A Level Chemistry H3 Organic Chemistry Quiz

A-Level Chemistry H3 Quiz - Organic Chemistry: Answer Key

Total Marks: 60

Section A: Multiple Choice (Questions 1–5)

1. A) The transition state of an exothermic step resembles the reactants more than the products.

• Explanation: The Hammond Postulate states that the transition state of a step resembles the species (reactant, product, or intermediate) that is closer in energy to it. For an exothermic step, the transition state is closer in energy to the reactants (the higher-energy species), so it resembles the reactants more.
• Common mistake: Students often confuse this with endothermic steps, where the transition state resembles the products.

[1 mark]

2. C) SN1

• Explanation: The rate law depends only on the concentration of the substrate (2-bromo-2-methylpropane) and not on the concentration of the nucleophile (OH⁻). This is characteristic of a unimolecular mechanism. Since the substrate is tertiary, it cannot undergo SN2 (which requires a backside attack and is hindered by bulky groups). The rate-determining step is the formation of the tertiary carbocation, which is slow and involves only the substrate. This is the SN1 mechanism.
• Note: E1 also has the same rate law, but the question asks for the mechanism consistent with the rate law. Since both SN1 and E1 share the same rate-determining step (carbocation formation), the best answer is SN1, as the question context is about substitution. However, a student could argue for both. The mark is for identifying the unimolecular mechanism.

[1 mark]

3. A) CH₃Br

• Explanation: SN2 reactions are favoured by less sterically hindered substrates. The rate of SN2 follows the order: methyl > primary > secondary > tertiary. CH₃Br is a methyl halide with the least steric hindrance around the electrophilic carbon, allowing the nucleophile to attack from the back side most easily.

[1 mark]

4. B) Zaitsev's rule

• Explanation: Zaitsev's rule states that in an elimination reaction, the more substituted alkene is the major product. This is because the more substituted alkene is more stable (due to hyperconjugation and inductive effects). The Hammond Postulate can be used to explain why the more stable alkene is favoured (the transition state resembles the alkene product), but the rule itself is Zaitsev's rule.

[1 mark]

5. B) (S)-2-methoxybutane

• Explanation: SN2 reactions proceed via a single concerted step with a backside attack. This leads to inversion of configuration at the stereocentre. (R)-2-bromobutane will be converted to (S)-2-methoxybutane. A racemic mixture would be formed via an SN1 mechanism.

[1 mark]

Section B: Short Answer Questions (Questions 6–12)

6. (a) The Hammond Postulate states that the transition state of a reaction step resembles the species (reactant, product, or intermediate) that is closest to it in energy. [1]

(b) Explanation: [3]

• The bromination of propane proceeds via a radical mechanism. The rate-determining step is the abstraction of a hydrogen atom by a bromine radical.
• This step is endothermic because the C–H bond broken is stronger than the H–Br bond formed.
• According to the Hammond Postulate, for an endothermic step, the transition state resembles the products (the alkyl radical).
• The secondary radical (from abstraction of a secondary H) is more stable than the primary radical (from abstraction of a primary H) due to hyperconjugation and inductive effects.
• Therefore, the transition state leading to the secondary radical is lower in energy than the transition state leading to the primary radical, making the secondary abstraction faster and more selective.

Marking Notes:

• 1 mark for identifying the step as endothermic.
• 1 mark for stating the transition state resembles the radical product.
• 1 mark for explaining that the more stable secondary radical leads to a lower-energy transition state and thus a faster reaction.

[4 marks]

7. (a) SN1 [1]

(b) Explanation: [3]

• The first step of the SN1 mechanism is the slow, rate-determining ionisation of the C–Br bond to form a planar tertiary carbocation (3-methylhexan-3-ylium) and a bromide ion.
• The carbocation is trigonal planar with an empty p-orbital perpendicular to the plane.
• The nucleophile (water) can attack the carbocation from either face of the planar ion with equal probability.
• Attack from the top face gives one enantiomer, and attack from the bottom face gives the other enantiomer.
• Since the starting material is a single enantiomer, (S)-3-bromo-3-methylhexane, and the intermediate is achiral, the product is formed as a racemic mixture (1:1 mixture of enantiomers).

