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A Level H1 Chemistry Organic Chemistry Quiz

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A-Level Chemistry H1 Quiz - Organic Chemistry

Name: _________________________ Class: _________________________ Date: _________________________ Score: ______ / 50

Duration: 45 minutes Total Marks: 50

Instructions:

Answer ALL questions in the spaces provided.
Show all working for calculation questions.
You may use a calculator and the Data Booklet.
The number of marks is given in brackets [ ] at the end of each question or part question.

Section A: Functional Groups, Nomenclature & Isomerism (12 marks)

1. The structural formula of a compound is shown below:

CH₃CH₂COCH₂CH₃

(a) Name the functional group present in this compound. [1]

(b) Give the IUPAC name of this compound. [1]

(c) Draw the structural formula of a functional group isomer of this compound that contains the same number of carbon atoms. [2]

2. Lactic acid has the structural formula CH₃CH(OH)COOH.

(a) Identify the two functional groups present in lactic acid. [2]

(b) Lactic acid exhibits optical isomerism. Explain what is meant by optical isomerism and identify the structural feature that gives rise to it in lactic acid. [3]

3. Compound X has the molecular formula C₄H₈.

(a) Draw the structural formulae of TWO straight-chain isomers of Compound X, each belonging to a different homologous series. Name each isomer. [3]

Isomer 1:

Name: _________________________

Isomer 2:

Name: _________________________

4. Define the term homologous series. [1]

5. Explain why isomers can have different physical properties, such as boiling point. [2]

Section B: Hydrocarbons & Reactions (14 marks)

6. Ethene reacts with bromine in an addition reaction.

(a) Write a balanced chemical equation for this reaction, using structural formulae. [2]

(b) State the colour change observed during this reaction. [1]

7. Propane (C₃H₈) and propene (C₃H₆) are both hydrocarbons.

(a) Describe a simple chemical test that can be used to distinguish between propane and propene. State the reagent used and the observation for each compound. [3]

(b) Propane undergoes complete combustion in excess oxygen. Write a balanced chemical equation for this reaction. [1]

(c) When propane undergoes incomplete combustion, carbon monoxide may be produced. Explain why carbon monoxide is dangerous to human health. [2]

8. The cracking of long-chain hydrocarbons produces shorter-chain alkanes and alkenes.

(a) Explain why cracking is an important industrial process. [2]

Section C: Oxygen-Containing Organic Compounds (14 marks)

9. Ethanol (C₂H₅OH) can be produced by fermentation of glucose.

(a) Write a balanced chemical equation for the fermentation of glucose to produce ethanol. [1]

(b) State TWO conditions required for fermentation. [2]

10. Ethanoic acid (CH₃COOH) is a weak acid.

(a) Write an equation for the dissociation of ethanoic acid in water, including state symbols. [2]

(b) Write the expression for the acid dissociation constant, Ka, of ethanoic acid. [1]

(c) A solution of ethanoic acid has a concentration of 0.10 mol dm⁻³ and a pH of 2.87. Calculate the value of Ka for ethanoic acid. [3]

11. Ethanoic acid reacts with ethanol in the presence of a concentrated sulfuric acid catalyst to form an ester.

(a) Name the ester produced in this reaction. [1]

(b) Write a balanced chemical equation for this reaction, using structural formulae. [2]

12. Describe a simple chemical test to distinguish between ethanol and ethanoic acid. [2]

Section D: Polymers & Integrated Organic Chemistry (10 marks)

13. Poly(ethene) is an addition polymer formed from ethene monomers.

(a) Draw the structural formula of poly(ethene), showing TWO repeat units. [2]

(b) Explain why poly(ethene) is non-biodegradable. [2]

(c) Suggest ONE environmental problem caused by the disposal of non-biodegradable polymers and describe ONE way this problem can be reduced. [2]

14. A student carried out the following reaction sequence:

Propan-1-ol → (Reaction 1) → Propanoic acid → (Reaction 2) → Ester

(a) Identify the reagent(s) and condition(s) required for Reaction 1. [2]

(b) Reaction 2 involves propanoic acid reacting with methanol. Draw the structural formula of the ester produced. [1]

15. What type of polymerisation occurs when ethene forms poly(ethene)? [1]

16. State the name of the small molecule eliminated during condensation polymerisation. [1]

17. Give ONE advantage of using biodegradable polymers over non-biodegradable polymers. [1]

18. Draw the structural formula of the monomer used to make poly(propene). [1]

19. Name the type of bond that links amino acids together in a protein. [1]

20. State ONE natural polymer and its monomer. [1]

END OF QUIZ

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Answers

A-Level Chemistry H1 Quiz - Organic Chemistry — Answer Key

Total Marks: 50

Section A: Functional Groups, Nomenclature & Isomerism (12 marks)

1. CH₃CH₂COCH₂CH₃

(a) Answer: Ketone / carbonyl group [1]

(b) Answer: Pentan-3-one [1] Marking note: Accept "3-pentanone". Must have correct numbering.

