OCR A-Level Chemistry Synthesis and analytical techniques (02), November 2020: Question 6

Alcohol Dehydration & Stereoisomerism

What this question tests

This question assesses your understanding of alcohol dehydration (elimination reactions using an acid catalyst like concentrated H₂SO₄ or H₃PO₄) and your ability to recognise when an organic product can exhibit stereoisomerism (specifically E/Z isomerism or optical isomerism).

Question 6

Dehydration of Alcohols and Stereoisomerism

✅ Correct Answer: D (heptan-4-ol)

When heptan-4-ol undergoes acid-catalysed dehydration (elimination of water), it forms hept-3-ene . This alkene possesses different groups on both carbons of the carbon-carbon double bond, allowing it to form both E and Z stereoisomers.

💡 Key Knowledge

Dehydration mechanism: Alcohols undergo elimination with an acid catalyst (heat under reflux with concentrated acid) to form alkenes and water.
E/Z Isomerism requirements: The C=C double bond must have two different groups attached to carbon 1 and two different groups attached to carbon 2.
Symmetrical secondary or tertiary alcohols often yield specific alkene regio- and stereoisomers upon elimination.

🧠 Exam Technique

Don't just guess! Draw out the structural formula of the alkene produced by each option before deciding. Systematically check each alkene for restricted rotation around the C=C bond and the presence of non-identical groups on each carbon atom.

❌ Common Errors

Assuming that because the starting alcohol is chiral or achiral, the product automatically inherits that property.
Forgetting that terminal alkenes (like pent-1-ene from pentan-1-ol) cannot exhibit E/Z isomerism because one carbon of the double bond has two identical hydrogen atoms attached.

Mark Breakdown: 1 mark available for selecting option D.

Detailed Breakdown of Options

Option A: 3-methylbutan-2-ol

Dehydration yields a mixture of alkenes (primarily 2-methylbut-2-ene and 3-methylbut-1-ene). 2-methylbut-2-ene has two identical methyl groups attached to one end of the C=C double bond, so it does not show E/Z isomerism.

Option B: pentan-1-ol

Dehydration yields pent-1-ene . Because carbon 1 of the double bond has two identical hydrogen atoms (-H and -H), it cannot form E/Z stereoisomers.

Option C: 2-methylhexan-2-ol

Dehydration gives 2-methylhex-1-ene and 2-methylhex-2-ene. Neither alkene satisfies the requirement for E/Z isomerism due to identical groups (-CH₃) on at least one of the double-bonded carbon atoms.

Option D: heptan-4-ol (Correct)

Structure: CH₃CH₂CH₂CH(OH)CH₂CH₂CH₃ .
Dehydration removes -OH and -H to form hept-3-ene ( CH₃CH₂CH=CHCH₂CH₂CH₃ ).
Both C3 and C4 have two different groups attached ( -H and -CH₂CH₂CH₃ vs -H and -CH₂CH₃ ), resulting in a mixture of E and Z stereoisomers.