Edexcel A Level Chemistry: Structure, Bonding and Introductory Organic Chemistry — Practice Questions

These are original questions written for Marlbridge, in the style and at the standard of the examination. They are not reproduced past-paper questions — examination boards hold copyright in their own papers. Use these alongside the official past papers available free from your board.

Related: Structure, Bonding and Organic Chemistry revision notes and the full study guide.

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Questions

1. Define a dative covalent bond and give one example, stating which atom provides both electrons. [3]

2. Explain what is meant by polarising power and polarisability, and use them to explain why aluminium chloride behaves covalently. [4]

3. Give the shape and bond angle of: PCl₅, SF₆, NH₄⁺, H₂O. [4]

4. Explain why CO₂ is non-polar but SO₂ is polar. [3]

5. Consider but-2-ene.

(a) State the two conditions required for E/Z isomerism and confirm that but-2-ene meets them. [3] (b) Explain why but-1-ene does not show E/Z isomerism. [2]

6. (This question draws on the Sₙ1/Sₙ2 mechanism distinction, which the specification states is tested in Unit 4, not Unit 1.) A single optically active enantiomer of 2-bromobutane is reacted with aqueous sodium hydroxide.

(a) Name the mechanism and the product. [2] (b) Explain what a chiral centre is and identify whether the product has one. [2] (c) The product is optically inactive. Explain what this reveals about the mechanism. [3]

7. (Unit 2, topic 10 content.) Rank C–F, C–Cl, C–Br and C–I by rate of nucleophilic substitution and explain the order. [3]

8. Explain why ice is less dense than liquid water. [3]

9. (Unit 2, topic 10 content.) 2-bromo-2-methylpropane is heated under reflux with ethanolic potassium hydroxide, instead of aqueous potassium hydroxide.

(a) Name the type of mechanism and give the name of the organic product. [2] (b) State the type of mechanism and the organic product that aqueous potassium hydroxide would give instead. [2]

10. Propene reacts with hydrogen bromide by electrophilic addition to give predominantly 2-bromopropane rather than 1-bromopropane. Explain this observation in terms of carbocation stability, referring to the mechanism by which the H⁺ and Br⁻ add across the double bond. [3]

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Answers

1. A covalent bond in which both electrons of the shared pair originate from the same atom [1] [1]. Example: NH₄⁺, where the nitrogen provides both electrons for the fourth bond [1].

2. Polarising power is the ability of a cation to distort an anion’s electron cloud, increasing with higher charge and smaller radius [1] [1]. Polarisability is how easily an anion’s electron cloud is distorted, increasing with larger radius [1]. Al³⁺ is small and triply charged, so it polarises the chloride ion heavily, drawing electron density into the bond and giving substantial covalent character [1].

3. PCl₅ — trigonal bipyramidal, 120° and 90° [1]. SF₆ — octahedral, 90° [1]. NH₄⁺ — tetrahedral, 109.5° [1]. H₂O — bent, 104.5° [1].

4. Both have polar bonds [1]. CO₂ is linear and symmetrical, so the dipoles cancel [1]. SO₂ is bent because sulfur has a lone pair, so the dipoles do not cancel and there is a net dipole [1].

5. (a) A C=C double bond preventing rotation [1], and two different groups on each of the double-bonded carbons [1]. In but-2-ene each carbon carries a CH₃ and an H, which are different [1]. (b) In but-1-ene the terminal carbon carries two hydrogen atoms [1], which are identical, so the second condition fails [1].

6. (a) Nucleophilic substitution [1]; butan-2-ol [1]. (b) A carbon bonded to four different groups [1]; butan-2-ol’s second carbon carries OH, H, CH₃ and C₂H₅ — four different groups, so yes [1]. (c) A racemic mixture must have formed [1], which means the intermediate was planar [1], allowing the nucleophile to attack with equal probability from either face — indicating an Sₙ1 mechanism via a carbocation [1].

7. C–I > C–Br > C–Cl > C–F [1]. The rate depends on the carbon–halogen bond enthalpy [1]; C–I is the weakest so it breaks most readily, while C–F is by far the strongest despite being the most polar [1].

8. Each water molecule can form four hydrogen bonds, arranged tetrahedrally around it [1]. In ice this produces a rigid, open lattice with holes between molecules [1]. On melting, the lattice partly collapses as molecules pack more closely, so the liquid is denser than the solid [1].

9. (a) Elimination [1]; the organic product is 2-methylpropene [1]. (b) Nucleophilic substitution [1]; the organic product would be 2-methylpropan-2-ol [1].

10. Protonation of propene can give either a secondary carbocation (H⁺ adds to the terminal carbon) or a primary carbocation (H⁺ adds to the middle carbon) [1]. The secondary carbocation is more stable, because the two alkyl groups attached to the positively charged carbon push electron density towards it, inductively stabilising the charge, whereas the primary carbocation has only one such alkyl group [1]. The reaction proceeds predominantly through the more stable secondary carbocation, so bromide attacks the middle carbon and 2-bromopropane is the major product [1].

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Where marks are usually lost

• Confusing polarising power with polarisability.
• Saying SO₂ is linear.
• Giving only one condition for E/Z isomerism.
• Explaining C–X reactivity by electronegativity.
• Explaining ice’s low density without mentioning the tetrahedral arrangement or the open lattice with holes.
• Mixing up which conditions give substitution and which give elimination — aqueous KOH substitutes to an alcohol, ethanolic KOH eliminates to an alkene.
• Explaining Markovnikov’s rule as “hydrogen goes to the carbon with more hydrogens” instead of via carbocation stability, which is the mechanism examiners require.
• Drawing curly arrows starting from an atom or a positive charge — they must always start from a bond or a lone pair, showing a pair of electrons moving.