Preparation of ketones

A ketone is a carbonyl compound in which the carbonyl carbon sits between two carbon groups, $\text{R–CO–R}'$ , with no hydrogen directly on that carbon. That one structural fact — carbonyl carbon fully flanked by carbon on both sides — is the thread that ties together every preparation route in this ISC Class 12 unit: oxidise a secondary alcohol, cleave a tetra- or tri-substituted alkene by ozonolysis, hydrate an internal alkyne, react an acid chloride with dialkyl cadmium, or add a Grignard reagent to a nitrile. The syllabus expects you to write the equation for each named route and to predict, from the starting material alone, whether the product is an aldehyde or a ketone. This article expands the five standard routes with a structural "why," one fully worked synthesis, a second worked route, and the traps examiners repeatedly test.

The core insight

A ketone forms whenever the reacting carbon ends up bonded to two carbon groups and no hydrogen, so the choice of route — alcohol oxidation, alkene ozonolysis, alkyne hydration, dialkyl cadmium, or Grignard-nitrile addition — is really a choice of which carbon skeleton already has that "two-carbon-neighbours" pattern built in.

The core idea

Oxidation of a secondary alcohol. A secondary alcohol has exactly one hydrogen on its carbinol carbon. Oxidising agents such as acidified potassium dichromate remove that one H to form the carbonyl, and then stop — there is no second H left on that carbon for the oxidant to abstract, so the reaction cannot run on to a carboxylic acid the way primary-alcohol oxidation does.

\text{R–CH(OH)–R}' \xrightarrow{K_2Cr_2O_7/H^+} \text{R–CO–R}'

where: $\text{R}$ and $\text{R}'$ are alkyl (or aryl) groups, and the carbinol carbon carries one H before oxidation and none after.

Ozonolysis of an alkene. When a carbon–carbon double bond has two alkyl substituents on the same carbon, ozonolysis followed by a reductive work-up cleaves the double bond and leaves a ketone at that carbon.

\text{R}_2\text{C=CR}'_2 \xrightarrow{(i)\ O_3\ (ii)\ Zn/H_2O} \text{R}_2\text{C=O} + \text{R}'_2\text{C=O}

where: each doubly-substituted alkene carbon becomes an independent carbonyl fragment; a carbon that instead carries only one alkyl group and one H yields an aldehyde on that side, not a ketone.

Hydration of an internal alkyne. Markovnikov addition of water across an internal (non-terminal) triple bond, catalysed by mercuric ion in acid, proceeds through an unstable enol that tautomerises to the more stable ketone.

\text{R–C≡C–R}' + H_2O \xrightarrow{Hg^{2+}/H_2SO_4} \text{R–CO–CH}_2\text{R}'

where: the OH adds to the more substituted alkyne carbon (Markovnikov), and only acetylene itself (a terminal, symmetric alkyne) gives an aldehyde instead of a ketone by this route.

Acid chloride with dialkyl cadmium, and nitrile with Grignard's reagent. These two routes both convert a carboxylic-acid-derived electrophile into a ketone in a single, controlled addition:

2\text{R–COCl} + \text{R}'_2\text{Cd} \rightarrow 2\text{R–CO–R}' + \text{CdCl}_2

\text{R–C≡N} + \text{R}'\text{MgX} \rightarrow \text{R–C(=NMgX)–R}' \xrightarrow{H_3O^+} \text{R–CO–R}' + \text{NH}_3

where: dialkyl cadmium is chosen deliberately because, once the ketone forms, it is far less reactive toward that ketone's own carbonyl than a Grignard reagent would be, so the reaction stops cleanly at the ketone; the Grignard-nitrile route instead relies on the nitrile carbon reacting only once, giving an isolable imine-magnesium salt that is hydrolysed afterwards. A sixth route, pyrolysis of a calcium carboxylate, gives ketones by dry distillation:

(\text{RCOO})_2\text{Ca} \xrightarrow{\Delta} \text{R–CO–R} + \text{CaCO}_3

Figure

A decision map showing which starting material and reagent lead to a ketone by each named route

Why it works

The reason ketones resist every one of these routes running "one step too far" is structural, and it is worth stating once, carefully, because the whole unit hinges on it. Once a carbonyl carbon is bonded to two carbon chains and one oxygen (as a double bond), there is no hydrogen left on that carbon. Oxidation reactions that convert alcohols and aldehydes to acids work by abstracting a C–H bond on the carbon being oxidised; a ketone carbon simply does not offer one. Cleaving a ketone into two smaller acids would instead require breaking a carbon–carbon bond, which needs conditions far harsher and less selective than the acidified dichromate used routinely in this unit — so under standard exam-level conditions, ketones are treated as oxidation dead-ends. The same logic explains why chemists deliberately swap reagents mid-route in two of the five methods above. A direct Grignard reagent added to an acid chloride would not stop at the ketone: the first addition gives the ketone, but the Grignard carbanion is itself nucleophilic enough to add a second time to that freshly formed carbonyl, over-shooting to a tertiary alcohol. Dialkyl cadmium is chosen specifically because it is a much weaker nucleophile — reactive enough to attack the acid chloride once, but too weak to attack the resulting ketone a second time. The Grignard-nitrile route avoids the same trap differently: the nitrile carbon can only accept one nucleophilic addition (nitrogen has nowhere else to bond), so the reaction is self-limiting even before hydrolysis releases the ketone.