Marking Notes:

• 1 mark for the planar carbocation intermediate.
• 1 mark for attack from both faces.
• 1 mark for the formation of a racemic mixture.

[4 marks]

8. (a) 2-methylprop-1-ene (also known as isobutene) [1]

(b) Mechanism: E2 [2]

• The substrate is a tertiary bromide, which is highly sterically hindered. This prevents the nucleophile (OH⁻) from performing a backside attack required for SN2.
• The base (OH⁻) is strong and will abstract a proton from a β-carbon, leading to the formation of a double bond and the expulsion of the bromide ion in a single concerted step.
• The major product is the most substituted alkene (Zaitsev's rule), which is 2-methylprop-1-ene.

Marking Notes:

• 1 mark for identifying E2.
• 1 mark for explaining the steric hindrance preventing SN2 and the strong base favouring elimination.

[3 marks]

9. (a) Rate = k[1-bromobutane][OH⁻] [1]

(b) Explanation: [2]

• The reaction to form butan-1-ol is an SN2 reaction.
• The rate law for an SN2 reaction is second order overall: first order with respect to the substrate and first order with respect to the nucleophile.
• If the concentration of sodium hydroxide is doubled, the rate of formation of butan-1-ol will double (increase by a factor of 2), because the rate is directly proportional to [OH⁻].

Marking Notes:

• 1 mark for stating the rate doubles.
• 1 mark for explaining the first-order dependence on [OH⁻].

[3 marks]

10. (a) Newman Projection (Anti): [2]

The anti conformation has the two largest groups (methyl groups) at a dihedral angle of 180°.

Image pending generation: diagram for Q10.

• 1 mark for correct staggered conformation.
• 1 mark for placing the methyl groups anti (180° apart).

(b) Explanation: [2]

• The anti conformation is more stable than the gauche conformation because the two bulky methyl groups are as far apart as possible (180° dihedral angle).
• This minimises steric strain (the repulsion between the electron clouds of the two bulky groups).
• In the gauche conformation, the methyl groups are 60° apart, leading to greater steric repulsion and higher energy.

Marking Notes:

• 1 mark for identifying the minimisation of steric strain.
• 1 mark for explaining the 180° separation of the methyl groups.

[4 marks]

11. (a) Rate = k[2-chlorobutane][base] [1]

(b) Explanation: [3]

• A bulky base, such as potassium tert-butoxide, is a strong base but a poor nucleophile due to its large, sterically hindered structure.
• The SN2 pathway requires the nucleophile to attack the electrophilic carbon from the back side. The bulky tert-butoxide ion cannot approach the carbon effectively due to steric hindrance.
• The E2 pathway requires the base to abstract a proton from a β-carbon. This proton is more accessible than the electrophilic carbon, so the bulky base can still perform this function.
• Therefore, the bulky base strongly favours the E2 pathway over the SN2 pathway.

Marking Notes:

• 1 mark for identifying the base as a poor nucleophile.
• 1 mark for explaining the steric hindrance to backside attack.
• 1 mark for explaining the accessibility of the β-hydrogen.

[4 marks]

12. (a) Regioselectivity is the preference for the formation of one structural isomer (regioisomer) over another when a reaction can produce two or more possible products. [1]

(b) Explanation: [2]

• The E2 reaction of 2-bromo-2-methylbutane can produce two alkenes: 2-methylbut-2-ene (more substituted) and 2-methylbut-1-ene (less substituted).
• Zaitsev's rule states that the more substituted alkene is the major product.
• 2-methylbut-2-ene is a trisubstituted alkene, while 2-methylbut-1-ene is a disubstituted alkene.
• Therefore, the major product is 2-methylbut-2-ene.

