IB DP Chemistry Reactivity 3.4: Electron-pair sharing: nucleophiles, electrophiles and substitution – 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 – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

This set checks Reactivity 3.4.1 to 3.4.5 against the IB’s DP Chemistry guide (2023 edition; first exams in 2025), and it suits every session from 2025. Every question is SL and HL content. Mechanism detail beyond describing how electron pairs move is not asked for, because the guide does not require it at SL for these understandings. Where masses are needed, use these Ar values: Br 79.90, C 12.01, Cl 35.45, H 1.01, I 126.90, N 14.01, O 16.00.

Before you start, you may want the study guide open. Other units sit on the IB DP chemistry hub; track progress with our checklist, and spot weak areas with a free diagnostic.

Questions

1. Explain what an electrophile is, and name one electrophile that carries no charge. [2]

2. Identify the nucleophile in each reaction and state whether it is neutral or negatively charged.

(a) CH₃CH₂CH₂Cl + CH₃O⁻ → CH₃CH₂CH₂OCH₃ + Cl⁻ [1] (b) (CH₃)₂CHI + H₂O → (CH₃)₂CHOH + HI [1] (c) CH₃CH₂Br + CH₃NH₂ → CH₃CH₂NHCH₃ + HBr [1]

3. Classify each species as a nucleophile or an electrophile: Cl⁻, NO₂⁺, NH₃, (CH₃)₃C⁺. [4]

4. This question is about heterolytic fission.

(a) Define heterolytic fission. [1] (b) Write an equation for the heterolytic fission of the C–Cl bond in 2-chloro-2-methylbutane, CH₃CH₂C(CH₃)₂Cl, and describe where the curly arrow starts and ends. [2] (c) State how the products would differ if the same bond broke homolytically. [1]

5. 1-Chlorohexane, CH₃(CH₂)₅Cl, undergoes nucleophilic substitution.

(a) Write an equation for its reaction with hydroxide ions. [1] (b) Write an equation for its reaction with cyanide ions and name the organic product. [2] (c) Write an equation for its reaction with excess ammonia. [2]

6. 2-Iodobutane reacts with aqueous hydroxide ions to form butan-2-ol. Describe and explain the movement of electron pairs in this reaction, and write the equation. [4]

7. Cycloheptene, C₇H₁₂, is a symmetrical cyclic alkene. Deduce equations, using molecular formulas, for its reaction with:

(a) chlorine [1] (b) hydrogen bromide [1] (c) water, stating the conditions used. [2]

8. Explain why alkenes are attacked by electrophiles, and why the reactions are classed as additions. [3]

9. A student has two colourless liquids, hexane and hex-1-ene. She shakes a sample of each with bromine water in a dark cupboard.

(a) Describe and explain what she sees with hex-1-ene. [2] (b) Describe and explain what she sees with hexane. [2]

10. Butan-2-ol can be made from but-2-ene, CH₃CH=CHCH₃, by two routes.

(a) Route 1 is a single step. Write the equation and state the conditions. [2] (b) Route 2 has two steps: but-2-ene reacts with hydrogen bromide, and the product then reacts with hydroxide ions. Write an equation for each step. [2] (c) Name the type of reaction in each step of route 2. [2] (d) Identify the electrophile in step 1 and the nucleophile in step 2 of route 2. [2]

11. A sample of 2.52 g of 2,3-dimethylbut-2-ene, (CH₃)₂C=C(CH₃)₂, reacts with excess hydrogen bromide. The mass of 2-bromo-2,3-dimethylbutane collected is 3.29 g.

(a) Explain why only one organic product forms. [1] (b) Write the equation for the reaction. [1] (c) Calculate the percentage yield. [4] (d) State the role of hydrogen bromide in this reaction. [1]

12. A chemist heats 3.75 g of 2-chlorobutane, CH₃CH₂CHClCH₃, with excess cyanide ions and isolates 2.80 g of the organic product.

