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A Level H2 Chemistry Organic Chemistry Quiz
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Questions
A-Level Chemistry H2 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.
- Use appropriate significant figures.
- You may use a Data Booklet.
Section A: Nomenclature and Isomerism (10 marks)
1. Draw the skeletal formula of 3-ethyl-2,2-dimethylhexane.
[2 marks]
2. 2-chlorobutane exhibits optical isomerism.
(a) Explain what is meant by the term chiral centre. [1 mark]
(b) Draw the two optical isomers of 2-chlorobutane using 3D wedge-and-dash notation. [2 marks]
3. A compound has the molecular formula C₄H₈O₂.
(a) Draw and name two functional group isomers of C₄H₈O₂ that are carboxylic acids. [2 marks]
(b) Draw and name one functional group isomer of C₄H₈O₂ that is an ester. [1 mark]
4. State the type of isomerism exhibited by but-1-ene and but-2-ene, and explain why they are isomers. [2 marks]
5. Define the term structural isomerism and give one example of a pair of structural isomers with the molecular formula C₃H₈O. [2 marks]
Section B: Reaction Mechanisms (14 marks)
6. 2-bromopropane reacts with aqueous sodium hydroxide via an S<sub>N</sub>2 mechanism.
(a) Outline the mechanism for this reaction, showing all curly arrows, relevant lone pairs, and charges. [3 marks]
(b) Draw the structure of the organic product and state its IUPAC name. [2 marks]
7. 2-bromo-2-methylpropane reacts with aqueous sodium hydroxide via an S<sub>N</sub>1 mechanism.
(a) Explain why this reaction proceeds via S<sub>N</sub>1 rather than S<sub>N</sub>2. [2 marks]
(b) Outline the mechanism for this reaction, clearly showing the structure of the intermediate. [3 marks]
8. Compare the rates of hydrolysis of 1-bromobutane, 2-bromobutane, and 2-bromo-2-methylpropane with aqueous sodium hydroxide. Explain your reasoning. [4 marks]
Section C: Organic Synthesis and Reactions (16 marks)
9. Ethene can be converted to ethanol via two different routes.
(a) State the reagents and conditions for the direct hydration of ethene to ethanol. [2 marks]
(b) State the reagents and conditions for the two-step conversion of ethene to ethanol via a halogenoalkane intermediate. [3 marks]
10. Propan-1-ol can be oxidised to propanoic acid.
(a) State the reagents and conditions required for this oxidation. [2 marks]
(b) Write a balanced equation for the reaction using [O] to represent the oxidising agent. [1 mark]
11. Propanoic acid reacts with ethanol in the presence of concentrated sulfuric acid.
(a) Name the organic product and state the type of reaction. [2 marks]
(b) Write a balanced equation for this reaction. [1 mark]
12. A student proposes the following two-step synthesis to convert ethanol to ethanoic acid:
Step 1: Ethanol → ethanal
Step 2: Ethanal → ethanoic acid
(a) State the reagents and conditions for Step 1 that would give a good yield of ethanal. [2 marks]
(b) Explain why the conditions in (a) are chosen rather than heating under reflux. [1 mark]
13. Ethene undergoes electrophilic addition with hydrogen bromide.
(a) Outline the mechanism for this reaction, showing all curly arrows and the structure of the intermediate. [2 marks]
Section D: Structure Determination and Analysis (10 marks)
14. An organic compound X has the molecular formula C₃H₆O.
Compound X reacts with 2,4-dinitrophenylhydrazine to form an orange precipitate, but does not react with Tollens' reagent.
(a) Identify the functional group present in X. [1 mark]
(b) Draw the structure of X. [1 mark]
(c) Write an equation for the reaction of X with 2,4-dinitrophenylhydrazine. You may represent 2,4-dinitrophenylhydrazine as R-NH-NH₂. [2 marks]
15. An alcohol Y with molecular formula C₄H₁₀O is oxidised by acidified potassium dichromate(VI) to form a ketone Z.
(a) Draw the structures of Y and Z. [2 marks]
(b) State what is observed during the oxidation of Y. [1 mark]
16. A hydrocarbon W decolourises bromine water. On complete combustion, 0.56 g of W produces 1.76 g of CO₂ and 0.72 g of H₂O.
