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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: 1 hour 15 minutes
Total Marks: 50
Instructions:
- Answer ALL questions in the spaces provided.
- Use of the Data Booklet is relevant to some questions.
- Show all working for calculation questions.
- Marks are indicated in brackets [ ].
- You are advised to spend no more than 1 hour 15 minutes on this quiz.
Section A: Multiple Choice (5 marks)
Circle the correct answer for each question.
1. Which of the following compounds exhibits geometrical isomerism?
A. 2-methylpropene
B. But-1-ene
C. But-2-ene
D. 2-methylbut-2-ene
[1 mark]
2. The reaction of ethanol with acidified potassium dichromate(VI) under reflux produces:
A. Ethanal
B. Ethanoic acid
C. Ethene
D. Ethyl ethanoate
[1 mark]
3. Which reagent and condition would convert propanenitrile to propylamine?
A. LiAlH₄ in dry ether
B. NaBH₄ in ethanol
C. H₂ with Ni catalyst, heat
D. Sn with concentrated HCl, heat
[1 mark]
4. The electrophile in the nitration of benzene is:
A. NO₃⁻
B. NO₂⁺
C. HNO₃
D. NO₂⁻
[1 mark]
5. Which of the following statements about the hydrolysis of halogenoalkanes is correct?
A. The rate of hydrolysis increases from chloroalkane to iodoalkane.
B. The reaction proceeds via an electrophilic substitution mechanism.
C. Tertiary halogenoalkanes react more slowly than primary halogenoalkanes.
D. Aqueous sodium hydroxide is not required for hydrolysis.
[1 mark]
Section B: Structured Questions (45 marks)
6. An organic compound A has the molecular formula C₄H₈O₂. Compound A reacts with sodium carbonate to produce a gas that turns limewater milky. Compound A also reacts with ethanol in the presence of concentrated sulfuric acid to form a sweet-smelling liquid B.
(a) Identify the functional group present in compound A. [1 mark]
(b) Draw the displayed formula of compound A. [1 mark]
(c) Name compound B and write a balanced equation for its formation from A and ethanol. [2 marks]
(d) Compound A can be reduced using lithium aluminium hydride in dry ether. Draw the structure of the organic product formed and name it. [2 marks]
7. Consider the following reaction scheme:
CH₃CH₂CH₂Br → CH₃CH₂CH₂NH₂
(C) (D)
(a) State the reagents and conditions required for this conversion. [2 marks]
(b) The reaction proceeds via an S_N2 mechanism. Using curly arrows, show the mechanism for this reaction. Include all relevant lone pairs, charges, and partial charges. [3 marks]
(c) Explain why compound C reacts via an S_N2 mechanism rather than an S_N1 mechanism. [2 marks]
8. Benzene undergoes electrophilic substitution reactions rather than electrophilic addition reactions.
(a) Explain why benzene does not readily undergo electrophilic addition reactions. [2 marks]
(b) Write an equation for the reaction of benzene with ethanoyl chloride in the presence of aluminium chloride. Name the organic product formed. [2 marks]
(c) Draw the mechanism for the reaction in part (b), showing the formation of the electrophile and the substitution step. Include all relevant curly arrows. [4 marks]
9. An organic compound E (C₃H₆O) gives a positive test with 2,4-dinitrophenylhydrazine but does not react with Tollens' reagent.
(a) Identify the functional group in compound E and draw its structure. [2 marks]
(b) Compound E can be reduced using sodium borohydride. Draw the structure of the product and name it. [2 marks]
(c) Describe a chemical test, including observations, to distinguish between compound E and propanal. [2 marks]
10. Amino acids are the building blocks of proteins.
(a) Draw the structure of the zwitterion of glycine (H₂NCH₂COOH). [1 mark]
(b) Explain why amino acids have relatively high melting points. [2 marks]
(c) When two molecules of glycine react together, a dipeptide is formed. Draw the structure of the dipeptide formed and identify the peptide linkage. [2 marks]
11. Consider the following organic compounds:
| Compound | Formula |
|---|---|
| F | CH₃CH₂COOH |
| G | CH₃COOCH₃ |
| H | CH₃CH₂CHO |
(a) Arrange compounds F, G, and H in order of increasing boiling point. Explain your reasoning. [3 marks]
(b) Compound F is a stronger acid than ethanol. Explain why, with reference to the stability of the conjugate base. [2 marks]
12. An organic compound J has the molecular formula C₄H₈. Compound J decolourises bromine water and exists as a pair of geometrical isomers.
