Carbonyl compounds · ⁨カルボニル化合物⁩

Aldehydes and ketones are carbonyl compounds 羰基化合物 — they contain the C=O carbonyl 羰基 group.

• in an aldehyde 醛 the carbonyl carbon is on the end of the chain (it also carries an H), written $\text{–CHO}$.
• in a ketone 酮 the carbonyl carbon is in the middle, between two other carbons.

Making aldehydes and ketones

Both are made by the oxidation 氧化 of an alcohol 醇 with acidified $\text{K}_2\text{Cr}_2\text{O}_7$ or $\text{KMnO}_4$:

• a primary alcohol, with distillation 蒸馏, gives an aldehyde.
• a secondary alcohol gives a ketone.

Reactions

• reduction 还原 with $\text{NaBH}_4$ or $\text{LiAlH}_4$ turns a carbonyl compound back into an alcohol (an aldehyde gives a primary alcohol; a ketone gives a secondary alcohol).
• reaction with hydrogen cyanide 氰化氢 ($\text{HCN}$), with $\text{KCN}$ as catalyst and heat, adds $\text{H}$ and $\text{CN}$ across the C=O to make a hydroxynitrile 羟基腈. This adds one carbon to the chain.

The mechanism: nucleophilic addition

The reaction with $\text{HCN}$ is a nucleophilic addition 亲核加成. The carbonyl carbon is slightly positive (oxygen pulls the electrons away). So:

1. the $\text{CN}^-$ ion (a nucleophile) attacks the slightly positive carbon.
2. this breaks the C=O double bond, leaving a negative oxygen ($\text{O}^-$).
3. the $\text{O}^-$ takes an $\text{H}^+$ (from $\text{HCN}$) to finish the hydroxynitrile.

Detecting any carbonyl

Add 2,4-DNPH reagent (2,4-dinitrophenylhydrazine). An orange precipitate confirms that the compound is an aldehyde or a ketone.

Telling an aldehyde from a ketone

Aldehydes are easily oxidised to carboxylic acids, but ketones are not. Two tests use this difference:

The iodoform test

If the compound has the $\text{CH}_3\text{CO}-$ group, warming it with alkaline aqueous iodine gives a pale yellow precipitate of tri-iodomethane 三碘甲烷 ($\text{CHI}_3$) and the ion $\text{RCO}_2^-$.

Worked example. A liquid gives an orange precipitate with 2,4-DNPH, gives no silver mirror with Tollens' reagent, and gives a yellow precipitate with alkaline aqueous iodine. Identify it. Take the tests one at a time, using each for exactly what it proves. The orange precipitate with 2,4-DNPH proves a carbonyl group is present - an aldehyde or a ketone, nothing else. No silver mirror with Tollens' rules out an aldehyde, so it must be a ketone. The yellow precipitate in the iodoform test proves a $\text{CH}_3\text{CO}-$ group next to the carbonyl. The simplest compound satisfying all three is propanone, $\text{CH}_3\text{COCH}_3$. Keep the roles straight: 2,4-DNPH finds any carbonyl, Tollens' separates aldehyde from ketone, and iodoform detects the methyl group beside the C=O.

• 2,4-DNPH gives an orange precipitate with any carbonyl — it tests for $\text{C}=\text{O}$.
• Tollens' (silver mirror) and Fehling's (brick-red) are positive for aldehydes only — this is how you tell aldehydes from ketones.
• Nucleophilic addition of HCN adds one carbon (→ hydroxynitrile); show the curly arrows.
• The iodoform (tri-iodomethane) test is positive for $\text{CH}_3\text{CO}-$ groups.