OCR A Level Chemistry A: Basic concepts of organic chemistry (H432) – 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.

Here you practise section 4.1.1, Basic concepts of organic chemistry, of the OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017. The questions test every learning outcome, 4.1.1(a) to (i), from Module 4, Core organic chemistry. H432 is the full A level. Questions 8, 10 and 11 are longer and draw on several outcomes at once.

Revise first with the study guide and the revision notes. Other topics are on the OCR chemistry hub; mark your progress on the printable checklist. For a check across the whole course, use the free diagnostics.

Relative atomic masses for the calculations: C = 12.0, H = 1.0, O = 16.0, Cl = 35.5.

Questions

1. Give the systematic name of each compound.

(a) CH₃CH₂C(CH₃)₂CH₂CH₂CH₃ [1] (b) CH₃CHClCH(CH₃)CH₂OH [1] (c) CH₃CH₂CH=C(CH₃)CH₃ [1]

2. Write the structural formula of each compound.

(a) 4-ethyl-2-methylheptane [1] (b) but-3-en-2-ol [1] (c) 1,1,2-trichloroethane [1]

3. For each compound, state whether it is aliphatic (but not alicyclic), alicyclic or aromatic, and whether it is saturated or unsaturated.

(a) methylcyclohexane [1] (b) ethylbenzene [1] (c) CH₃CH₂C≡CH [1] (d) cyclopentene [1]

4.

(a) Define the term homologous series. [2] (b) State what is meant by a functional group. [1] (c) Give the general formula of an alkyl group and the name of the alkyl group with four carbon atoms. [1]

5.

(a) A chloroalkane CₙH₂ₙ₊₁Cl has Mᵣ = 120.5. Find its molecular formula. [2] (b) Predict the molecular formula of the alcohol in the series CₙH₂ₙ₊₁OH that has twelve carbon atoms. [1] (c) Show that every alkene with one C=C bond has the same percentage by mass of carbon, and give that value to three significant figures. [2]

6. Ethyl propanoate is an ester. Its molecule has a CH₃CH₂ group bonded to a C=O carbon, which is bonded through an O atom to a CH₂CH₃ group.

(a) State the molecular formula of ethyl propanoate. [1] (b) Deduce the empirical formula. [1] (c) Write its structural formula, showing the ester group the way the specification expects. [1] (d) State which atoms a skeletal formula of ethyl propanoate leaves out. [1]

7. Heptane, C₇H₁₆, has several structural isomers.

(a) Explain the term structural isomers. [1] (b) Five isomers of C₇H₁₆ have a longest chain of exactly five carbon atoms. Name all five. [5]

8. Alcohol P belongs to the series CₙH₂ₙ₊₁OH and has Mᵣ = 102.0.

(a) Find the molecular formula of P. [2] (b) Three isomers of P have an unbranched carbon chain. Give the structural formula and name of each. [3] (c) A student names one of these “hexan-4-ol”. Explain the error and state the correct name. [1] (d) Give the structural formula of 2-methylpentan-2-ol and state, with a reason, whether it is a structural isomer of P. [2]

9.

(a) Describe homolytic fission of a covalent bond. [1] (b) Write an equation for the homolytic fission of the C–C bond in ethane. [1] (c) The C–I bond in 2-iodo-2-methylbutane, CH₃CH₂C(CH₃)₂I, breaks heterolytically. Write an equation for this fission, show the dipole on the bond, and explain which atom receives the bonding pair. [3] (d) From this list, identify every radical: Br•, H⁺, (CH₃)₃C•, CN⁻, Cl⁻. [1]

10. Ammonia reacts with bromomethane in one step to form CH₃NH₃⁺ and Br⁻.

(a) State what a curly arrow represents. [1] (b) Show the dipole on the C–Br bond and explain its origin. [2] (c) A student’s mechanism has two arrows. Arrow 1 starts on an H atom of NH₃ and ends on the carbon atom. Arrow 2 starts on the Br atom and ends on the carbon atom. Identify the error in each arrow and describe the correct arrow. [4]

