Difficulty: Intermediate | Prerequisites: Part 1 of these notes (carbonyl overview, reversible vs irreversible reactions), familiarity with nucleophilic addition mechanism
This second set of notes covers the specific nucleophile classes that attack carbonyls: hydrogen nucleophiles (hydride donors such as NaBH₄ and borane) and oxygen nucleophiles (water, forming hydrates). These are core reactions in the carbonyl chemistry unit and will appear repeatedly through the rest of Organic Chemistry II. You should understand what makes a reaction reversible or irreversible (covered in Part 1) before working through this material.
Hydride reagents (NaBH₄, BH₃) deliver H⁻ to the carbonyl carbon, reducing aldehydes and ketones to alcohols. Water adds across the carbonyl to form geminal diols (hydrates), but this equilibrium usually favours the carbonyl form for ketones and the hydrate form only for certain aldehydes. Knowing the selectivity and reversibility of each reagent is the practical takeaway.
NaBH₄ (sodium borohydride)
A mild reducing agent that delivers hydride (H⁻) to the carbonyl carbon of aldehydes and ketones. It is selective: it reduces C=O but does not reduce C=C double bonds or esters under normal conditions. Think of it as a gentle, targeted hydrogen delivery system. Used in protic solvents such as methanol (CH₃OH).
Borane (BH₃)
A stronger, less selective reducing agent that can reduce carbonyls as well as carboxylic acids and other functional groups. It delivers hydride through a different mechanism (coordination to the oxygen first, then hydride transfer). In simple terms, it is a more aggressive version of NaBH₄.
Hydride (H⁻)
A hydrogen atom carrying a pair of electrons, making it a nucleophile. It is not a free species in solution; it is delivered by a reagent (NaBH₄, LiAlH₄, BH₃). Think of it as hydrogen acting as an electron donor rather than a proton.
Geminal diol (hydrate)
The product of water adding across a carbonyl: two hydroxyl groups on the same carbon (gem-diol). For most ketones, the equilibrium lies far to the carbonyl side, so the hydrate is only a minor species in solution. In simple terms, it is a carbonyl that has "soaked up" a water molecule.
Aldehyde vs ketone reactivity
Aldehydes are more reactive toward nucleophilic addition than ketones for two reasons: less steric hindrance (one H vs two R groups flanking the carbonyl) and less electron donation into the carbonyl carbon (alkyl groups are weakly electron-donating, stabilising the partial positive charge). Think of aldehydes as more "exposed" and more "hungry" for electrons.
NaBH₄ reduction of aldehydes and ketones
NaBH₄ is used in a protic solvent (typically CH₃OH or ethanol).
The borohydride ion (BH₄⁻) delivers one of its four hydrides (H⁻) to the electrophilic carbonyl carbon. The mechanism involves a 2-electron process.
The result is an alkoxide intermediate, which is protonated by the solvent to give the alcohol product.
Aldehydes give primary alcohols. Ketones give secondary alcohols.
NaBH₄ is selective: it does not reduce isolated C=C bonds, esters, or carboxylic acids under standard conditions. This makes it very useful when you need to reduce only the carbonyl in a multifunctional molecule.
Borane (BH₃) reduction
Borane is an electrophilic reducing agent. It first coordinates to the lone pair on the carbonyl oxygen, then delivers hydride to the carbon.
BH₃ is less selective than NaBH₄. It can reduce carboxylic acids and amides in addition to aldehydes and ketones.
It also hydrates alkenes (hydroboration), so if your molecule has a C=C double bond, BH₃ may react with both functional groups.
The hydride as a nucleophile
H⁻ from NaBH₄ is a "hydrogen nucleophile" in the lecture's classification. It is not a proton (H⁺) and it is not a hydrogen atom (H·). It carries a lone pair.
After hydride delivery and protonation, the net transformation is: carbonyl (C=O) becomes alcohol (C–OH with an H on the carbon).
Conversion of the alcohol product
The lecture notes show that once you have reduced a ketone to an alcohol, you can convert it further. For example, treating the resulting alcohol with HBr or PBr₃ gives an alkyl bromide. This is a common synthetic sequence: reduce the carbonyl, then substitute the OH.
