OCR A-Level Chemistry Synthesis and analytical techniques (02), June 2019: Question 13

1 mark · Medium difficulty · Multiple Choice

Identify which of the given alcohols could be formed by the reduction of a carbonyl compound with NaBH4.

Practise this question

Question

Multiple choice question 13 asks which of three compounds (1: 2-Methylpentan-2-ol, 2: 2-Methylpentan-1-ol, 3: 3-Methylpentan-2-ol) could be formed when a carbonyl compound is reacted with NaBH4. Options A, B, C, and D give combinations of these statements, followed by an answer box and [1] mark allocation.
Question text

13 A carbonyl compound is reacted with NaBH4.

Which compound(s) could be formed?

1 2-Methylpentan-2-ol

2 2-Methylpentan-1-ol

3 3-Methylpentan-2-ol

A 1, 2 and 3

B Only 1 and 2

C Only 2 and 3

D Only 1

Your answer

[1]

Mark scheme

Show the mark scheme Mark scheme for question 13 indicating the correct answer is C.

13 C 1 AO1.2

How to answer it

Reduction of Carbonyl Compounds with NaBH₄

What this question tests

This question assesses your understanding of the reduction of carbonyl compounds (aldehydes and ketones) using aqueous sodium tetrahydridoborate(III) ( NaBH₄ ). You need to apply knowledge of functional group transformations, carbon skeleton structures, and isomerism to deduce which specific alcohols can be synthesized from a single unknown starting carbonyl compound.

Question 13

Full Worked Solution & Breakdown

✅ Correct Answer: C (Only 2 and 3)

The correct option is C because reducing a carbonyl compound with NaBH₄ can yield 2-methylpentan-1-ol (from 2-methylpentanal, an aldehyde) and 3-methylpentan-2-ol (from 3-methylpentan-2-one, a ketone). Compound 1 ( 2-methylpentan-2-ol ) is a tertiary alcohol and cannot be formed by reducing any carbonyl compound containing a straight chain of 5 carbons with a methyl branch at position 2.

💡 Key Knowledge

  • NaBH₄ acts as a reducing agent, supplying hydride ions ( H⁻ ).
  • Aldehydes reduce to form primary alcohols.
  • Ketones reduce to form secondary alcohols.
  • Tertiary alcohols cannot be produced by reducing carbonyl compounds because adding a hydride to a ketone carbonyl carbon always leaves at least one hydrogen attached.

🧠 Exam Technique

Draw out the skeletal or structural formula of each candidate molecule provided in the statements:

  • Statement 1: 2-methylpentan-2-ol is a tertiary alcohol (the -OH is on C2, which also holds a methyl group and an ethyl/propyl chain). Reject immediately!
  • Statement 2: 2-methylpentan-1-ol is a primary alcohol, formed by reducing 2-methylpentanal. (Valid)
  • Statement 3: 3-methylpentan-2-ol is a secondary alcohol, formed by reducing 3-methylpentan-2-one. (Valid)

❌ Common Errors

Students often lose marks on multiple-choice questions like this by assuming NaBH₄ can turn any alcohol skeleton into a product, or by failing to classify alcohols (primary, secondary, tertiary) correctly. Remember: tertiary alcohols are a dead giveaway for incorrect options in carbonyl reduction questions!

Mark Scheme Note: 1 mark awarded for selecting option C (AO1.2 - Demonstrating knowledge and understanding of organic reaction types).

Topics

Module 6: Organic chemistry and analysis · 6.1 Aromatic compounds, carbonyls and acids

Question and mark scheme from the OCR A-Level Chemistry examination, Synthesis and analytical techniques (02), June 2019. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.