11. Compound W contains 24.4% carbon, 3.4% hydrogen and 72.2% chlorine by mass. Its Mᵣ is 147.5.

(a) Calculate the empirical formula of W. [3] (b) Deduce the molecular formula of W. [1] (c) W is one of five structural isomers with this molecular formula, all on an unbranched three-carbon chain. Name all five. [5] (d) State whether W is saturated or unsaturated, and give a reason. [1]

Answers

1. (a) Six-carbon chain, two methyls on C3: 3,3-dimethylhexane [1] (b) Four-carbon chain containing C–OH, numbered from the OH end: 3-chloro-2-methylbutan-1-ol [1] (c) Number from the end nearest the C=C: 2-methylpent-2-ene [1] Examiner insight: In (b), numbering from the CH₃ end gives the substituents the same locants, {2, 3}, but puts the OH on C4; the suffix group takes priority, so “2-chloro-3-methylbutan-4-ol” is unlikely to gain credit.

2. (a) CH₃CH(CH₃)CH₂CH(C₂H₅)CH₂CH₂CH₃ [1] (b) CH₂=CHCH(OH)CH₃ [1] (c) CHCl₂CH₂Cl [1] Examiner insight: A structural formula must be unambiguous; C₄H₈O for (b) is a molecular formula and describes several compounds, so it does not answer the question.

3. (a) Alicyclic, saturated [1] (b) Aromatic, unsaturated [1] (c) Aliphatic, unsaturated (C≡C) [1] (d) Alicyclic, unsaturated (C=C in the ring) [1] Examiner insight: Here each mark needs both parts right; calling cyclopentene aromatic because it is a ring is a common way to lose (d).

4. (a) A series of organic compounds with the same functional group [1], each successive member differing by CH₂ [1] (b) The group of atoms responsible for the characteristic reactions of a compound [1] (c) CₙH₂ₙ₊₁; butyl [1] Examiner insight: “Similar chemical properties” is not a substitute for “the same functional group”; the specification’s definition names the functional group and the CH₂ difference.

5. (a) Mᵣ = 14n + 1 + 35.5, so 14n + 36.5 = 120.5 [1]; n = 6, C₆H₁₃Cl [1] (b) 2(12) + 1 = 25 hydrogens on the chain: C₁₂H₂₅OH [1] (c) % C = (12n ÷ 14n) × 100, and n cancels [1]; 85.7% [1] Examiner insight: On a “show that” part, quoting 85.7% from one example such as ethene does not show it holds for every member; the algebra with n is the point.

6. (a) C₅H₁₀O₂ [1] (b) The ratio 5 : 10 : 2 has no common factor, so C₅H₁₀O₂ [1] (c) CH₃CH₂COOCH₂CH₃ [1] (d) The hydrogen atoms bonded to carbon in the alkyl chains [1] Examiner insight: For (b), dividing by 2 gives non-whole numbers, which is not a formula; when no common factor exists, the empirical formula equals the molecular formula.

7. (a) Compounds with an identical molecular formula whose structural formulae differ [1] (b) 2,2-dimethylpentane [1]; 2,3-dimethylpentane [1]; 2,4-dimethylpentane [1]; 3,3-dimethylpentane [1]; 3-ethylpentane [1] Examiner insight: Names such as “2-ethylpentane” or “4,4-dimethylpentane” describe chains numbered or chosen wrongly; each is really another listed isomer or a hexane, so it adds nothing to the list.