NaBH₄ reduction followed by treatment with a nitrile (shown in the notes as NaBH₄/CH₃OH then NaCN) illustrates sequential functional group transformations.
Hydration of carbonyls
Water acts as the nucleophile, adding across the C=O to form a geminal diol (two OH groups on the same carbon).
This reaction is reversible. The equilibrium position depends on the substrate.
Aldehyde vs ketone: which hydrates more?
Aldehydes hydrate more readily than ketones. Two factors:
Steric: aldehydes have one H and one R on the carbonyl carbon, leaving more room for the nucleophile. Ketones have two R groups, creating more steric crowding in the tetrahedral product.
Electronic: each alkyl group donates electron density into the carbonyl carbon (inductive effect), making ketone carbonyls less electrophilic. Aldehydes, with only one alkyl group, have a more electrophilic carbon.
Formaldehyde (H₂C=O) exists almost entirely as its hydrate in water. Acetone (CH₃COCH₃) has a negligible amount of hydrate at equilibrium.
Energy comparison
The lecture highlights that aldehydes are higher in energy relative to their hydrates than ketones are. This means the thermodynamic driving force for hydration is greater for aldehydes.
Ketones are already more stabilised (lower in energy) by their two alkyl groups, so they gain less stability from adding water.
Acid-catalysed hydration
Under acidic conditions (cat. H₃O⁺), the carbonyl oxygen is protonated first. This makes the carbon even more electrophilic and susceptible to attack by water.
After water attacks, a proton transfer gives the geminal diol.
Base-catalysed hydration
Under basic conditions (cat. OH⁻), hydroxide ion is the nucleophile rather than water. OH⁻ is a stronger nucleophile than H₂O.
After hydroxide attacks the carbonyl carbon, the resulting alkoxide is protonated by water to give the diol.
Hemiacetal formation (preview)
When an alcohol (ROH) rather than water adds to a carbonyl, the product is a hemiacetal (one OR and one OH on the same carbon). This follows the same pattern and is also reversible. It will be covered in detail in later lectures.
NaBH₄ reduction
RCHO + NaBH₄ (CH₃OH) → RCH₂OH (primary alcohol)
R₂C=O + NaBH₄ (CH₃OH) → R₂CHOH (secondary alcohol)
Hydration equilibrium
RCHO + H₂O ⇌ RCH(OH)₂ (geminal diol) — equilibrium favours hydrate for formaldehyde, carbonyl for most other aldehydes and all ketones
R₂C=O + H₂O ⇌ R₂C(OH)₂ — equilibrium strongly favours the ketone
Alcohol to alkyl halide (downstream)
R–OH + HBr → R–Br + H₂O
R–OH + PBr₃ → R–Br + (HO)₃P
NaBH₄ reduction is used routinely in pharmaceutical labs to reduce ketone intermediates to alcohol products during drug synthesis. It is mild enough to leave other sensitive functional groups intact, which is why it is preferred over stronger reagents like LiAlH₄ in many contexts. Hydration equilibria matter in biochemistry: formaldehyde's tendency to exist as its hydrate is part of why it is so reactive (and toxic) in biological systems, cross-linking proteins through its carbonyl form.
Students often confuse NaBH₄ and LiAlH₄. NaBH₄ is milder and used in protic solvents. LiAlH₄ is much more reactive, reacts violently with water, and must be used in anhydrous ether solvents. They are not interchangeable.
Students sometimes think hydride (H⁻) is the same as a proton (H⁺). They are opposites. Hydride is a nucleophile (electron-rich). A proton is an electrophile (electron-poor).
A common error is assuming that all carbonyl hydrations are negligible. Formaldehyde is almost fully hydrated in water. The generalisation "ketones do not hydrate" is broadly correct, but "carbonyls do not hydrate" is too sweeping.
Students sometimes forget that BH₃ can also do hydroboration of alkenes. If a substrate has both a C=O and a C=C, BH₃ may react with both, and you need to account for that.
⚠️ Expect a question asking you to choose between NaBH₄ and LiAlH₄ for a selective reduction. Know what each one does and does not reduce.
