Edexcel International A Level Chemistry: Organic chemistry: general principles and halogenoalkanes (YCH11) – 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.

This set covers Topic 10A: General principles and Topic 10B: Halogenoalkanes of the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry specification (YCH11), Issue 1, September 2017, outcomes 10.1 to 10.14, including Core Practicals 5 and 6. It is International AS (IAS) content in Unit 2: Energetics, Group Chemistry, Halogenoalkanes and Alcohols. Learn the content first from the study guide and the revision notes, and track progress on the printable checklist. More resources are on the course hub.

Data for this set: H = 1.0, C = 12.0, N = 14.0, O = 16.0, Cl = 35.5, Br = 79.9, Ag = 107.9. Use these mean bond enthalpies in kJ mol⁻¹: C–F 467, C–Cl 346, C–Br 290, C–I 228.

Questions

1. This question is about general principles.

(a) State the meaning of the term nucleophile. [1] (b) Describe what happens to the bonding electrons in heterolytic fission and state what type of species is formed. [2]

2. Classify each reaction as addition, elimination, substitution, oxidation, reduction, hydrolysis or polymerisation. Give one label for each.

(a) C₃H₈ + 5O₂ → 3CO₂ + 4H₂O [1] (b) n CH₂=CHCH₃ → poly(propene) [1] (c) CH₃CH₂CH₂CH₂I + H₂O → CH₃CH₂CH₂CH₂OH + HI [1] (d) CH₃CH₂CHBrCH₃ + KOH (ethanol) → CH₃CH=CHCH₃ + KBr + H₂O [1]

3. Three isomers of C₅H₁₁Cl are shown.

A: CH₃CH₂CH(CH₃)CH₂Cl B: (CH₃)₂CHCHClCH₃ C: (CH₃)₂CClCH₂CH₃

Name each isomer and state whether it is a primary, secondary or tertiary halogenoalkane. [6]

4. 1-chloro-3-methylbutane, (CH₃)₂CHCH₂CH₂Cl, is reacted separately with four reagents.

(a) For KOH(aq), name the organic product and give a balanced equation. [2] (b) Name the organic product with KOH dissolved in ethanol and state the role of the hydroxide ion. [2] (c) Name the organic product with KCN in ethanol and state why this reaction is useful in synthesis. [2] (d) State the conditions for the reaction with ammonia and name the organic product. [2]

5. Describe the mechanism for the reaction of 1-bromo-2-methylbutane, CH₃CH₂CH(CH₃)CH₂Br, with excess ammonia. Include dipoles, curly arrows, lone pairs and the final products. [5]

6. 2-chloropentane, CH₃CHClCH₂CH₂CH₃, is heated under reflux with ethanolic potassium hydroxide.

(a) Name all three alkenes that can form. [3] (b) Explain why the alkenes include two that differ in the position of the C=C. [1]

7. A student added equal amounts of three primary halogenoalkanes to silver nitrate in ethanol at 40 °C and timed the first appearance of a precipitate.

Halogenoalkane	Time / s
1-chloro-4-methylpentane	1800
1-bromo-4-methylpentane	240
1-iodo-4-methylpentane	30

(a) Calculate 1/time for each compound. [2] (b) Calculate how many times faster the iodo compound reacts than the bromo compound. [1] (c) Explain the trend in rate using the bond enthalpy data. [2] (d) State why the polarity of the C–X bond does not explain this trend. [1]

8. You are asked to compare the rates of hydrolysis of the primary, secondary and tertiary isomers of C₄H₉I using aqueous silver nitrate in ethanol.

(a) State two variables you would control. [2] (b) Explain why ethanol is used. [1] (c) Predict the order of rate, fastest first. [1] (d) State one safety precaution and its reason. [1]

9. In Core Practical 6, 11.10 g of 2-methylpropan-2-ol was shaken with excess concentrated hydrochloric acid. After purification, 9.85 g of 2-chloro-2-methylpropane was collected.