8. (a) Mᵣ = 14n + 18 = 102.0 [1]; n = 6, C₆H₁₃OH [1] (b) CH₃CH₂CH₂CH₂CH₂CH₂OH, hexan-1-ol [1]; CH₃CH(OH)CH₂CH₂CH₂CH₃, hexan-2-ol [1]; CH₃CH₂CH(OH)CH₂CH₂CH₃, hexan-3-ol [1] (c) The chain is numbered from the wrong end; the OH must get the lowest number, so it is hexan-3-ol [1] (d) (CH₃)₂C(OH)CH₂CH₂CH₃ and it is a structural isomer: same molecular formula C₆H₁₄O, different structure [1] [1] Examiner insight: In (b), drawing hexan-1-ol a second time with the OH on the far end looks like a new isomer but is the same compound; naming each structure exposes the repeat.

9. (a) Each bonding atom receives one electron from the shared pair, forming two radicals [1] (b) CH₃–CH₃ → CH₃• + CH₃• [1] (c) CH₃CH₂C(CH₃)₂–I with Cδ+ and Iδ− [1]; CH₃CH₂C(CH₃)₂I → CH₃CH₂C⁺(CH₃)₂ + I⁻ [1]; iodine takes both electrons because it is more electronegative than carbon [1] (d) Br• and (CH₃)₃C• [1] Examiner insight: In (b), writing CH₃⁺ and CH₃⁻ describes heterolytic fission; the question asks for homolytic, and the dots on both products are what show it.

10. (a) The movement of a pair of electrons [1] (b) Cδ+–Brδ− [1]; bromine is more electronegative than carbon, so it attracts the bonding pair more strongly [1] (c) Arrow 1: an H atom has no lone pair to donate [1]; the arrow should start from the lone pair on N and end on the δ+ carbon [1]. Arrow 2: it points the wrong way and starts on an atom [1]; it should start from the C–Br bond and end on the Br atom [1] Examiner insight: Examiners look at exactly where each arrow tail sits; a tail that starts near the N but not on the lone pair or a charge can read as starting from the atom.

11. (a) C: 24.4 ÷ 12.0 = 2.03; H: 3.4 ÷ 1.0 = 3.4; Cl: 72.2 ÷ 35.5 = 2.03 [1]; ratio 1 : 1.67 : 1 [1]; multiply by 3: C₃H₅Cl₃ [1] (b) Empirical formula mass = 36.0 + 5.0 + 106.5 = 147.5, equal to Mᵣ, so C₃H₅Cl₃ [1] (c) 1,1,1-trichloropropane [1]; 1,1,2-trichloropropane [1]; 1,1,3-trichloropropane [1]; 1,2,2-trichloropropane [1]; 1,2,3-trichloropropane [1] (d) Saturated: only single C–C bonds; the formula fits CₙH₂ₙ₊₂ with three H replaced by Cl [1] Examiner insight: Rounding 1.67 to 2 in (a) gives CH₂Cl, which does not fit the Mᵣ of 147.5; a ratio ending near .33 or .67 is a signal to multiply by 3.

Where marks are usually lost

• Picking a chain that is not the longest, as in questions 1 and 7.
• Numbering from the end that suits the substituents instead of the OH or C=C, as in question 8.
• Writing prefixes in number order rather than alphabetical order.
• Giving a molecular formula when a structural formula is asked for.
• Calling an alicyclic alkene aromatic.
• Leaving “same functional group” or “differ by CH₂” out of the homologous series definition.
• Listing the same isomer twice under two names.
• Dropping the dots on radicals or writing them as ions.
• Starting a curly arrow on an atom, or pointing it away from where the pair goes.
• Rounding a mole ratio such as 1.67 instead of scaling it.

Next steps

• Recap the tables and rules: revision notes
• Worked examples and explanations: study guide
• Empirical formula practice: Amount of substance practice
• Bond polarity: Bonding and structure practice
• Check the whole course: all free 10-minute diagnostics
• More OCR chemistry topics: course hub
• Printable checklist: OCR A Level Chemistry checklist
• Book a free trial class: /trial/

Official syllabus

OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017, section 4.1.1 Basic concepts of organic chemistry, learning outcomes (a) to (i). Published by OCR.