AQA A-Level Chemistry Paper 3, June 2018: Question 11
1 mark · Medium difficulty · Multiple Choice
Identify which statement applies to both 2-methylbutan-1-ol and 2-methylbutan-2-ol.
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Question text
11 Which statement is correct about both 2-methylbutan-1-ol and 2-methylbutan-2-ol?
[1 mark]
A They can be formed by alkaline hydrolysis of esters.
B They can be oxidised by reaction with acidified
potassium dichromate(VI).
C They can be formed by hydration of 2-methylbut-2-ene.
D They have four peaks in their 13C NMR spectra.
Mark scheme
Show the mark scheme
11 A
How to answer it
Comparing Isomeric Alcohols: Reactions and ¹³C NMR
This question evaluates your synoptic understanding of organic chemistry across multiple modules:
- Ester Hydrolysis: Base-catalysed (alkaline) cleavage of esters yielding carboxylate salts and alcohols.
- Oxidation of Alcohols: Distinguishing primary (1°), secondary (2°), and tertiary (3°) alcohols using acidified potassium dichromate(VI).
- Electrophilic Addition: Products formed by the hydration of unsymmetrical alkenes.
- Carbon-13 (¹³C) NMR: Identifying non-equivalent carbon environments and molecular symmetry.
Identifying the Shared Chemical Property
Analysis of 2-methylbutan-1-ol and 2-methylbutan-2-ol
✅ Correct Answer: A
They can be formed by alkaline hydrolysis of esters.
Mark scheme: Option A = 1 mark.
Alkaline hydrolysis (saponification) of an ester with aqueous NaOH yields a sodium carboxylate salt and the corresponding alcohol:
R-COO-R' + NaOH → R-COO⁻Na⁺ + R'-OH
Because any alcohol moiety can theoretically form an ester, both primary and tertiary alcohols are released upon alkaline hydrolysis of their corresponding esters.
💡 Key Knowledge: Molecular Structures
Drawing out the structures immediately reveals the classification differences:
- 2-methylbutan-1-ol:
CH₃-CH₂-CH(CH₃)-CH₂OH
• Classification: Primary (1°) alcohol.
• ¹³C environments: 5 distinct carbons (no plane of symmetry) → 5 peaks. - 2-methylbutan-2-ol:
CH₃-CH₂-C(CH₃)₂(OH)
• Classification: Tertiary (3°) alcohol.
• ¹³C environments: 4 distinct environments (the two methyl groups attached to C2 are chemically equivalent) → 4 peaks.
❌ Why the Distractors are Incorrect
- B is false: 2-methylbutan-2-ol is a tertiary alcohol. Tertiary alcohols have no hydrogen atom on the carbon carrying the -OH group, so they cannot be oxidised by acidified K₂Cr₂O₇ under standard conditions.
- C is false: Hydration of 2-methylbut-2-ene ( CH₃-C(CH₃)=CH-CH₃ ) adds H and OH across the double bond.
- Major product: 2-methylbutan-2-ol (via tertiary carbocation).
- Minor product: 3-methylbutan-2-ol (via secondary carbocation).
- 2-methylbutan-1-ol cannot be formed because the C=C double bond is between carbons 2 and 3, not 1 and 2.
- D is false: While 2-methylbutan-2-ol has 4 peaks due to symmetry between the two C2 methyl groups, 2-methylbutan-1-ol lacks this symmetry and has 5 peaks in its ¹³C NMR spectrum.
🧠 Exam Technique & Strategy
- Disprove quickly: Look for classic organic tests first. Spotting that 2-methylbutan-2-ol is a tertiary alcohol instantly eliminates B in under 5 seconds.
- Alkene hydration positioning: Check double bond positions. 2-methylbut-2-ene has the unsaturation between C2 and C3. An alcohol with an -OH on C1 (a butan-1-ol) is impossible to obtain without rearrangement. This instantly eliminates C.
- Symmetry check for ¹³C NMR: For 2-methylbutan-1-ol, trace every carbon: C1 (-CH₂OH), C2 (-CH-), C2-branch (-CH₃), C3 (-CH₂-), C4 (-CH₃). Every single carbon experiences a unique chemical environment, giving 5 signals, eliminating D.
Topics
Organic Chemistry · 3.3.4 Alkenes · 3.3.5 Alcohols · 3.3.9 Carboxylic Acids and Derivatives · 3.3.15 Nuclear Magnetic Resonance Spectroscopy
Question and mark scheme from the AQA A-Level Chemistry examination, Paper 3, June 2018. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.