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.

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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

$CH_3CHO + HCN \longrightarrow CH_3CH(OH)CN$CH3​CHO+HCN⟶CH3​CH(OH)CN

Ethanal plus HCN gives 2-hydroxypropanenitrile. Similarly:

$CH_3COCH_3 + HCN \longrightarrow (CH_3)_2C(OH)CN$CH3​COCH3​+HCN⟶(CH3​)2​C(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.

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:

1. From the lone pair on C of CN⁻ to the carbonyl C.
2. From the C=O π bond onto the O atom.
3. 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.

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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

$CH_3CHO + 2[H] \longrightarrow CH_3CH_2OH$CH3​CHO+2[H]⟶CH3​CH2​OH

$CH_3COCH_3 + 2[H] \longrightarrow (CH_3)_2CHOH$CH3​COCH3​+2[H]⟶(CH3​)2​CHOH

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.

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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).

$R-CHO + 2[Ag(NH_3)_2]^+ + 3OH^- \longrightarrow R-COO^- + 2Ag + 4NH_3 + 2H_2O$R−CHO+2[Ag(NH3​)2​]++3OH−⟶R−COO−+2Ag+4NH3​+2H2​O

At A-Level you can write:

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

3.2 Fehling's (and Benedict's) Solution

What it is: A deep blue solution of Cu²⁺ complexed with tartrate (Fehling's) or citrate (Benedict's) ions in alkali.

Procedure: Warm with the unknown in a water bath.

Observations:

• Aldehyde: Blue solution turns to a brick-red precipitate of copper(I) oxide (Cu₂O).
• Ketone: No change — stays blue.

What is happening: Cu²⁺ is reduced to Cu⁺ (as Cu₂O) while the aldehyde is oxidised to a carboxylate.

$R-CHO + 2Cu^{2+} + 5OH^- \longrightarrow R-COO^- + Cu_2O + 3H_2O$R−CHO+2Cu2++5OH−⟶R−COO−+Cu2​O+3H2​O

Historical note: Benedict's solution is the standard chemistry teaching reagent in British schools; Fehling's is older and uses two separate solutions mixed immediately before use. OCR accepts either as a test for aldehydes.

3.3 Acidified Potassium Dichromate(VI)

Orange K₂Cr₂O₇ / H₂SO₄ is a general oxidising agent.

Observations:

• Aldehyde: Orange → green (Cr₂O₇²⁻ → Cr³⁺) on warming. The aldehyde is oxidised to a carboxylic acid.
• Ketone: No colour change.

This is not a definitive test because primary and secondary alcohols also give the same colour change. But combined with Tollens' or Fehling's it tells you the compound is an aldehyde.

3.4 Summary Table: Aldehyde vs Ketone Tests

Test	Reagent	Aldehyde result	Ketone result
Tollens'	[Ag(NH₃)₂]⁺	Silver mirror	No change
Fehling's	Cu²⁺ in alkali	Brick-red precipitate Cu₂O	No change (stays blue)
Acidified dichromate	K₂Cr₂O₇ / H₂SO₄	Orange → green	No change
2,4-DNP (next section)	2,4-dinitrophenylhydrazine	Yellow/orange precipitate	Yellow/orange precipitate (both carbonyls)

graph TD
  A[Unknown compound] --> B{2,4-DNP: orange ppt?}
  B -->|No| C[Not a carbonyl]
  B -->|Yes| D[Carbonyl confirmed]
  D --> E{Tollens: silver mirror?}
  E -->|Yes| F[Aldehyde]
  E -->|No| G[Ketone]

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4. 2,4-Dinitrophenylhydrazine (2,4-DNP / Brady's Reagent)

This is the definitive test for the carbonyl group itself — it tells you that a C=O is present, but does not distinguish aldehyde from ketone (both react).

4.1 Procedure

Add a few drops of the unknown to Brady's reagent (2,4-DNP dissolved in methanol/sulfuric acid).

4.2 Observation

• Aldehyde or ketone present: A bright yellow, orange or red precipitate forms immediately.
• No carbonyl: No precipitate.

4.3 What is happening — condensation reaction

The NH₂ of 2,4-DNPH attacks the δ+ C of the C=O, then water is eliminated to give a 2,4-dinitrophenylhydrazone.

$R_2C=O + H_2N-NH-C_6H_3(NO_2)_2 \longrightarrow R_2C=N-NH-C_6H_3(NO_2)_2 + H_2O$R2​C=O+H2​N−NH−C6​H3​(NO2​)2​⟶R2​C=N−NH−C6​H3​(NO2​)2​+H2​O

This is a nucleophilic addition-elimination (sometimes called a condensation). The yellow/orange colour comes from the extended conjugation through the aromatic ring.

