IB Chemistry SLTopic 6 — Electron TransferPaper 1 & 2Organic~11 min read
Reducing Carboxylic Acids, Aldehydes and Ketones
Every arrow on the alcohol oxidation diagram can be run backwards. The destinations are already familiar — what is new is choosing the reagent, and knowing which of the two will refuse to stop where you want it to.
LiAlH4 is the strong reagent: it reduces carboxylic acids, aldehydes and ketones.
LiAlH4 is used in dry ether, followed by dilute acid.
NaBH4 is milder and safer, used in water or alcohol, but it cannot reduce carboxylic acids.
Both work by supplying the nucleophilic hydride ion, H–.
Write the reducing agent as [H] in equations, and still balance them.
Reducing a carboxylic acid with LiAlH4 goes all the way to the alcohol — the aldehyde cannot be isolated.
One ladder, two directions
Nothing new has to be learned about where these reactions end up. The sequence you built for oxidation works just as well read downwards, and the same structural rule applies: a primary alcohol sits below an aldehyde, a secondary alcohol below a ketone.
The right-hand column has no top rung, and that is the same fact as before: a ketone has no hydrogen on its carbonyl carbon, so there is nothing left to oxidise.
Two reagents, one of them fussy
Both reagents deliver a hydride ion, H– — a hydrogen atom carrying a lone pair, which is a nucleophile. The difference is how vigorously they hand it over.
Only one cell in this grid is a cross, and it is the one worth remembering: sodium borohydride will not touch a carboxylic acid.
LiAlH4
NaBH4
Name
lithium aluminium hydride
sodium borohydride
Conditions
anhydrous, usually dry ether, then dilute acid
aqueous or alcoholic solution
Reduces
carboxylic acids, aldehydes, ketones
aldehydes and ketones only
Handling
reacts violently with water
much safer and easier to use
Active species
the hydride ion, H–
the hydride ion, H–
The reason LiAlH4 must be kept dry is the same reason it is powerful. The hydride ion is a strong base as well as a nucleophile, so it will pull a proton off water instantly, giving hydrogen gas and destroying the reagent before it can reach your organic molecule. Sodium borohydride holds its hydride more tightly, which is why it survives in water — and also why it cannot manage a carboxylic acid.
Count the [H] carefully. An aldehyde or ketone needs two, because one hydrogen goes to the carbon and one to the oxygen. A carboxylic acid needs four, because it is two steps down the ladder and a molecule of water is lost on the way.
The aldehyde you cannot catch
Here is the point examiners like. Reducing a carboxylic acid with LiAlH4 does pass through the aldehyde — but the aldehyde is more easily reduced than the acid was, so the moment it appears it is reduced again. There is no way to stop the reaction in the middle.
The second step is the distillation you met on the previous page — oxidise the alcohol and take the aldehyde out of the flask before it can go any further.
WORKED EXAMPLE
Give the organic product when each compound is treated with LiAlH4 in dry ether, followed by dilute acid. (a) propanoic acid (b) propanal (c) propanone
(a) propanoic acidA carboxylic acid is reduced past the aldehyde and does not stop.propan-1-ol(b) propanalAn aldehyde has its carbonyl carbon at the end of the chain.propan-1-ol(c) propanoneA ketone has its carbonyl carbon in the middle, so the OH ends up in the middle too.propan-2-olNotice (a) and (b) give the same product. That is exactly why the aldehyde cannot be isolated from the reduction of the acid.
WORKED EXAMPLE
A mixture contains butanal and butanoic acid. State a reagent that would reduce only the butanal, and write the equation for that reduction using [H].
Choosing the reagentNaBH₄ reduces aldehydes and ketones but is not strong enough for carboxylic acids.sodium borohydride, NaBH₄, in aqueous or alcoholic solutionThe equationCH₃CH₂CH₂CHO + 2[H] → CH₃CH₂CH₂CH₂OHTwo [H]: one adds to the carbon, one to the oxygen of the C=O group. No water is formed, so nothing else appears on the right.
WORKED EXAMPLE
Describe how ethanal could be prepared starting from ethanoic acid, naming the reagents and conditions for each step.
Why not do it directly?LiAlH₄ would reduce the acid straight past the aldehyde to ethanol, and the aldehyde cannot be caught.Step 1 — reduce all the wayLiAlH₄ in dry ether, then dilute acid → ethanolStep 2 — oxidise back up, carefullyacidified K₂Cr₂O₇, distilling the product offDistillation is essential in step 2. Reflux and you would simply arrive back at the ethanoic acid you started with.
💡 Exam tip
Match reagent to substrate: only LiAlH4 will reduce a carboxylic acid.
Quote the conditions: dry ether then dilute acid for LiAlH4, aqueous or alcoholic for NaBH4.
Count the [H]: two for an aldehyde or ketone, four for a carboxylic acid.
Remember the water when reducing a carboxylic acid, or the equation will not balance.
Say that both reagents supply the hydride ion, H–, if asked how they work.
For an aldehyde from an acid, give the two-step route and explain why one step will not do.
⚠️ Common mix-up
Using NaBH4 on a carboxylic acid. It will not work.
Claiming LiAlH4 stops at the aldehyde. It does not, and cannot be made to.
Reducing a ketone to a primary alcohol. The carbonyl is in the middle of the chain, so the product is secondary.
Writing 2[H] for a carboxylic acid instead of 4[H].
Using LiAlH4 in water. It reacts with water violently, which is why dry ether is specified.
Confusing [H] with H2. The square brackets stand for the reducing agent, not hydrogen gas.
Up next: Reducing Unsaturated Compounds — that last mix-up is a useful hint. There is a reduction that really does use hydrogen gas, and it works on carbon–carbon double bonds rather than carbonyls.
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