Aldehydes & Ketones

Think about a carpenter nailing two boards together. One nail holds them, but you could still pull them apart if you tried. Now imagine one board has a magnet built into the wood and the nail is partially charged — one end positive, one end negative. That is essentially the carbonyl group: a carbon–oxygen double bond (C=O) where the electrons are not shared equally. Oxygen, far more electronegative, pulls the shared electrons toward itself, leaving the carbon with a partial positive charge (δ+) and the oxygen with a partial negative charge (δ–).

The geometry matters for understanding reactivity. The carbonyl carbon is sp²-hybridised, meaning it sits at the centre of a flat, trigonal-planar arrangement with bond angles of about 120°. This flat geometry means there is no steric "roof" blocking the carbonyl from above and below — it is wide open for attack. The π bond (the second bond in C=O) sits above and below this flat plane and is relatively easy to break, because π bonds are weaker than σ bonds. These two facts together — the partial positive charge on carbon, and the exposed π system — explain why the carbonyl group is such a reactive electrophile.

Aldehydes (RCHO) have a hydrogen atom attached to the carbonyl carbon. Ketones (RCOR') have two carbon groups. This difference is not cosmetic. The hydrogen in an aldehyde is smaller and less electron-donating than an alkyl group, so the carbonyl carbon of an aldehyde remains more δ+ and more accessible — making aldehydes more reactive than ketones in nearly every nucleophilic addition reaction.

Naming aldehydes and ketones follows the same logic as all IUPAC naming: find the longest chain that includes the functional group, number it to give the functional group the lowest possible number, then name the substituents. The key difference between the two classes is the ending and the numbering rule.

For aldehydes, replace the terminal “-e” of the alkane name with -al. Because the aldehyde group is always at the end of the chain (by definition — the C=O is attached to H, so it must be terminal), the aldehyde carbon is always C-1. You never need to write the number. A five-carbon chain with an aldehyde at C-1 is pentanal. A four-carbon chain is butanal. Common names you must know: methanal = formaldehyde, ethanal = acetaldehyde, propanal = propionaldehyde.

For ketones, replace “-e” with -one. Now you do need a number, because the carbonyl can be at different positions along the chain. Always number from the end that gives the carbonyl the lower locant. A six-carbon ketone with C=O at C-3 is hexan-3-one. Common names: propanone = acetone, butanone = methyl ethyl ketone (MEK). The “-one” suffix also appears in many drug names (testosterone, cortisone, progesterone) — all are ketone-containing steroids.

The carbonyl group is polar, but it cannot donate a hydrogen bond (no O–H or N–H bond on the molecule itself). This places aldehydes and ketones between two extremes: they cannot hydrogen-bond to themselves the way alcohols do, but they can accept hydrogen bonds from water. The result is a compound that has higher boiling points than comparable alkanes (because of dipole–dipole interactions between the polar C=O groups), but lower boiling points than comparable alcohols (no H-bond donors).

Small aldehydes and ketones (up to about C-4) are fully miscible with water, because the C=O can act as a hydrogen-bond acceptor with water molecules. Acetone is one of the most important lab solvents precisely because it is polar enough to dissolve many organic compounds while also mixing with water. Beyond C-6 or so, the non-polar hydrocarbon chain dominates and water solubility drops sharply.

Almost every reaction of aldehydes and ketones involves the same fundamental event: a nucleophile attacks the electrophilic carbonyl carbon. Picture a moth flying toward a lamp — the negatively charged (or electron-rich) nucleophile is attracted to the partial positive charge on the carbonyl carbon. When it attacks, the π bond breaks. The electrons from the C=O π bond move to the oxygen, giving it a full negative charge (an alkoxide). An acid step then protonates the oxygen, and the product is an alcohol with the nucleophile now attached to what was the carbonyl carbon.

This mechanism — nucleophilic addition — is fundamentally different from electrophilic addition (which alkenes undergo). In nucleophilic addition, the nucleophile strikes first. The flat, trigonal geometry of the carbonyl makes the carbon accessible from above and below, which is why these reactions work well. This same mechanism explains hydride reduction, acetal formation, hemiacetal formation, and cyanohydrin synthesis.

Reduction of a carbonyl is simply adding hydrogen across the C=O. The nucleophile here is a hydride ion (H⁻), delivered by a reducing agent. Think of it as using a "hydrogen donor" to convert the C=O into a C–OH. Two reducing agents are used, and choosing between them matters clinically and practically.

Sodium borohydride (NaBH₄) is a mild, selective reagent. It reduces aldehydes and ketones to their corresponding alcohols but does NOT reduce less-reactive groups like carboxylic acids, esters, or amides. It can be used in water or alcohol solvents. In the lab and in pharmaceutical synthesis, NaBH₄ is the go-to for selectively reducing a ketone in a complex molecule without touching other functional groups.

Lithium aluminium hydride (LiAlH₄) is a powerful reducing agent — it reduces almost everything, including carboxylic acids, esters, and amides. It reacts violently with water and must be used in dry ether. LiAlH₄ is used when you need to reduce stubborn functional groups, but its lack of selectivity is a disadvantage in multistep synthesis.

