Aldehydes, Ketones and Carboxylic Acids • Topic 2 of 3

Reactions of Aldehydes & Ketones

Because the carbonyl carbon is electron-poor (δ+), the central reaction of aldehydes and ketones is nucleophilic addition. A nucleophile attacks the planar carbon, which rehybridises from sp2 to sp3, and the oxygen accepts the electron pair to give a tetrahedral alkoxide that is then protonated.

Important nucleophilic additions

(a) HCN adds to give a cyanohydrin (R2C(OH)CN), a useful route to α-hydroxy acids. (b) NaHSO3 gives a crystalline bisulphite addition product used to purify and separate aldehydes/methyl ketones. (c) Alcohols add (with dry HCl) to give hemiacetals and then acetals [R–CH(OR')2], a common way of protecting the carbonyl group. (d) Ammonia derivatives (NH2–Z) add and then lose water to give C=N compounds: hydroxylamine → oximes, hydrazine → hydrazones, phenylhydrazine → phenylhydrazones, and 2,4-DNP → orange 2,4-dinitrophenylhydrazones (a test for the carbonyl group).

Relative reactivity: aldehyde vs ketone

Aldehydes are more reactive than ketones towards nucleophilic addition for two reasons: (i) steric — an aldehyde has only one bulky group plus a small H, so the nucleophile reaches the carbon easily; (ii) electronic — the two alkyl groups of a ketone are electron-donating (+I), reducing the δ+ on carbon and stabilising the reactant. Hence the order is HCHO > CH3CHO > CH3COCH3.

Oxidation and reduction

Aldehydes are easily oxidised to carboxylic acids (even by mild reagents); ketones resist oxidation and break only under vigorous conditions. Reduction to alcohols uses NaBH4 or LiAlH4 (or H2/Ni): aldehyde → 1° alcohol, ketone → 2° alcohol. Reduction all the way to the –CH2– (alkane) is done by Clemmensen reduction (Zn-Hg/conc. HCl, acidic) or Wolff–Kishner reduction (NH2NH2 then KOH/glycol, basic).

Reactions due to α-hydrogen

The H on the carbon next to C=O (the α-H) is acidic because the resulting carbanion (enolate) is resonance-stabilised. Aldol condensation: two molecules of an aldehyde/ketone bearing α-H combine in dilute base to give a β-hydroxy carbonyl (an aldol), which on heating loses water to give an α,β-unsaturated product. A cross-aldol uses two different carbonyl partners (useful only when the choice of products is controlled, e.g. one partner has no α-H).

Cannizzaro reaction

Aldehydes with no α-H (e.g. HCHO, C6H5CHO) undergo self oxidation–reduction (disproportionation) in concentrated alkali: one molecule is reduced to an alcohol and another oxidised to the carboxylate salt.

Tests for distinction

Tollens' test: aldehydes reduce ammoniacal AgNO3 to a silver mirror; ketones do not. Fehling's test: aliphatic aldehydes give a red-brown Cu2O precipitate; aromatic aldehydes and ketones do not. Iodoform test: compounds with a CH3CO– group (or CH3CH(OH)–) give a yellow CHI3 precipitate with I2/NaOH — positive for acetaldehyde and methyl ketones, negative for other aldehydes/ketones.

Nucleophilic addition: Nu attacks the delta-plus carbonyl carbonCOδ+δ−Nu:−→CO−Nu→H+R–C(OH)(Nu)sp² C → tetrahedral sp³ alkoxide → protonation
Solved Examples
1
Worked Example
Why is acetaldehyde more reactive than acetone towards HCN addition?
Solution
  1. Nucleophilic addition is favoured by a more δ+ and less hindered carbonyl carbon.
  2. Acetone has two electron-donating CH3 groups (+I) that lower δ+ and crowd the carbon.
  3. Acetaldehyde has only one CH3 and a small H, so it is less hindered and more electrophilic.

Answer: Acetaldehyde is more reactive due to lower steric hindrance and a higher δ+ on its carbonyl carbon.