(CH₃)₃COH + HCl → (CH₃)₃CCl + H₂O

(a) Calculate the percentage yield. [4] (b) The density of 2-chloro-2-methylpropane is 0.85 g cm⁻³. State which layer in the separating funnel contains the product. [1] (c) State the purpose of washing with sodium hydrogencarbonate solution and of adding anhydrous sodium sulfate. [2] (d) The boiling point of the product is 51 °C and that of the alcohol is 82 °C. State which fraction should be collected in the final distillation. [1]

10. Liquid Y is a halogenoalkane, CₙH₂ₙ₊₁X. A sample of Y was warmed with NaOH(aq), acidified with dilute nitric acid, and silver nitrate solution was added. A cream precipitate formed. It did not dissolve in dilute ammonia but did dissolve in concentrated ammonia.

(a) Explain why the mixture was acidified before adding silver nitrate. [1] (b) Identify X. [1] (c) 2.055 g of Y gave 2.817 g of the dried precipitate. Deduce the molecular formula of Y. [3] (d) Y reacts faster with silver nitrate in ethanol than any of its structural isomers. Give the structural formula and name of Y. [2]

11. This question is about bond properties and synthesis.

(a) The C–F bond is the most polar carbon–halogen bond, yet fluoroalkanes are almost unreactive towards nucleophiles. Explain this. [2] (b) Suggest a two-step route from propene to 2-methylpropanenitrile, (CH₃)₂CHCN. Give the reagent and conditions for each step, the intermediate, and the type of each reaction. [4] (c) Starting from 6.30 g of propene, each step has a yield of 80%. Calculate the mass of 2-methylpropanenitrile formed. [3]

Answers

1. (a) An electron-pair donor: it uses a lone pair to make a new covalent bond [1] (b) Both bonding electrons go to one atom [1]; ions form, a positive and a negative ion (giving electrophiles or nucleophiles) [1] Examiner insight: “Donates electrons” alone is vague; say a lone pair or electron pair, and that a covalent bond forms.

2. (a) Oxidation (combustion) [1] (b) Polymerisation [1] (c) Hydrolysis (substitution is also correct) [1] (d) Elimination [1] Examiner insight: When a question asks for one label, give one; a list of two that includes a wrong term may lose the mark.

3. A: 1-chloro-2-methylbutane [1], primary [1] B: 2-chloro-3-methylbutane [1], secondary [1] C: 2-chloro-2-methylbutane [1], tertiary [1] Examiner insight: For B both directions give locants 2 and 3, so chloro, first alphabetically, takes the lower number; 3-chloro-2-methylbutane is wrong.

4. (a) 3-methylbutan-1-ol [1]; (CH₃)₂CHCH₂CH₂Cl + KOH → (CH₃)₂CHCH₂CH₂OH + KCl [1] (b) 3-methylbut-1-ene [1]; OH⁻ acts as a base (removes H⁺) [1] (c) 4-methylpentanenitrile, (CH₃)₂CHCH₂CH₂CN [1]; it increases the length of the carbon chain by one carbon [1] (d) Ammonia dissolved in ethanol, heated in a sealed tube (under pressure) [1]; 3-methylbutylamine (3-methylbutan-1-amine) [1] Examiner insight: Name the nitrile by counting the CN carbon; “3-methylbutanenitrile” here describes the wrong chain length.

5. Dipole shown, Cδ+–Brδ− [1] Curly arrow starting at the nitrogen lone pair of NH₃ and ending at the δ+ carbon [1] Curly arrow from the C–Br bond to Br, giving Br⁻ [1] Intermediate CH₃CH₂CH(CH₃)CH₂NH₃⁺ shown [1] Second NH₃ lone pair removes H⁺ (arrow from N–H bond to N), giving CH₃CH₂CH(CH₃)CH₂NH₂ and NH₄⁺ (NH₄Br overall) [1] Examiner insight: An arrow that starts from the nitrogen atom with no lone pair drawn is ambiguous; draw the lone pair and start the arrow on it.

6. (a) Pent-1-ene [1]; (E)-pent-2-ene [1]; (Z)-pent-2-ene [1] (b) H⁺ can be removed from either carbon next to the C–Cl carbon (carbon 1 or carbon 3) [1] Examiner insight: Treat the E and Z isomers as separate products; giving “pent-2-ene” once only accounts for one of them.