(a) Draw and name the two geometrical isomers of compound J. [2 marks]
(b) Compound J can be polymerised. Draw a section of the polymer chain showing two repeat units. [2 marks]
(c) Explain why the polymer formed from compound J is not biodegradable. [1 mark]
13. Explain why phenol is more reactive than benzene towards electrophilic substitution. [2 marks]
14. Describe the mechanism of the acid-catalysed hydrolysis of ethyl ethanoate. Include all relevant curly arrows and the structures of the products. [4 marks]
15. An organic compound K (C₂H₄O₂) reacts with sodium hydrogen carbonate to produce effervescence. Identify compound K and write a balanced equation for its reaction with sodium hydrogen carbonate. [2 marks]
16. Explain why tertiary halogenoalkanes undergo S_N1 hydrolysis more readily than primary halogenoalkanes. [2 marks]
17. Describe the bonding in benzene, with reference to sigma and pi bonds. [3 marks]
18. Draw the structures of the two optical isomers of alanine (CH₃CH(NH₂)COOH). [2 marks]
19. Explain why amides are weaker bases than amines. [2 marks]
20. An organic compound L has the molecular formula C₃H₈O. Compound L does not react with acidified potassium dichromate(VI). Identify compound L and explain your reasoning. [2 marks]
END OF QUIZ
Check your work carefully.
Answers
A-Level Chemistry H2 Quiz - Organic Chemistry - ANSWER KEY
Total Marks: 50
Section A: Multiple Choice (5 marks)
1. C. But-2-ene [1 mark]
But-2-ene has different groups on each carbon of the C=C bond (CH₃ and H on each carbon), allowing cis/trans isomerism. 2-methylpropene and but-1-ene have two identical groups on one carbon of the double bond; 2-methylbut-2-ene has two CH₃ groups on one carbon.
2. B. Ethanoic acid [1 mark]
Primary alcohols are oxidised to aldehydes then carboxylic acids. Under reflux with excess oxidising agent, ethanol is fully oxidised to ethanoic acid.
3. A. LiAlH₄ in dry ether [1 mark]
Nitriles are reduced to primary amines using lithium aluminium hydride (a strong reducing agent). NaBH₄ is not strong enough; H₂/Ni reduces nitriles but requires different conditions; Sn/HCl is used for nitro compounds.
4. B. NO₂⁺ [1 mark]
The nitronium ion (NO₂⁺) is the electrophile generated from HNO₃ and H₂SO₄ in the nitration of benzene.
5. A. The rate of hydrolysis increases from chloroalkane to iodoalkane. [1 mark]
The C-X bond strength decreases down Group 17 (C-I is weakest), making iodoalkanes hydrolyse fastest. The mechanism is nucleophilic substitution, tertiary halogenoalkanes react faster via S_N1, and aqueous NaOH is required.
Section B: Structured Questions (45 marks)
6. (a) Carboxylic acid / carboxyl group (-COOH) [1 mark]
Reaction with Na₂CO₃ producing CO₂ confirms the presence of a carboxylic acid group.
(b) Displayed formula of butanoic acid: CH₃CH₂CH₂COOH (or structural formula showing all bonds) [1 mark]
(c) Compound B is ethyl butanoate. [1 mark]
Equation: CH₃CH₂CH₂COOH + C₂H₅OH ⇌ CH₃CH₂CH₂COOC₂H₅ + H₂O [1 mark]
Accept reversible arrow with H₂SO₄ catalyst indicated.
(d) Product: CH₃CH₂CH₂CH₂OH (butan-1-ol) [1 mark for structure, 1 mark for name]
LiAlH₄ reduces carboxylic acids to primary alcohols.
7. (a) Excess concentrated ammonia in ethanol, heated under pressure / in a sealed tube. [2 marks]
Accept: NH₃(conc) in ethanol, heat.
(b) Mechanism (S_N2):
- Curly arrow from lone pair on N of NH₃ to the carbon attached to Br [1 mark]
- Curly arrow from C-Br bond to Br (showing Br leaving) [1 mark]
- Transition state shown with partial bonds, Br with δ−, N with δ+ [1 mark]
- Product: CH₃CH₂CH₂NH₃⁺ Br⁻ (then deprotonation by NH₃ to give amine)
(c) Compound C is a primary halogenoalkane. [1 mark]
The carbon attached to Br is relatively unhindered, allowing backside attack by the nucleophile. The primary carbocation that would form in an S_N1 mechanism is too unstable. [1 mark]
8. (a) Benzene has a delocalised π-electron system which is highly stable (aromatic stabilisation). [1 mark]
Electrophilic addition would disrupt this delocalisation and destroy the aromatic stability, requiring a large activation energy. Substitution preserves the aromatic ring. [1 mark]
(b) Equation: C₆H₆ + CH₃COCl → C₆H₅COCH₃ + HCl [1 mark]
Product: phenylethanone (acetophenone) [1 mark]
(c) Mechanism:
- Formation of electrophile: CH₃COCl + AlCl₃ → CH₃CO⁺ + AlCl₄⁻ [1 mark]
- Curly arrow from benzene ring to CH₃CO⁺ [1 mark]
- Formation of arenium ion (positive charge delocalised in ring) [1 mark]
- Curly arrow from C-H bond to ring, restoring aromaticity; AlCl₄⁻ removes H⁺ [1 mark]
- Products: C₆H₅COCH₃ + HCl + AlCl₃ (regenerated)
9. (a) Compound E is a ketone (positive 2,4-DNP test; negative Tollens' test). [1 mark]
Structure: CH₃COCH₃ (propanone) [1 mark]
(b) Reduction product: CH₃CH(OH)CH₃ (propan-2-ol) [1 mark for structure, 1 mark for name]
(c) Test: Add Tollens' reagent (ammoniacal silver nitrate) and warm. [1 mark]
Observation: Propanal gives a silver mirror; propanone gives no reaction. [1 mark]
Accept: Fehling's solution (propanal gives red ppt; propanone no reaction).
