Carbonyl mechanisms: nucleophilic addition, aldol and Grignard
Every carbonyl reaction starts from the same fact: the C=O carbon is electron-poor and the oxygen is electron-rich. A nucleophile attacks the carbon, the π electrons fold onto oxygen, and a tetrahedral alkoxide forms. What happens next is decided by one question — is there a leaving group on that carbon? If no, the alkoxide is simply protonated (addition). If yes, the oxygen kicks back down and expels it (acyl substitution).
The one fact the whole chapter rests on
Oxygen is more electronegative than carbon, so the C=O bond is strongly polarised: partial positive on carbon, partial negative on oxygen. Written as a resonance structure, that is C⁺–O⁻. The consequence is that the carbonyl carbon is permanently hungry for electrons, and its oxygen is permanently available as a place to park a negative charge.
Those two properties generate the entire chapter. A nucleophile attacks carbon; the π electrons retreat onto oxygen, which can hold the charge comfortably; you now have a tetrahedral alkoxide intermediate. Everything after that is bookkeeping.
The fork: addition or substitution?
Look at what else is attached to the carbonyl carbon.
| Carbonyl | Attached groups | Leaving group? | Outcome |
|---|---|---|---|
| Aldehyde | H and R | No | Addition — protonate the alkoxide |
| Ketone | R and R' | No | Addition |
| Acid chloride | R and Cl | Yes, excellent | Acyl substitution, very fast |
| Ester | R and OR' | Yes, moderate | Acyl substitution |
| Amide | R and NR'₂ | Poor | Substitution, but sluggish — usually needs acid or base and heat |
Reactivity runs acid chloride > anhydride > ester > amide, and the order is exactly the leaving-group ability of the group attached. That is the whole trend; you do not need to memorise it separately.
Nucleophilic addition, step by step
Hydride reduction (NaBH₄ or LiAlH₄)
- A hydride is delivered to the carbonyl carbon. Arrow from the B–H or Al–H bond to the carbon.
- The C=O π electrons fold onto oxygen, giving a tetrahedral alkoxide.
- Aqueous workup protonates the alkoxide to an alcohol.
NaBH₄ is mild and reduces aldehydes and ketones but leaves esters and acids essentially alone. LiAlH₄ is aggressive and reduces esters, acids and amides too. If a question gives you a molecule with both a ketone and an ester and asks you to reduce only the ketone, NaBH₄ is the answer it wants.
Grignard addition
A Grignard reagent is a carbon nucleophile. The C–Mg bond is so polarised that the carbon behaves as a carbanion, which makes it the standard way to build a carbon–carbon bond.
- The carbanion carbon attacks the carbonyl carbon.
- π electrons fold onto oxygen — tetrahedral alkoxide.
- H₃O⁺ workup gives the alcohol.
Product class follows the starting carbonyl: formaldehyde gives a primary alcohol, any other aldehyde gives a secondary alcohol, a ketone gives a tertiary alcohol. With an ester you get a double addition, because the first addition expels the alkoxide to give a ketone, which is more reactive than the ester was — so a second equivalent adds immediately and you end up with a tertiary alcohol bearing two identical R groups.
Grignards and protic solvents do not mix. A carbanion is an extremely strong base. Any O–H or N–H in the flask — water, alcohol, carboxylic acid, terminal alkyne — protonates it instantly and destroys the reagent. If a question hands you a Grignard and a substrate containing an OH, the Grignard is consumed by the OH first.
Enolates: the other end of the carbonyl
So far the carbonyl carbon has been the electrophile. But the carbon next door — the α carbon — has an unusual property too: its protons are acidic, with a pKa around 20 for a ketone, because the resulting anion is stabilised by resonance onto the carbonyl oxygen.
That anion is the enolate, and it is a nucleophile. So a carbonyl compound can act as an electrophile at the carbonyl carbon and a nucleophile at the α carbon — which is what makes carbonyls able to react with themselves.
The aldol reaction
- Enolate formation. Hydroxide removes an α proton; the resulting carbanion is resonance-stabilised onto oxygen.
- Addition. The enolate's α carbon attacks the carbonyl carbon of a second molecule.
- Protonation. Workup gives the aldol product: a β-hydroxy aldehyde or ketone. The name is literal — it contains an aldehyde and an alcohol.
Heat, or stronger base, pushes it one step further. The aldol condensation eliminates water from the β-hydroxy carbonyl to give an α,β-unsaturated carbonyl, which is conjugated and therefore more stable. If a question says "with heat" or draws a C=C next to the C=O, it wants the condensation product.
Crossed aldols need control. Mixing two different enolisable carbonyls gives four products. Exam questions avoid this by making one partner non-enolisable (benzaldehyde, formaldehyde) or by forming one enolate completely with LDA before adding the electrophile.
Acetals, imines and the protecting-group trick
Two additions in a row, with dehydration in between, give the condensation products:
- Alcohol + aldehyde/ketone, acid catalysed → hemiacetal → acetal. Acetals are stable to base and to nucleophiles, and hydrolyse back in aqueous acid. That reversibility is why they are the standard protecting group for a ketone you need to survive a Grignard.
- Primary amine + aldehyde/ketone → carbinolamine → imine (Schiff base). With a secondary amine you cannot form the C=N, so the mechanism takes the alternative route and gives an enamine.
Mistakes that cost marks
- Forgetting the workup. Grignard and hydride reactions give an alkoxide. The alcohol only exists after the H₃O⁺ written on the second arrow, and a mechanism that stops at the alkoxide is incomplete.
- Adding one equivalent to an ester. Esters take two equivalents of Grignard or LiAlH₄, because the intermediate ketone or aldehyde is more reactive than the starting material.
- Giving the aldol addition product when the question heated it. Heat means condensation — eliminate water and conjugate.
- Deprotonating the wrong position. Only α protons are acidic. A proton two carbons away from the carbonyl is an ordinary, unreactive C–H.
- Running a Grignard next to a free OH or NH. Protect it first, or the reagent never reaches the carbonyl.
How Organic Chemistry AI helps here
Carbonyl problems are long — several steps, a workup, and often a decision about how many equivalents add. A solve numbers the steps so you can see exactly where the tetrahedral intermediate forms and what happens to it, and it names the reaction type so you can check your classification before you check the product. The flashcard decks include carbonyl chemistry — nucleophilic acyl substitution, acetal formation, aldol addition and aldol condensation are all in the bundled library, offline.
Flashcard decks cover carbonyl chemistry — acyl substitution, acetal formation, aldol addition and condensation — with the scheme drawn on the answer side.
Frequently asked
Why does an ester take two equivalents of Grignard?
The first addition gives a tetrahedral alkoxide that expels the OR group, producing a ketone. That ketone is more electrophilic than the ester it came from, so a second equivalent adds before you can stop it. The result is a tertiary alcohol with two identical R groups from the Grignard.
When do I get the aldol addition product and when the condensation?
Mild conditions and low temperature stop at the β-hydroxy carbonyl (addition). Heat, or stronger base, eliminates water to give the conjugated α,β-unsaturated carbonyl (condensation). A question that says 'with heat' or 'Δ' wants the condensation product.
Why use an acetal as a protecting group?
Because the protection is easy to undo. An acetal is inert to bases, nucleophiles, Grignards and hydrides, so a ketone hidden as an acetal survives a reaction aimed at another part of the molecule — and aqueous acid at the end brings the ketone straight back.
Untangle a multi-step carbonyl problem
Photograph the scheme and read what happens at the tetrahedral intermediate, step by step.
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