IB Chemistry HLTopic 6 — Electron TransferPaper 1 & 2Organic~9 min read
Reducing Carboxylic Acids, Aldehydes & Ketones
The last page climbed a ladder: alcohol to aldehyde to carboxylic acid. This page climbs back down it. Same rungs, opposite direction, and only two reagents to know — one that reduces everything and one that is a bit more polite about it.
📚 What you need to know
Reduction is written with [H], the reducing agent, just as oxidation used [O].
LiAlH4 (lithium aluminium hydride) is strong: it reduces acids, aldehydes and ketones. Used in dry ether, then dilute acid.
NaBH4 (sodium borohydride) is milder: aldehydes and ketones only, and it works in water or alcohol.
Both work by delivering the hydride ion, H−, which attacks the carbon of the C=O group.
You cannot stop a carboxylic acid at the aldehyde with LiAlH4.
The ladder, going down
The aldehyde in the top row is real but cannot be isolated. LiAlH4 reduces it the moment it appears, so the reaction runs straight through to the alcohol.
The two reagents
Reagent
Reduces
Conditions
Notes
LiAlH4
Carboxylic acids, aldehydes, ketones
Dry ether, then dilute acid added afterwards
Very reactive; reacts violently with water, so the solvent must be anhydrous
NaBH4
Aldehydes and ketones only
Aqueous or alcoholic solution
Milder and much safer to handle; will not touch a carboxylic acid
Both reagents work the same way underneath: they hand over a hydride ion, H−. That is a hydrogen with two electrons, so it is a nucleophile, and it attacks the slightly positive carbon of the C=O bond. This is why the reaction is a reduction — that carbon is gaining electron density.
Why LiAlH4 needs dry ether: the hydride ion is a powerful base as well as a nucleophile. Put it in water and it grabs a proton straight away, producing hydrogen gas and destroying the reagent before it can do the job. The dilute acid goes in only at the end, once the reduction is complete.
Ketone to secondary alcohol
RCOR’ + 2[H] → RCH(OH)R’
Notice that only the carboxylic acid produces water. That is the extra oxygen being removed, and it is also why that step needs four [H] rather than two.
Worked examples
WORKED EXAMPLE
Propanone is treated with NaBH4 in aqueous solution. Name the product and write the equation.
Step 1: Identify the starting materialPropanone, CH3COCH3, is a ketone.Step 2: Ketones reduce to secondary alcoholsNaBH4 is fine here — it handles ketones.CH3COCH3 + 2[H] → CH3CH(OH)CH3Step 3: Name itPropan-2-olthe OH lands on the carbon that used to hold the C=O, so it is carbon 2
WORKED EXAMPLE
A student wants butanal from butanoic acid. Explain why a single reduction will not work, and suggest a two-step route.
Step 1: Try the direct routeOnly LiAlH4 is strong enough to touch a carboxylic acid, and it does not stop halfway.Butanoic acid + 4[H] → butan-1-ol + H2OStep 2: Accept the overshoot and go back upButan-1-ol is a primary alcohol, so mild oxidation with distillation gives the aldehyde.Butan-1-ol + [O] → butanal + H2OReduce all the way with LiAlH4, then oxidise back up by distillingdown two rungs then up one — awkward, but it is the only route at this level
WORKED EXAMPLE
A molecule contains both a ketone group and a carboxylic acid group. Suggest a reagent that reduces only the ketone, and explain your choice.
Step 1: Compare what each reagent attacksLiAlH4 would reduce both groups. NaBH4 cannot reduce carboxylic acids.Step 2: Choose the milder oneNaBH4 in aqueous or alcoholic solutionStep 3: State the outcomeThe C=O of the ketone becomes an alcohol; the COOH group is untouched.NaBH4, because it is selective for aldehydes and ketones“the weaker reagent is the more useful one” is a common theme in organic synthesis
💡 Exam tip
Use [H] and get the number right: 4[H] for a carboxylic acid, 2[H] for an aldehyde or ketone.
Remember the water only appears when a carboxylic acid is reduced.
Quote conditions: LiAlH4 in dry ether, followed by dilute acid. NaBH4 in water or alcohol.
If asked to choose a reagent, justify it by saying what the other one would also attack.
You are expected to know reagents and general conditions, but not exact temperatures.
Watch the product class: aldehydes give primary alcohols, ketones give secondary ones.
⚠ Common mix-up
Using NaBH4 on a carboxylic acid. It cannot do it, and questions test exactly this.
Trying to stop a LiAlH4 reduction at the aldehyde. It runs straight through.
Writing 2[H] for a carboxylic acid. It needs 4[H].
Using LiAlH4 in water. It reacts with water violently, which is why dry ether is specified.
Saying a ketone reduces to a primary alcohol. The carbon already has two R groups, so it can only be secondary.
Forgetting that this is redox. The carbon is gaining electron density, and its oxidation number falls.
Up next: Reducing Unsaturated Compounds — the same word, reduction, but now hydrogen is added straight across a C=C or C≡C bond rather than to a carbonyl.
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