How a Silver Mirror Reveals an Aldehyde

The polar carbonyl group invites nucleophiles to attack its carbon, and the hydrogens next to it are unusually acidic. Those two features explain almost every reaction of aldehydes and ketones — from depositing silver mirrors to building new carbon–carbon bonds.

This part covers nucleophilic addition and reactivity, addition reactions, reductions and tests for aldehydes, and reactions that depend on $\alpha$-hydrogen.

A nucleophile attacks the electrophilic carbonyl carbon from above or below its plane, the carbon changes from trigonal planar to tetrahedral, and the alkoxide formed picks up a proton; aldehydes react faster than ketones because they are less crowded and have fewer electron-releasing groups reducing the carbon's positive charge.

Mechanism:

- Step 1 — the nucleophile attacks the carbonyl carbon, forming a tetrahedral alkoxide intermediate
- Step 2 — the alkoxide oxygen captures H${}^{+}$ to give the addition product

Reactivity: methanal > other aldehydes > ketones

- Steric effect — two alkyl groups in a ketone crowd the approaching nucleophile
- Electronic effect — alkyl groups release electrons, lowering the carbon's positive charge
- Aromatic aldehydes such as benzaldehyde are less reactive than aliphatic ones, because ring resonance also reduces that charge

Worked example. Ordering four compounds by reactivity towards nucleophiles:

$$\text{methanal}>\text{ethanal}>\text{propanone}>\text{butanone}$$

Each step adds bulk or electron release around the carbonyl carbon, so each is attacked more slowly than the one before.

An everyday example. A crowded auto-rickshaw stand lets a new passenger in more slowly than an empty one — just as a bulkier ketone slows an incoming nucleophile.

The substance. Attack can happen from either face of the flat carbonyl group, so a new chiral centre usually forms as a racemic mixture.

**Aldehydes and ketones add HCN to form cyanohydrins, NaHSO${}_3$ to form crystalline adducts, alcohols to form hemiacetals and acetals, ammonia derivatives to form C=N compounds with loss of water, and Grignard reagents to form alcohols after hydrolysis.

The addition reactions:

- HCN** — cyanohydrins; a base catalyst supplies the CN${}^{-}$ nucleophile
- Sodium hydrogensulphite — crystalline adducts that acid or alkali convert back, useful for purifying carbonyl compounds
- Alcohols — with dry HCl, aldehydes form hemiacetals and then acetals; ketones with ethane-1,2-diol form cyclic ketals
- Ammonia derivatives — hydroxylamine gives an oxime, hydrazine a hydrazone, 2,4-dinitrophenylhydrazine an orange 2,4-DNP derivative, and semicarbazide a semicarbazone
- Grignard reagents — methanal gives primary, other aldehydes secondary, and ketones tertiary alcohols

Worked example. Propanone with hydroxylamine:

$${\text{(CH}}_3{)}_2\text{C=O}+{\text{H}}_2\text{NOH}\to {\text{(CH}}_3{)}_2\text{C=NOH}+{\text{H}}_2\text{O}$$

Mass balances: $58+33=73+18=91$ g mol${}^{-1}$ on each side.

An everyday example. The orange precipitate with 2,4-DNP reagent is the standard laboratory test for any aldehyde or ketone.

The substance. Ammonia derivatives add and then eliminate water, so the final product is a condensation product, not a simple adduct.

**NaBH${}_4$ or LiAlH${}_4$ reduces aldehydes to primary and ketones to secondary alcohols; Clemmensen and Wolff-Kishner reductions convert C=O straight to CH${}_2$; and aldehydes, being easily oxidised, give a silver mirror with Tollens' reagent and a red-brown precipitate with Fehling's solution.

Reductions:

- To alcohols** — NaBH${}_4$, LiAlH${}_4$, or H${}_2$ with Ni, Pt or Pd
- Clemmensen — zinc amalgam and conc. HCl turn C=O into CH${}_2$
- Wolff-Kishner — forming the hydrazone, then heating with KOH in ethylene glycol, also gives CH${}_2$

Tests for aldehydes:

- Tollens' reagent, ammoniacal silver nitrate, deposits a silver mirror:

$$\text{RCHO}+2[{\text{Ag(NH}}_3{)}_2{]}^{+}+3{\text{OH}}^{-}\to {\text{RCOO}}^{-}+2\text{Ag}+2{\text{H}}_2\text{O}+4{\text{NH}}_3$$

- Fehling's solution — aliphatic aldehydes give a **red-brown precipitate of Cu${}_2$O; aromatic aldehydes do not respond
- Ketones give neither test under these conditions

Worked example.** Butanal and butanone share the formula C${}_4$H${}_8$O ($M=72$ g mol${}^{-1}$), but only butanal gives a silver mirror. Oxidising $0.72$ g ($0.010$ mol) of butanal deposits $2\times 0.010\times 108=2.16$ g of silver.

An everyday example. Glucose in a sweet gives a red precipitate with Fehling's solution because its open-chain form contains an aldehyde group.

The substance. Fehling's test separates aliphatic from aromatic aldehydes — benzaldehyde gives a silver mirror but no red precipitate.

**Aldehydes and ketones with an $\alpha$-hydrogen undergo aldol condensation in dilute alkali, joining two molecules into a $\beta$-hydroxy carbonyl compound that loses water to form an $\alpha ,\beta$-unsaturated one; aldehydes without $\alpha$-hydrogen instead undergo the Cannizzaro reaction in concentrated alkali, giving an alcohol and a carboxylate salt.

Why $\alpha$-hydrogens are acidic.** The carbonyl group withdraws electrons, and the anion formed on losing an $\alpha$-hydrogen is stabilised by resonance.

Aldol condensation:

$$2{\text{CH}}_3\text{CHO}\stackrel{\text{dil. NaOH}}{\to }{\text{CH}}_3{\text{CH(OH)CH}}_2\text{CHO}\stackrel{\text{heat}}{\to }{\text{CH}}_3\text{CH=CHCHO}+{\text{H}}_2\text{O}$$

Ethanal gives 3-hydroxybutanal, which dehydrates to but-2-enal.

Cross-aldol condensation. Two different carbonyl compounds that both have $\alpha$-hydrogens give four products; the reaction is most useful when one partner, such as benzaldehyde, has none.

Cannizzaro reaction — one molecule is oxidised and one reduced:

$$2\text{HCHO}+\text{conc. KOH}\to {\text{CH}}_3\text{OH}+\text{HCOOK}$$

Uses: methanal as formalin and in resins; ethanal for making ethanoic acid; propanone and butanone as solvents; benzaldehyde, vanillin and cinnamaldehyde in flavourings and perfumes.

Worked example. Of ethanal, propanone, methanal, benzaldehyde and 2,2-dimethylpropanal, only ethanal and propanone have $\alpha$-hydrogens and give aldol condensation; the three aldehydes without them undergo the Cannizzaro reaction.

An everyday example. The vanilla flavour in ice cream is vanillin, one of the aldehydes on that list of uses.

The substance. **The presence or absence of an $\alpha$-hydrogen decides the path** — aldol with it, Cannizzaro without it.