4.4 Using 2,4-DNP to identify an unknown carbonyl

The 2,4-DNP derivative of each carbonyl has a specific, sharp melting point. In the days before spectroscopy, organic chemists would:

1. Add unknown + 2,4-DNP → precipitate.
2. Filter off the solid.
3. Recrystallise from ethanol.
4. Measure the melting point and compare to a data book.

A match (to within 1 °C) confirmed the identity of the starting carbonyl. This is still a classic OCR practical exam question — be ready to describe the recrystallisation and melting-point determination steps.

Exam Tip: The full procedure is: add to Brady's → filter the precipitate → recrystallise from hot solvent → cool, filter, wash with ice-cold solvent → dry → determine melting point → compare with data book values.

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5. The Iodoform Test — A Specialty Carbonyl Test

The iodoform test is the most discriminating of the classical carbonyl tests. It identifies the structural unit CH₃-CO-R (a methyl ketone, where R is H, alkyl or aryl) — plus ethanol and any secondary alcohol with a CH₃-CH(OH)- arrangement, because those are oxidised in situ to methyl carbonyls under the reaction conditions.

5.1 Reagent

I₂ in aqueous NaOH (sometimes written as NaOI / sodium iodate(I)). Warm gently.

5.2 Observation

A pale yellow crystalline precipitate of triiodomethane (CHI₃, "iodoform") with a distinctive antiseptic / "TCP" smell. Positive: yellow ppt forms. Negative: no precipitate.

5.3 What gives a positive result?

Compound type	Example	Positive?
Methyl ketone CH₃-CO-R	Propanone CH₃COCH₃, butan-2-one CH₃COCH₂CH₃	Yes
Ethanal CH₃CHO	CH₃CHO	Yes
Methyl secondary alcohol CH₃-CH(OH)-R	Propan-2-ol, butan-2-ol	Yes (oxidised in situ)
Ethanol CH₃CH₂OH	CH₃CH₂OH	Yes (oxidised in situ to CH₃CHO)
Other carbonyls	Propanal, pentan-3-one	No
Tertiary alcohols, methanol, propan-1-ol	(CH₃)₃COH, CH₃OH, CH₃CH₂CH₂OH	No

5.4 Mechanism in outline

1. The methyl C–H bonds α to the C=O are acidic because the resulting carbanion is stabilised by delocalisation into the carbonyl.
2. NaOH deprotonates a methyl α-H. The resulting enolate is iodinated by I₂.
3. Steps 1–2 repeat twice more until all three methyl H atoms are replaced by I: now we have a –CO–CI₃ group.
4. The CI₃⁻ group is an unusually good leaving group; OH⁻ attacks the carbonyl carbon, kicks out the CI₃⁻, and gives the carboxylate.
5. CI₃⁻ picks up an H⁺ from water → CHI₃ (the yellow precipitate).

Net: CH₃-CO-R + 3I₂ + 4NaOH → CHI₃ + R-COONa + 3NaI + 3H₂O.

5.5 Why this test is useful

The iodoform test was the standard way to distinguish methyl ketones from other ketones before NMR was widely available. It is still useful for narrowing down an unknown — a positive iodoform tells you the carbonyl is of the CH₃-CO-R type. Combined with the molecular formula and the 2,4-DNP / Tollens' / Fehling's results, an unknown carbonyl can usually be identified from C₃ to about C₆.

Exam Tip: Iodoform-positive = compound contains either a CH₃-CO-R group, or a CH₃-CH(OH)-R group (oxidised to a methyl carbonyl by the alkaline iodine), or it is ethanol/ethanal.

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6. LiAlH₄ — A Stronger Reducing Agent

LiAlH₄ (lithium tetrahydridoaluminate(III), often called "LAH") is a stronger reducing agent than NaBH₄. The OCR specification mentions it as a comparison rather than requiring its full mechanism, but you should know what it does:

Functional group	NaBH₄	LiAlH₄
Aldehyde → 1° alcohol	Yes	Yes
Ketone → 2° alcohol	Yes	Yes
Carboxylic acid → 1° alcohol	No	Yes
Ester → 1° alcohol + 1° alcohol	No	Yes
Amide → amine	No	Yes
C=C (alkene)	No	No
Nitrile R–CN → amine R–CH₂NH₂	No	Yes