AQA A-Level Chemistry: Introduction to organic chemistry (7405) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover section 3.3.1 Introduction to organic chemistry of the AQA AS and A-level Chemistry specification (7404/7405), version 1.2, July 2026, for AS and A-level exams June 2016 onwards: 3.3.1.1 Nomenclature, 3.3.1.2 Reaction mechanisms and 3.3.1.3 Isomerism. It is all AS content, so nothing here is A-level only. The section is examined on Paper 2 at AS and A-level, and A-level Paper 3 can test any content.

Learn the ideas first in the study guide and the revision notes. The AQA A-Level Chemistry hub and the printable checklist cover the whole course. Use Ar values H = 1.0, C = 12.0, O = 16.0 where needed.

Questions

1. Define each term.

(a) Structural isomers [1] (b) General formula [1]

2. State three characteristics of a homologous series. [3]

3. Compound Z contains 62.1% carbon, 10.3% hydrogen and 27.6% oxygen by mass. Its Mr is 116.

(a) Calculate the empirical formula of Z. [3] (b) Deduce the molecular formula of Z. [2] (c) Z is an unbranched carboxylic acid. Name Z and give its structural formula. [2]

4. Give the IUPAC name of each compound.

(a) CH₃CH₂CHBrCH₃ [1] (b) (CH₃)₂CHCH₂CHO [1] (c) CH₃CH=CHCH₂CH₂CH₃ [1] (d) CH₃C(CH₃)(OH)CH₂CH₃ [1]

5. Give the structural formula of each compound.

(a) 1,2-dichloropropane [1] (b) pentan-3-one [1] (c) 2,2-dimethylbutanoic acid [1]

6. Each name below is wrong. Give the correct IUPAC name and explain the error.

(a) 3-methylbutane [2] (b) 2-ethylpropan-1-ol [2]

7. Cyclohexane reacts with bromine in ultraviolet light to form bromocyclohexane, C₆H₁₁Br, by a free-radical mechanism.

(a) Write an equation for the initiation step. [1] (b) Write equations for the two propagation steps. [2] (c) Write an equation for a termination step that forms a hydrocarbon with twelve carbon atoms. [1] (d) State what the dot in a radical such as Br• represents. [1]

8. This question is about curly arrows.

(a) A hydroxide ion reacts with 2-bromopropane to form propan-2-ol and a bromide ion in a single step that needs two curly arrows. For each arrow, state where it starts and where it ends. [4] (b) State what a curly arrow represents. [1] (c) In the first step of the reaction between propene and hydrogen bromide, a new bond forms between a carbon atom and a hydrogen atom. State where the curly arrow that forms this bond starts, and explain why it does not start from a lone pair. [2]

9. There are four structural isomers of C₄H₉Cl.

(a) Name the four isomers. [4] (b) From your answer, identify one pair of chain isomers and one pair of position isomers. [2]

10. This question is about E–Z isomerism.

(a) Explain why but-2-ene has E–Z isomers but but-1-ene does not. [2] (b) An isomer of 1-chloro-2-methylbut-1-ene has this structure:

    Cl       CH₂CH₃
      \     /
       C = C
      /     \
     H       CH₃

(i) Use the CIP priority rules to identify the higher-priority group on each carbon of the C=C. Explain your choices. [3] (ii) State whether this is the E or the Z isomer. [1]

11. A 0.350 g sample of hydrocarbon W contains 0.300 g of carbon. The Mr of W is 70.0.

(a) Calculate the empirical formula and the molecular formula of W. [3] (b) W is an alkene with an unbranched chain. Name the two possible structures of W. [2] (c) One of these structures has E–Z isomers. Name it and draw both stereoisomers, labelling each E or Z. [3] (d) Name a cyclic compound that is a functional group isomer of W. [1]

12. This question is about skeletal formulas.

(a) The skeletal formula of compound P is a hexagon with an OH group on one corner. Give the molecular formula and the name of P. [2] (b) The skeletal formula of compound Q is a zigzag of five lines, with =O on the second carbon from one end. Give the name and the molecular formula of Q. [2] (c) State the type of isomerism shown by P and Q. [1]

Answers

1. (a) Compounds with the same molecular formula but a different structural formula [1]. (b) An algebraic formula that applies to every member of a homologous series, e.g. CₙH₂ₙ₊₂ [1]. [2] Examiner insight: “Same formula, different structure” is too vague for (a); say which formula is the same and which differs.

2. Same general formula [1]; successive members differ by CH₂ [1]; same functional group so similar chemical properties, or a gradual trend in physical properties [1]. [3] Examiner insight: “Similar properties” with no word “chemical” is usually too vague — physical properties change gradually along the series.

3. (a) Moles: C = 62.1/12.0 = 5.175; H = 10.3/1.0 = 10.3; O = 27.6/16.0 = 1.725 [1]. Divide by 1.725: C = 3.00, H = 5.97, O = 1.00 [1]. Empirical formula C₃H₆O [1]. (b) Empirical formula mass = 58.0, and 116/58.0 = 2 [1]; molecular formula C₆H₁₂O₂ [1]. (c) Hexanoic acid [1]; CH₃CH₂CH₂CH₂CH₂COOH [1]. [7] Examiner insight: Show the division by Ar; an error in (a) can usually be carried forward into (b), but only if the working is visible.

4. (a) 2-bromobutane [1] (b) 3-methylbutanal [1] (c) hex-2-ene [1] (d) 2-methylbutan-2-ol [1] [4] Examiner insight: For (b), numbering starts at the CHO carbon, so “2-methylbutanal” counts from the wrong end and earns nothing.