Marking Notes:

• 1 mark for identifying 2-methylbut-2-ene.
• 1 mark for explaining it is the more substituted alkene.

[3 marks]

Section C: Extended Response Questions (Questions 13–20)

13. (a) SN1 [1]

(b) Explanation: [3]

• The reaction of 2-iodo-2-methylpropane with iodide ion proceeds via an SN1 mechanism.
• The rate-determining step is the ionisation of the C–I bond to form a tertiary carbocation, (CH₃)₃C⁺.
• Tertiary carbocations are very stable due to the +I effect of three methyl groups and hyperconjugation.
• The high stability of the tertiary carbocation lowers the activation energy for its formation, making the reaction fast.
• In contrast, 1-iodobutane would react via an SN2 mechanism. The primary carbon is not sterically hindered, but the formation of a primary carbocation (required for SN1) is highly unfavourable. The SN2 reaction is slow because the iodide ion is a relatively weak nucleophile in acetone.

Marking Notes:

• 1 mark for identifying the tertiary carbocation intermediate.
• 1 mark for explaining its stability (inductive effect/hyperconjugation).
• 1 mark for contrasting with the primary substrate.

(c) Stereochemical outcome: The reaction of 1-iodobutane with radioactive iodide ion (¹³¹I⁻) via an SN2 mechanism results in inversion of configuration at the carbon atom. However, since 1-iodobutane is not a chiral molecule (the carbon bearing the iodine is not a stereocentre), the product is simply 1-iodobutane with the radioactive isotope incorporated. The stereochemical outcome is not observable in this case because the molecule is achiral. If it were a chiral primary halide, the product would have the inverted configuration. [1]

[5 marks]

14. (a) The rate-determining step is the first step: the ionisation of (CH₃)₃CCl to form the tert-butyl carbocation, (CH₃)₃C⁺, and Cl⁻. This is the step with the higher activation energy (Ea1). [1]

(b) Explanation: [3]

• The first step (ionisation) is endothermic because it involves breaking a strong C–Cl bond to form an unstable carbocation.
• The second step (attack of water on the carbocation) is exothermic because a new, strong C–O bond is formed.
• According to the Hammond Postulate:
  • For the endothermic first step, the transition state (TS1) resembles the products (the carbocation and chloride ion), which are higher in energy.
  • For the exothermic second step, the transition state (TS2) resembles the reactants (the carbocation and water), which are higher in energy than the products.
• Since TS1 resembles the high-energy carbocation intermediate, it is more "product-like" (relative to the reactants of that step). TS2 resembles the carbocation (the reactant of that step), so it is more "reactant-like".

Marking Notes:

• 1 mark for identifying the first step as endothermic.
• 1 mark for identifying the second step as exothermic.
• 1 mark for correctly applying the Hammond Postulate to both steps.

(c) Stereochemical outcome: The reaction proceeds via a planar carbocation intermediate. The nucleophile (water) can attack from either face of the planar ion, leading to a racemic mixture of products (if the starting material is a single enantiomer). [1]

[5 marks]

15. (a) 2-methylprop-1-ene (CH₃)₂C=CH₂ [1]

(b) Explanation: [3]

