Reduction of carbonyl compounds

The carbonyl carbon in $C=O$ is electrophilic — oxygen pulls electron density away from carbon, leaving the carbon short of electrons and open to attack by an electron-rich species. Reduction exploits exactly this: a source of hydrogen adds across the double bond, and the product is always the corresponding alcohol.

where: $R\text{–CHO}$ is an aldehyde and $R\text{–CH}_2\text{OH}$ is the primary alcohol formed when one hydrogen adds to the carbonyl carbon and one to the oxygen.

where: $R_2C=O$ is a ketone (two carbon-based groups on the carbonyl carbon) and $R_2CH\text{–OH}$ is the secondary alcohol formed the same way. Three reagents carry out this addition, and they are not interchangeable:

• Catalytic hydrogenation passes $H_2$ gas over the carbonyl compound in the presence of finely divided Ni, Pt, or Pd. The metal surface adsorbs and dissociates $H_2$ into individual hydrogen atoms, which then add across the C=O bond. It works on both aldehydes and ketones but needs the metal catalyst and usually heat — $H_2$ gas alone does nothing to a carbonyl group.
• Lithium aluminium hydride ($LiAlH_4$) is a powerful hydride donor: the $AlH_4^-$ ion delivers $H^-$ directly to the carbonyl carbon. It is strong enough to reduce aldehydes, ketones, carboxylic acids, and even esters all the way to alcohols. Because it reacts violently — sometimes explosively — with water and other protic solvents, it must be used in dry ether, with the alcohol only released afterwards in a separate aqueous acid work-up step.
• Sodium borohydride ($NaBH_4$) is a gentler hydride donor. It reduces aldehydes and ketones efficiently and can even be used directly in protic solvents like ethanol or water, which makes it far more convenient than $LiAlH_4$. Its trade-off is reach: it is not nucleophilic enough to attack the less electrophilic carbonyl carbon of an ester or carboxylic acid, so those groups pass through a $NaBH_4$ reduction largely untouched.

In every hydride case, the mechanism is the same two-step pattern: hydride adds to the carbonyl carbon to give an alkoxide, and the alkoxide is protonated — by solvent or on acid work-up — to give the alcohol.

where: $H^-$ is the hydride ion supplied by $LiAlH_4$ or $NaBH_4$, $R_2CH\text{–O}^-$ is the intermediate alkoxide, and $H^+$ (from water or dilute acid) is the proton that finishes the job.

There is no single formula to derive here — this is descriptive reactivity, and the pattern follows directly from carbonyl electronics. The oxygen of $C=O$ is far more electronegative than carbon, so the carbon end of the double bond carries a partial positive charge and is genuinely electron-poor. Any species that can offer electron density — molecular hydrogen dissociated on a metal surface, or a hydride ion sitting on aluminium or boron — is drawn to that carbon and bonds to it, and the electrons that were in the $\pi$ bond of C=O move fully onto oxygen, generating an alkoxide (or, in catalytic hydrogenation, the alcohol directly once the second hydrogen adds to oxygen). The difference between $LiAlH_4$ and $NaBH_4$ is a difference in how tightly each hydride is held and how eagerly it will attack a weaker electrophile. Aluminium is less electronegative than boron, so the $Al$–$H$ bonds in $AlH_4^-$ are more polarised and the hydride is released more readily — $LiAlH_4$ is the stronger, more reactive nucleophile. An ester's carbonyl carbon is a comparatively poor electrophile in the first place, because the attached $-OR$ group donates electron density into the carbonyl by resonance, partly cancelling the pull of the oxygen; only a strongly nucleophilic hydride source can still force an addition there. $NaBH_4$'s hydride, held more tightly by the more electronegative boron, simply is not reactive enough to attack that weakened electrophile, so esters and acids pass through a $NaBH_4$ treatment essentially unreacted while aldehydes and ketones — genuinely electrophilic carbonyls — are reduced smoothly. This is also why $LiAlH_4$ must stay away from water: it is such an aggressive hydride donor that it reduces the O–H bond of water itself, releasing hydrogen gas violently, so it is always used in dry ether with the alcohol liberated only afterwards, in a deliberate aqueous acid work-up.

Reduce propanal ($CH_3CH_2CHO$) using (a) catalytic hydrogenation and (b) $NaBH_4$, then compare with reducing ethyl acetate ($CH_3COOC_2H_5$) using $LiAlH_4$, and explain why $NaBH_4$ could not do the ester step.

Step 1 — Catalytic hydrogenation of propanal. Hydrogen gas is passed over the aldehyde with a Ni or Pt catalyst; $H_2$ adds across the C=O bond:

Step 2 — $NaBH_4$ reduction of propanal. A hydride ion is delivered to the carbonyl carbon, and the alkoxide intermediate is protonated on aqueous work-up:

Both routes give propan-1-ol, a primary alcohol, since propanal is an aldehyde — the reagent choice does not change the class of product, only how vigorously it must be applied.

Step 3 — $LiAlH_4$ reduction of the ester. Ethyl acetate's carbonyl carbon carries the $-OC_2H_5$ leaving group. $LiAlH_4$ is strong enough to add hydride twice, expelling that leaving group as a second alkoxide along the way:

so the single ester is cleaved into two molecules of ethanol — one from the acyl portion, one from the alkoxy portion that left as ethoxide and was then protonated.

Step 4 — Why $NaBH_4$ cannot do Step 3. The ester carbonyl is a weaker electrophile than an aldehyde or ketone carbonyl, because the $-OR$ group feeds electron density back into the carbonyl carbon by resonance. $NaBH_4$'s hydride, held more tightly by boron, is not nucleophilic enough to attack that weakened centre, so an ester (or a carboxylic acid) is essentially untouched by $NaBH_4$ under normal conditions — only $LiAlH_4$ has the strength to drive the reduction through. Final answer: propanal gives propan-1-ol with either reagent, while ethyl acetate needs $LiAlH_4$ specifically, giving two equivalents of ethanol.

Competitive stretch

If asked to reduce acetone, $(CH_3)_2C=O$, with $LiAlH_4$, the logic is identical to Step 2 but the product class changes: a ketone gives a secondary alcohol, $(CH_3)_2CHOH$ (propan-2-ol), not a primary one — the aldehyde/ketone rule (primary versus secondary product) applies regardless of which reducing agent, mild or strong, is chosen.

• Assuming any reducing agent handles any carbonyl compound equally — $NaBH_4$ reduces aldehydes and ketones but not esters or carboxylic acids, while $LiAlH_4$ reduces all four.
• Forgetting that $LiAlH_4$ reacts violently with water and other protic solvents — it is always used in dry ether, with a separate aqueous acid work-up step needed to free the alcohol.
• Mixing up which carbonyl class gives which alcohol class — aldehyde gives primary alcohol, ketone gives secondary alcohol; simple carbonyl reduction never produces a tertiary alcohol (that needs Grignard addition to a ketone instead).
• Writing catalytic hydrogenation without the catalyst or conditions — $H_2$ gas alone does not touch a carbonyl group; Ni, Pt, or Pd and usually heat are required.

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