Worked example

Problem. Prepare butan-2-one ( $\text{CH}_3\text{COCH}_2\text{CH}_3$ ) from butan-2-ol, and separately prepare acetone ( $\text{CH}_3\text{COCH}_3$ ) from acetonitrile using a Grignard reagent, explaining at each stage why the reaction stops cleanly at the ketone.

Step 1. Classify butan-2-ol, $\text{CH}_3\text{CH(OH)CH}_2\text{CH}_3$ , as a secondary alcohol: its carbinol carbon carries exactly one H, flanked by $\text{CH}_3$ and $\text{CH}_2\text{CH}_3$ .

Step 2. Oxidise with acidified potassium dichromate:

\text{CH}_3\text{CH(OH)CH}_2\text{CH}_3 \xrightarrow{K_2Cr_2O_7/H_2SO_4,\ \Delta} \text{CH}_3\text{COCH}_2\text{CH}_3

The single carbinol H is removed to form the $\text{C=O}$ ; no second H remains on that carbon, so the oxidant cannot push the reaction further to an acid — over-oxidation, a real risk with primary alcohols, does not occur here.

Step 3. Turn to the nitrile route: react acetonitrile with methylmagnesium iodide (from $\text{CH}_3\text{I}$ and Mg in dry ether):

\text{CH}_3\text{C≡N} + \text{CH}_3\text{MgI} \rightarrow \text{CH}_3\text{–C(=NMgI)–CH}_3

The Grignard carbon adds to the electrophilic nitrile carbon exactly once, giving an isolable metalated imine (ketimine salt) rather than over-adding as it would to a ketone.

Step 4. Hydrolyse the imine salt with dilute acid:

\text{CH}_3\text{–C(=NMgI)–CH}_3 \xrightarrow{H_3O^+} \text{CH}_3\text{COCH}_3 + \text{NH}_3 + \text{Mg(OH)I}

Answer: butan-2-ol gives butan-2-one by simple secondary-alcohol oxidation, and acetonitrile plus methylmagnesium iodide, followed by acid hydrolysis, gives acetone — both routes terminate at the ketone because the electrophilic carbon can accept only the reactivity available to it before becoming carbon-flanked and hydrogen-free.

Competitive stretch

Given hex-3-ene, $\text{CH}_3\text{CH}_2\text{CH=CHCH}_2\text{CH}_3$ , check substitution before predicting the product: each double-bond carbon here carries one H and one alkyl group ( $\text{CH}_3\text{CH}_2$ –), not two alkyl groups. Ozonolysis with $\text{O}_3$ then $\text{Zn/H}_2\text{O}$ therefore cleaves it to two molecules of an aldehyde, $2\ \text{CH}_3\text{CH}_2\text{CHO}$ (propanal), not a ketone — a useful reminder that ozonolysis gives a ketone fragment only when the alkene carbon bears two alkyl substituents and no vinylic H, exactly as tested in the quiz below.

Common traps

Confusing the dialkyl cadmium route (deliberately unreactive toward the ketone product) with a direct Grignard addition to an acid chloride, which over-adds to a tertiary alcohol.
Forgetting that ozonolysis gives a ketone fragment only from a fully-substituted (no vinylic H) alkene carbon; a mono-substituted alkene carbon gives an aldehyde fragment instead.
Assuming secondary alcohols can be over-oxidised to acids like primary alcohols — they cannot, since the required second C–H bond is simply absent after the first oxidation.
Mixing up which alkyne hydration gives a ketone (any internal alkyne) versus an aldehyde (acetylene itself only).

Ask the tutor: if a specific ketone synthesis problem — picking the right starting material or explaining why one route beats another — is tripping you up, ask the AI tutor to walk the retrosynthesis with you step by step.

Why can't a primary alcohol be used to make a ketone directly?

A primary alcohol's carbinol carbon carries two hydrogens, not one. Oxidation removes the first H to give an aldehyde, but the aldehyde carbon still has one H left, so further oxidation removes that too, driving the product all the way to a carboxylic acid rather than stopping at a stable carbonyl flanked by two carbons. Only a secondary alcohol, with exactly one carbinol H, halts naturally at the ketone stage.

Why does the Grignard-nitrile route need a separate hydrolysis step while the alcohol oxidation does not?

The Grignard addition to a nitrile produces a magnesium-bound imine salt, $\text{R–C(=NMgX)–R}'$ , not the ketone itself — nitrogen is still attached to the carbon in place of oxygen. Only aqueous acid workup exchanges that nitrogen-magnesium linkage for a carbonyl oxygen, releasing ammonia and freeing the ketone. Alcohol oxidation, by contrast, transforms the alcohol oxygen directly into the carbonyl oxygen in one redox step, so no separate hydrolysis is needed.

How do I quickly decide, on sight, which route to use for a given target ketone?

Look at what starting material or fragment you are handed. A secondary alcohol already in hand points straight to dichromate oxidation. A tetra- or tri-substituted alkene points to ozonolysis, since it cleaves precisely at the double bond. An acid chloride, if a clean ketone (not a tertiary alcohol) is the goal, points to dialkyl cadmium rather than a direct Grignard. A nitrile plus a Grignard reagent is the route when you need to build a new carbon-carbon bond at the future carbonyl carbon.