17B.6-17B.8 - Carbonyl Compounds And Reactions

Recognising Aldehydes And Ketones

A carbonyl group is a carbon–oxygen double bond; the atoms attached to its carbon distinguish aldehydes from ketones.

In this lesson the word "carbonyl" means aldehydes and ketones, not every organic molecule that happens to contain a $\ce{C=O}$ bond. Carboxylic acids, esters and acyl chlorides also contain a carbonyl group, but their reactions are assigned to neighbouring lessons.

The assessed distinction is the position and bonding environment of the carbonyl carbon.

In a displayed formula, an aldehyde has the pattern $\ce{R-C(=O)-H}$. In a condensed formula, the end group is usually written as $\ce{CHO}$, not as $\ce{COH}$; writing $\ce{COH}$ can be confused with an alcohol group. A ketone has the pattern $\ce{R-C(=O)-R'}$, where both groups attached to the carbonyl carbon are carbon-containing groups.

The name can help, but it is not the evidence by itself. Names ending in -al are aldehydes and names ending in -one are ketones, but structure questions often expect you to identify the group from a displayed, skeletal or semi-displayed structure.

Common-error contrast

$\ce{CH3CH2CHO}$ and $\ce{CH3COCH3}$ both have molecular formula $\ce{C3H6O}$, but they are not the same functional group. In $\ce{CH3CH2CHO}$, the carbonyl carbon is bonded to H, so the compound is an aldehyde. In $\ce{CH3COCH3}$, the carbonyl carbon is bonded to two carbon groups, so the compound is a ketone.

Properties From The Polar C=O Bond

The $\ce{C=O}$ bond is polar because oxygen is more electronegative than carbon. The carbonyl carbon is electron-poor, $\ce{C}^{\delta+}$, and the oxygen is electron-rich, $\ce{O}^{\delta-}$, with lone pairs on oxygen. That polarity explains both physical properties and later reaction mechanisms.

The simple aldehydes and ketones considered here do not form intermolecular hydrogen bonds with each other because their molecules do not contain an $\ce{O-H}$, $\ce{N-H}$ or F-H bond. They can act as hydrogen-bond acceptors, but they cannot donate a hydrogen bond to another aldehyde or ketone molecule.

This gives a useful comparison:

The same carbonyl oxygen can form hydrogen bonds with water molecules because water can donate $\ce{O-H}$ hydrogen bonds. Small aldehydes and ketones are therefore soluble in water, but solubility decreases as the non-polar hydrocarbon chain becomes a larger fraction of the molecule.

Worked example

Explain why propanone is soluble in water but has a lower boiling temperature than propan-1-ol.

Propanone contains a polar $\ce{C=O}$ bond and the oxygen has lone pairs, so water molecules can hydrogen-bond to the carbonyl oxygen. That helps propanone dissolve. Propanone molecules cannot hydrogen-bond to each other because propanone has no $\ce{O-H}$ bond, whereas propan-1-ol molecules can hydrogen-bond to each other. More energy is needed to separate the alcohol molecules, so propan-1-ol has the higher boiling temperature.

Oxidation Tests For Aldehydes

Aldehydes are more readily oxidised than ketones. In this specification, aldehydes can be oxidised to carboxylic acids, and the oxidising agent may be represented as $\ce{[O]}$:

$\ce{RCHO + [O] -> RCOOH}$

For ethanal:

$\ce{CH3CHO + [O] -> CH3COOH}$

That equation is not asking you to write the full redox chemistry of the oxidising reagent. It is showing that the aldehyde carbon gains oxygen character and becomes a carboxylic acid carbon. The reactions of carboxylic acids themselves belong to the next lesson.

Three named oxidising tests are important.

Fehling’s and Benedict’s tests here concern simple aliphatic aldehydes. Aromatic aldehydes such as benzaldehyde can give a negative result; Tollens’ reagent is the appropriate distinguishing test for those examples.

Ketones are not usually oxidised by these reagents under the conditions used for these tests. This is why the observations can distinguish an aldehyde from a ketone after a general carbonyl test has shown that a $\ce{C=O}$ compound is present.

Worked example

An unknown liquid gives an orange precipitate with 2,4-DNPH. A fresh sample gives a silver mirror with Tollens' reagent. Deduce the functional group and explain the evidence.

The 2,4-DNPH result shows that the unknown contains an aldehyde or ketone carbonyl group. The silver mirror shows that the unknown is oxidised by Tollens' reagent, which is a positive aldehyde test. Therefore the unknown is an aldehyde, not a ketone.

Guided practice

Complete the reasoning for this test sequence: compound X gives a positive 2,4-DNPH test but no change with acidified dichromate(VI).

Answer: X contains an aldehyde or ketone because 2,4-DNPH is positive. The lack of orange-to-green change with acidified dichromate(VI) means it is not oxidised like an aldehyde under these conditions. X is therefore likely to be a ketone.

Fehling's, Benedict's and Tollens' reagents are alkaline. Their immediate organic oxidation product is a carboxylate ion, not the free acid; acidification gives the carboxylic acid. The simplified $\ce{RCHO + [O] -> RCOOH}$ equation summarises oxidation, while an ionic Tollens' equation shows the alkaline conditions explicitly:

Use fresh, separate small samples for different tests. Warm flammable carbonyl liquids in a water bath, with eye protection. Tollens' reagent must be freshly prepared and disposed of promptly under the laboratory method; do not store it or let silver-containing residues dry.

Reduction To Alcohols

Lithium tetrahydridoaluminate, also called lithium aluminium hydride or $\ce{LiAlH4}$, reduces aldehydes and ketones in dry ether. The reducing agent can be represented as $\ce{[H]}$ in equations.

The product depends on the carbonyl family.

The dry ether condition matters because $\ce{LiAlH4}$ reacts violently with water. In a full practical route, a careful work-up is needed after reduction, but the assessed reagent-condition pair here is lithium tetrahydridoaluminate in dry ether.

The $\ce{[H]}$ notation is a bookkeeping shortcut. It does not mean bubbling hydrogen gas through the mixture. The useful structural idea is that the $\ce{C=O}$ group becomes $\ce{C-OH}$, and the carbonyl carbon gains hydrogen.

Worked example

Predict the organic product when butan-2-one is reduced using lithium tetrahydridoaluminate in dry ether.

Butan-2-one is a ketone: $\ce{CH3COCH2CH3}$. Reduction of a ketone gives a secondary alcohol, so the $\ce{C=O}$ group becomes $\ce{CH(OH)}$.

The product is butan-2-ol. The carbon skeleton has not changed; only the carbonyl group has been reduced.