AQA A-Level Chemistry Paper 3, June 2022: Question 3

15 marks · Medium difficulty · State/Explain/Describe

Identify the mechanisms, isomers, draw the elimination mechanism, represent enantiomers in 3D, and explain the relative rates of hydrolysis for halogenoalkanes reacting with sodium hydroxide and silver nitrate.

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Question text

03 Under suitable conditions, 2-bromobutane reacts with sodium hydroxide to produce a

mixture of five products, A, B, C, D and E.

Products A, B and C are alkenes.

A is a structural isomer of B and C.

A does not exhibit stereoisomerism.

B and C are a pair of stereoisomers.

Products D and E are alcohols.

D and E are a pair of enantiomers.

03.1 Give the names of the two concurrent mechanisms responsible for the formation of

the alkenes and the alcohols.

[2 marks]

Mechanism to form alkenes

Mechanism to form alcohols

03.2 Define the term stereoisomers.

[2 marks]

03.3 Deduce the name of isomer A.

Explain why A does not exhibit stereoisomerism.

[2 marks]

Name

Explanation

03.4 Outline the mechanism for the reaction of 2-bromobutane with sodium hydroxide to

form alkene A.

[3 marks]

03.5 Deduce the name of isomer B and the name of isomer C.

Explain the origin of the stereoisomerism in B and C.

[2 marks]

Names

Explanation

03.6 Draw 3D representations of enantiomers D and E to show how their structures are

related.

[2 marks]

03.7 A student compares the rates of hydrolysis of 1-chlorobutane, 1-bromobutane and

1-iodobutane.

The suggested method is:

• add equal volumes of the three halogenoalkanes to separate test tubes

• add equal volumes of aqueous silver nitrate to each test tube

• record the time taken for a precipitate to appear in each test tube.

State and explain the order in which precipitates appear.

[2 marks]

Order in which precipitates appear

Explanation

Mark scheme

Show the mark scheme

Question Answers Additional Comments/Guidelines Mark

(for alkenes) elimination Allow base elimination 1

Not nucleophilic elimination

03.1

(for alcohols) nucleophilic substitution 1

(2 x AO1)

(Different molecules/compounds with the) same (molecular and) 1

structural formula

03.2

Different spatial arrangement of atoms Allow different spatial arrangement of 1

bonds/groups (2 x AO1)

A = but-1-ene Not butene 1

two groups/atoms/Hs the same on one of the C=C carbons Allow two groups/atoms/Hs the same on first C 1

03.3 Not two groups the same on one side of C=C

Ignore references to no chiral carbon

Ignore ‘priority’ i.e. 2 groups with the same (1 x AO1,

priority… gets M2 for ‘2 groups the same–A-LEVEL CHEMISTRY–…’ – JUNE 1 x AO3)2022

H H If wrong halogenoalkane used then max 2/3

M1 lone pair on O, negative charge (anywhere) and 1

H3CCH2 C C H curly arrow from lone pair to H on carbon 1

M2 Not if (covalent) NaOH / additional arrows to or

M3 +

Br H from NaOH / additional arrows to or from Na

- M2 curly arrow from C(1)–H to C(1)–C(2) 1

M2 is standalone from M1

Allow ecf if H on carbon 3 attacked in M1 for curly 15

arrow from C(3)-H to C(2)–C(3)

Not as ecf if H on carbon 2 attacked in M1 for curly

arrow from C(2)-H

M3 Curly arrow from C–Br to Br (mark is 1

independent)

03.4 Not if any additional arrows / incorrect polarity (3 x AO2)

or formal charges on C–Br

Allow ecf for mechanism to form but-2-ene from

03.3

Allow E1 mechanism

M1 curly arrow from C–Br bond to the Br

M2 curly arrow from lone pair on O of OH– to a

correct H on the correct C adjacent to C+ on

the carbocation

M3 curly arrow from a correct C–H bond to a

correct C–C bond

penalise M1 for any additional arrow(s) to/from the

Br to/from anything else

penalise M2 for any additional arrow(s) on NaOH

– A-LEVEL CHEMISTRY – –

16 Z-but-2-ene AND E-but-2-ene allow ‘cis’/‘trans’ and B and C either way round 1

Allow E/Z but-2-ene, cis/trans but-2-ene

03.5 lack of/restricted/no (free) rotation around C=C/double bond Allow C=C/double bond cannot rotate 1

(1 x AO1,

1 x AO3)

OH OH M1 any correct 2D or 3D structure of butan-2-ol 1

Allow C2H5

C C M2 must show at least one wedge bond and one 1

CH3 H3C dash bond in each structure from the chiral C and

H3CH2C CH2CH3

H H any bonds in the plane cannot be at 180º to each (1 x AO2,

other 1 x AO3)

03.6

second structure could be drawn as mirror image of

first or with same orientation of bonds and two

groups swapped round

Allow ECF for second structure from incorrect first

structure, providing molecule is chiral

Silver iodide then silver bromide then silver chloride Allow yellow then cream then white 1

