6.1.2 Carbonyl Compounds

Heat the aldehyde under reflux with acidified potassium dichromate(VI). The aldehyde is fully oxidised to the corresponding carboxylic acid; the carbon skeleton stays the same.

For example, ethanal gives ethanoic acid. In an equation you can show the oxidising agent as [O]: CH₃CHO + [O] → CH₃COOH. Acidified dichromate changes from orange to green as it is reduced.

NaBH₄ reduces the C=O group to C–OH. An aldehyde therefore gives a primary alcohol, while a ketone gives a secondary alcohol.

Keep every carbon in the same place and change only the carbonyl carbon: RCHO → RCH₂OH and RCOR′ → RCH(OH)R′. If you are given a skeletal structure, redraw that same skeleton before changing C=O to CH–OH.

OCR commonly writes the reagent as NaCN(aq)/H⁺(aq). CN⁻ adds to the carbonyl carbon and the carbonyl oxygen becomes –OH after protonation.

The product contains –OH and –CN on the same carbon, and the carbon chain has gained one carbon because the carbon of CN⁻ is now part of the molecule. Check that the C=O has disappeared in your product.

Put δ⁺ on the carbonyl carbon and δ⁻ on oxygen. Treat the nucleophile as H⁻: the first curly arrow goes from H⁻ to the carbonyl carbon, while a second arrow moves the C=O π pair onto oxygen.

You now have an alkoxide intermediate with O⁻ and a new C–H bond. Protonate O⁻ with water to form the alcohol. The hydride attack must happen before the protonation step.

Start with the same C=O dipole. The curly arrow comes from the carbon end of CN⁻ to the δ⁺ carbonyl carbon, and the C=O π pair moves onto oxygen.

The intermediate has O⁻ and a new C–C bond to CN. Protonate O⁻ with H⁺ or water to give the hydroxynitrile. This mechanism is worth knowing as a carbon-chain extension as well as a carbonyl reaction.

Add 2,4-dinitrophenylhydrazine (2,4-DNP/2,4-DNPH) to the sample. Aldehydes and ketones give a yellow/orange precipitate, so a positive result tells you that a carbonyl group is present.

This test does not tell you whether the carbonyl is an aldehyde or a ketone. OCR does not require the reaction equation or the structure of the 2,4-DNP derivative.

A particular aldehyde or ketone forms a solid 2,4-DNP derivative with a characteristic melting point. Measure the derivative’s melting point and compare it with reference values to identify the original carbonyl compound.

A close match supports the identification; a broad or depressed melting range suggests the solid may be impure. The useful number here is the melting point of the derivative, not the boiling point of the original liquid.

Warm the sample gently with Tollens’ reagent, ammoniacal silver nitrate. An aldehyde gives a silver mirror (or a grey/silver deposit of silver); a ketone gives no reaction under the test conditions.

Use 2,4-DNP first if the job is simply to detect a carbonyl. Tollens’ reagent is the follow-up test that distinguishes an aldehyde from a ketone.

During a positive Tollens’ test, the aldehyde is oxidised to a carboxylic acid while silver ions in the reagent are reduced to Ag(s). The deposited silver produces the mirror.

The redox directions are a useful check: aldehyde gains oxygen overall; Ag⁺ gains electrons and becomes silver metal. OCR accepts [O] in equations involving Tollens’ reagent.