(a) Calculate the empirical formula of W. [2 marks]
(b) Given that the relative molecular mass of W is 56, determine its molecular formula. [1 mark]
17. Describe a simple chemical test to distinguish between propanal and propanone, including the reagents and expected observations. [2 marks]
18. An organic compound Q has the molecular formula C₂H₄O₂. It reacts with sodium carbonate to produce a gas that turns limewater milky. Identify Q and write a balanced equation for its reaction with sodium carbonate. [2 marks]
19. State the reagents and conditions required to convert benzene to nitrobenzene, and name the type of reaction. [2 marks]
20. Explain why phenol is more acidic than ethanol, using resonance structures to support your answer. [2 marks]
END OF QUIZ
Answers
A-Level Chemistry H2 Quiz - Organic Chemistry - ANSWER KEY
Section A: Nomenclature and Isomerism
1. Skeletal formula of 3-ethyl-2,2-dimethylhexane
[2 marks]
CH3
|
CH3-C-CH2-CH-CH2-CH3
| |
CH3 CH2
|
CH3
Marking:
- Correct longest chain (hexane backbone) [1]
- Correct substituent positions (2,2-dimethyl, 3-ethyl) [1]
Accept any clear skeletal representation showing correct connectivity.
2. (a) A chiral centre is a carbon atom bonded to four different groups/atoms. [1 mark]
(b) Optical isomers of 2-chlorobutane:
[2 marks]
CH3 CH3
| |
Cl-C-H H-C-Cl
| |
C2H5 C2H5
Wedge-and-dash: one isomer with Cl as wedge and H as dash; other isomer with Cl as dash and H as wedge.
Marking: One mark for each correct 3D representation showing non-superimposable mirror images.
3. (a) Carboxylic acid isomers of C₄H₈O₂:
- Butanoic acid: CH₃CH₂CH₂COOH [1]
- 2-methylpropanoic acid: (CH₃)₂CHCOOH [1]
(b) Ester isomer:
- Methyl propanoate: CH₃CH₂COOCH₃ [1]
(Accept ethyl ethanoate: CH₃COOCH₂CH₃)
4. But-1-ene and but-2-ene are position isomers. [1]
They have the same molecular formula (C₄H₈) and the same functional group (C=C), but the position of the double bond differs (carbon-1 vs carbon-2). [1]
5. Structural isomerism occurs when compounds have the same molecular formula but different structural formulae (different arrangement of atoms). [1]
Example for C₃H₈O: Propan-1-ol (CH₃CH₂CH₂OH) and propan-2-ol (CH₃CH(OH)CH₃) [1]
(Accept any valid pair, e.g., propan-1-ol and methoxyethane.)
Section B: Reaction Mechanisms
6. (a) S<sub>N</sub>2 mechanism for 2-bromopropane + NaOH:
[3 marks]
- Curly arrow from OH⁻ lone pair to the carbon attached to Br [1]
- Curly arrow from C-Br bond to Br (leaving group departure) [1]
- Transition state shown with partial bonds, or correct product with Br⁻ leaving [1]
(b) Product: Propan-2-ol, CH₃CH(OH)CH₃ [1]
IUPAC name: Propan-2-ol [1]
7. (a) 2-bromo-2-methylpropane is a tertiary halogenoalkane. [1]
The bulky methyl groups cause steric hindrance, preventing backside attack by the nucleophile (required for S<sub>N</sub>2). The tertiary carbocation formed in S<sub>N</sub>1 is stabilised by the +I effect of three alkyl groups. [1]
(b) S<sub>N</sub>1 mechanism:
[3 marks]
Step 1 (slow): (CH₃)₃CBr → (CH₃)₃C⁺ + Br⁻ [1]
- Curly arrow from C-Br bond to Br
Step 2 (fast): (CH₃)₃C⁺ + OH⁻ → (CH₃)₃COH [1] - Curly arrow from OH⁻ lone pair to C⁺
- Carbocation intermediate clearly shown [1]
8. Rate order: 2-bromo-2-methylpropane > 2-bromobutane > 1-bromobutane [1]
- 2-bromo-2-methylpropane (tertiary) reacts via S<sub>N</sub>1; rate depends only on halogenoalkane concentration; tertiary carbocation is most stable. [1]
- 2-bromobutane (secondary) can react via both S<sub>N</sub>1 and S<sub>N</sub>2; intermediate rate. [1]
- 1-bromobutane (primary) reacts via S<sub>N</sub>2; least steric hindrance but requires bimolecular collision; slowest rate. [1]
Section C: Organic Synthesis and Reactions
9. (a) Direct hydration of ethene:
Reagents: Steam (H₂O) [1]
Conditions: H₃PO₄ catalyst, 300°C, 60 atm [1]
(b) Two-step route via halogenoalkane:
Step 1: Ethene + HBr → Bromoethane. Reagents: HBr(g), room temperature. [1]
Step 2: Bromoethane + NaOH(aq) → Ethanol. Reagents: NaOH(aq), heat under reflux. [1]
(Accept HCl and chloroethane as alternative.) [1]
10. (a) Oxidation of propan-1-ol to propanoic acid:
Reagents: K₂Cr₂O₇/H₂SO₄ (acidified potassium dichromate(VI)) [1]
Conditions: Heat under reflux [1]
(b) CH₃CH₂CH₂OH + 2[O] → CH₃CH₂COOH + H₂O [1]
11. (a) Product: Ethyl propanoate [1]
Type of reaction: Esterification / Condensation [1]
(b) CH₃CH₂COOH + CH₃CH₂OH ⇌ CH₃CH₂COOCH₂CH₃ + H₂O [1]
(Accept reversible arrow; H₂SO₄ written above arrow.)