(c) Answer: Pentanal: CH₃CH₂CH₂CH₂CHO [2] Marking note: 1 mark for correct structural formula showing aldehyde group; 1 mark for correct functional group isomer (aldehyde, not another ketone or alcohol with different C count). Also accept other functional group isomers with 5 carbons (e.g., unsaturated alcohols, but aldehyde is the most direct).

2. Lactic acid: CH₃CH(OH)COOH

(a) Answer: Hydroxyl group (–OH) / alcohol AND carboxyl group (–COOH) / carboxylic acid [2] Marking note: 1 mark for each functional group correctly identified.

(b) Answer: Optical isomerism occurs when a molecule has a chiral centre (a carbon atom bonded to four different groups) and exists as two non-superimposable mirror images (enantiomers). [1] The two enantiomers rotate plane-polarised light in opposite directions. [1] In lactic acid, the central carbon atom (C-2) is bonded to four different groups: –H, –OH, –COOH, and –CH₃, making it a chiral centre. [1] Marking note: Must mention chiral centre/asymmetric carbon and non-superimposable mirror images for full marks.

3. Compound X: C₄H₈

Answer: Isomer 1: CH₃CH₂CH=CH₂ (but-1-ene) [1.5] Isomer 2: Cyclobutane OR CH₃CH=CHCH₃ (but-2-ene) [1.5] Marking note: 1 mark for each correct structural formula; 0.5 marks for each correct name. The two isomers must belong to different homologous series (alkene and cycloalkane, or positional isomers of alkenes are acceptable if clearly different series is not possible — accept but-1-ene and but-2-ene as different homologous series is not strictly required here, but ideally one alkene and one cycloalkane).

4. Answer: A homologous series is a family of organic compounds with the same functional group and similar chemical properties, where each successive member differs by a –CH₂– group. [1]

5. Answer: Isomers have different structural arrangements of atoms, leading to differences in molecular shape and surface area. [1] This affects the strength of intermolecular forces (e.g., van der Waals forces, dipole-dipole interactions, hydrogen bonding), resulting in different boiling points. [1]

Section B: Hydrocarbons & Reactions (14 marks)

6. Ethene + bromine

(a) Answer: CH₂=CH₂ + Br₂ → CH₂Br–CH₂Br [2] Marking note: 1 mark for correct reactants with structural formulae; 1 mark for correct product (1,2-dibromoethane).

(b) Answer: The reddish-brown/orange colour of bromine is decolourised (turns colourless). [1]

(c) Answer: Alkenes contain a carbon-carbon double bond (C=C), which consists of one sigma (σ) bond and one pi (π) bond. [1] The pi bond is weaker and more exposed, making it susceptible to attack by electrophiles. Addition reactions break only the pi bond while the sigma bond remains intact. [1] Alkanes contain only carbon-carbon single bonds (C–C), which are strong sigma bonds. Breaking these bonds requires more energy and occurs via substitution reactions under more vigorous conditions (e.g., UV light). [1] Marking note: Must reference the presence of the C=C double bond and the nature of the pi bond.

7. Propane vs propene

(a) Answer: Reagent: Bromine water (or bromine in organic solvent). [1] Propane: No visible change; the reddish-brown/orange colour of bromine remains. [1] Propene: The reddish-brown/orange colour of bromine is decolourised immediately. [1] Marking note: Accept acidified potassium manganate(VII) as alternative reagent (propane: no change/purple remains; propene: purple decolourised and brown precipitate forms).

(b) Answer: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O [1]

(c) Answer: Carbon monoxide binds irreversibly to haemoglobin in red blood cells, forming carboxyhaemoglobin. [1] This prevents haemoglobin from carrying oxygen around the body, leading to oxygen deprivation (hypoxia) which can be fatal. [1]

8. Cracking

(a) Answer: Cracking converts long-chain hydrocarbons (which have low demand/high supply) into shorter-chain alkanes and alkenes which are in higher demand. [1] The shorter-chain alkanes are used as fuels (petrol), while the alkenes are important feedstocks for the petrochemical industry (making polymers, solvents, etc.). [1]

Section C: Oxygen-Containing Organic Compounds (14 marks)

9. Fermentation of glucose

(a) Answer: C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂ [1]

(b) Answer: 1. Yeast (as a source of enzymes/zymase). [1] 2. Anaerobic conditions (absence of oxygen) / warm temperature (approximately 30–37°C). [1] Marking note: Accept any two valid conditions.

