Aldehydes, Ketones and Carboxylic Acids
The carbonyl group and beyond — the polar \(\ce{C=O}\) that invites nucleophiles, the α-hydrogen that builds carbon chains, and the resonance that makes carboxylic acids truly acidic
- The structure of the carbonyl group and the nucleophilic addition mechanism.
- Why aldehydes are more reactive than ketones toward nucleophiles.
- Key additions — HCN, bisulphite, acetals, and the ammonia derivatives (2,4-DNP test).
- The aldol and Cannizzaro reactions, and the haloform reaction.
- Tollens' and Fehling's tests for aldehydes; reduction by Clemmensen and Wolff–Kishner.
- Why carboxylic acids are acidic, and how they form their many derivatives.
The Carbonyl Group
The carbonyl group \(\ce{C=O}\) is the heart of this chapter. Carbon is \(sp^2\) and planar, and because oxygen is far more electronegative, the bond is strongly polar: carbon carries \(\delta^+\) and oxygen \(\delta^-\). That electron-poor carbon is exactly what a nucleophile attacks. An aldehyde has the group at the chain end (\(\ce{-CHO}\)); a ketone has it between two carbons (\(\ce{>C=O}\)).
A ketone has two alkyl groups whose \(+I\) effect reduces the carbon's \(\delta^+\), and which crowd the approaching nucleophile. An aldehyde has only one, so it is both more electrophilic and less hindered.
Preparation of Carbonyls
Aldehydes and ketones are made chiefly by controlled oxidation of alcohols, by ozonolysis, or from acid derivatives. The trick for aldehydes is to stop oxidation before the acid stage.
| Method | Reaction | Gives |
|---|---|---|
| Oxidation of alcohol | \(\ce{1^\circ ->[PCC] RCHO};\ \ce{2^\circ -> ketone}\) | aldehyde / ketone |
| Ozonolysis | \(\ce{C=C ->[O3][Zn/H2O]}\) carbonyls | both |
| Rosenmund reduction | \(\ce{RCOCl + H2 ->[Pd/BaSO4] RCHO}\) | aldehyde |
| Friedel–Crafts acylation | \(\ce{ArH + RCOCl ->[AlCl3] ArCOR}\) | aryl ketone |
Nucleophilic Addition
The signature reaction of the carbonyl is nucleophilic addition: a nucleophile attacks the electrophilic carbon, the \(\pi\) electrons shift onto oxygen, and the resulting alkoxide is protonated.
| Nucleophile | Product | Use |
|---|---|---|
| \(\ce{HCN}\) | cyanohydrin | chain extension |
| \(\ce{NaHSO3}\) | bisulphite adduct | purification of carbonyls |
| alcohol (dry \(\ce{HCl}\)) | hemiacetal → acetal | protecting group |
| \(\ce{RMgX}\) | alcohol | C–C bond formation |
Addition–Elimination & the Carbonyl Test
Ammonia derivatives add to the carbonyl and then eliminate water, giving compounds with a \(\ce{C=N}\) bond. The most useful is 2,4-dinitrophenylhydrazine (2,4-DNP), whose orange precipitate is a classic test for any aldehyde or ketone.
| Reagent | Product |
|---|---|
| hydroxylamine \(\ce{NH2OH}\) | oxime |
| hydrazine \(\ce{NH2NH2}\) | hydrazone |
| 2,4-DNP | 2,4-dinitrophenylhydrazone (orange ppt) |
| semicarbazide | semicarbazone |
α-Hydrogen: The Aldol Reaction
A hydrogen on the carbon next to the carbonyl (the \(\alpha\)-hydrogen) is weakly acidic, because the carbanion it leaves is stabilised by the carbonyl. A base can remove it, generating a nucleophilic carbon that attacks another carbonyl — the aldol reaction.
Two molecules with \(\alpha\)-H combine to a \(\beta\)-hydroxy carbonyl (the "aldol"), which on heating loses water to an \(\alpha,\beta\)-unsaturated carbonyl. It is a key carbon–carbon bond-forming reaction — and it requires at least one \(\alpha\)-hydrogen.
Cannizzaro & Haloform Reactions
Two more named reactions hinge on structure. Aldehydes with no \(\alpha\)-hydrogen cannot do the aldol, so with strong base they disproportionate instead — the Cannizzaro reaction. Methyl ketones undergo the haloform reaction, the basis of the iodoform test.
One molecule is oxidised (to the carboxylate) and another reduced (to the alcohol). Benzaldehyde, also lacking \(\alpha\)-H, behaves the same way.
