Aldehydes, Ketones and Carboxylic Acids · Lesson 4 of 15
Nucleophilic Addition to the Carbonyl Group
“Follow electron movement and predict cyanohydrins, hydrogensulphite adducts and Grignard alcohols.”
• Explain the steps and geometry change of nucleophilic addition. • Compare steric and electronic effects on reactivity. • Predict HCN, hydrogensulphite and Grignard addition products. • Distinguish reaction conditions from hydrolytic work-up.
A polar π bond can accept an electron pair
In an aldehyde or ketone, the carbonyl carbon is electrophilic. A negatively charged nucleophile donates a lone pair to that carbon while the C=O π electron pair moves onto oxygen. These two electron movements belong together: the π bond must open as the new σ bond forms. Carbon changes from approximately planar sp² to tetrahedral sp³, and the intermediate is an alkoxide. A proton donor then protonates oxygen to give an alcohol-type addition product. The original carbonyl oxygen remains in the molecule.
A curved electron-pair arrow begins at a lone pair or bond and ends where that pair is used. It is different from a reaction arrow between species. The nucleophile approaches a face of the planar carbonyl at an oblique angle, avoiding occupied orbitals; it is not constrained to a perfectly perpendicular path. If a new stereogenic carbon is formed from an achiral planar substrate with achiral reagents, attack on either face can give a racemic pair. A flat condensed formula alone does not imply one enantiomer.
Steric access and electrophilicity act together
For similar substrates, methanal reacts readily because it has no alkyl groups. Aldehydes generally react faster than ketones: replacing H by a carbon group both crowds the incoming nucleophile and donates electron density that reduces carbonyl electrophilicity. Bulky ketones are especially resistant. The common order HCHO > ordinary aliphatic aldehydes > comparable ketones is a useful tendency, not a universal ordering that ignores substituents or reaction conditions.
In benzaldehyde, conjugation with the phenyl ring reduces the electrophilicity of the carbonyl compared with a simple aliphatic aldehyde such as propanal. A para nitro group withdraws electron density and increases electrophilicity, whereas a para methyl group tends to lower it. Thus a representative increasing addition-reactivity order is acetophenone < 4-methylbenzaldehyde < benzaldehyde < 4-nitrobenzaldehyde. Steric crowding and electronic donation both explain why 2,2,6-trimethylcyclohexanone forms cyanohydrin less readily than cyclohexanone.
Problem
Compare ethanal, propanone, methyl tert-butyl ketone and di-tert-butyl ketone toward HCN.
- 1.Ethanal has one H and one small methyl group at C=O. Propanone has two methyl groups, which are more donating and more crowded.
- 2.Replacing methyl by tert-butyl progressively blocks access. Increasing reactivity is di-tert-butyl ketone < methyl tert-butyl ketone < propanone < ethanal. This comparison applies to the specified reaction, not to every possible property.
HCN adds an OH and a one-carbon nitrile group
Pure HCN reacts slowly because it supplies little of the stronger nucleophile CN⁻. A small amount of cyanide under suitable controlled conditions promotes reaction. The carbon end of CN⁻ bonds to carbonyl carbon, making an alkoxide; the alkoxide takes a proton from HCN or another proton donor. The product has OH and CN on the same carbon and is called a cyanohydrin. The net reaction is R₂CO + HCN ⇌ R₂C(OH)CN. It adds one carbon, which is useful because later hydrolysis of CN can produce an α-hydroxy acid.
Problem
Predict the product of ethanal with HCN.
- 1.Start from CH₃–CH=O. Attach CN to the carbonyl carbon, change C=O to C–O⁻ and then protonate O. The product is CH₃CH(OH)CN, 2-hydroxypropanenitrile.
- 2.The former aldehyde H stays attached, and the nitrile carbon raises the total from two to three. The new carbon with OH has four different substituents; without stereoselective control, both enantiomers can form.