• The tertiary bromide, (CH₃)₃CBr, has only one type of β-hydrogen (all β-carbons are equivalent methyl groups). Therefore, elimination can only produce a single alkene: 2-methylprop-1-ene.
• The primary bromide, (CH₃)₂CHCH₂Br, has two different types of β-hydrogens:
  • β-hydrogens on the CH group (C2) → produces 2-methylprop-1-ene.
  • β-hydrogens on the terminal CH₃ group (C3) → would produce 2-methylprop-2-ene, but this is the same as 2-methylprop-1-ene after renumbering. Wait, let's re-examine.
  • Actually, (CH₃)₂CHCH₂Br has β-hydrogens on C1 (the CH group) and on the two equivalent methyl groups (C3 and C4). Abstraction from C1 gives (CH₃)₂C=CH₂ (2-methylprop-1-ene). Abstraction from a methyl group gives CH₂=C(CH₃)-CH₃, which is also 2-methylprop-1-ene. So, in this specific case, only one product is formed.
  • Let's use a better example: 2-bromobutane. It has β-hydrogens on C1 and C3, giving but-1-ene and but-2-ene. The question should be corrected. Let's assume the question meant 1-bromobutane, which gives but-1-ene and but-2-ene.
• Correction for the question: The question states that (CH₃)₂CHCH₂Br gives two products: 2-methylprop-1-ene and 2-methylprop-2-ene. This is incorrect. It gives only one product. The question should be revised to use 1-bromobutane or 2-bromobutane. For the purpose of this answer key, we will answer based on the intended concept.

Revised Explanation (assuming the question meant 1-bromobutane):

• The tertiary bromide, (CH₃)₃CBr, has only one type of β-hydrogen, so only one alkene (2-methylprop-1-ene) can be formed.
• 1-bromobutane has two different types of β-hydrogens (on C2 and C3), leading to two possible alkenes: but-1-ene and but-2-ene.

Marking Notes:

• 1 mark for identifying the single product from the tertiary bromide.
• 2 marks for explaining the different types of β-hydrogens.

(c) Explanation: [2]

• The tertiary bromide is more likely to undergo E1 elimination.
• E1 involves the formation of a carbocation intermediate. Tertiary carbocations are very stable, making the E1 pathway feasible.
• Primary carbocations are highly unstable, so the primary bromide cannot undergo E1. It must proceed via the E2 mechanism, which does not involve a carbocation intermediate.

Marking Notes:

• 1 mark for identifying the tertiary bromide.
• 1 mark for justifying with carbocation stability.

[6 marks]

16. (a) Major organic product: The reaction of a vicinal dibromide (2,3-dibromobutane) with zinc dust is a dehalogenation reaction, which removes the two bromine atoms and forms a double bond. The product is but-2-ene. [2]

(b) Explanation: [2]

• The starting material is (2R,3R)-2,3-dibromobutane. The two bromine atoms are on adjacent carbons.
• The reaction with zinc is a syn-elimination (or anti-elimination depending on the mechanism). For acyclic compounds, the elimination is typically anti-periplanar.
• For (2R,3R), the two bromine atoms are on opposite sides of the molecule (anti). The anti-periplanar elimination of bromine atoms from (2R,3R) gives the E-isomer of but-2-ene.
• The product is (E)-but-2-ene.

Marking Notes:

• 1 mark for identifying but-2-ene.
• 1 mark for identifying the E-isomer and explaining the anti-periplanar geometry.

[4 marks]

17. (a) SN2 [1]

(b) Rate = k[1-bromopropane][CN⁻] [1]

(c) Explanation: [1]

• The rate of reaction is directly proportional to the concentration of the nucleophile (CN⁻).
• If the concentration of sodium cyanide is doubled, the rate of reaction will double.

(d) Stereochemical outcome: The SN2 reaction proceeds via a backside attack, leading to inversion of configuration at the electrophilic carbon. Since 1-bromopropane is achiral, the product is achiral, but the mechanism is confirmed by the inversion of configuration in chiral substrates. [1]

[4 marks]

18. (a) Step 1 Mechanism: SN1 (The alcohol is protonated by HBr, water leaves to form a secondary carbocation, which is then attacked by Br⁻. However, secondary carbocations can rearrange. In this case, a hydride shift could occur to form a more stable tertiary carbocation. The major product would be 2-bromo-2-methylpropane. The question should be revised to avoid this complication. Let's assume the question is simplified and the major product is 1-bromo-2-methylpropane via SN1 without rearrangement, which is not realistic. For the purpose of this answer key, we will state SN1.) [1]

(b) Step 2 Mechanism: E2 (The substrate is a primary bromide, and the base is strong. E2 is favoured over SN2 with a strong base, especially at higher temperatures.) [1]

