressures of five organic compounds are measured at 20 °C. The vapour pressures of compounds A, B and C are NOT shown in the table.
| Compound | IUPAC name | Molar mass (g·mol⁻¹) | Vapour pressure (kPa) at 20 °C |
|---|---|---|---|
| A | Hexane | 86 | ? |
| B | 2-Methylpentane | 86 | ? |
| C | 2,2-Dimethylbutane | 86 | ? |
| D | Pentan-1-ol | 88 | 0,22 |
| E | Pentanal | 86 | 3,4 |
- aDefine the term vapour pressure. (2)
- bThe vapour pressures of compounds A, B and C are given below in random order: 16,2 kPa 34,0 kPa 24,5 kPa Write down the vapour pressure of compound C. (1)
- cFully explain your answer to QUESTION (b). (3)
- dConsider compounds D and E. Which compound has the HIGHER boiling point? (1)
- eFully explain the large difference between the vapour pressures of compounds D and E. (4)
Hint
Among chain isomers, more branching means a smaller surface area; then compare the TYPE of intermolecular force for D and E.
Worked answer
a) Vapour pressure is the pressure exerted by a vapour in equilibrium with its liquid in a closed system (at a given temperature). ✓✓
b) 34,0 kPa ✓
c) A, B and C are chain isomers with the same molar mass, so the comparison depends on branching. C is the MOST branched, giving it the smallest surface area over which molecules make contact ✓, so it has the WEAKEST London (dispersion) forces ✓. The least energy is needed for its molecules to escape the liquid, so C has the HIGHEST vapour pressure ✓ (and A, unbranched, has the lowest: 16,2 kPa; B: 24,5 kPa).
d) D (pentan-1-ol) ✓ — its very low vapour pressure (0,22 kPa) shows the strongest intermolecular forces, so it boils at the highest temperature.
e) D and E have comparable molar masses, so the difference is due to the TYPE of intermolecular force ✓. D (an alcohol, O–H group) has hydrogen bonds between molecules ✓, while E (an aldehyde) has only dipole–dipole forces (plus London forces) ✓. Hydrogen bonds are stronger, so more energy is needed for molecules of D to escape the liquid — far fewer molecules enter the vapour phase, giving D the much lower vapour pressure ✓.
Modelled on Nov 2024 Paper 2, Question 3 — same skills, new scenario.
Question 3
In which ONE of the following pure liquids are the molecules held together by HYDROGEN BONDS?
Hint
Hydrogen bonding requires a hydrogen atom bonded directly to O (or N or F).
Worked answer
D. Butanoic acid contains an O–H group, so a strongly positive H on one molecule is attracted to a lone pair on the O of a neighbouring molecule — hydrogen bonding. A (butanal) and B (butan-2-one) have polar C=O groups but no O–H, so their strongest force is dipole–dipole. C (ethyl methanoate, an ester) also has no H bonded to O, so it cannot hydrogen-bond with itself. Modelled on Nov 2024 Paper 2, Question 1.1 — same skill, new scenario.
Question 4
The aldehydes form a homologous series. Which ONE of the following general formulas represents this series?
Hint
Test the formula on the simplest members, e.g. methanal (CH₂O) and ethanal (C₂H₄O).
Worked answer
B. Methanal is CH₂O and ethanal is C₂H₄O — both fit CₙH₂ₙO, the general formula for aldehydes (and ketones). A (CₙH₂ₙ₊₂O) is the general formula of the alcohols. C would describe a compound with an extra degree of unsaturation. D (CₙH₂ₙO₂) is the general formula of carboxylic acids and esters. Modelled on Nov 2024 Paper 2, Question 1.2 — same skill, new scenario.
Question 5
The letters A to F in the table below represent six organic compounds.
| Letter | Condensed structural formula |
|---|---|
| A | CH₃CH(CH₃)CH₂CH₂OH |
| B | CH₃C≡CCH₃ |
| C | CH₃COOCH₂CH₂CH₃ |
| D | C(CH₃)₄ |
| E | CH₃CH₂CH₂CH₂CHO |
| F | CH₃CH₂CH₂COCH₃ |
Use the information in the table to answer the questions that follow.
- aDefine the term functional group. (2)
- bWrite down the IUPAC name of compound A. (3)
- cWrite down the IUPAC name of compound F. (2)
- dWrite down the LETTER of the compound that is an aldehyde, and the NAME of the functional group of aldehydes. (2)
- eCompound C is prepared from a carboxylic acid and an alcohol. Write down the IUPAC NAMES of the carboxylic acid and the alcohol used. (2)
- fDefine the term chain isomer and write down the IUPAC name of ONE chain isomer of compound D. (4)
- gWrite down the general formula of the homologous series to which compound B belongs. (1)
Hint
Number each parent chain from the end nearest the first substituent or functional group, and remember that an ester's name starts with the alkyl group from the alcohol.
