• Aldehydes have the carbonyl group at the end of the chain (bonded to at least one H); ketones have it in the middle (bonded to two alkyl groups).
• Both have a polarised C=O bond (δ+ carbon, δ− oxygen) — this drives almost everything they do.
• Carbonyls can't hydrogen bond with each other (no O–H or N–H), so their boiling points sit below the equivalent alcohol — but small ones dissolve in water via hydrogen bonding with water molecules.
• Aldehydes are easily oxidised to carboxylic acids; ketones resist oxidation. This difference is the basis of three classic distinguishing tests: acidified K₂Cr₂O₇, Tollens' reagent, and Fehling's solution.
• Both are reduced by LiAlH₄: aldehydes → primary alcohols, ketones → secondary alcohols.
• Both undergo nucleophilic addition with HCN, extending the carbon chain by one and forming hydroxynitriles.
• 2,4-DNPH detects any carbonyl (orange precipitate); the iodoform test specifically detects methyl ketones (and ethanal) with a yellow precipitate.

The C=O group — and why position matters

Aldehydes and ketones are both built around the carbonyl functional group, C=O. That's the shared feature. The difference between them comes down to where that carbonyl sits on the carbon chain and what it's attached to.

Think of it like this: the carbonyl carbon needs two more bonds besides the double bond to oxygen. In an aldehyde, one of those two remaining bonds is always to a hydrogen atom (R–CHO). Because of this, the carbonyl carbon can never be "buried" in the middle of a chain — it's always forced to be at the very end, which is why it's always carbon number 1 and you never need to write the "1" in the name (e.g. it's "propanal," not "propan-1-al").

In a ketone, both of the remaining bonds go to alkyl (R) groups — there's no carbonyl-hydrogen. That means the carbonyl carbon needs a carbon chain on both sides, so it's physically impossible for it to sit at the end. It's always somewhere in the middle.

Quick reference table

Why carbonyls behave the way they do in water and when heated

Oxygen is more electronegative than carbon, so it pulls electron density towards itself across the C=O double bond. This creates a permanent dipole: the carbon becomes slightly positive (δ+) and the oxygen slightly negative (δ−). This single fact — the polarised carbonyl — is the root cause of nearly every physical and chemical property in this whole topic, so it's worth really internalising.

Boiling points: carbonyls vs. alcohols

Because carbonyls have a permanent dipole, molecules of aldehyde or ketone attract each other through permanent dipole-dipole forces plus the usual London (dispersion) forces. That's it though — there's no O–H or N–H bond anywhere in a plain aldehyde or ketone, so carbonyl molecules cannot hydrogen bond with each other.

Compare that to an alcohol, which has an O–H bond and so hydrogen bonds extensively with itself. Hydrogen bonds are much stronger than dipole-dipole forces, so it takes more energy (a higher temperature) to separate alcohol molecules. That's why every carbonyl in the table below boils at a noticeably lower temperature than its "equivalent" alcohol (same number of carbons).

Solubility in water

Here's the twist: even though carbonyls can't hydrogen bond with each other, the oxygen atom still has lone pairs. That means a carbonyl molecule can hydrogen bond with something that has a δ+ hydrogen already built in — like water. The δ− oxygen on the carbonyl accepts a hydrogen bond from the δ+ hydrogen of a water molecule.

This is why small aldehydes and ketones (like methanal, ethanal, propanone) dissolve readily in water. But as the hydrocarbon chain gets longer, two things work against solubility:

• The long non-polar hydrocarbon "tail" can't hydrogen bond with water at all — it just sits there.
• That tail actually disrupts the existing hydrogen-bonded network within the water itself, without offering anything to replace it.

So solubility becomes a tug-of-war: the strength of the potential carbonyl···water hydrogen bond has to overcome (a) the London forces holding carbonyl molecules to each other, and (b) the hydrogen bonding holding water molecules to each other. For long-chain carbonyls, the non-polar part of the molecule wins, and the compound won't dissolve.

Aldehydes oxidise. Ketones don't. That's the whole story.

This is arguably the single most exam-important fact in this chapter: aldehydes can be oxidised to carboxylic acids; ketones essentially cannot be oxidised at all under normal lab conditions. Why? Because oxidation of a carbonyl requires a hydrogen atom directly attached to the carbonyl carbon to be available for removal — and aldehydes have exactly that (R–CHO), while ketones don't (R–CO–R′, no H on the carbonyl carbon).

To force a ketone to oxidise you'd need an extremely powerful oxidising agent, and even then it would have to break a C–C bond to do it — a destructive reaction that doesn't happen under standard test conditions. So in practice: no colour change with a ketone = negative test, every time, across all three tests below.

3a. Acidified potassium dichromate(VI), K₂Cr₂O₇ / H₂SO₄

Heat the aldehyde gently with acidified potassium dichromate(VI) solution.

3b. Tollens' reagent — the "silver mirror" test

Tollens' reagent is made by adding aqueous ammonia to silver nitrate solution, forming the colourless silver(I) complex ion [Ag(NH₃)₂]⁺ (also called ammoniacal silver nitrate). Gently warm — don't boil — the sample with Tollens' reagent.

If an aldehyde is present, it gets oxidised to a carboxylic acid, and in doing so it reduces the Ag⁺ ions all the way down to solid metallic silver, Ag. This silver deposits as a thin, shiny coating on the inside of the test tube — the famous "silver mirror."

3c. Fehling's solution (or Benedict's solution)

Fehling's solution contains blue Cu²⁺ ions dissolved in sodium hydroxide (Benedict's is identical chemistry but uses sodium carbonate instead). Gently warm the sample with the solution.