Hydrogensulphite addition is reversible
Sodium hydrogensulphite, NaHSO₃, forms α-hydroxysulphonate addition compounds: R₂CO + HSO₃⁻ ⇌ R₂C(OH)SO₃⁻. Sulphur is bonded to the former carbonyl carbon; this is not an alkyl sulphate with a new C–O–S connection. Many aldehydes and accessible ketones form these adducts, but steric crowding can make addition unfavourable for other ketones. Their ionic nature and suitable crystallisation behaviour permit separation or purification. Treating an isolated adduct with dilute acid or alkali can regenerate the carbonyl compound.
A test result therefore has a scope: failure to isolate an adduct does not prove there is no carbonyl group, and a positive adduct test does not distinguish all aldehydes from all ketones. In analytical reasoning, combine it with oxidation and other evidence.
Grignard reagents make a new carbon–carbon bond
In RMgX the carbon group behaves as a strongly basic carbon nucleophile. In dry ether it adds to an aldehyde or ketone to make a magnesium alkoxide. Only after addition is complete is water or dilute acid introduced to release the alcohol. Water present initially consumes RMgX to give RH, so “RMgX/H₃O⁺ together” is not a valid description of the reaction sequence.
| Starting carbonyl | After R′MgX then acidic work-up | Alcohol class |
|---|---|---|
| HCHO | R′CH₂OH | Primary |
| RCHO, R not H | RCH(OH)R′ | Secondary |
| RCOR″ | RC(OH)(R′)R″ | Tertiary |
Free OH, COOH or other proton donors can consume the reagent by acid–base reaction; other reactive groups such as nitro or additional carbonyl groups can cause unwanted reaction. Functional-group compatibility must be checked. A carboxylic acid does not behave like an aldehyde just because it contains C=O: its acidic proton usually quenches an ordinary Grignard reagent.
Problem
React cyclohexanecarbaldehyde with PhMgBr then H₃O⁺.
- 1.Write the substrate as C₆H₁₁–CHO; the aldehyde carbon is outside the ring. Phenyl bonds to that carbon, oxygen becomes OH, and its original H remains.
- 2.The product is C₆H₁₁CH(OH)C₆H₅, a secondary alcohol. It is not a tertiary cyclohexanol, because the ring carbon is not the original carbonyl carbon.
Quick Quiz
Quiz
Where must the electron-pair arrow from CN⁻ end?
Explanation: CN⁻ forms a new carbon–carbon bond with electrophilic carbon.
Which has OH and CN on the same carbon?
Explanation: HCN adds across C=O to give this connectivity.
Which product follows propanone plus ethylmagnesium bromide then acid?
Explanation: Retain two methyl groups at the former carbonyl carbon and add ethyl plus OH.
Why must the Grignard addition precede acidic work-up?
Explanation: The strong base reacts with proton donors to form a hydrocarbon.
What happens to carbon geometry during nucleophilic addition?
Explanation: A new σ bond forms as the π bond opens.
Practice Problems
- Rank butanone, propanone, propanal and ethanal in increasing addition reactivity.
- Draw condensed products of propanone with HCN and NaHSO₃.
- Explain why cyclohexanone and 2,2,6-trimethylcyclohexanone differ in cyanohydrin yield.
- Choose a carbonyl and Grignard reagent for 1-phenylethanol.
- Why is a free COOH group incompatible with a simple Grignard alcohol synthesis?
A useful order is butanone < propanone < propanal < ethanal. Ketones have two donating carbon groups and greater crowding; the larger ethyl-containing member is less accessible than its methyl analogue. This is a qualitative ranking for comparable conditions.
Key Takeaways
• Nucleophilic addition opens the carbonyl π bond and forms a tetrahedral intermediate. • Electron-pair arrows start at electron sources. • Steric and electronic effects jointly control reactivity. • Cyanohydrins have OH and CN on the same carbon. • Hydrogensulphite adducts can regenerate carbonyl compounds. • Grignard addition forms a C–C bond before aqueous work-up. • Proton donors and other reactive groups limit organomagnesium chemistry.