Carbonyls: Reactions and Tests

Now that you know the structure of aldehydes and ketones, we can look at their chemistry. This is one of the most heavily examined areas of A-Level Chemistry because it ties together nucleophilic addition, reduction, oxidation, and several characteristic test-tube reactions. Master this lesson and you will have a huge head-start on the rest of the organic course.

This lesson covers the OCR A-Level Chemistry A (H432) specification point 6.1.2 (c)–(d): reactions of carbonyl compounds and chemical tests to distinguish aldehydes from ketones.

1. Nucleophilic Addition of HCN

Because the carbonyl carbon is δ⁺, it is attacked by nucleophiles. Hydrogen cyanide (HCN) provides the cyanide ion, CN⁻, which is one of the strongest carbon-based nucleophiles you meet at A-Level. The reaction produces a hydroxynitrile (sometimes called a cyanohydrin).

1.1 Overall equation

$CH_3CHO + HCN \longrightarrow CH_3CH(OH)CN$

Ethanal plus HCN gives 2-hydroxypropanenitrile. Similarly:

$CH_3COCH_3 + HCN \longrightarrow (CH_3)_2C(OH)CN$

Propanone gives 2-hydroxy-2-methylpropanenitrile.

1.2 Conditions

The reaction is slow with HCN alone because HCN is a weak acid and gives very little free CN⁻. In practice you use:

KCN or NaCN to provide CN⁻ directly, plus
Dilute H₂SO₄ (or HCl) to give H⁺ for the second step.

This mixture generates both the nucleophile and the proton source needed.

1.3 Mechanism (nucleophilic addition)

Step 1 — attack: the CN⁻ ion uses its lone pair on carbon to attack the δ+ carbonyl carbon. The C=O π bond breaks and both electrons move onto oxygen.

          CN(-)                        CN
           |                           |
  R  \ /          slow     R \ /
      C = O   ---------->     C  - O(-)
  H  /                     H /

Step 2 — protonation: the alkoxide (O⁻) is strongly basic and immediately grabs a proton from H⁺ (or from H₂O in the medium), giving the neutral hydroxynitrile.

Curly arrows you must draw:

From the lone pair on C of CN⁻ to the carbonyl C.
From the C=O π bond onto the O atom.
From a lone pair on O⁻ to H⁺.

Key Insight: This creates a new C–C bond — it is a chain-extension reaction. A 3-carbon carbonyl becomes a 4-carbon hydroxynitrile. We will see this used strategically in Lesson 11.

1.4 A hazard and a chirality point

HCN is highly toxic — any question asking for a practical hazard is looking for this point.
When CN⁻ attacks the planar carbonyl carbon, it can approach from either face. For an unsymmetrical aldehyde or ketone (like ethanal) this produces a racemic mixture of two enantiomers, because the carbonyl carbon becomes a new chiral centre.

2. Reduction with NaBH₄

Sodium tetrahydridoborate(III), NaBH₄, delivers a hydride ion (H⁻) which acts as a nucleophile on the carbonyl carbon. The overall effect is to reduce the C=O to a C–OH.

Starting material	Reduced to
Aldehyde R–CHO	Primary alcohol R–CH₂OH
Ketone R–CO–R'	Secondary alcohol R–CH(OH)–R'

2.1 Equations

$CH_3CHO + 2[H] \longrightarrow CH_3CH_2OH$

$CH_3COCH_3 + 2[H] \longrightarrow (CH_3)_2CHOH$

At A-Level you write [H] to represent the hydrogen added by the reducing agent. Two hydrogens are added in total — one to C, one to O.

2.2 Conditions

NaBH₄ in water or methanol, usually at room temperature.
Alternative: LiAlH₄ in dry ether (more aggressive; not required by OCR but sometimes mentioned).

2.3 Mechanism

Step 1: H⁻ (from NaBH₄) attacks the δ+ carbonyl C. The C=O π bond breaks and the electrons go onto oxygen, giving an alkoxide.

Step 2: The alkoxide is protonated by water (or methanol) to give the alcohol.

This is another nucleophilic addition — mechanistically the same pattern as HCN addition, just with H⁻ as the nucleophile.

Exam Tip: NaBH₄ reduces carbonyls but not C=C double bonds. So if you see an aldehyde with an alkene elsewhere in the molecule, the alkene is untouched.

3. Oxidation and Tests to Distinguish Aldehydes from Ketones

Aldehydes are easily oxidised to carboxylic acids because they have an H on the carbonyl carbon that can be replaced by an OH. Ketones, with no such H, resist oxidation under normal lab conditions. This difference is the basis of three classical test-tube reactions.

3.1 Tollens' Reagent — The Silver Mirror Test

What it is: A solution of silver nitrate in aqueous ammonia containing the complex ion [Ag(NH₃)₂]⁺.

Procedure: Warm gently (≈60 °C) in a water bath. Do not boil — explosive silver fulminate can form if solutions are kept too long.

Observations:

Aldehyde: A shiny silver mirror deposits on the inside of the test tube.
Ketone: No change.

What is happening: The Ag⁺ in [Ag(NH₃)₂]⁺ is reduced to Ag⁰ (metallic silver) as the aldehyde is oxidised to a carboxylic acid (or its ammonium salt under the basic conditions).

$R-CHO + 2[Ag(NH_3)_2]^+ + 3OH^- \longrightarrow R-COO^- + 2Ag + 4NH_3 + 2H_2O$

At A-Level you can write:

$R-CHO + [O] \longrightarrow R-COOH$

Lesson 2: Carbonyls — Reactions and Tests