12. (a) Step 1 (ethanol → ethanal):
Reagents: K₂Cr₂O₇/H₂SO₄ [1]
Conditions: Distil off the ethanal as it forms (immediate distillation) [1]
(b) Ethanal is volatile (b.p. 21°C). Distillation removes it from the oxidising mixture before further oxidation to ethanoic acid occurs. Heating under reflux would cause over-oxidation. [1]
13. (a) Electrophilic addition of HBr to ethene:
[2 marks]
- Curly arrow from C=C π-bond to H of HBr [1]
- Curly arrow from H-Br bond to Br, forming Br⁻
- Carbocation intermediate: CH₃CH₂⁺ shown [1]
- Second step: Curly arrow from Br⁻ lone pair to C⁺ forming CH₃CH₂Br
Section D: Structure Determination and Analysis
14. (a) X reacts with 2,4-DNPH (carbonyl present) but not Tollens' reagent (not an aldehyde), so X is a ketone. [1]
(b) Propanone: CH₃COCH₃ [1]
(c) CH₃COCH₃ + R-NH-NH₂ → CH₃C(CH₃)=N-NH-R + H₂O [2]
(One mark for correct organic product structure, one mark for balanced equation with water.)
15. (a) Y is oxidised to a ketone, so Y must be a secondary alcohol.
C₄H₁₀O secondary alcohol: Butan-2-ol, CH₃CH(OH)CH₂CH₃ [1]
Z (ketone): Butanone, CH₃COCH₂CH₃ [1]
(b) Observation: Orange solution turns green. [1]
16. (a) Mass of C in CO₂ = (12/44) × 1.76 = 0.48 g
Mass of H in H₂O = (2/18) × 0.72 = 0.08 g
Mass of C + H = 0.56 g, so W is a hydrocarbon (no other elements). [1]
| Element | Mass (g) | Moles | Ratio |
|---|---|---|---|
| C | 0.48 | 0.48/12 = 0.04 | 1 |
| H | 0.08 | 0.08/1 = 0.08 | 2 |
Empirical formula: CH₂ [1]
(b) Empirical formula mass = 12 + 2 = 14
n = 56/14 = 4
Molecular formula: C₄H₈ [1]
17. To distinguish between propanal and propanone:
Reagents: Tollens' reagent (ammoniacal silver nitrate) [1]
Observations: Propanal forms a silver mirror; propanone shows no visible change. [1]
(Accept Fehling's solution: propanal gives brick-red precipitate; propanone no change.)
18. Q is ethanoic acid (CH₃COOH). [1]
Equation: 2CH₃COOH + Na₂CO₃ → 2CH₃COONa + CO₂ + H₂O [1]
19. Reagents: Concentrated nitric acid (HNO₃) and concentrated sulfuric acid (H₂SO₄) [1]
Conditions: Heat under reflux at 50–55°C.
Type of reaction: Electrophilic substitution (nitration). [1]
20. Phenol is more acidic than ethanol because the phenoxide ion (C₆H₅O⁻) is stabilised by resonance. [1]
The negative charge on oxygen is delocalised into the benzene ring (show resonance structures with charge delocalised to ortho and para positions). In ethanol, the ethoxide ion (C₂H₅O⁻) has no such resonance stabilisation; the negative charge is localised on oxygen. [1]
END OF ANSWER KEY