(a) Give an equation for this substitution. [1] (b) Give the IUPAC name of the nitrile formed. [1] (c) Identify the nucleophile and the leaving group. [2] (d) Calculate the percentage yield. [4]

Answers

1. A species that accepts an electron pair from its partner (the nucleophile) to form a new covalent bond [1]. Neutral example: Br₂, HBr, HCl or the δ+ carbon of a halogenoalkane [1]. [2] Examiner insight: “A positive ion” is not a definition; an answer has to mention accepting an electron pair to describe what an electrophile does.

2. (a) CH₃O⁻ (methoxide ion), negatively charged [1] (b) H₂O, neutral [1] (c) CH₃NH₂ (methylamine), neutral [1] [3] Examiner insight: Look for the atom that ends up bonded to carbon with its own lone pair; naming the halogenoalkane as the nucleophile scores nothing.

3. Cl⁻: nucleophile [1]. NO₂⁺: electrophile [1]. NH₃: nucleophile [1]. (CH₃)₃C⁺: electrophile [1]. [4] Examiner insight: Charge is a guide, not the rule: decide by whether the species has a pair to give (nucleophile) or room to accept one (electrophile).

4. (a) A covalent bond breaks with both bonding electrons going to one fragment, forming two ions [1] (b) CH₃CH₂C(CH₃)₂Cl → CH₃CH₂C⁺(CH₃)₂ + Cl⁻ [1]; the curly arrow begins on the C–Cl bond and its head points at chlorine [1] (c) Homolytic fission would give two radicals (one electron each), not ions [1] [4] Examiner insight: Arrows drawn from the carbon, or from the plus sign, claim electrons come from somewhere they do not, so they earn nothing.

5. (a) CH₃(CH₂)₅Cl + OH⁻ → CH₃(CH₂)₅OH + Cl⁻ [1] (b) CH₃(CH₂)₅Cl + CN⁻ → CH₃(CH₂)₅CN + Cl⁻ [1]; heptanenitrile [1] (c) CH₃(CH₂)₅Cl + 2NH₃ → CH₃(CH₂)₅NH₂ + NH₄Cl: hexan-1-amine as the organic product [1]; 2NH₃ and NH₄Cl to balance [1] [5] Examiner insight: The nitrile name counts the CN carbon, so six carbons plus one gives heptanenitrile; “hexanenitrile” loses the naming mark.

6. Iodine is more electronegative than carbon, so the carbon bonded to iodine carries δ+ [1]. A lone pair on the oxygen of OH⁻ is donated to this δ+ carbon, forming a C–O bond [1]. The C–I bonding pair moves onto the iodine atom and the bond breaks; iodide is the leaving group [1]. CH₃CH₂CHICH₃ + OH⁻ → CH₃CH₂CH(OH)CH₃ + I⁻ [1] [4] Examiner insight: “Describe and explain” needs the reason (bond polarity), not just the arrows; an answer that only lists what moves gives no explanation.

7. (a) C₇H₁₂ + Cl₂ → C₇H₁₂Cl₂ [1] (b) C₇H₁₂ + HBr → C₇H₁₃Br [1] (c) C₇H₁₂ + H₂O → C₇H₁₃OH [1]; steam with an acid catalyst such as phosphoric acid, heated [1] [4] Examiner insight: Check that each product formula equals alkene plus reagent; a missing hydrogen in the HBr or water product is the most common slip.

8. The C=C bond contains a pi bond whose electrons lie above and below the plane of the molecule, giving a region of high electron density [1]. This electron-rich region attracts electron-poor species and donates a pair to them, so electrophiles attack [1]. The pi bond breaks and the whole reagent bonds to the two carbons with nothing removed, so the product is a single saturated molecule: addition [1]. [3] Examiner insight: Saying “a double bond is strong” works against you; the point is high electron density, and the pi bond is the part that breaks.

9. (a) Bromine water turns from orange-brown to colourless [1]. The electron-rich C=C bond reacts with bromine by electrophilic addition, which needs no light, so the bromine is used up [1]. (b) No colour change; the mixture stays orange-brown [1]. Hexane contains only C–C and C–H bonds, strong and hardly polar; it can only react by radical substitution, which needs UV light to break the Br–Br bond homolytically [1]. [4] Examiner insight: Write “colourless”, not “clear”: a clear liquid can still be coloured.

