OxfordAQA A-Level Chemistry: Introduction to organic chemistry (9620) – 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 OxfordAQA International AS and A-level Chemistry (9620) specification, Version 5.3, for International AS and A-level exams May/June 2020 onwards: 3.3.1.1 Nomenclature, 3.3.1.2 Reaction mechanisms and 3.3.1.3 Isomerism. It is entirely International AS content; nothing is International A-level only. Unit 2 (Organic 1 and Physical 1) is where the specification assesses it, and questions there can also use amount of substance (3.1.2).

Study the ideas first in the study guide and revision notes. Use Ar values H = 1.0, C = 12.0, Cl = 35.5, Br = 79.9.

Questions

1. This question is about definitions.

(a) Define the term stereoisomers. [1] (b) Apart from containing the same functional group, state two features shared by the members of a homologous series. [2]

2. 1,4-Dibromobutane has the structural formula BrCH₂CH₂CH₂CH₂Br.

(a) Give its molecular formula. [1] (b) Give its empirical formula. [1] (c) Describe its skeletal formula. [1]

3. Compound R contains 17.8% carbon, 3.0% hydrogen and 79.2% bromine by mass. Its Mr is 201.8.

(a) Calculate the empirical formula of R. [3] (b) Deduce the molecular formula of R. [1] (c) R has an unbranched chain with one bromine atom on each end carbon. Name R. [1]

4. Give the IUPAC name of each compound.

(a) CH₃CH₂CH(CH₃)CH₂CH₂CH₃ [1] (b) CH₂=CHCH₂CH(CH₃)CH₃ [1] (c) CH₃CH(OH)CH₂CH(CH₃)CH₃ [1] (d) (CH₃)₃CCHO [1]

5. This question is about drawing and reading formulas.

(a) Give the structural formula of 2-bromo-3-methylbutane. [1] (b) Give the structural formula of 2-methylpentanenitrile. [1] (c) A skeletal formula shows a hexagon with one double line in the ring and Br on the corner next to the double bond. Give the molecular formula of this compound. [1]

6. Each name below breaks an IUPAC rule. Give the correct name and state the rule broken.

(a) 2-ethylbutane [2] (b) 2-methylbutan-3-ol [2]

7. Dichloromethane, CH₂Cl₂, reacts with chlorine in ultraviolet light to form trichloromethane, CHCl₃, by a free-radical mechanism.

(a) Give the equation for the step that forms chlorine radicals. [1] (b) Write equations for the two steps that together convert CH₂Cl₂ into CHCl₃ and regenerate a chlorine radical. [2] (c) Two •CHCl₂ radicals meet. Write the equation for this step and name the organic product. [2]

8. This question is about curly arrows.

(a) Iodoethane reacts with cyanide ions: CH₃CH₂I + CN⁻ → CH₃CH₂CN + I⁻. Two curly arrows are needed. Give the start point and the end point of each. [4] (b) A water molecule bonds to the carbocation (CH₃)₃C⁺. State where the curly arrow starts and ends, and give the formula of the species formed. [2]

9. Hexane, C₆H₁₄, has several structural isomers.

(a) Name the five structural isomers of C₆H₁₄, including hexane. [5] (b) State the type of structural isomerism they show. [1]

10. For each alkene, state whether it has E–Z isomers and give a reason.

(a) hex-3-ene [1] (b) 3-ethylpent-2-ene [1] (c) 1-bromoprop-1-ene [1] (d) 2-methylpent-2-ene [1]

11. An alkene has this structure.

    Br       CH(CH₃)₂
      \     /
       C = C
      /     \
    Cl       CH₂CH₂CH₃

(a) Apply the CIP priority rules to decide which group ranks higher on each carbon of the double bond, explaining each decision. [3] (b) Deduce whether the structure shown is the E isomer or the Z isomer. [1]

12. Compound X contains 53.0% carbon, 7.7% hydrogen and 39.3% chlorine by mass. Its Mr is 90.5.

(a) Show that the molecular formula of X is C₄H₇Cl. [3] (b) X has an unbranched chain, a C=C between carbon 1 and carbon 2, and its chlorine atom on carbon 1. Name X. [1] (c) Draw the Z isomer of X, using the CIP rules to justify your answer. [3] (d) Name a position isomer of X that has no E–Z isomers, and explain why it has none. [2]

Answers

1. (a) Compounds that share a structural formula but have their atoms arranged differently in space [1]. (b) Any two, one mark each: same general formula; successive members differ by CH₂; similar chemical reactions; gradual trend in physical properties such as boiling point [2]. [3] Examiner insight: A stereoisomer definition built on “molecular formula” describes isomers in general; the key point is the same structural formula.

2. (a) C₄H₈Br₂ [1] (b) C₂H₄Br (divide by 2) [1] (c) A zigzag of three lines (four carbons, counting both ends), with Br written on each end [1]. [3] Examiner insight: An empirical formula must be fully simplified; C₄H₈Br₂ given for (b) only repeats (a) rather than answering the question.

3. (a) Moles: C = 17.8 / 12.0 = 1.483; H = 3.0 / 1.0 = 3.0; Br = 79.2 / 79.9 = 0.991 [1]. Divide by 0.991: C = 1.50, H = 3.03, Br = 1.00 [1]. Multiply by 2: C₃H₆Br₂ [1]. (b) Empirical formula mass = 36.0 + 6.0 + 159.8 = 201.8, equal to Mr, so the molecular formula is C₃H₆Br₂ [1]. (c) 1,3-dibromopropane [1]. [5] Examiner insight: Rounding 1.50 to 2 gives a wrong formula; a ratio ending in .5 is a signal to double every value.

