Organic Chemistry · Alcohols and Phenols
Alcohols from Carbonyl Compounds: Reduction
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Reduction of a carbonyl compound adds hydrogen across the C=O double bond: the carbon gains a Hydride An H- equivalent — a hydrogen with an extra electron pair Full entry → (H-) and the oxygen gains a proton, converting the planar sp2 carbonyl carbon into a tetrahedral sp3 alcohol carbon. The product depends on the starting carbonyl: aldehydes give primary alcohols, ketones give secondary alcohols, and — with the more powerful reagent — esters and carboxylic acids also give primary alcohols. Two reagents dominate: sodium borohydride (NaBH4), a mild hydride source that works in water or methanol and reduces only aldehydes and ketones among common groups, and lithium aluminum hydride (LiAlH4), a powerful hydride source for anhydrous ether that also reduces esters, acids, amides, and nitriles.
Why this matters
Reduction is how chemists — and cells — convert carbonyls into alcohols. Biologically, enzymes such as alcohol dehydrogenase use the coenzyme NADH to deliver hydride to carbonyls (for example, converting pyruvate to lactate). In the lab, NaBH4 and LiAlH4 are the workhorses of pharmaceutical synthesis, and the Chemoselectivity The ability of a reagent to react with one functional group and not others Full entry → of NaBH4 (it leaves C=C bonds, esters, and nitro groups untouched) lets a chemist reduce one group without damaging others. Reduction is also the direct reverse of oxidation (Topic 7).
The college version
Core Concepts
What "reduction" means here
In organic chemistry, reduction means the addition of hydrogen (or gain of electron density). The carbonyl carbon is δ+ because oxygen pulls electrons away; hydride, with its lone pair, is attracted to it. Mechanism in words: the hydride's electron pair forms a new C–H bond at the carbonyl carbon, and the π electrons of the C=O move onto the oxygen, creating an Alkoxide R–O-, the anionic oxygen left after hydride addition Full entry → (R–CH2–O-). Adding water or dilute acid (the workup) protonates the alkoxide to give the alcohol. The carbonyl carbon drops from oxidation state +1 (aldehyde) or +2 (ketone) to −1 (primary alcohol) or 0 (secondary alcohol).
Sodium borohydride: the mild reagent
NaBH4 reduces aldehydes and ketones cleanly in methanol, ethanol, or water — protic solvents it tolerates because it reacts with them only slowly. It does not reduce esters, carboxylic acids, amides, or nitriles, and it leaves C=C double bonds alone. This selectivity is its greatest asset: an α,β-unsaturated ketone is reduced only at the C=O (1,2-reduction), preserving the alkene. One NaBH4 can deliver up to four hydrides, so the stoichiometry is roughly one mole of borohydride per four moles of ketone.
Lithium aluminum hydride: the powerful reagent
LiAlH4 reduces aldehydes, ketones, esters, carboxylic acids, amides, and nitriles — essentially every carbonyl. It reacts violently with water and alcohols, so it must be used in anhydrous diethyl ether or THF and quenched carefully. Esters consume two hydrides: the first addition gives an aldehyde, the second reduces it to a primary alcohol. Because the ester's alkoxy group (–OR) is expelled as an alkoxide, an ester gives two alcohols.
Chemoselectivity: which functional groups survive
A single molecule often contains several reducible groups. NaBH4 lets you reduce a ketone while leaving an ester, a nitro group, or a halogen untouched; LiAlH4 is far less choosy. Neither reagent reduces isolated C=C bonds (that requires catalytic hydrogenation, Chapter 8), so an alkene survives.
Stereochemistry of ketone reduction
The carbonyl is planar, and simple hydride reagents attack either face with equal probability: an achiral ketone gives a Racemic mixture Equal amounts of both enantiomers Full entry → of alcohol enantiomers — a textbook illustration of why a chiral environment (reagent, catalyst, or substrate) is needed to make one enantiomer preferentially. In a ring, facial selectivity appears: reduction of 4-tert-butylcyclohexanone gives predominantly trans-4-tert-butylcyclohexanol (OH equatorial) — hydride approaches the more accessible face.
How It Works / Step-by-Step Process
- Identify the carbonyl: aldehyde, ketone, ester, or carboxylic acid — this fixes the product's substitution.
- Choose the reagent: NaBH4 for aldehydes/ketones; LiAlH4 when esters or acids must be reduced.
- Add the hydride: H- attacks the carbonyl carbon; the C=O π pair moves to oxygen, forming the alkoxide.
- Work up with water or dilute acid to protonate the alkoxide (with LiAlH4, quench excess reagent carefully). If the carbonyl is cyclic or chiral, predict the major stereoisomer from face accessibility and product stability.