Allow iodide/AgI then bromide/AgBr then

chloride/AgCl

Allow iodo(butane) then bromo(butane) then

03.7 chloro(butane)

Ignore iodine then bromine then chlorine

bond strength C−I < C−Br < C−Cl Ignore incorrect formulae 1

Allow carbon-halogen bond strength decreases (2 x AO3)

down the group / from Cl to I

How to answer it

Reactions & Isomerism of 2-Bromobutane

📌 What this question tests

This question assesses core Organic Chemistry principles across Halogenoalkanes and Isomerism:

Distinguishing between elimination and nucleophilic substitution mechanisms.
Standard definitions and molecular requirements for stereoisomerism (both E/Z geometric isomerism and optical enantiomers).
Drawing the curved arrow mechanism for base-catalysed elimination without regiospecific errors.
Drawing correct 3D tetrahedral wedge-and-dash diagrams for optical isomers.
Explaining relative rates of hydrolysis using carbon–halogen bond enthalpy rather than bond polarity.

Question 03.1 • 2 Marks

Concurrent Mechanisms Forming Alkenes and Alcohols

Identification of competitive reaction pathways of halogenoalkanes

✅ Correct Answer

Mechanism to form alkenes: Elimination (allow 'base elimination')

Mechanism to form alcohols: Nucleophilic substitution

💡 Key Knowledge

OH⁻ as a Base: In alcoholic/ethanolic conditions and high temperatures, OH⁻ acts as a proton acceptor (base), favouring elimination to yield alkenes.
OH⁻ as a Nucleophile: In aqueous conditions and warm temperatures, OH⁻ attacks the electron-deficient carbon, favouring nucleophilic substitution to yield alcohols.

❌ Common Errors & Examiner Commentary

Do NOT write "nucleophilic elimination": There is no such mechanism name in A-Level Chemistry. Elimination is purely base-induced here.
Forgetting the full name "nucleophilic substitution" (writing just "substitution" often forfeits the mark if the syllabus specifies mechanism classification).

Mark scheme: 1 mark for 'elimination', 1 mark for 'nucleophilic substitution'. (2 × AO1)

Question 03.2 • 2 Marks

Definition of Stereoisomers

Precise textbook definition required

✅ Correct Answer

Molecules/compounds with the same structural formula...

...but with a different spatial arrangement of atoms (or bonds/groups in space).

🧠 Exam Technique

Always state both halves of the definition:
1. Same structural formula (or same molecular and structural formula).
2. Different spatial arrangement of atoms.
Avoid using vague terms like "different shape" or "different 3D look". Always write "spatial arrangement".

Mark scheme: M1 for 'same (molecular and) structural formula'. M2 for 'different spatial arrangement of atoms / bonds / groups'. (2 × AO1)

Question 03.3 • 2 Marks

Identifying Isomer A & Why It Lacks Stereoisomerism

Deducing structural isomerism vs geometric isomerism in alkenes

✅ Correct Answer

Name of isomer A: but-1-ene

Explanation: There are two identical groups/atoms (two H atoms) on one of the C=C carbon atoms (specifically C1).

🧠 Exam Technique

For an alkene to exhibit E/Z isomerism, each carbon atom of the C=C double bond must be attached to two different atoms or groups.
In but-1-ene ( CH₂=CH-CH₂-CH₃ ), carbon 1 has two hydrogens. Swapping them produces an identical molecule.

❌ Common Errors

Writing just "butene" without the locant number "1" — this scores 0.
Saying "two groups on the same side of the C=C" — this is a description of cis-isomers, NOT why stereoisomerism is absent!
Referring to a lack of a chiral centre. Stereoisomerism in alkenes is geometric (E/Z), not optical.

Mark scheme: M1 for 'but-1-ene'. M2 for 'two groups/atoms/Hs the same on one of the C=C carbons'. (1 × AO1, 1 × AO3)

Question 03.4 • 3 Marks

Mechanism: Formation of But-1-ene

Base-catalysed elimination mechanism from 2-bromobutane

✅ Correct Arrow Movements (E2 Pathway)

Arrow 1 (M1): From a lone pair on the oxygen of :OH⁻ to one of the H atoms on Carbon-1 (the terminal -CH₃ group).
Arrow 2 (M2): From the C(1)–H bond to the C(1)–C(2) single bond (forming the C=C double bond).
Arrow 3 (M3): From the C(2)–Br bond to the Br atom (breaking the bond to release Br⁻).

📐 How to Draw the Mechanism Clearly

Structure to draw:
Draw C(1) fully displayed as H–CH₂– attached to –CH(Br)–CH₂–CH₃ .

1. Place :OH⁻ near an H on C1. Arrow starts strictly at the lone pair on O and points to that H.
2. Arrow starts strictly on the middle of the C(1)–H bond and curves to the C(1)–C(2) bond.
3. Arrow starts on the C(2)–Br bond and ends directly on the Br atom.

