Why Fehling's Solution Turns Red With Ethanal but Not With Acetone

The smell of cinnamon, the flavour of vanilla and the solvent in nail polish remover all come from compounds with a carbonyl group, C=O. When that group sits at the end of a chain with a hydrogen attached, the compound is an aldehyde; when it sits between two carbons, it is a ketone. That small difference decides which tests they pass.

This lesson covers naming and preparing aldehydes and ketones by standard methods, and the chemical tests that distinguish them.

Aldehydes are named with the suffix -al and ketones with -one, and they are prepared mainly by oxidising or dehydrogenating alcohols, by ozonolysis of alkenes, by hydrating alkynes, and by named reactions of acid chlorides, nitriles and aromatic compounds.

Naming:

- $\mathrm{HCHO}$ is methanal and $\mathrm{C}{\mathrm{H}}_3\mathrm{CHO}$ is ethanal
- $\mathrm{C}{\mathrm{H}}_3\mathrm{COC}{\mathrm{H}}_3$ is propanone and $\mathrm{C}{\mathrm{H}}_3\mathrm{COC}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_3$ is butanone
- ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{CHO}$ is benzaldehyde and ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COC}{\mathrm{H}}_3$ is acetophenone
- The aldehyde carbon is always carbon 1, while a ketone's carbonyl gets the lowest locant, as in pentan-2-one

From alcohols:

- Oxidation — primary alcohols give aldehydes with pyridinium chlorochromate (PCC); secondary alcohols give ketones with acidified ${\mathrm{K}}_2\mathrm{C}{\mathrm{r}}_2{\mathrm{O}}_7$
- Dehydrogenation — alcohol vapour passed over copper at 573 K

From hydrocarbons:

- Ozonolysis of alkenes — ozone, then zinc and water, splits the double bond into two carbonyl compounds
- Hydration of alkynes — ethyne gives ethanal; other alkynes give ketones

From acid chlorides, nitriles and aromatic compounds:

- Rosenmund reduction — $\mathrm{RCOCl}+{\mathrm{H}}_2\stackrel{Pd/BaS{\mathrm{O}}_4}{\to }\mathrm{RCHO}+\mathrm{HCl}$
- Stephen reaction — nitriles with $\mathrm{SnC}{\mathrm{l}}_2$ and HCl, then hydrolysis, give aldehydes
- Nitriles with Grignard reagents, then hydrolysis, give ketones: $\mathrm{C}{\mathrm{H}}_3\mathrm{CN}$ with ${\mathrm{C}}_2{\mathrm{H}}_5\mathrm{MgBr}$ gives butanone
- Friedel-Crafts acylation — benzene with ethanoyl chloride and $\mathrm{AlC}{\mathrm{l}}_3$ gives acetophenone
- Etard reaction — toluene with chromyl chloride gives benzaldehyde

Worked example — ozonolysis. An alkene ${\mathrm{C}}_5{\mathrm{H}}_{10}$ gives propanone and ethanal. Joining the two carbonyl carbons with a double bond gives $(\mathrm{C}{\mathrm{H}}_3{)}_2\mathrm{C}=\mathrm{CHC}{\mathrm{H}}_3$, which is 2-methylbut-2-ene.

An everyday example. Nail polish remover sold at cosmetics counters across India is largely propanone, which industry produces alongside phenol from cumene.

The substance. Acidified dichromate usually overshoots a primary alcohol to the carboxylic acid — to stop at the aldehyde, use a milder reagent such as PCC or distil the aldehyde out as it forms.

Aldehydes are easily oxidised, so they reduce Tollens' reagent to a silver mirror and Fehling's solution to a red precipitate, while ordinary ketones give neither; the iodoform and 2,4-DNP tests help classify carbonyl compounds further.

Why aldehydes respond. An aldehyde has a hydrogen on its carbonyl carbon, so it is readily oxidised to a carboxylic acid; a ketone would need a carbon-carbon bond to break.

Tollens' test — ammoniacal silver nitrate:

$$\mathrm{RCHO}+2[\mathrm{Ag}(\mathrm{N}{\mathrm{H}}_3{)}_2{]}^{+}+3O{\mathrm{H}}^{-}\to \mathrm{RCO}{\mathrm{O}}^{-}+2Ag+2{\mathrm{H}}_2\mathrm{O}+4N{\mathrm{H}}_3$$

All aldehydes, aliphatic and aromatic, give a silver mirror; ketones do not.

Fehling's test — deep blue alkaline copper(II) tartrate:

$$\mathrm{RCHO}+2C{\mathrm{u}}^{2+}+5O{\mathrm{H}}^{-}\to \mathrm{RCO}{\mathrm{O}}^{-}+\mathrm{C}{\mathrm{u}}_2\mathrm{O}+3{\mathrm{H}}_2\mathrm{O}$$

Aliphatic aldehydes give a red-brown precipitate of copper(I) oxide, but aromatic aldehydes such as benzaldehyde and ketones do not.

Other tests:

- Iodoform test — ethanal and methyl ketones, with a $\mathrm{C}{\mathrm{H}}_3\mathrm{CO}-$ group, give a yellow precipitate, which separates propanone from pentan-3-one
- 2,4-Dinitrophenylhydrazine — gives an orange-red precipitate with both aldehydes and ketones, confirming a carbonyl group
- Schiff's reagent — regains its pink colour with aldehydes

Worked example — ethanal, propanone and benzaldehyde:

- Fehling's: only ethanal gives a red precipitate
- Tollens': ethanal and benzaldehyde give silver mirrors; propanone does not
- Iodoform: ethanal and propanone give yellow precipitates; benzaldehyde does not

An everyday example. Benedict's solution, a gentler relative of Fehling's, is used in school biology practicals to detect reducing sugars such as glucose, whose open-chain form has an aldehyde group.

The substance. Fructose is a ketone yet reduces Tollens' and Fehling's reagents — in alkaline solution it rearranges into a form containing an aldehyde group.