7. (a) Chloro 5.56 × 10⁻⁴ s⁻¹; bromo 4.17 × 10⁻³ s⁻¹ [1]; iodo 3.33 × 10⁻² s⁻¹ [1] (b) 240 ÷ 30 = 8.0 times faster [1] (c) Bond enthalpy falls from C–Cl (346) to C–Br (290) to C–I (228 kJ mol⁻¹) [1]; the weaker bond breaks more easily, so hydrolysis is faster [1] (d) C–Cl is the most polar, so polarity would predict the chloro compound to react fastest, which is the opposite of what is seen [1] Examiner insight: Quote the bond enthalpy values in the comparison; a statement that “C–I is weak” without comparing all three bonds gives a thinner answer.

8. (a) Any two from: temperature (water bath); volume and concentration of silver nitrate; volume of ethanol; amount of each halogenoalkane [2] (b) Halogenoalkanes do not mix with water; ethanol is a common solvent for both [1] (c) Tertiary > secondary > primary [1] (d) Heat with a water bath, not a flame, because ethanol and halogenoalkanes are flammable [1] Examiner insight: “Same amount” is too loose; name the quantity, for example the same volume of silver nitrate solution of the same concentration.

9. (a) M(C₄H₁₀O) = 74.0 and M(C₄H₉Cl) = 92.5 g mol⁻¹ [1] n(alcohol) = 11.10 ÷ 74.0 = 0.1500 mol [1] Theoretical mass = 0.1500 × 92.5 = 13.88 g [1] Yield = 9.85 ÷ 13.88 × 100 = 71.0% [1] (b) The upper layer, because the product is less dense than water [1] (c) NaHCO₃ neutralises (removes) remaining acid [1]; anhydrous sodium sulfate removes water (dries the product) [1] (d) The fraction boiling at about 51 °C [1] Examiner insight: Keep at least three significant figures in intermediate values; early rounding can shift the final yield.

10. (a) To remove excess OH⁻ ions, which would react with Ag⁺ and form a precipitate that hides the result [1] (b) Bromine (bromide ions, AgBr precipitate) [1] (c) n(AgBr) = 2.817 ÷ 187.8 = 0.01500 mol, which equals n(Y) [1] M(Y) = 2.055 ÷ 0.01500 = 137.0 g mol⁻¹ [1] 14n + 1.0 + 79.9 = 137.0 gives n = 4, so C₄H₉Br [1] (d) (CH₃)₃CBr [1]; 2-bromo-2-methylpropane [1] Examiner insight: Show the link from moles of precipitate to moles of Y; a molecular formula stated without the mole ratio gives the examiner no method to credit.

11. (a) The C–F bond has the highest bond enthalpy (467 kJ mol⁻¹) [1]; so it is very hard to break, and bond strength outweighs polarity [1] (b) Step 1: HBr (at room temperature) gives 2-bromopropane as the major product [1]; electrophilic addition [1] Step 2: KCN in ethanol, heated under reflux [1]; nucleophilic substitution [1] (c) n(propene) = 6.30 ÷ 42.0 = 0.150 mol [1] Overall yield = 0.80 × 0.80 = 0.64 [1] Mass = 0.150 × 0.64 × 69.0 = 6.62 g [1] Examiner insight: Combine step yields by multiplying; adding or averaging the two percentages gives an overall yield that is too high.

Where marks are usually lost

• Writing “KOH” without “aqueous” or “in ethanol”, so the conditions do not fix the product.
• Calling ethanolic hydroxide a nucleophile; in elimination it acts as a base.
• Missing lone pairs, dipoles or the second ammonia molecule in mechanisms.
• Using polarity to explain why iodoalkanes hydrolyse fastest.
• Counting nitrile chains without the CN carbon.
• Forgetting to acidify with dilute nitric acid before the silver nitrate test.
• Placing the Core Practical 6 product in the lower layer.
• Rounding moles too early in yield calculations.

Next steps

• Review the revision notes and the study guide for questions you found hard.
• Find more Edexcel resources on the course hub and tick off outcomes on the checklist.
• Take one of the free 10-minute diagnostics.
• Book a free trial class for help with organic mechanisms.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry (XCH11/YCH11) specification, Issue 1, September 2017, Pearson Education Limited – Unit 2, Topic 10: Organic Chemistry: Halogenoalkanes, Alcohols and Spectra, parts 10A General principles and 10B Halogenoalkanes, outcomes 10.1 to 10.14.