Difficulty: Intermediate | Prerequisites: LT5 hydride reduction basics, polar reaction classification, curved-arrow mechanism fundamentals, electronegativity concepts.
This topic sits at the heart of organic chemistry II: how do you reduce a carbonyl group, and how do you control which carbonyl gets reduced when a molecule has more than one? Hydride reagents (LiAlH4, NaBH4, DIBAL) are the main tools for turning carbonyls into alcohols (or, in some cases, stopping at the aldehyde stage). Before tackling this material, you should be comfortable with polar basic reaction types, curved-arrow mechanisms, and the idea that nucleophiles attack electrophilic carbons. If those feel shaky, revisit LT5 first.
Hydride reagents deliver H⁻ to electrophilic carbonyl carbons, reducing them to alcohols. LiAlH4 (LAH) is the stronger reagent and reduces every carbonyl functional group. NaBH4 is milder and chemoselective, reducing only the most reactive carbonyls (aldehydes, ketones, acid chlorides, anhydrides) while leaving esters and amides untouched.
Hydride (H⁻)
A hydrogen atom carrying a negative charge, acting as a nucleophile. In simple terms, it is a hydrogen that donates its pair of electrons to an electrophilic carbon.
Lithium aluminium hydride (LiAlH4 / LAH)
A strong reducing agent that delivers H⁻ to any carbonyl functional group (aldehydes, ketones, esters, acid chlorides, anhydrides, amides). Think of it as the "heavy-duty" hydride source: it reduces everything.
Sodium borohydride (NaBH4)
A milder, chemoselective reducing agent. It only reduces functional groups more reactive than an ester (aldehydes, ketones, acid chlorides, anhydrides). Think of it as the "gentle" hydride: it leaves esters and amides alone.
DIBAL (diisobutylaluminium hydride)
A bulky aluminium hydride that can reduce an ester to an aldehyde when used at low temperature (−78 °C). In simple terms, DIBAL lets you stop the reduction partway, something LAH cannot do.
Chemoselectivity
The preference of a reagent for one functional group over another in the same molecule. In simple terms, when a molecule has two different reducible groups, a chemoselective reagent picks only one.
Irreversible nucleophile
A nucleophile whose conjugate acid has a very high pKa (H⁻ has pKaH ≈ 35), meaning the reverse reaction is thermodynamically unfavourable. Once it adds, it stays.
Tetrahedral intermediate
The sp3 intermediate formed when a nucleophile adds to a carbonyl carbon. In simple terms, the flat C=O becomes a four-cornered carbon bearing the nucleophile and an alkoxide.
Aqueous workup (H3O⁺ quench)
The acid wash step after a hydride reduction that protonates the alkoxide intermediate to give the final alcohol product. Think of it as "finishing" the reaction by adding a proton source.
Organic reactions fall into five types: polar basic, polar acidic, polar neutral, radical, and pericyclic. Hydride reductions are polar basic reactions. The reagent dictates which type occurs.
Irreversible nucleophiles (H⁻, carbanions) are strong bases with very high conjugate acid pKa values (pKaH of H⁻ ≈ 35). Once they add to a carbonyl, the reaction does not reverse under normal conditions. This is the defining feature that separates them from weaker, reversible nucleophiles like water or alcohols.
An aprotic solvent (THF, diethyl ether) is required because these reagents would react with protic solvents (water, alcohols) before ever reaching the carbonyl.
The hydride ion (H⁻) acts as a nucleophile. Its filled orbital (HOMO, a sigma bond) attacks the empty orbital (LUMO, the pi-star of C=O) on the electrophilic carbonyl carbon.
Step 1: Nucleophilic addition. H⁻ attacks the carbonyl carbon, forming a tetrahedral alkoxide intermediate.
Step 2: Aqueous workup. H3O⁺ protonates the alkoxide to give the alcohol product.
With aldehydes, this is a straightforward addition: one equivalent of hydride, one primary alcohol out.
Aldehydes are more reactive than esters toward nucleophilic addition. The reason is resonance: in an ester, the lone pair on oxygen donates into the carbonyl, making the carbonyl carbon less electrophilic. Aldehydes lack this extra resonance stabilisation.
When LAH reacts with an ester:
The first hydride adds to the ester carbonyl, forming a tetrahedral intermediate.
The leaving group (alkoxide, pKa ≈ 16) departs, collapsing the intermediate to an aldehyde. This step is reversible in the sense that the tetrahedral intermediate can lose the leaving group.
The aldehyde is more reactive than the original ester, so a second hydride immediately attacks.
The final product after workup is a primary alcohol.
The reduction cannot stop at the aldehyde stage with LAH because LAH has more than one hydride per molecule and the aldehyde intermediate is more reactive than the starting ester.
Amides are the exception. When an amide is treated with LAH:
The first hydride adds, forming a tetrahedral intermediate.
The nitrogen-containing leaving group departs, generating an imine intermediate (not an aldehyde).
A second hydride reduces the imine to an amine.
After workup, the product is an amine, not an alcohol.
So: every carboxylic acid derivative except amides gives an alcohol with LAH. Amides give amines.
The difference in reactivity comes down to how polar the metal-hydrogen bond is.
In NaBH4: the B-H bond has an electronegativity difference (ΔEN) of about 0.2 (B = 2.04, H = 2.2). The bond is less polar, so the hydride carries less negative charge. Less reactive.
In LiAlH4: the Al-H bond has ΔEN ≈ 0.6 (Al = 1.61, H = 2.2). The bond is more polar, so the hydride carries more negative charge. More reactive.
Because NaBH4 delivers a "weaker" hydride, it is chemoselective: it only reduces functional groups that are more reactive than an ester. Those groups are aldehydes, ketones, acid chlorides, and anhydrides.