2
Worked Example
Write the aldol product and the final α,β-unsaturated product when ethanal undergoes aldol condensation.
Solution
  1. Dilute NaOH removes an α-H of one CH3CHO to give the enolate.
  2. The enolate adds to the carbonyl C of a second CH3CHO → aldol: CH3CH(OH)CH2CHO (3-hydroxybutanal).
  3. On heating it loses water → CH3CH=CHCHO (but-2-enal).

Answer: Aldol = 3-hydroxybutanal; final product = but-2-enal (after –H2O).

3
Worked Example
Benzaldehyde is treated with concentrated NaOH. What happens and why?
Solution
  1. Benzaldehyde has no α-hydrogen, so it cannot do an aldol.
  2. With concentrated alkali it undergoes the Cannizzaro reaction (disproportionation).
  3. One molecule is reduced to benzyl alcohol; the other is oxidised to sodium benzoate.

Answer: Cannizzaro reaction: 2 C6H5CHO → C6H5CH2OH + C6H5COONa.

4
Worked Example
Which of these give a positive iodoform test: ethanal, propanal, propan-2-ol, acetone? Explain.
Solution
  1. Iodoform forms from a CH3CO– group or a CH3CH(OH)– group (which is first oxidised to it).
  2. Ethanal (CH3CHO) and acetone (CH3COCH3) have CH3CO– → positive.
  3. Propan-2-ol (CH3CH(OH)CH3) is oxidised to acetone → positive; propanal lacks the CH3CO group → negative.

Answer: Ethanal, acetone and propan-2-ol give positive; propanal is negative.

5
Worked Example
Distinguish between propanal and propanone using a chemical test.
Solution
  1. Propanal is an aldehyde (oxidisable); propanone is a ketone (not easily oxidised).
  2. Add Tollens' reagent (ammoniacal AgNO3) and warm.
  3. Propanal gives a silver mirror; propanone gives no reaction.

Answer: Tollens' test — propanal forms a silver mirror, propanone does not. (Fehling's also works for the aliphatic aldehyde.)

6
Worked Example
Name the reagents for converting acetone to propane (the –CH2– level) under acidic and under basic conditions.
Solution
  1. Reducing C=O fully to CH2 needs Clemmensen or Wolff–Kishner.
  2. Acidic: Zn-Hg amalgam with concentrated HCl (Clemmensen reduction).
  3. Basic: hydrazine (NH2NH2) then KOH in ethylene glycol, heat (Wolff–Kishner).

Answer: Clemmensen (Zn-Hg/HCl) in acid; Wolff–Kishner (NH2NH2/KOH) in base — both give propane.

Key Points

  • The key reaction is nucleophilic addition: Nu attacks the δ+ carbon (sp2 → sp3); examples are HCN (cyanohydrin), NaHSO3, alcohols (acetals), and NH2Z derivatives (oximes/hydrazones).
  • Aldehydes > ketones in reactivity (less steric hindrance and greater δ+); aldehydes oxidise easily, ketones do not.
  • Reduction: NaBH4/LiAlH4 → alcohols; Clemmensen (Zn-Hg/HCl) and Wolff–Kishner (NH2NH2/KOH) → alkanes (–CH2–).
  • Compounds with α-H undergo aldol condensation (base) → β-hydroxy carbonyl → α,β-unsaturated product; those with no α-H undergo Cannizzaro (disproportionation) in conc. alkali.
  • Distinction: Tollens (silver mirror) and Fehling (red Cu2O) are positive for aldehydes; iodoform is positive for CH3CO– / CH3CH(OH)– compounds.
Quick Check — 4 MCQs
Tap an option to check your answer0 / 4
Q1.Which compound will NOT give the Cannizzaro reaction?
Explanation: Cannizzaro needs an aldehyde with NO α-H. CH3CHO has α-H, so it does aldol instead.
Q2.A positive iodoform test is given by:
Explanation: Acetophenone (C6H5COCH3) has a CH3CO– group, so it gives iodoform; the others do not.
Q3.Aldehydes are more reactive than ketones in nucleophilic addition mainly because of:
Explanation: One small H and one R group make the aldehyde less hindered and more electrophilic.
Q4.Reduction of a ketone with NaBH4 gives a:
Explanation: NaBH4 reduces C=O to C–OH; a ketone yields a secondary alcohol.
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