5. (a) CH₂ClCHClCH₃ [1] (b) CH₃CH₂COCH₂CH₃ [1] (c) CH₃CH₂C(CH₃)₂COOH [1] [3] Examiner insight: A structural formula must be unambiguous; writing C₆H₁₂O₂ for (c) is a molecular formula and gains no credit.

6. (a) 2-methylbutane [1]. The chain must be numbered from the end nearer the branch, giving the lower locant [1]. (b) 2-methylbutan-1-ol [1]. The longest chain containing the C–OH has four carbons, not three; the “ethyl” group is part of the main chain [1]. [4] Examiner insight: Drawing the structure from the wrong name first, then renaming it, is the safest route; the explanation mark needs the rule, not just the new name.

7. (a) Br₂ → 2Br• [1] (b) Br• + C₆H₁₂ → C₆H₁₁• + HBr [1]; C₆H₁₁• + Br₂ → C₆H₁₁Br + Br• [1] (c) 2C₆H₁₁• → C₁₂H₂₂ [1] (d) An unpaired electron [1]. [5] Examiner insight: A radical written without its dot is usually not credited, and curly arrows are not needed in these steps.

8. (a) Arrow 1 starts from a lone pair on the oxygen of OH⁻ [1] and ends at the carbon bonded to bromine (the δ+ carbon) [1]. Arrow 2 starts from the C–Br bond [1] and ends at the bromine atom, which leaves as Br⁻ [1]. (b) The movement of a pair of electrons [1]. (c) It starts from the C=C double bond [1]. The electron pair for the new bond comes from a covalent bond (the double bond), not from a lone pair, because propene has no lone pairs [1]. [7] Examiner insight: An arrow drawn from the minus sign of OH⁻ instead of from a lone pair is a common way to lose the first mark in (a).

9. (a) 1-chlorobutane [1]; 2-chlorobutane [1]; 1-chloro-2-methylpropane [1]; 2-chloro-2-methylpropane [1]. (b) Chain isomers: 1-chlorobutane and 1-chloro-2-methylpropane (or 2-chlorobutane and 2-chloro-2-methylpropane) [1]. Position isomers: 1-chlorobutane and 2-chlorobutane (or 1-chloro-2-methylpropane and 2-chloro-2-methylpropane) [1]. [6] Examiner insight: Naming one compound twice under different names (for example 3-chlorobutane for 2-chlorobutane) does not count as a fourth isomer.

10. (a) In but-2-ene each carbon of the C=C has two different groups (H and CH₃), and there is restricted rotation about the C=C [1]. In but-1-ene, carbon 1 has two identical groups (H and H), so no E–Z isomers [1]. (b)(i) Left carbon: Cl has a higher atomic number (17) than H (1), so Cl is higher priority [1]. Right carbon: CH₃ and CH₂CH₃ are both attached through carbon, a tie [1]; at the next atoms CH₂CH₃ has (C, H, H) and CH₃ has (H, H, H), so CH₂CH₃ is higher priority [1]. (ii) Cl and CH₂CH₃ are on the same side: Z [1]. [6] Examiner insight: In (b)(i) the tie at the first atom must be stated before comparing further out; “ethyl is bigger” does not earn the mark.

11. (a) Mass of H = 0.350 − 0.300 = 0.050 g; moles C = 0.300/12.0 = 0.0250, H = 0.050/1.0 = 0.050 [1]. Ratio 1 : 2, empirical formula CH₂ [1]. 70.0/14.0 = 5, molecular formula C₅H₁₀ [1]. (b) pent-1-ene [1]; pent-2-ene [1]. (c) Pent-2-ene [1].

   H₃C       CH₂CH₃            H₃C       H
      \     /                     \     /
       C = C                       C = C
      /     \                     /     \
     H       H                   H       CH₂CH₃
       (Z)                          (E)

Z isomer with CH₃ and CH₂CH₃ on the same side [1]; E isomer with them on opposite sides [1]. (d) Cyclopentane (or methylcyclobutane) [1]. [9] Examiner insight: In (c) each drawing needs the correct label; two correct drawings with the labels swapped lose both drawing marks.

12. (a) C₆H₁₂O [1]; cyclohexanol [1]. (b) Hexan-2-one [1]; C₆H₁₂O [1]. (c) Functional group isomerism (same molecular formula, different functional groups) [1]. [5] Examiner insight: A five-line zigzag has six carbons, because both line ends count; giving pentan-2-one loses the name mark.

Where marks are usually lost

• Not showing the mole ratio step in empirical formula work, so no method credit can be given.
• Losing count of carbons in a skeletal formula by missing the line ends.
• Numbering from the wrong end, or adding a locant to -al or -oic acid names.
• Choosing a “longest chain” that leaves out the functional group carbon.
• Writing radicals without dots, or adding curly arrows to free-radical steps.
• Starting curly arrows from atoms or charges rather than from lone pairs or bonds.
• Counting the same isomer twice under two names.
• Ranking CIP groups by size or mass rather than by atomic number.
• Explaining E–Z isomerism without mentioning restricted rotation about the C=C.

Next steps

• Introduction to organic chemistry revision notes
• Introduction to organic chemistry study guide
• AQA A-Level Chemistry exam preparation
• AQA A-Level Chemistry hub
• Printable checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

AQA, AS and A-level Chemistry specification (7404/7405), version 1.2, July 2026, for AS and A-level exams June 2016 onwards: section 3.3.1 Introduction to organic chemistry (3.3.1.1 to 3.3.1.3).