(c) Answer: Ethanol is toxic to yeast. [1] When the ethanol concentration reaches approximately 15%, the yeast enzymes become denatured and fermentation stops. [1]

10. Ethanoic acid

(a) Answer: CH₃COOH(aq) ⇌ CH₃COO⁻(aq) + H⁺(aq) [2] Marking note: 1 mark for correct equation; 1 mark for equilibrium arrow (⇌) and state symbols (aq).

(b) Answer: Ka = [CH₃COO⁻][H⁺] / [CH₃COOH] [1]

(c) Answer: [H⁺] = 10⁻²·⁸⁷ = 1.35 × 10⁻³ mol dm⁻³ [1] Since [CH₃COO⁻] = [H⁺] = 1.35 × 10⁻³ mol dm⁻³ [0.5] [CH₃COOH] ≈ 0.10 – 1.35 × 10⁻³ ≈ 0.09865 mol dm⁻³ [0.5] Ka = (1.35 × 10⁻³)² / 0.09865 = 1.85 × 10⁻⁵ mol dm⁻³ [1] Marking note: Accept 1.8 × 10⁻⁵ to 1.9 × 10⁻⁵. Must show working for full marks.

11. Esterification

(a) Answer: Ethyl ethanoate [1]

(b) Answer: CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O [2] Marking note: 1 mark for correct reactants; 1 mark for correct products and equilibrium arrow.

(c) Answer: Esters are used as flavourings/fragrances in food and perfumes / as solvents (e.g., in nail varnish remover) / as plasticisers. [1] Marking note: Accept any valid use.

12. Answer: Add sodium carbonate (or sodium hydrogencarbonate) solution. [1] Ethanoic acid will react to produce effervescence (CO₂ gas evolved), while ethanol will not react. [1] Marking note: Accept other valid tests, e.g., litmus paper (ethanoic acid turns blue litmus red, ethanol no change).

Section D: Polymers & Integrated Organic Chemistry (10 marks)

13. Poly(ethene)

(a) Answer: –[CH₂–CH₂]–[CH₂–CH₂]– or a clear diagram showing two ethene units linked by single bonds with the repeat unit in brackets and subscript 'n'. [2] Marking note: 1 mark for correct repeat unit; 1 mark for showing two repeat units with correct bonding.

(b) Answer: Poly(ethene) has a non-polar carbon-carbon backbone with strong C–C and C–H bonds. [1] Microorganisms (bacteria/fungi) lack enzymes capable of breaking these strong, non-polar bonds, so the polymer cannot be broken down naturally in the environment. [1]

(c) Answer: Environmental problem: Accumulation in landfills/littering leading to visual pollution and harm to wildlife (e.g., animals ingesting or becoming entangled in plastic waste). [1] Solution: Recycling poly(ethene) to produce new products / developing biodegradable alternatives (e.g., bioplastics from plant starch) / incineration with energy recovery. [1] Marking note: Accept any valid problem and corresponding solution.

14. Reaction sequence

(a) Answer: Reagent: Acidified potassium dichromate(VI) (K₂Cr₂O₇/H₂SO₄) [1] Condition: Heat under reflux [1] Marking note: Accept acidified potassium manganate(VII) as alternative oxidising agent.

(b) Answer: CH₃CH₂COOCH₃ (methyl propanoate) [1]

15. Answer: Addition polymerisation [1]

16. Answer: Water (H₂O) [1] Marking note: Accept hydrogen chloride (HCl) if referring to specific condensation polymers, but water is the most common.

17. Answer: Biodegradable polymers can be broken down by microorganisms, reducing long-term waste accumulation in landfills and environmental pollution. [1]

18. Answer: CH₂=CHCH₃ (propene) [1]

19. Answer: Peptide bond (or amide bond) [1]

20. Answer: Protein (natural polymer) and amino acids (monomer) / Starch and glucose / Cellulose and glucose / Natural rubber and isoprene. [1] Marking note: Accept any valid natural polymer and its correct monomer.

END OF ANSWER KEY