4. (a) 3-methylhexane [1] (b) 4-methylpent-1-ene [1] (c) 4-methylpentan-2-ol [1] (d) 2,2-dimethylpropanal [1] [4] Examiner insight: In (c) the –OH sets the numbering, so “2-methylpentan-4-ol” is not a correct name even though the methyl locant looks lower.

5. (a) CH₃CHBrCH(CH₃)CH₃ [1] (b) CH₃CH₂CH₂CH(CH₃)CN [1] (c) C₆H₉Br (cyclohexene is C₆H₁₀; one H replaced by Br) [1] [3] Examiner insight: In (b) the CN carbon is C1 of the five-carbon stem, so only four more chain carbons follow it; writing CN on the end of a five-carbon chain adds a carbon.

6. (a) 3-methylpentane [1]. The longest chain has five carbons; the “ethyl” carbons are part of the main chain [1]. (b) 3-methylbutan-2-ol [1]. The –OH group must get the lowest possible locant [1]. [4] Examiner insight: The question asks for the rule as well as the corrected name, so state the rule in words rather than leaving the new name to speak for itself.

7. (a) Cl₂ → 2Cl• [1] (b) Cl• + CH₂Cl₂ → •CHCl₂ + HCl [1]; •CHCl₂ + Cl₂ → CHCl₃ + Cl• [1] (c) 2•CHCl₂ → CHCl₂CHCl₂ [1]; 1,1,2,2-tetrachloroethane [1]. [5] Examiner insight: Check that each equation balances and that every radical carries its dot; curly arrows are not required here, so do not spend time on them.

8. (a) Arrow 1 starts from the lone pair on the carbon of CN⁻ [1] and ends at the carbon bonded to iodine (the δ+ carbon) [1]. Arrow 2 starts from the C–I bond [1] and ends at the iodine atom, which leaves as I⁻ [1]. (b) From a lone pair on the oxygen of H₂O to the positive carbon [1]; forms (CH₃)₃COH₂⁺ [1]. [6] Examiner insight: Drawing the first arrow from the nitrogen, or from the minus sign, does not show where the new C–C bond comes from; the carbon lone pair forms it.

9. (a) hexane [1]; 2-methylpentane [1]; 3-methylpentane [1]; 2,2-dimethylbutane [1]; 2,3-dimethylbutane [1]. (b) Chain isomerism [1]. [6] Examiner insight: “4-methylpentane” or “2-ethylbutane” are repeats of isomers already listed under a wrong name and add nothing.

10. (a) Yes: each C=C carbon carries H and CH₂CH₃ [1]. (b) No: carbon 3 carries two identical CH₂CH₃ groups [1]. (c) Yes: carbon 1 carries Br and H; carbon 2 carries CH₃ and H [1]. (d) No: carbon 2 carries two identical CH₃ groups [1]. [4] Examiner insight: Give the yes or no together with a reason tied to the groups on a named carbon of the C=C; a bare yes or no answers only half the question.

11. (a) Left carbon: Br (atomic number 35) beats Cl (17) [1]. Right carbon: both groups attach through C, a tie [1]; next atoms are (C, C, H) for CH(CH₃)₂ and (C, H, H) for CH₂CH₂CH₃, so CH(CH₃)₂ has the higher priority [1]. (b) Br and CH(CH₃)₂ are on the same side: Z [1]. [4] Examiner insight: State the tie before comparing further out; “the branched group is bigger” is not a CIP reason.

12. (a) Moles: C = 53.0 / 12.0 = 4.417; H = 7.7 / 1.0 = 7.7; Cl = 39.3 / 35.5 = 1.107 [1]. Divide by 1.107: C = 3.99, H = 6.96, Cl = 1.00, so the empirical formula is C₄H₇Cl [1]. Its mass is 48.0 + 7.0 + 35.5 = 90.5 = Mr, so the molecular formula is C₄H₇Cl [1]. (b) 1-chlorobut-1-ene [1]. (c) Carbon 1: Cl (17) beats H (1) [1]. Carbon 2: CH₂CH₃ (C, 6) beats H [1].

    Cl       CH₂CH₃
      \     /
       C = C
      /     \
     H       H
   (Z)-1-chlorobut-1-ene

Cl and CH₂CH₃ drawn on the same side [1]. (d) 2-chlorobut-1-ene (or 3-chlorobut-1-ene or 4-chlorobut-1-ene) [1]; carbon 1 of the C=C carries two H atoms, so there are no E–Z isomers [1]. [9] Examiner insight: In a “show that” the final formula is already given, so show every stage: moles, ratio and empirical formula mass compared with Mr.

Where marks are usually lost

• Rounding a ratio such as 1.50 instead of doubling every value.
• Giving an empirical formula that can still be simplified.
• Leaving the CN carbon out of a nitrile’s stem, or adding an extra one.
• Numbering an alcohol from the end nearer a branch instead of the –OH.
• Repeating an isomer under a second, wrong name.
• Writing radicals without dots or radical equations that do not balance.
• Starting a curly arrow on a charge or the wrong atom of an ion.
• Deciding CIP priority by size, mass or number of atoms.
• Labelling a drawing Z when the higher-priority groups are on opposite sides.

Next steps

• Revision notes: introduction to organic chemistry
• Study guide: introduction to organic chemistry
• Amount of substance
• OxfordAQA A-Level Chemistry hub
• Printable checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

OxfordAQA, International AS and A-level Chemistry (9620) specification, Version 5.3, for International AS and A-level exams May/June 2020 onwards: section 3.3.1 Introduction to organic chemistry (International AS), points 3.3.1.1 to 3.3.1.3.