10. (a) CH₃CH=CHCH₃ + H₂O → CH₃CH₂CH(OH)CH₃ [1]; steam with an acid catalyst (for example phosphoric acid) [1] (b) Step 1: CH₃CH=CHCH₃ + HBr → CH₃CH₂CHBrCH₃ [1]. Step 2: CH₃CH₂CHBrCH₃ + OH⁻ → CH₃CH₂CH(OH)CH₃ + Br⁻ [1] (c) Step 1: electrophilic addition [1]. Step 2: nucleophilic substitution [1] (d) Step 1: HBr (its δ+ hydrogen) [1]. Step 2: the hydroxide ion, OH⁻ [1] [8] Examiner insight: Both words of each reaction type are needed: “addition” alone or “substitution” alone does not say who attacks.

11. (a) The alkene is symmetrical: both carbons of the C=C carry the same groups, so H and Br can add only one way round [1] (b) (CH₃)₂C=C(CH₃)₂ + HBr → (CH₃)₂CBrCH(CH₃)₂ [1] (c) M(C₆H₁₂) = 84.18 g mol⁻¹, so n = 2.52 / 84.18 = 0.0299 mol [1]. M(C₆H₁₃Br) = 165.09 g mol⁻¹ [1]. Theoretical mass = 0.0299 × 165.09 = 4.94 g [1]. Percentage yield = (3.29 / 4.94) × 100 = 66.6% [1] (d) Electrophile [1] [7] Examiner insight: Show the moles and the theoretical mass on separate lines, so your method is visible to the examiner even if the final percentage goes wrong.

12. (a) CH₃CH₂CHClCH₃ + CN⁻ → CH₃CH₂CH(CN)CH₃ + Cl⁻ [1] (b) 2-methylbutanenitrile [1] (c) Nucleophile: the cyanide ion, CN⁻ [1]. Leaving group: the chloride ion, Cl⁻ [1] (d) M(C₄H₉Cl) = 92.58 g mol⁻¹, so n = 3.75 / 92.58 = 0.0405 mol [1]. M(C₅H₉N) = 83.15 g mol⁻¹ [1]. Theoretical mass = 0.0405 × 83.15 = 3.37 g [1]. Percentage yield = (2.80 / 3.37) × 100 = 83.1% [1] [8] Examiner insight: The nitrile has one more carbon than the chloroalkane, so its molar mass must be worked out from C₅H₉N; reusing the C₄ formula gives the wrong theoretical mass.

Where marks are usually lost

• Defining nucleophiles and electrophiles by charge instead of by donating or accepting an electron pair.
• Naming the halogenoalkane as the nucleophile; it supplies the δ+ carbon and is the electrophile.
• Not recognising that the alkene’s C=C bond acts as the nucleophile in electrophilic addition.
• Curly arrows that start at an atom or a positive charge instead of at a lone pair or a bond.
• Fish-hook arrows or radical products in a heterolytic fission answer.
• Missing HX or NH₄X when the nucleophile is neutral, leaving the equation unbalanced.
• Naming nitriles without counting the CN carbon.
• Calling halogen addition to an alkene a substitution, or stating the hydration conditions without a catalyst.
• Writing “clear” instead of “colourless” for decolourised bromine water.
• Rounding moles too early in yield questions; keep at least three significant figures through the working.

Next steps

• Recap with the revision notes
• Revisit any weak area in the study guide
• Check yourself with all free 10-minute diagnostics
• See where this unit sits in the Reactivity 3 overview
• Browse other units via the IB DP chemistry course hub
• Book a free trial class
• Mark your progress on the printable checklist

Official syllabus

Source: the Chemistry guide for the IB Diploma Programme (International Baccalaureate, published February 2023; first assessment 2025). Every question targets Reactivity 3.4 understandings 3.4.1 to 3.4.5, which are SL and HL content.