Worked answer
a) A bond or an atom or a group of atoms that determine(s) the physical and chemical properties of a group of organic compounds. ✓✓
b) 3-methylbutan-1-ol ✓✓✓ (longest chain 4 C with the OH on C1; methyl on C3.)
c) Pentan-2-one ✓✓ (5-C chain, carbonyl C on position 2, bonded to two C atoms → ketone.)
d) E ✓; formyl group ✓.
e) Ethanoic acid ✓ and propan-1-ol ✓ (the acyl part CH₃CO– comes from ethanoic acid; the –OCH₂CH₂CH₃ comes from propan-1-ol).
f) Chain isomers: compounds with the same molecular formula but different types of chains (different branching). ✓✓ Compound D is 2,2-dimethylpropane (C₅H₁₂), so ONE chain isomer is pentane ✓✓ (or 2-methylbutane).
g) CₙH₂ₙ₋₂ ✓ (compound B, but-2-yne, is an alkyne).
Modelled on Nov 2025 Paper 2, Question 2 — same skills, new scenario.
Question 6
Two unlabelled bottles in a school laboratory contain compounds P and Q. Both compounds have the molecular formula C₃H₆O₂ and are STRAIGHT-CHAIN molecules, but they belong to two DIFFERENT homologous series.
The structural formula of compound Q is shown in the diagram.
The vapour pressures of the two compounds at 25 °C are:
| Compound | Vapour pressure at 25 °C |
|---|---|
| P | 0,5 kPa |
| Q | 65,0 kPa |
- aDefine the term functional isomers. (2)
- bIdentify the TWO homologous series to which compounds P and Q belong. (2)
- cWrite down the STRUCTURAL FORMULA of compound P. (2)
- dWrite down the IUPAC name of compound Q. (2)
- eWhich compound, P or Q, has the higher boiling point? Explain the answer fully by referring to the TYPE of intermolecular forces in each compound, their relative STRENGTH and the ENERGY involved. (4)
This question has a diagram, shown in the app.
Hint
The diagram shows Q contains the –COO– link between two carbon groups; the isomer from the other series must contain –COOH. A much lower vapour pressure signals much stronger intermolecular forces.
Worked answer
a) Functional isomers are compounds with the same molecular formula ✓ but different functional groups (they belong to different homologous series). ✓
b) P: carboxylic acids ✓ · Q: esters ✓
c) Compound P is propanoic acid. Full structural formula: a 3-carbon chain, CH₃–CH₂–COOH, with EVERY bond shown — all C–H bonds, the C=O double bond and the O–H bond on the last carbon. ✓✓ (1 mark for the correctly drawn –COOH group, 1 mark for the rest of the molecule.)
d) Methyl methanoate ✓✓
e) P ✓. P (a carboxylic acid) has hydrogen bonding between its molecules, while the strongest intermolecular force between Q (ester) molecules is dipole–dipole forces. ✓ Hydrogen bonds are STRONGER than dipole–dipole forces ✓ — consistent with P's much lower vapour pressure. MORE ENERGY is therefore needed to overcome the intermolecular forces in P, so P boils at a higher temperature. ✓
Modelled on Nov 2025 Paper 2, Question 3 — same skills, new scenario.
Question 7
Butan-2-one is used as a solvent in some paints. The functional group present in butan-2-one is the … group.
Hint
First decide to which homologous series the suffix -one points.
Worked answer
D. Butan-2-one is a ketone (suffix -one). Ketones contain the carbonyl group, C=O, bonded to two other carbon atoms. A: hydroxyl (−OH) belongs to alcohols. B: carboxyl (−COOH) belongs to carboxylic acids. C: formyl (−CHO) belongs to aldehydes.
Modelled on Nov 2025 Paper 2, Question 1.1 — same skills, new scenario.
Question 8
Consider the following statements about any two members of the same homologous series:
(i) They have the same general formula.
(ii) They have the same molecular formula.
(iii) They contain the same functional group.
Which of these statements are ALWAYS true?
Hint
Members of a series differ by CH₂ — what stays the same?
Worked answer
B. Members of the same homologous series share the same general formula (i) and the same functional group (iii). Consecutive members differ by a CH₂ unit, so their molecular formulas differ — (ii) is false. A omits (iii); C and D wrongly include (ii).
Modelled on Nov 2025 Paper 2, Question 1.3 — same skills, new scenario.