AQA A-Level Chemistry Paper 2, 2018: Question 2

7 marks · Medium difficulty · State/Explain/Describe

Deduce the IUPAC name, nucleophilic substitution mechanism details, and structures of intermediate compounds in an organic synthesis pathway starting from a bromoalcohol.

Practise this question

Question

Question 02 presents a reaction pathway starting with CH2(OH)CH(CH3)CH2Br reacting with NaOH (Reaction 1) to form CH2(OH)CH(CH3)CH2OH, which undergoes Reaction 2 to form Compound Y (C4H6O2), and then Reaction 3 to form Compound Z. Part 02.1 asks for the IUPAC name of CH2(OH)CH(CH3)CH2Br (1 mark). Part 02.2 asks to explain why the halogenoalkane is attacked by the nucleophile in Reaction 1 (3 marks). Part 02.3 states that the infrared spectrum of Compound Y shows an absorption in the range 1680–1750 cm⁻¹ and asks to draw the displayed formula of Compound Y (1 mark). Part 02.4 states that Compound Z has the empirical formula C3H4NO, asking for the structure of Compound Z and the reagent for Reaction 3 (2 marks).
Question text

02 Halogenoalkanes are useful compounds in synthesis. A reaction pathway is shown.

Reaction 1

CH2(OH)CH(CH3)CH2Br CH2(OH)CH(CH3)CH2OH

NaOH

Reaction 2

Reaction 3

Compound Z Compound Y

C4H6O2

02.1 Give the IUPAC name for CH2(OH)CH(CH3)CH2Br

[1 mark]

02.2 Reaction 1 occurs via a nucleophilic substitution mechanism.

Explain why the halogenoalkane is attacked by the nucleophile in this reaction.

[3 marks]

02.3 The infrared spectrum of Compound Y shows a significant absorption in the range

1680–1750 cm–1

Draw the displayed formula of Compound Y.

[1 mark]

02.4 Compound Z has the empirical formula C3H4NO

Give the structure of Compound Z.

Suggest the reagent for Reaction 3.

[2 marks]

Structure

Reagent for Reaction 3

Mark scheme

Show the mark scheme Mark scheme for Question 02: 02.1 requires '3-bromo-(2)-methylpropan-1-ol ONLY' (1 mark). 02.2 awards M1 for 'Bromine is more electronegative than carbon', M2 for 'C is partially positive / electron deficient', and M3 for 'Lone/electron pair (on the nucleophile) donated to the partially positive carbon' (3 marks total). 02.3 shows the displayed formula of methylpropanedial: O=CH-CH(CH3)-CH=O with all bonds shown (1 mark). 02.4 awards 1 mark for the structure of Compound Z, showing the dinitrile-diol NC-CH(OH)-CH(CH3)-CH(OH)-CN, and 1 mark for 'KCN & (dil) acid' or 'HCN' as the reagent for Reaction 3.

Question Answers Mark Additional Comments/Guidance

3-bromo-(2)-methylpropan-1-ol ONLY 3 and 1 are essential, 2 may be omitted, but any

02.1 1 other number here is wrong

Ignore hyphens and commas

Bromine is more electronegative than carbon M1 Allow difference in electronegativity if polarity of

bond shown

C is partially positive / electron deficient M2 M2 and M3 can be awarded from diagram that

02.2 shows nucleophilic attack

Lone/electron pair (on the nucleophile) donated to the partially M3 Allow lone pair attracted to / attacks the partially

positive carbon positive carbon

Must be displayed with all bonds shown

02.3 1

– – –

1 Not need be displayed

See General Marking instructions section 3.12 for

penalties for incorrectly drawn bonds such as

1 C HO or C NC etc.

02.4

KCN & (dil) acid Allow HCN

12 Allow Ignore alcoholic solvents

H H

O Penalise conc. HCl, H2SO4 or any HNO3

O

NC C CH2 CH2 C CN C CH CH C

HO OH from H H

Total 7

How to answer it

Organic Synthesis: Halogenoalkanes, Carbonyls & Mechanism Theory

🎯 What This Question Tests

This question assesses multi-step organic reaction pathways, IUPAC nomenclature rules, theoretical understanding of nucleophilic substitution mechanisms, infrared spectroscopy interpretation, and carbonyl addition reactions:

  • IUPAC Nomenclature: Numbering priority of functional groups (-ol over halogen/alkyl prefixes).
  • Mechanism Theory: Explaining nucleophilic attack in halogenoalkanes using electronegativity, bond polarity (Cδ+—Brδ-), and electron pair donation.
  • Oxidation of Diols & IR Spectroscopy: Interpreting C=O stretching absorptions (1680–1750 cm⁻¹) and drawing strict displayed formulas.
  • Nucleophilic Addition (Cyanohydrin Formation): Deducing structures using empirical formulas and identifying reagents (KCN / dilute acid).
Question 02.1 • 1 Mark

IUPAC Name for CH₂OHCH(CH₃)CH₂Br

✅ Correct Answer

3-bromo-2-methylpropan-1-ol

Mark Scheme Notes:
• The locants 3 and 1 are mandatory.
• "2" may be omitted (as 3-bromomethylpropan-1-ol has only one possible position for the methyl group), but if any number other than 2 is written, it is marked incorrect.
• Hyphens and commas are ignored, but spelling and priority must be exact.