❌ Examiner Traps to Avoid

Targeting the wrong carbon: Removing an H from Carbon-3 produces but-2-ene, not isomer A (but-1-ene). The question explicitly asks for alkene A!
Arrow starting points: Never start arrows from the negative charge itself unless the lone pair is clearly drawn next to it. Start directly from the lone pair.
Covalent NaOH: Do not draw a covalent Na–OH bond with arrows coming from Na or the bond; use free :OH⁻ .

Mark scheme: M1 (lone pair on :OH⁻ to C1–H), M2 (C1–H bond to C1–C2 bond), M3 (C2–Br bond to Br). (3 × AO2)

Question 03.5 • 2 Marks

Isomers B and C: Stereoisomerism in But-2-ene

Nomenclature and structural causes of geometric isomerism

✅ Correct Answer

Names of isomers B and C:
Z-but-2-ene AND E-but-2-ene
(Allow cis-but-2-ene and trans-but-2-ene)

Explanation:
Restricted rotation (or no free rotation / lack of rotation) around the C=C double bond.

💡 Key Knowledge

The π-bond (pi bond) forms from sideways overlap of p-orbitals above and below the plane of the carbon atoms.
Rotating the molecule about the double bond would require breaking this π-bond, which has a high activation energy at room temperature. Thus, rotation is restricted.

Mark scheme: M1 for both 'Z-but-2-ene AND E-but-2-ene' (or cis/trans). M2 for 'restricted/lack of rotation around C=C double bond'. (1 × AO1, 1 × AO3)

Question 03.6 • 2 Marks

3D Representations of Enantiomers D and E

Displaying tetrahedral geometry and non-superimposable mirror images

✅ Correct Representation Requirements

The central asymmetric carbon atom (C2 of butan-2-ol) must have 4 different groups: –H , –OH , –CH₃ , and –CH₂CH₃ (or –C₂H₅ ).
Both enantiomers must be drawn in 3D: each must show two normal in-plane bonds, one wedge (coming forward), and one dashed/hatched bond (going back).
Structure 2 must be the clear mirror image of Structure 1 (or have two groups swapped).

🧠 Exam Technique for 3D Tetrahedral Drawings

Standard Template:
• Draw a central 'C'.
• Draw two solid lines at ~109.5° (e.g., one going straight up to –OH , one down-left to –CH₂CH₃ ).
• Draw a solid wedge pointing down-right to –CH₃ .
• Draw a hashed/dashed line pointing slightly between them to –H .
• Draw a vertical dashed mirror line and reflect the exact positions for the second enantiomer!

❌ Common Errors

Bonds in the plane drawn at 180°: A square-planar looking cross (+) will instantly lose the 3D mark (M2). Bonds in the plane must look tetrahedral (~109.5°).
Drawing butan-1-ol instead of butan-2-ol (butan-1-ol is not chiral and cannot have enantiomers).
Attaching the ethyl group through the wrong atom (e.g., writing C–H₂C–CH₃ with the bond pointing to H instead of C).

Mark scheme: M1 for correct connectivity of butan-2-ol. M2 for correct 3D stereochemical representation showing mirror imagery (at least one wedge and one dash per structure). (1 × AO2, 1 × AO3)

Question 03.7 • 2 Marks

Rates of Hydrolysis: 1-Chlorobutane vs 1-Bromobutane vs 1-Iodobutane

Trend in reactivity of halogenoalkanes with aqueous silver nitrate

✅ Correct Answer

Order in which precipitates appear:
Silver iodide first, then silver bromide, then silver chloride.
(Allow: 1-iodobutane > 1-bromobutane > 1-chlorobutane, or yellow then cream then white)

Explanation:
Bond strength / bond enthalpy order: C–I < C–Br < C–Cl
(The C–I bond is the weakest and requires the least energy to break).

💡 Underlying Chemistry: Bond Enthalpy vs Polarity

Students often think bond polarity determines the rate (since C–Cl is the most polar and C has the greatest δ+). This is incorrect!

Bond enthalpy is the overriding factor:

Going down Group 7, atomic radius increases.
The shared pair of electrons is further from the halogen nucleus, so the C–X bond becomes longer and weaker.
Lower activation energy means faster hydrolysis, releasing X⁻ ions quickest for 1-iodobutane to form the AgI precipitate first.

❌ Common Errors

Saying "chlorine is more electronegative so C–Cl reacts fastest" — completely backwards! Bond enthalpy dominates over bond polarity.
Stating "iodine then bromine then chlorine" instead of naming the precipitate (silver iodide) or the haloalkane (1-iodobutane). Halogen elements (I₂, Br₂, Cl₂) are not formed!

Mark scheme: M1 for correct order (AgI first, then AgBr, then AgCl). M2 for explanation linked to bond strength/enthalpy (C–I < C–Br < C–Cl). (2 × AO3)

Topics

Organic Chemistry · 3.3.1 Introduction to Organic Chemistry · 3.3.3 Halogenoalkanes · 3.3.7 Optical Isomerism

Question and mark scheme from the AQA A-Level Chemistry examination, Paper 3, June 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.