Common Confusions
| Do Not Confuse | With | Difference |
|---|---|---|
| NaBH₄ scope | LiAlH₄ scope | NaBH₄ reduces aldehydes/ketones only; LiAlH₄ also reduces esters, acids, amides, nitriles |
| Reduction of a C=O | Hydrogenation of a C=C | Hydride reagents add H to the carbonyl carbon; H2/metal adds H,H across alkenes |
| Aldehyde reduction product | Ketone reduction product | Aldehydes → primary alcohols; ketones → secondary alcohols |
| Reduction | Oxidation | Reduction adds H / gains electron density; oxidation removes H / adds C–O bonds (Topic 7) |
| Ester reduction products | One alcohol | An ester gives two alcohols: the acyl-side alcohol plus the alkoxy alcohol |
| Racemic product | No reaction | A racemate means reduction happened with no facial selectivity (planar carbonyl, achiral reagent) |
| Alkoxide | Alcohol | The direct product is the alkoxide; the alcohol appears only after the workup protonates it |
Eli explains
The same idea, in plain words
Explain it like I’m 10
A carbonyl is a carbon double-bonded to an oxygen — imagine a seesaw with the carbon at the top. Reduction pushes that seesaw down: you add a hydrogen to the carbon and a hydrogen to the oxygen, the double bond becomes a single bond, and the molecule ends up with an OH handle — it has become an alcohol. Sodium borohydride is the gentle helper that does this to aldehydes and ketones without touching anything else; lithium aluminum hydride is the strong helper that can also do it to acids and esters, but it is so powerful you must keep water away.
Worked example
Example 1: Stoichiometry of a NaBH₄ reduction
Problem: How many grams of NaBH4 (molar mass 37.83 g/mol) reduce 0.500 mol of cyclohexanone, assuming one borohydride delivers four hydrides?
The mole ratio is 1 mol NaBH4 : 4 mol ketone. Write the formula before substituting:
n(NaBH4) = n(ketone)4 = 0.500 mol4 = 0.125 mol
Convert moles to grams with dimensional analysis:
0.125 mol NaBH4 × 37.83 g NaBH41 mol NaBH4 = 4.73 g NaBH4
Answer: 4.73 g of sodium borohydride.
Example 2: Ester reduction gives two alcohols
Problem: What forms when methyl benzoate, C6H5CO2CH3, is treated with excess LiAlH4 in ether, then worked up?
Step 1 — First hydride: H- adds to the ester carbonyl carbon; the –OCH3 group leaves as methoxide, giving the aldehyde benzaldehyde, C6H5CHO.
Step 2 — Second hydride: H- adds to the aldehyde carbon, giving the alkoxide C6H5CH2O-.
Step 3 — Workup: Protonation gives benzyl alcohol, C6H5CH2OH (acyl side), and the methoxide becomes methanol, CH3OH. One ester, two alcohols.
Example 3: Chemoselectivity on an enone
Problem: 4-Phenyl-3-buten-2-one, C6H5CH=CHCOCH3, is treated with one equivalent of NaBH4 in methanol. Which functional group reacts, and what is the product?
The reagent attacks only the carbonyl carbon (1,2-reduction); the C=C is untouched because hydride does not add to double bonds. The product is 4-phenyl-3-buten-2-ol, C6H5CH=CHCH(OH)CH3 — the alkene survives (reducing it would require H2/Pd). A classic demonstration of reagent-controlled chemoselectivity.
Key takeaways
- Aldehyde + NaBH4 or LiAlH4 → primary alcohol; ketone + either → secondary alcohol.
- NaBH4: mild, works in protic solvents, reduces only aldehydes/ketones among common groups.
- LiAlH4: powerful, requires anhydrous conditions, also reduces esters, acids, amides, nitriles.
- Ester + LiAlH4 → two alcohols (acyl-side primary alcohol plus the alkoxy alcohol).
- Hydride attacks the carbonyl carbon; isolated C=C bonds are not reduced by these reagents.
- An achiral ketone reduced by an achiral reagent gives a racemic alcohol — no facial preference.
- The alkoxide becomes the alcohol only after the aqueous workup.
Check yourself
6 review questions from the chapter. Try each one, then open the answer.
What are the products of reducing an aldehyde and a ketone with NaBH₄?
Show answer
Aldehyde → primary alcohol; ketone → secondary alcohol.
Why must LiAlH₄ reactions be run in anhydrous ether while NaBH₄ works in methanol or water?
Show answer
LiAlH₄ reacts violently with protic solvents, so it requires anhydrous ether/THF; NaBH₄ reacts only slowly with them, so protic solvents are fine.
An enone, C6H5CH=CHCOCH3, is treated with NaBH₄. What is reduced, and what survives?
Show answer
The carbonyl is reduced (1,2-reduction) to give 4-phenyl-3-buten-2-ol; the C=C bond survives because hydride reagents do not reduce isolated or conjugated alkenes.
How many moles of NaBH₄ are needed to reduce 2.00 mol of cyclohexanone (four hydrides per borohydride)?
Show answer
n(NaBH4) = 2.00/4 = 0.500 mol — one mole of NaBH₄ per four moles of ketone.
Which two alcohols form when methyl benzoate is reduced with excess LiAlH₄?
Show answer
Benzyl alcohol (acyl side) and methanol (methoxy group).
Why does reduction of an achiral ketone give a racemic mixture of alcohols?
Show answer
The planar carbonyl is attacked from either face with equal probability, giving equal amounts of both enantiomers.
Study tools & related lessonsKey vocabulary · Related
Key vocabulary
- Hydride
- An H- equivalent — a hydrogen with an extra electron pair
- Alkoxide
- R–O-, the anionic oxygen left after hydride addition
- Sodium borohydride (NaBH₄)
- Mild hydride donor that reduces aldehydes/ketones in protic solvents
- Lithium aluminum hydride (LiAlH₄)
- Powerful hydride donor requiring anhydrous ether
- Chemoselectivity
- The ability of a reagent to react with one functional group and not others
- Racemic mixture
- Equal amounts of both enantiomers
Sources & references
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