(c) Major organic product of Step 2: The E2 elimination of 1-bromo-2-methylpropane gives 2-methylprop-1-ene as the major product. [1]

(d) Explanation: [2]

• The E2 reaction of 1-bromo-2-methylpropane can produce two alkenes: 2-methylprop-1-ene (more substituted) and 2-methylprop-2-ene (which is the same as 2-methylprop-1-ene after renumbering). Actually, it only gives one product. Let's use a different example. If Step 1 gave 2-bromobutane, then Step 2 would give but-2-ene (major) and but-1-ene (minor).
• Assuming Step 1 gives 2-bromobutane: The E2 elimination of 2-bromobutane can produce but-1-ene and but-2-ene.
• Zaitsev's rule states that the more substituted alkene is the major product.
• But-2-ene is a disubstituted alkene, while but-1-ene is a monosubstituted alkene.
• Therefore, but-2-ene is the major product.

Marking Notes:

• 1 mark for identifying the more substituted alkene.
• 1 mark for applying Zaitsev's rule.

[5 marks]

19. (a) SN2 and E2 [1]

(b) Stereochemical outcome of SN2 product: The SN2 reaction proceeds via a backside attack, leading to inversion of configuration at the stereocentre. (R)-3-bromohexane will produce (S)-3-methoxyhexane. [1]

(c) Explanation: [2]

• The E2 reaction of (R)-3-bromohexane can produce two alkenes: hex-2-ene and hex-3-ene.
• Hex-2-ene can exist as E/Z isomers (cis/trans) because the double bond has two different groups on each carbon.
• Hex-3-ene is symmetrical and does not exhibit E/Z isomerism.
• The formation of hex-2-ene will give a mixture of (E)-hex-2-ene and (Z)-hex-2-ene.

(d) Explanation: [2]

• A bulky base, such as potassium tert-butoxide, is a strong base but a poor nucleophile.
• It will favour the E2 pathway over the SN2 pathway.
• The bulky base cannot easily perform the backside attack required for SN2 due to steric hindrance.
• It can, however, abstract a β-hydrogen, which is more accessible, leading to the elimination product.

Marking Notes:

• 1 mark for identifying E2.
• 1 mark for explaining the steric hindrance to SN2.

[6 marks]

20. (a) Order with respect to (CH₃)₃CBr: [1]

• Comparing Experiments 1 and 2: [(CH₃)₃CBr] doubles (0.10 → 0.20), while [OH⁻] is constant (0.10). The rate doubles (1.2 × 10⁻⁴ → 2.4 × 10⁻⁴).
• Therefore, the order with respect to (CH₃)₃CBr is 1.

Order with respect to OH⁻: [1]

• Comparing Experiments 1 and 3: [OH⁻] doubles (0.10 → 0.20), while [(CH₃)₃CBr] is constant (0.10). The rate stays the same (1.2 × 10⁻⁴).
• Therefore, the order with respect to OH⁻ is 0.

(b) Rate equation: Rate = k[(CH₃)₃CBr] [1]

(c) Mechanism: The rate equation is first order with respect to the substrate and zero order with respect to the nucleophile. This is consistent with a unimolecular mechanism. Since the substrate is tertiary, it is likely SN1 (or E1). The rate-determining step is the formation of the tertiary carbocation. [2]

(d) Calculation of k: [2]

• Using Experiment 1: Rate = k[(CH₃)₃CBr]
• 1.2 × 10⁻⁴ mol dm⁻³ s⁻¹ = k × 0.10 mol dm⁻³
• k = (1.2 × 10⁻⁴) / 0.10 = 1.2 × 10⁻³ s⁻¹
• Units: s⁻¹ (since the overall order is 1, the units of k for a first-order reaction are s⁻¹).

Marking Notes:

• 1 mark for the correct calculation.
• 1 mark for the correct units.

[7 marks]

END OF ANSWER KEY