💡 Key Knowledge

  • Principal Functional Group: The alcohol group (-OH) has higher priority than halogen (-bromo) or alkyl (-methyl) substituents.
  • Chain Numbering: Number the 3-carbon chain starting from the end closest to the suffix group:
    • C1 = —CH₂OH
    • C2 = —CH(CH₃)—
    • C3 = —CH₂Br
  • Alphabetical Order: Substituents precede the root alphabetically: bromo before methyl.

❌ Common Errors & Pitfalls

  • Numbering from the wrong end: Writing 1-bromo-2-methylpropan-3-ol (fails to give the principal -OH group the lowest locant number).
  • Incorrect alphabetical order: Writing 2-methyl-3-bromopropan-1-ol (bromo must come before methyl).
Question 02.2 • 3 Marks

Why Halogenoalkanes Are Attacked by Nucleophiles

Explaining the driving force of the SN reaction

✅ Correct Answer (Mark Scheme Breakdown)

  • M1: Bromine is more electronegative than carbon (or a clear polarity difference is stated).
  • M2: Carbon is partially positive (electron-deficient / Cδ+).
  • M3: A lone pair / electron pair (on the nucleophile) is donated to (or attacks / is attracted to) the partially positive carbon.

🧠 Exam Technique: 3-Point Explanations

Always structure your nucleophilic attack explanations with this logical 3-step sequence:

  1. Electronegativity comparison: "Halogen is more electronegative than carbon..."
  2. Resulting bond polarity: "...creating a polar bond where carbon is Cδ+ (electron-deficient)."
  3. Action of the nucleophile: "...the nucleophile donates a lone pair of electrons to this Cδ+ atom."
Examiner note: M2 and M3 can be awarded from an accurately annotated mechanism sketch showing partial charges and a curly arrow originating from a lone pair on :OH⁻ to Cδ+.

❌ Common Misconceptions

  • Omitting the word "lone" or "electron" pair — simply saying "the nucleophile is donated" scores zero for M3.
  • Stating that bromine is a "good leaving group" instead of explaining why the carbon is attacked in the first place.
  • Describing the bond as ionic rather than polar covalent.
Question 02.3 • 1 Mark

Displayed Formula of Compound Y (C₄H₆O₂)

Deducing the oxidation product from IR spectroscopy

✅ Correct Displayed Formula

Compound Y is 2-methylpropanedial. Both primary alcohol groups in the starting diol are oxidized to aldehyde groups.

H | H - C - H | O = C-C-C = O | | | H H H
Mark Rule: Must be fully displayed. Every single bond (including C—H and C=O double bonds) must be drawn explicitly.

📐 Structural Deduction

  1. Starting Material: Reaction 1 produces the diol CH₂(OH)CH(CH₃)CH₂OH . Both —OH groups are primary alcohols.
  2. IR Evidence: Peak at 1680–1750 cm⁻¹ corresponds uniquely to a carbonyl C=O bond.
  3. Formula Match: The diol formula is C₄H₁₀O₂. Oxidation removes 4 hydrogens (-4H) to yield C₄H₆O₂, which represents two C=O aldehyde groups at both ends.

❌ Trap Alert: "Displayed" vs "Structural" Formula

A frequent mark-loser in AQA exams is drawing shorthand groups like —CHO or —CH₃ . In a displayed formula, every bond between every atom must be drawn out as an individual line (e.g. C=O must show a double bond, and each C—H on the methyl branch must be shown separately).

Question 02.4 • 2 Marks

Structure of Compound Z & Reagents for Reaction 3

Nucleophilic addition to form a dicyanohydrin

✅ Correct Answer

Structure of Compound Z:

H CH₃ H | | | NC - C - C - C - CN | | | OH H OH

Reagent for Reaction 3:

KCN and (dilute) acid  OR  HCN

Marking Notes:
• Structure does not need to be displayed (structural/semi-displayed allowed).
• Bonds must link correctly: penalise bonds drawn as C—HO or C—NC.
• If Y was incorrectly identified as butanedial, the corresponding dicyanohydrin is allowed via error-carried-forward (ECF).

📐 Molecular vs. Empirical Formula Deduction

  1. Reaction 3 adds cyanide (—CN) across carbonyl groups (nucleophilic addition).
  2. Both aldehyde groups add HCN, giving a di-hydroxynitrile (dicyanohydrin).
  3. Molecular Formula: C₄H₆O₂ + 2 HCN = C₆H₈N₂O₂
  4. Empirical Formula Check: Divide by 2:
    C₆H₈N₂O₂ ÷ 2 = C₃H₄NO (exactly matches the question prompt!)

❌ Common Errors & Penalties

  • Reagent Penalties: Concentrated acids ( conc. HCl or conc. H₂SO₄ ) or any nitric acid ( HNO₃ ) are heavily penalised because they react destructively or hydrolyse nitriles.
  • Connectivity Errors: Writing the bond connected to the oxygen of OH or the nitrogen of CN (e.g. C—NC is an isonitrile, not a nitrile). Always ensure the C—C and C—O bonds are cleanly drawn.

Topics

Organic Chemistry · 3.3.1 Introduction to Organic Chemistry · 3.3.3 Halogenoalkanes · 3.3.5 Alcohols · 3.3.6 Organic Analysis · 3.3.8 Aldehydes and Ketones

Question and mark scheme from the AQA A-Level Chemistry examination, Paper 2, 2018. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.