3.8.2Organic Chemistry
Reduction of Carbonyls — Oxford AQA International A Level Chemistry Revision Notes & Topical Questions
5 min read 6 parts Lesson 30 of 55
Part 1 of 6Reagent and Products
Carbonyl compounds are reduced by adding two hydrogen atoms across the C=O bond.
Reducing agent
- Sodium tetrahydridoborate(III), NaBH₄, in aqueous or alcoholic solution acts as a source of the hydride ion, H⁻.
Aldehydes
- Reduction of an aldehyde gives a primary alcohol: CH₃CHO + 2[H] → CH₃CH₂OH.
Ketones
- Reduction of a ketone gives a secondary alcohol: CH₃COCH₃ + 2[H] → CH₃CH(OH)CH₃.
Selectivity
- NaBH₄ reduces C=O but not C=C, because the hydride is attracted to the δ+ carbon and repelled by the electron-rich double bond.
Reverse reaction
- Oxidising the alcohols with acidified dichromate regenerates the carbonyl compounds, linking the two reactions.
Part 2 of 6The Mechanism
The reaction is a nucleophilic addition in two steps.
1
The C=O bond is polar because oxygen is more electronegative, so the carbon is δ+ and the oxygen is δ−.
2
The hydride ion uses its lone pair to attack the δ+ carbon, and the C=O π electrons move onto oxygen.
3
This forms a negatively charged alkoxide intermediate containing a new C–H bond.
4
The negative oxygen picks up a proton from water or the solvent, forming the O–H bond of the alcohol.
5
Curly arrows run from the hydride lone pair to carbon, from the C=O bond to oxygen, and from the oxygen lone pair to H⁺.
Part 3 of 6Stereochemistry
Stereochemistry
Planar carbonyl
- The carbonyl carbon and its three attached atoms lie in one plane, so the hydride can attack from either face.
Chiral products
- Reducing an unsymmetrical ketone such as butanone gives butan-2-ol, which has a chiral centre.
Racemic mixture
- Attack from each face is equally likely,
- so equal amounts of both enantiomers form and the product does not rotate plane-polarised light.
Symmetrical products
- Reducing propanone or ethanal gives alcohols that are not chiral, so no optical isomers form.
Part 4 of 6Linking to Other Reactions
Linking to Other Reactions
Compared with oxidation
- Oxidation of alcohols removes hydrogen, whereas reduction of carbonyls adds hydrogen, so they are opposite processes.
Writing equations
- [H] represents hydrogen from the reducing agent and equations are balanced by including two [H] per carbonyl group.
Safer reagent
- NaBH₄ can be used in aqueous solution and is easier to handle than more reactive hydride reagents.
Checking the product
- The alcohol formed can be confirmed by the loss of the C=O peak and the appearance of a broad O–H peak in the infrared spectrum.
Part 5 of 6Worked Example
Question Write the equation, using [H], for the reduction of butanone and explain why the product is optically inactive.
- Butanone is the ketone CH₃COCH₂CH₃, so it is reduced to a secondary alcohol.
- Add 2[H]: CH₃COCH₂CH₃ + 2[H] → CH₃CH(OH)CH₂CH₃, which is butan-2-ol.
- Carbon 2 is bonded to four different groups (H, OH, CH₃ and C₂H₅), so it is a chiral centre.
- The planar C=O is attacked equally from both faces, so a 50:50 racemic mixture forms and its optical effects cancel.
Answer
- CH₃COCH₂CH₃ + 2[H] → CH₃CH(OH)CH₂CH₃;
- the product is a racemic mixture of butan-2-ol enantiomers, so it is optically inactive.
Part 6 of 6Exam Checklist & Tips
3.8.2 – Key Points for the Exam
- Reduction Gain of electrons or, for organic compounds, addition of hydrogen.
- Hydride ion The H⁻ ion, a nucleophile with a lone pair.
- Nucleophilic addition A reaction in which a nucleophile attacks a δ+ carbon and adds across a double bond.
- Racemic mixture An equal mixture of two enantiomers that has no effect on plane-polarised light.
Common Mistakes
- Saying NaBH₄ reduces C=C bonds; it reduces C=O but leaves C=C unchanged.
- Reducing a ketone to a primary alcohol, when ketones give secondary alcohols.
- Showing the hydride attacking the oxygen atom instead of the δ+ carbon.
- Forgetting that the intermediate carries a negative charge on oxygen before protonation.
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