Making aldehydes and ketones

Start from what you can buy

There is no single way to make an aldehyde or a ketone. A chemist looks at what is on the shelf and picks the method that fits. The starting materials are alcohols, hydrocarbons and the derivatives of carboxylic acids.

The map shows the routes in one picture. We take them one family at a time.

From alcohols: oxidation and dehydrogenation

A secondary alcohol is oxidised to a ketone. The usual oxidising agents are acidified potassium dichromate, acidified potassium permanganate or chromium trioxide in aqueous acid (the Jones reagent). The ketone does not oxidise further under mild conditions.

$\text{CH}_{3}\text{CH}(\text{OH})\text{CH}_{3} \xrightarrow{\text{K}_{2}\text{Cr}_{2}\text{O}_{7}\ \text{H}_{2}\text{SO}_{4}} \text{CH}_{3}\text{COCH}_{3}$

A primary alcohol gives an aldehyde, but a strong aqueous oxidant carries on to the carboxylic acid. There are two ways round this.

1. Use PCC, pyridinium chlorochromate, in dichloromethane. It has no water, so the aldehyde stops.
2. Distil the aldehyde out of the mixture as it forms. The aldehyde has no hydrogen bonds, so it boils at a lower temperature than the alcohol.

$\text{CH}_{3}\text{CH}_{2}\text{OH} \xrightarrow{\text{PCC}} \text{CH}_{3}\text{CHO}$

Dehydrogenation is the industrial route. The vapour of the alcohol is passed over copper at $573\ \text{K}$. A primary alcohol gives an aldehyde, a secondary alcohol gives a ketone, and a tertiary alcohol loses water to give an alkene.

$\text{CH}_{3}\text{CH}_{2}\text{OH} \xrightarrow{\text{Cu}\ 573\ \text{K}} \text{CH}_{3}\text{CHO} + \text{H}_{2}$

A tertiary alcohol has no hydrogen on the carbon with the $\text{OH}$ group. So it cannot be oxidised to a carbonyl compound under these conditions.

Look at the green curve. At $k_2/k_1 = 5$ only about $13\ \%$ of the alcohol is ever in the flask as aldehyde. At $0.05$ it is about $85\ \%$. This is why PCC works, and why a chemist who uses dichromate must distil the aldehyde away at once. The model has two first-order steps. It is a simple picture, but it shows the logic.

Practise with new numbers

From hydrocarbons

1. Hydration of alkynes. Ethyne adds water in the presence of mercury(II) sulphate and dilute sulphuric acid at $333\ \text{K}$. The product is the enol, and it changes to ethanal at once. Other terminal alkynes give methyl ketones, because water adds by Markovnikov's rule. An unsymmetrical internal alkyne gives a mixture of two ketones.

$\text{CH}_{3}\text{C}≡\text{CH} \xrightarrow{\text{HgSO}_{4}\ \text{H}_{2}\text{SO}_{4}\ 333\ \text{K}} \text{CH}_{3}\text{COCH}_{3}$

2. Hydroboration, then oxidation. A terminal alkyne reacts with a hindered borane and then with alkaline hydrogen peroxide. Water adds the other way (anti-Markovnikov), so the product is an aldehyde. Hex-1-yne gives hexanal.
3. Ozonolysis of alkenes. Ozone adds across the double bond. Zinc and water (or dimethyl sulphide) then split the ozonide into two carbonyl compounds. Each carbon of the double bond becomes a carbonyl carbon. A carbon with a hydrogen gives an aldehyde, and a carbon with two carbon groups gives a ketone.

$\text{CH}_{3}\text{CH}{=}\text{C}(\text{CH}_{3})_{2} \xrightarrow{1.\ \text{O}_{3}\ 2.\ \text{Zn}\ \text{H}_{2}\text{O}} \text{CH}_{3}\text{CHO} + \text{CH}_{3}\text{COCH}_{3}$

From acyl chlorides: Rosenmund reduction

An acyl chloride, $\text{RCOCl}$, is reduced with hydrogen over palladium on barium sulphate. The catalyst is partly poisoned with a little quinoline or sulphur, so it does not carry on to the alcohol.