Here is a question every TMU examiner asks: how do you tell an aldehyde from a ketone? The answer is oxidation. Aldehydes are easily oxidised; ketones resist mild oxidants. This difference in oxidation behaviour forms the basis of three classic tests — Tollens', Fehling's, and Benedict's — all of which give a positive result with aldehydes (and some sugars) and a negative result with ketones.

The logic is straightforward: in an aldehyde, the carbonyl carbon (C=O) already has one hydrogen attached. Oxidation removes that hydrogen and adds an oxygen, converting –CHO into –COOH. In a ketone, there is no such hydrogen on the carbonyl carbon — both substituents are carbon groups — so mild oxidants cannot attack. The ketone just sits there unchanged.

Tollens' test: Silver-ammonia complex [Ag(NH₃)₂]⁺ in alkaline solution is used. When an aldehyde is present, it reduces Ag⁺ to metallic silver (Ag⁰), which deposits as a bright mirror on the inner wall of a clean test tube. The aldehyde is oxidised to a carboxylate. Ketones do not reduce Ag⁺ — no mirror forms.

Fehling's test uses an alkaline solution of copper(II) ions complexed with tartrate (deep blue solution). Aldehydes reduce Cu²⁺ to Cu⁺, which precipitates as brick-red Cu₂O. Ketones give no precipitate (solution remains blue). Benedict's reagent works identically (Cu²⁺ complexed with citrate); it is the version used clinically to test urine for glucose (an aldehyde sugar). A colour change from blue → green → yellow → orange → red indicates increasing glucose concentration.

Imagine an alcohol walking up to an aldehyde and shaking hands. They form a hemiacetal — a half-way product. If a second alcohol joins in and the first "hand" lets go (the OH leaves), you get an acetal. This two-step process is reversible: under aqueous acid, an acetal slowly breaks back into the aldehyde and two alcohol molecules.

More precisely: a hemiacetal has the formula R–CH(OH)(OR′). It has one OH and one OR group on the same carbon. Acetals are R–CH(OR′)₂ — two alkoxy groups, no OH, no carbonyl. The forward reaction (aldehyde → acetal) is catalysed by acid and driven by removal of water. The reverse (acetal → aldehyde) occurs in dilute aqueous acid — the acetal is simply hydrolysed.

This chemistry is not just a textbook curiosity. It is central to carbohydrate chemistry. When glucose cyclises from its open-chain form (with a free aldehyde) to its ring form, it forms a hemiacetal intramolecularly — the C-5 OH attacks the C-1 aldehyde. The resulting α and β anomers of glucose are simply different configurations of this hemiacetal carbon (called the anomeric carbon). Glycosidic bonds in disaccharides and polysaccharides are acetals — one OH of a second sugar attacks the hemiacetal, displacing water.

The aldol reaction is how nature builds carbon chains. It is the reaction that plants use to assemble sugars and that your cells use in fatty acid synthesis. The key requirement is an α-hydrogen — a hydrogen sitting on the carbon immediately next to the carbonyl. In dilute base, that hydrogen is pulled off, generating a carbanion (called an enolate). The enolate is a nucleophile. It attacks the carbonyl of a second aldehyde or ketone molecule, forming a new C–C bond. The product is a β-hydroxy carbonyl compound — the "aldol" (both an aldehyde and an alcohol, hence the name).

The aldol product can then undergo elimination (dehydration) if heated. The β-OH and an α-H leave as water, forming an α,β-unsaturated carbonyl — a compound with a C=C conjugated with the C=O. This two-step process (aldol addition then dehydration) is called an aldol condensation. The product is thermodynamically stable because the double bond is conjugated with the carbonyl.

The iodoform test is one of the most specific tests in organic chemistry — it detects a very particular structural feature: a methyl group (CH₃) directly attached to a carbonyl. The reagent is iodine in alkaline solution (I₂ / NaOH, also written as I₂ / KOH). A positive result is an immediate yellow precipitate of iodoform (CHI₃), which also has a distinctive "antiseptic" smell (iodoform was historically used as an antiseptic).

The mechanism involves three successive iodinations of the methyl group, followed by cleavage of the C–C bond by hydroxide. The methyl ketone CH₃COR first gets all three H's on the methyl group replaced by I (forming CI₃COR), then OH⁻ attacks the carbonyl to break the C–CI₃ bond, releasing CHI₃ and the carboxylate RCOO⁻. The yellow CHI₃ precipitate confirms the presence of a methyl ketone.

Most aldol and oxidation reactions require α-hydrogens — the hydrogen on the carbon next to the carbonyl. Benzaldehyde (C₆H₅CHO) has no α-carbon at all; formaldehyde (HCHO) has no α-carbon either. When you treat these aldehydes with concentrated NaOH, they cannot form an enolate and cannot undergo aldol. Instead, they do something unusual: they react with each other. One molecule acts as the oxidant (gets oxidised to a carboxylate), and the other acts as the reductant (gets reduced to an alcohol). This self-disproportionation is the Cannizzaro reaction.

These questions reflect the types and difficulty of questions that appear in TMU past papers (2019–2022) for this topic. Work through them yourself before checking the answers.