Oxidation, Reduction & the Silver Mirror
Aldehydes are easily oxidised to acids; ketones strongly resist (there is no \(\ce{C-H}\) on the carbonyl carbon to remove). Two gentle oxidising tests exploit this to distinguish aldehydes from ketones.
| Test / reagent | Aldehyde | Ketone |
|---|---|---|
| Tollens' (ammoniacal \(\ce{AgNO3}\)) | silver mirror | no reaction |
| Fehling's (\(\ce{Cu^2+}\)) | red \(\ce{Cu2O}\) ppt | no reaction |
| \(\ce{LiAlH4}/\ce{NaBH4}\) | 1° alcohol | 2° alcohol |
| Clemmensen (\(\ce{Zn-Hg}/\ce{HCl}\)) | alkane | alkane |
| Wolff–Kishner (\(\ce{NH2NH2}/\ce{KOH}\)) | alkane | alkane |
Carboxylic Acids: Acidity
Carboxylic acids (\(\ce{-COOH}\)) are the most acidic of the common organic compounds — more acidic than phenols and alcohols. The reason, once again, is the conjugate base: the carboxylate ion spreads its negative charge equally over two oxygen atoms by resonance.
Because a carboxylic acid is stronger than carbonic acid, it liberates \(\ce{CO2}\) from \(\ce{NaHCO3}\) (brisk effervescence) — a test phenol fails. Electron-withdrawing groups raise acidity further: \(\ce{ClCH2COOH} > \ce{CH3COOH}\), and formic acid is stronger than acetic.
Reactions of Carboxylic Acids
Beyond salt formation, the \(\ce{-COOH}\) group is the gateway to a whole family of acid derivatives — esters, acid chlorides, anhydrides and amides — each obtained by replacing the \(\ce{-OH}\).
| Reaction | Reagent | Product |
|---|---|---|
| Salt formation | \(\ce{NaOH}\), \(\ce{NaHCO3}\) | carboxylate |
| Esterification (Fischer) | \(\ce{R'OH}/\ce{H+}\) | ester |
| Acid chloride | \(\ce{SOCl2}\) | \(\ce{RCOCl}\) |
| Amide | \(\ce{NH3}\), then \(\Delta\) | \(\ce{RCONH2}\) |
| Reduction | \(\ce{LiAlH4}\) | 1° alcohol |
| HVZ (α-halogenation) | \(\ce{X2}/\text{red P}\) | α-halo acid |
Putting It to Work
Problem. Arrange \(\ce{HCHO},\ \ce{CH3CHO},\ \ce{CH3COCH3}\) by reactivity toward nucleophilic addition.
Solution. Fewer/smaller alkyl groups → more \(\delta^+\), less hindrance:
Problem. How would you chemically distinguish acetaldehyde from acetone?
Solution. Only the aldehyde is oxidised by Tollens'/Fehling's:
Problem. Give the aldol product of two acetaldehyde molecules.
Solution. α-carbon of one adds to the carbonyl of the other:
Problem. Will acetaldehyde or benzaldehyde undergo the Cannizzaro reaction? Why?
Solution. Cannizzaro needs no α-H; acetaldehyde has α-H:
Problem. Arrange acetic acid, phenol and ethanol by acidity, and explain.
Solution. Compare the stability of the conjugate base:
Problem. Which gives a positive iodoform test: propan-2-ol or propan-1-ol?
Solution. Iodoform needs a \(\ce{CH3-CH(OH)-}\) group:
Chapter Summary
Polar \(\ce{C=O}\); aldehydes more reactive than ketones (electronic + steric).
Nucleophilic addition of HCN, bisulphite, alcohols, Grignard; 2,4-DNP detects carbonyls.
Aldol condensation forms C–C bonds; needs at least one α-H.
Cannizzaro disproportionation (HCHO, benzaldehyde); haloform from methyl ketones.
Tollens' silver mirror and Fehling's red \(\ce{Cu2O}\); ketones don't respond.
Carboxylate resonance makes \(\ce{-COOH}\) strongly acidic; \(\ce{NaHCO3}\) test; many derivatives.
Problems
For each item, first decide whether it concerns the carbonyl's electrophilic carbon, its α-hydrogen, or the acidity of \(\ce{-COOH}\) — then apply the relevant idea. Difficulty rises down the list.
- Why is the carbonyl carbon electrophilic, and why are aldehydes more reactive than ketones?
- Write the Rosenmund reduction and state what it produces.
- Give the mechanism (in words) of nucleophilic addition of HCN to acetaldehyde.
- What does the 2,4-DNP test detect, and what does a positive result look like?
- Write the aldol condensation of acetaldehyde and name the product.
- Why does benzaldehyde undergo the Cannizzaro reaction but acetaldehyde does not?
- Describe Tollens' and Fehling's tests and what they distinguish.
- Compare Clemmensen and Wolff–Kishner reductions and when each is used.
- Which compounds give a positive iodoform test? Give two examples.
- Explain, using resonance, why carboxylic acids are more acidic than phenols.
- How would you distinguish acetic acid from phenol with a single reagent?
- Starting from acetic acid, show how to make (a) an ester, (b) an acid chloride, (c) an amide.