$\text{RCOCl} + \text{H}_{2} \xrightarrow{\text{Pd}\text{-}\text{BaSO}_{4}} \text{RCHO} + \text{HCl}$

An acyl chloride and a dialkylcadmium give a ketone: $2\,\text{RCOCl} + \text{R'}_2\text{Cd} \to 2\,\text{RCOR'} + \text{CdCl}_2$. A lithium dialkylcuprate does the same thing, and neither reagent attacks the ketone that forms.

From nitriles: Stephen reduction

A nitrile, $\text{RCN}$, is reduced with tin(II) chloride and hydrochloric acid. This gives an imine hydrochloride. Boiling water then turns it into the aldehyde. The reaction is the Stephen reduction.

$\text{RCN} + _{2}[\text{H}] + \text{HCl} \xrightarrow{\text{SnCl}_{2}} \text{RCH}{=}\text{NH}\cdot \text{HCl} \xrightarrow{\text{H}_{3}\text{O}+} \text{RCHO}$

Diisobutylaluminium hydride, DIBAL-H, also reduces a nitrile to an aldehyde after water is added.

A nitrile and a Grignard reagent give a ketone. The Grignard reagent adds to the carbon of the $\text{C}≡\text{N}$ group. The imine salt is hydrolysed by dilute acid.

$\text{RCN} + \text{R}'\text{MgX} \longrightarrow \text{RC}(\text{R}'){=}\text{NMgX} \xrightarrow{\text{H}_{3}\text{O}+} \text{RCOR}'$

From esters

An ester is reduced to an aldehyde by DIBAL-H in a cold solvent, then water. The temperature is very low, about $-78\ ^\circ\text{C}$, so the aldehyde does not react further. Warm DIBAL-H, or lithium aluminium hydride, would give the alcohol.

Dry distillation of calcium salts

Heating calcium ethanoate gives propanone and calcium carbonate. Heating a mixture of calcium methanoate and calcium ethanoate gives ethanal. This is an old method, but it is in the syllabus.

Try it in Python

The program has a small table of alcohols. For each one it prints what PCC makes and what hot acidified dichromate makes.

Common mistakes

• Using dichromate to make an aldehyde. Under reflux it goes on to the acid. Use PCC, or distil the product out.
• Writing the Rosenmund catalyst as plain palladium. It is palladium on barium sulphate, poisoned.
• Reading Stephen's reaction as a hydrolysis of the nitrile to an acid. Here tin(II) chloride reduces the nitrile.
• Hydrating ethyne and expecting a ketone. Water adds, and the product is an aldehyde, ethanal. Only the other terminal alkynes give ketones.
• Oxidising a tertiary alcohol. There is no carbonyl product.

Remember these

• Secondary alcohol, oxidant: ketone. Primary alcohol: PCC for the aldehyde, dichromate for the acid.
• Cu at $573\ \text{K}$: dehydrogenation. Ethyne with Hg$^{2+}$ and acid: ethanal.
• Rosenmund: acyl chloride, H$_2$, Pd-BaSO$_4$. Stephen: nitrile, SnCl$_2$ and HCl, then water.
• Grignard reagent with a nitrile: a ketone. DIBAL-H with an ester: an aldehyde.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Which reagent turns hexan-1-ol into hexanal and stops there?

2. Which alkyne gives butanone on hydration?

3. Name the starting material for the Rosenmund synthesis of propanal.

4. What does ozonolysis of $\text{CH}_{3}\text{CH}_{2}\text{CH}{=}\text{CHCH}_{2}\text{CH}_{3}$ followed by zinc and water give?

References

• OpenStax, Chemistry 2e, Chapter 20, section 20.3 Aldehydes, Ketones, Carboxylic Acids, and Esters.
• NCERT, Chemistry Part II for Class XII, Chapter 8, the section on the preparation of aldehydes and ketones.