Diisobutylaluminium hydride (DIBAL) is a bulky aluminium hydride. At low temperature (−78 °C), it can reduce an ester to an aldehyde and stop there. The bulky isobutyl groups and cold temperature stabilise the tetrahedral intermediate so it does not collapse to release a leaving group before workup. On aqueous workup, the aldehyde is released.
This is the reagent you reach for when you need an aldehyde from an ester.
Aldehyde + LiAlH4: Aldehyde → (1. LiAlH4, 2. H3O⁺) → Primary alcohol
Ketone + LiAlH4: Ketone → (1. LiAlH4, 2. H3O⁺) → Secondary alcohol
Ester + LiAlH4: Ester → (1. LiAlH4, 2. H3O⁺) → Primary alcohol (two hydrides consumed; passes through aldehyde intermediate)
Amide + LiAlH4: Amide → (1. LiAlH4, 2. H3O⁺) → Amine (passes through imine intermediate)
Ketone + NaBH4: Ketone → (NaBH4, EtOH) → Secondary alcohol
Ester + DIBAL: Ester → (1. DIBAL, −78 °C, 2. workup) → Aldehyde
Electronegativity values to remember: H = 2.2, B = 2.04, Al = 1.61. The larger the ΔEN between the metal and hydrogen, the more polar the M-H bond, the more reactive the hydride.
The LiAlH4 reduction of p-isobutylacetophenone is an actual step in the industrial synthesis of ibuprofen, one of the most widely used over-the-counter painkillers. Chemoselectivity matters in pharmaceutical manufacturing because drug molecules often carry multiple functional groups, and reducing the wrong one destroys the product.
Students often think NaBH4 can reduce esters. It cannot. NaBH4 only reduces functional groups more reactive than an ester (aldehydes, ketones, acid chlorides, anhydrides).
Students sometimes assume that LAH reduction of an ester stops at the aldehyde. It does not, because the aldehyde intermediate is more reactive than the starting ester, and LAH has multiple hydrides available.
Students confuse the product of amide reduction with LAH. Amides give amines, not alcohols. Every other carboxylic acid derivative gives an alcohol.
Students forget the aqueous workup step. The direct product of hydride addition is an alkoxide ion, not a neutral alcohol. You must protonate it with H3O⁺.
⚠️ You will be asked to predict products of LiAlH4 and NaBH4 reductions and to draw their mechanisms. Know both steps (nucleophilic addition, then workup).
⚠️ Chemoselectivity is a favourite exam topic. If a molecule has both a ketone and an ester, NaBH4 reduces only the ketone. LAH reduces both.
⚠️ Expect a question asking you to compare LiAlH4, NaBH4, and DIBAL, and to choose the correct reagent for a given transformation.
⚠️ The electronegativity argument for why LAH is more reactive than NaBH4 (Al-H bond is more polar than B-H bond) is a commonly tested reasoning question.
⚠️ IR spectroscopy data may appear: the disappearance of a carbonyl stretch (~1700 cm⁻¹) and the appearance of a broad O-H stretch (~3300 cm⁻¹) confirms reduction to an alcohol.
True or false: NaBH4 can reduce an ester to an alcohol.
Fill in the blank: The direct product of hydride addition to a carbonyl is a(n) ______, which must be protonated during workup.
True or false: LiAlH4 reduction of an amide yields an alcohol.
Fill in the blank: DIBAL at −78 °C reduces an ester to a(n) ______.
True or false: The Al-H bond in LiAlH4 is less polar than the B-H bond in NaBH4.
Answers: 1. False. 2. Alkoxide. 3. False (it yields an amine). 4. Aldehyde. 5. False (Al-H is more polar).
Q: What product is formed when a ketone is treated with (1) LiAlH4, (2) H3O⁺?
A: A secondary alcohol.
Q: You have a molecule with both a ketone and an ester group. You want to reduce only the ketone. Which reagent do you use, and why?
A: NaBH4. It is chemoselective and only reduces functional groups more reactive than an ester. The ketone is more reactive than the ester, so NaBH4 reduces the ketone and leaves the ester intact.
Q: Why does LiAlH4 reduction of an ester not stop at the aldehyde stage?
A: Because the aldehyde intermediate is more reactive than the starting ester, and LAH still has available hydrides. The second reduction occurs immediately.
Q: Explain, using electronegativity, why LiAlH4 is a stronger reducing agent than NaBH4.
A: Aluminium (EN = 1.61) is less electronegative than boron (EN = 2.04). The Al-H bond therefore has a larger electronegativity difference (ΔEN ≈ 0.6) than the B-H bond (ΔEN ≈ 0.2), making the hydride in LAH more negatively charged and more nucleophilic.
Q: What reagent and conditions would you use to convert an ester directly to an aldehyde?
A: DIBAL at −78 °C, followed by aqueous workup. The low temperature and bulky substituents prevent the reduction from proceeding past the aldehyde.
This connects to LT5 (where hydride reduction was first introduced as a transformation) and to LT6/LT12/LT18 (synthesis problems that require choosing the right reducing agent). The concept of chemoselectivity here also applies when you learn about Grignard reagents and Gilman reagents in the next part of LT8, where the same "which functional group reacts?" logic drives reagent choice.
Hydride reduction, LiAlH4, LAH, NaBH4, sodium borohydride, lithium aluminium hydride, DIBAL, diisobutylaluminium hydride, DIBAL-H, carbonyl reduction, aldehyde reduction, ketone reduction, ester reduction, amide reduction, chemoselectivity, chemoselective reduction, irreversible nucleophile, tetrahedral intermediate, aqueous workup, H3O+ quench, polar basic reaction, nucleophilic addition to carbonyl, CHEM 212, LT8, organic chemistry II, Penn State