⚠️ Aldehyde vs ketone reactivity toward nucleophilic addition is a classic exam concept. Be ready to explain both the steric and electronic reasons.
⚠️ You may be asked to draw the mechanism of acid-catalysed or base-catalysed hydration of a carbonyl. Know both pathways.
⚠️ The equilibrium position for hydration (aldehyde vs ketone, formaldehyde vs acetone) is commonly tested. Be prepared to predict which direction the equilibrium lies and explain why.
⚠️ Multi-step synthesis questions may ask you to reduce a carbonyl to an alcohol (NaBH₄) and then convert the alcohol to a halide (HBr or PBr₃). Recognise the two-step pattern.
True or False: NaBH₄ will reduce an ester to an alcohol under standard conditions.
Fill in the blank: Aldehydes are more reactive toward nucleophilic addition than ketones because of _______ and _______ effects.
True or False: Formaldehyde exists primarily as its geminal diol (hydrate) in aqueous solution.
True or False: Borane (BH₃) is more selective than NaBH₄.
Fill in the blank: In acid-catalysed hydration, the carbonyl _______ is protonated first to activate the carbon toward nucleophilic attack.
Answers: 1. False (NaBH₄ is too mild for esters). 2. Steric and electronic. 3. True. 4. False (BH₃ is less selective). 5. Oxygen.
Q: What is the product when cyclohexanone is treated with NaBH₄ in methanol, followed by aqueous workup?
A: Cyclohexanol (a secondary alcohol). The hydride adds to the carbonyl carbon, and the solvent protonates the alkoxide.
Q: Explain why acetone (propanone) exists almost entirely as the ketone in water, while formaldehyde exists mostly as the hydrate.
A: Acetone has two methyl groups flanking the carbonyl. These create steric strain in the tetrahedral hydrate product and donate electron density to the carbonyl carbon, making it less electrophilic. Formaldehyde has two hydrogens, so there is minimal steric strain in the hydrate and no electron-donating alkyl groups to stabilise the carbonyl.
Q: You have a molecule with both a ketone and an isolated alkene. You want to reduce only the ketone. Which reagent do you choose, and why?
A: NaBH₄. It selectively reduces the C=O without touching the C=C. BH₃ would react with the alkene via hydroboration.
Q: Draw the mechanism of base-catalysed hydration of an aldehyde (RCHO).
A: Hydroxide (OH⁻) attacks the electrophilic carbonyl carbon. The pi bond breaks, and the electrons move onto the oxygen, forming an alkoxide. The alkoxide is protonated by water to give the geminal diol, regenerating OH⁻.
Q: A student claims that reducing a ketone with NaBH₄ gives a tertiary alcohol. What is wrong with this claim?
A: Reduction of a ketone with NaBH₄ gives a secondary alcohol, not tertiary. The hydride replaces the C=O pi bond with a C–H bond, converting the sp2 carbon to sp3 with one H, one OH, and two R groups. A tertiary alcohol would require three R groups and no H on the carbinol carbon.
Hydride reductions connect back to the oxidation reactions in Part 1: oxidation converts alcohols to carbonyls, and reduction converts carbonyls back to alcohols. These are inverse transformations, and knowing both directions is essential for retrosynthesis. Hydration equilibria foreshadow hemiacetal and acetal formation (next lecture topic in most courses), where alcohols rather than water add to the carbonyl. The selectivity of NaBH₄ vs LiAlH₄ will come up again when you study carboxylic acid derivatives (esters, amides), where LiAlH₄ reduces them but NaBH₄ does not.
Hydrogen nucleophile, hydride donor, NaBH₄, sodium borohydride, LiAlH₄, lithium aluminium hydride, borane, BH₃, carbonyl reduction, ketone reduction, aldehyde reduction, geminal diol, hydrate, hydration equilibrium, acid-catalysed hydration, base-catalysed hydration, aldehyde reactivity, ketone reactivity, steric effects, electronic effects, selective reduction, HBr, PBr₃, alcohol to halide, hemiacetal (preview), formaldehyde hydrate