An aldehyde reduces the blue Cu²⁺ ions to Cu⁺ ions, which precipitate out as a brick-red solid, copper(I) oxide, Cu₂O.

LiAlH₄ turns carbonyls into alcohols

Where oxidation only worked on aldehydes, reduction works on both aldehydes and ketones. The reducing agent is lithium tetrahydridoaluminate (also called lithium aluminium hydride), LiAlH₄, used in dry ether as the solvent (it reacts violently with water, so the solvent must be anhydrous).

LiAlH₄ works by generating a hydride ion, :H⁻, which acts as a nucleophile. This hydride ion attacks the δ+ carbonyl carbon — meaning this reduction is actually a specific case of nucleophilic addition (see next section).

Why the carbonyl carbon is a target — and how to draw the mechanism

Remember the polarisation from Section 2? The δ+ carbon of the C=O group is electron-poor, which makes it attractive to anything carrying a lone pair of electrons it can grab — i.e., a nucleophile. This is the basis of a whole category of carbonyl reactions called nucleophilic addition.

The classic example in this specification is addition of hydrogen cyanide, HCN (usually generated in situ from acidified KCN, since HCN itself is a toxic gas). The nucleophile here is the cyanide ion, :CN⁻.

The two-step mechanism

1. Step 1 — Attack: The lone pair on the carbon of :CN⁻ attacks the δ+ carbonyl carbon. As the new C–C bond forms, the C=O π bond breaks, pushing both electrons onto the oxygen. This creates a negatively charged intermediate — an alkoxide ion (O⁻) with a −CN group now attached to the former carbonyl carbon.
2. Step 2 — Protonation: The negatively charged oxygen quickly grabs a proton (H⁺) from the solvent (water, dilute acid, or leftover HCN), giving a neutral –OH group.

Why does this reaction matter beyond the mechanism itself? Because it's a rare and valuable way to increase the chain length by exactly one carbon — the cyanide ion brings its own carbon into the product. That makes it genuinely useful in organic synthesis routes, not just a mechanism to memorise.

Naming the product

The product is called a hydroxynitrile. The nitrile group (–C≡N) is the higher-priority functional group, so it gets the main suffix "-nitrile" and is numbered as carbon 1. The hydroxyl group is present but not the priority group, so instead of the usual "-ol" suffix, it's named with the prefix "hydroxy-."

6a. 2,4-Dinitrophenylhydrazine (2,4-DNPH) — "is there a carbonyl here at all?"

2,4-DNPH is a reagent that reacts with any carbonyl compound — it doesn't distinguish aldehyde from ketone, it just confirms that a C=O group is present at all. This matters because other functional groups (carboxylic acids, esters) that also technically contain a carbon-oxygen double bond do not give a positive result — so a positive 2,4-DNPH test specifically confirms an aldehyde or ketone.

The reaction is a condensation reaction: the carbonyl compound reacts with the –NH₂ group of 2,4-DNPH, and a small molecule (water, H₂O) is lost as the two molecules join together.

A neat extra step: once you've confirmed a carbonyl is present, you can purify the orange precipitate by recrystallisation and then measure its melting point. Comparing this to a table of known literature melting points lets you identify exactly which aldehyde or ketone you started with — each one forms a derivative with a unique, sharp melting point.

6b. The Iodoform Test — spotting methyl ketones specifically

This test is more selective: it only gives a positive result for compounds containing a CH₃CO– group (a "methyl ketone"). By coincidence, ethanal also happens to contain this exact group (CH₃CHO has a CH₃CO- fragment), so it gives a positive result too, even though it's an aldehyde, not a ketone.

The reagent is iodine dissolved in alkali (I₂/NaOH), heated with the sample. The reaction happens in two stages:

1. Halogenation: all three hydrogens of the –CH₃ group are substituted by iodine atoms, forming a –CI₃ group.
2. Hydrolysis: the alkaline solution then hydrolyses this intermediate, breaking off the –CI₃ group as a precipitate of tri-iodomethane (iodoform), CHI₃ — a distinctive yellow solid — while the rest of the molecule becomes a carboxylate salt.

Bonus context: iodination under acidic conditions

If you instead treat a ketone like propanone with iodine under acidic conditions (rather than alkaline), a completely different reaction happens: only one hydrogen is substituted, giving iodopropanone.

This product is colourless and forms slowly, and since the brown/yellow colour of iodine fades steadily as the reaction proceeds, this reaction is a favourite for kinetics practicals — you can monitor the colour change with a colorimeter and work out reaction orders for propanone, iodine, and H⁺ by varying their concentrations.

• I can explain why an aldehyde's carbonyl group is always at the end of the chain, and a ketone's is always in the middle.
• I can name and draw the structures of methanal, ethanal, propanal, propanone, and pentan-2-one.
• I understand why the C=O bond is polarised and can label the δ+ and δ− atoms.
• I can explain why carbonyl boiling points are lower than the equivalent alcohol's.
• I can explain why small carbonyls dissolve in water but large ones don't, using intermolecular forces.
• I know why aldehydes oxidise but ketones don't (the H on the carbonyl carbon).
• I can describe the method and observation for all three oxidation tests: K₂Cr₂O₇, Tollens', and Fehling's.
• I can write the reduction equation for an aldehyde and a ketone using LiAlH₄, and name the alcohol products.
• I can draw the full curly-arrow mechanism for nucleophilic addition of HCN to a carbonyl.
• I can name hydroxynitrile products correctly, using the hydroxy- prefix.
• I know that 2,4-DNPH detects any carbonyl (orange ppt) and can identify the compound via melting point.
• I can describe the iodoform test, including which structural feature triggers a positive result.
• I know that certain alcohols (R-CH(OH)-CH₃) also give a positive iodoform test, and why.