Reduce a ketone to a secondary alcohol
Reduce an internal C=O with NaBH₄ or LiAlH₄ while retaining the carbonyl position.
Section 17 · Carbonyl compounds
AS Level; Cambridge International Chemistry 9701, 2025–2027 syllabus. Distinguish aldehydes from ketones by structure and evidence; state reagent, condition and observation. Written practice is self-checked
Before you begin
- Use carbonyl structures, alcohol oxidation and reduction from Topics 13 and 16, nucleophiles and curly arrows, and the organic reaction patterns from Topic 15.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What class of alcohol forms when a ketone is reduced?
What you will learn
- Use NaBH₄ or LiAlH₄
- Convert C=O to C–OH
- Classify the product as secondary
The ketone carbonyl becomes an alcohol
NaBH₄ or LiAlH₄ reduces a ketone to a secondary alcohol. The C=O becomes C–OH at the same carbon; carbon–carbon connectivity is retained.
Butanone + 2[H] → butan-2-ol. The product OH-bearing carbon remains attached to two carbon groups.
Use the starting position
Keep the ketone's carbonyl position when naming the alcohol. Reduction does not rearrange the carbon skeleton in this syllabus pathway.
If the ketone is symmetrical, equivalent chain directions can give the same alcohol name. Avoid shifting OH to an end carbon unless the starting carbonyl was there.
Pause and explain
What happens to the ketone C=O during reduction?
Worked example
Predict the product from pentan-3-one and classify it.
Show the worked solution
- Retain the five-carbon skeleton.
- Reduce C=O at carbon 3 to C–OH.
- That carbon has two carbon neighbours, giving a secondary alcohol.
Answer Pentan-3-one + 2[H] → pentan-3-ol; secondary alcohol, using NaBH₄ or LiAlH₄.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Give the product and reagent for reducing propanone.
Give me a hint
Keep the carbonyl at C2.
Compare my reasoning
- Propanone is a three-carbon ketone.
- Convert C=O to C–OH.
- Name the secondary alcohol.
Propanone + 2[H] → propan-2-ol; NaBH₄ or LiAlH₄.
Look for these in your work
- Product correct
- Position retained
- Reagent stated
Reduce 3-methylbutan-2-one and name the product.
Give me a hint
Keep the carbonyl at C2 and the methyl branch at C3.
Compare my reasoning
- The starting molecule contains five carbon atoms.
- Convert the C2 carbonyl to C2–OH.
- The product is 3-methylbutan-2-ol.
3-methylbutan-2-one → 3-methylbutan-2-ol; NaBH₄ or LiAlH₄.
Look for these in your work
- Branch retained
- OH position correct
- Name correct
Why is the product from a ketone not a primary alcohol?
Give me a hint
Count carbon groups bonded to the carbonyl carbon.
Compare my reasoning
- A ketone carbonyl carbon is bonded to two carbons.
- Reduction does not remove either carbon group.
- The resulting C–OH carbon is secondary.
It retains two carbon neighbours after reduction, so the alcohol is secondary.
Look for these in your work
- Carbon neighbours counted
- Connectivity retained
- Class justified
Common mistakes
- Calling the product primary
- Changing the carbonyl position
- Using NaOH instead of a reducing agent
What can you explain now?
What class of alcohol forms when a ketone is reduced?
Compare with the explanation
Secondary
The former carbonyl carbon remains attached to two carbon groups.
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Make it stick
How do ketone position and carbon neighbours determine the reduction product's name and class?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.