10. (a) Zwitterion of glycine: ⁺H₃NCH₂COO⁻ [1 mark]
(b) Amino acids exist as zwitterions in the solid state. [1 mark]
Strong ionic/electrostatic attractions between the ⁺NH₃ and COO⁻ groups of neighbouring molecules require significant energy to overcome, resulting in high melting points. [1 mark]
(c) Dipeptide structure: H₂NCH₂CONHCH₂COOH [1 mark for structure]
Peptide linkage: -CONH- (amide group) [1 mark]
11. (a) Order of increasing boiling point: G < H < F [1 mark]
- G (CH₃COOCH₃): ester, only permanent dipole-dipole and van der Waals' forces [1 mark]
- H (CH₃CH₂CHO): aldehyde, permanent dipole-dipole and van der Waals' forces (similar Mr to G but more polar)
- F (CH₃CH₂COOH): carboxylic acid, forms hydrogen bonds (dimers), strongest intermolecular forces [1 mark]
(b) In CH₃CH₂COOH, the conjugate base CH₃CH₂COO⁻ is stabilised by resonance/delocalisation of the negative charge over two oxygen atoms. [1 mark]
In ethanol, the conjugate base CH₃CH₂O⁻ has the negative charge localised on one oxygen atom, making it less stable. The more stable conjugate base corresponds to the stronger acid. [1 mark]
12. (a) Compound J is but-2-ene (CH₃CH=CHCH₃).
- (E)-but-2-ene / trans-but-2-ene: CH₃ groups on opposite sides [1 mark]
- (Z)-but-2-ene / cis-but-2-ene: CH₃ groups on same side [1 mark]
(b) Polymer: poly(but-2-ene)
CH₃ H CH₃ H
| | | |
—[C—C]—[C—C]—
| | | |
H CH₃ H CH₃
[2 marks for correct repeat unit structure]
(c) The polymer has a carbon-carbon backbone with no functional groups that can be hydrolysed. [1 mark]
Accept: The polymer is not susceptible to enzymatic degradation / lacks hydrolysable linkages.
13. The lone pair on the oxygen atom in phenol is delocalised into the benzene ring. [1 mark]
This increases the electron density of the ring, making it more susceptible to electrophilic attack compared to benzene. [1 mark]
14. Mechanism of acid-catalysed hydrolysis of ethyl ethanoate:
- Protonation of carbonyl oxygen by H⁺ [1 mark]
- Nucleophilic attack by water at the carbonyl carbon, curly arrow from O of H₂O to C [1 mark]
- Proton transfer and loss of ethanol, curly arrow showing C-O bond breaking [1 mark]
- Deprotonation to give ethanoic acid; products: CH₃COOH and C₂H₅OH [1 mark]
15. Compound K is ethanoic acid (CH₃COOH). [1 mark]
Equation: CH₃COOH + NaHCO₃ → CH₃COONa + H₂O + CO₂ [1 mark]
16. Tertiary halogenoalkanes form relatively stable tertiary carbocations. [1 mark]
The stability of the tertiary carbocation lowers the activation energy for the S_N1 pathway, whereas primary carbocations are too unstable to form readily. [1 mark]
17. Each carbon in benzene uses three sp² hybrid orbitals to form sigma bonds: two C-C and one C-H. [1 mark]
The remaining p orbital on each carbon overlaps sideways to form a delocalised π system above and below the ring. [1 mark]
This delocalisation gives benzene extra stability (aromaticity). [1 mark]
18. Optical isomers of alanine:
- Draw tetrahedral carbon with CH₃, NH₂, COOH, H; one with NH₂ on left, one with NH₂ on right (mirror images). [1 mark each]
19. In amides, the lone pair on nitrogen is delocalised into the carbonyl group (resonance with C=O). [1 mark]
This makes the lone pair less available for protonation compared to amines, where the lone pair is localised on nitrogen. [1 mark]
20. Compound L is propan-2-ol (CH₃CH(OH)CH₃) or a tertiary alcohol (not possible with C₃H₈O) – actually propan-2-ol is a secondary alcohol; it does react with acidified dichromate. Correction: Compound L is methoxyethane (CH₃OC₂H₅) or propan-2-ol? Wait, C₃H₈O with no reaction with dichromate must be an ether: methoxyethane. [1 mark]
Ethers do not undergo oxidation with acidified dichromate because they lack the O-H group necessary for oxidation. [1 mark]
END OF ANSWER KEY