Reduction, oxidation and the tests for aldehydes and ketones

Reduction to alcohols

Adding hydrogen to C=O gives an alcohol. An aldehyde gives a primary alcohol and a ketone gives a secondary alcohol.

• Sodium borohydride, NaBH$_4$, in ethanol or water. It is mild and reduces aldehydes and ketones.
• Lithium aluminium hydride, LiAlH$_4$, in dry ether, then dilute acid. It is stronger and reduces more groups. It reacts with water, so the solvent must be dry.
• Hydrogen over nickel, platinum or palladium. It also reduces a C=C bond in the same molecule.

$\text{CH}_{3}\text{COCH}_{2}\text{CH}_{3} \xrightarrow{1.\ \text{NaBH}_{4}\ 2.\ \text{H}_{3}\text{O}+} \text{CH}_{3}\text{CH}(\text{OH})\text{CH}_{2}\text{CH}_{3}$

Both hydrides work the same way. The hydride ion, $\text{H}^-$, is the nucleophile that adds to the carbon.

Reduction to hydrocarbons

The C=O group can be turned into $\text{CH}_{2}$. There are two classical methods. They suit different molecules.

1. Clemmensen reduction. Zinc amalgam, $\text{Zn-Hg}$, and concentrated hydrochloric acid, heated. It needs a molecule that survives strong acid.
2. Wolff-Kishner reduction. Hydrazine, $\text{NH}_2\text{NH}_2$, makes a hydrazone. Heating with potassium hydroxide in ethylene glycol, at about $473\ \text{K}$, gives the alkane and nitrogen gas. It needs a molecule that survives strong base.

$\text{C}_{6}\text{H}_{5}\text{COCH}_{3} \xrightarrow{\text{Zn}\text{-}\text{Hg}\ \text{conc.}\ \text{HCl}} \text{C}_{6}\text{H}_{5}\text{CH}_{2}\text{CH}_{3}$

The choice is made by what else is in the molecule. An acid-sensitive group needs Wolff-Kishner, and a base-sensitive group needs Clemmensen.

Oxidation: why an aldehyde is not a ketone

An aldehyde carries a hydrogen on the carbonyl carbon. Replacing this hydrogen with an $\text{OH}$ group gives a carboxylic acid. So an aldehyde is oxidised very easily, even by mild oxidising agents. A ketone has no such hydrogen. Oxidising it must break a carbon-carbon bond, and this needs strong conditions, such as hot concentrated nitric acid or potassium permanganate. The result is a mixture of acids with fewer carbons. Cyclohexanone gives hexanedioic acid, which is used to make nylon-6,6.

This difference gives us tests.

Tollens' test. Tollens' reagent is ammoniacal silver nitrate, which has the complex ion $[\text{Ag}(\text{NH}_{3})_{2}]^{+}$ in alkali. Warm it with an aldehyde, and silver forms as a silver mirror.

$\text{RCHO} + _{2}[\text{Ag}(\text{NH}_{3})_{2}]^{+} + _{3}\text{OH}^{-} \longrightarrow \text{RCOO}^{-} + _{2}\text{Ag} + _{4}\text{NH}_{3} + _{2}\text{H}_{2}\text{O}$

Fehling's test. Fehling's solution has copper(II) ions held in alkaline tartrate. Heat it with an aliphatic aldehyde, and a red-brown precipitate of copper(I) oxide forms.

$\text{RCHO} + _{2}\text{Cu}^{2+} + _{5}\text{OH}^{-} \longrightarrow \text{RCOO}^{-} + \text{Cu}_{2}\text{O} + _{3}\text{H}_{2}\text{O}$

Aromatic aldehydes, such as benzaldehyde, do not give Fehling's test, but they do give Tollens' test. Ketones give neither.

Iodoform test. Warm the compound with iodine and sodium hydroxide, which gives sodium hypoiodite. A yellow precipitate of iodoform, $\text{CHI}_3$, and a characteristic antiseptic smell show a positive result. The test is given by compounds that have the group $\text{CH}_{3}\text{CO}\text{-}$ attached to carbon or hydrogen, and by alcohols with $\text{CH}_{3}\text{CH}(\text{OH})\text{-}$, which are first oxidised.

$\text{CH}_{3}\text{COCH}_{3} + _{3}\text{I}_{2} + _{4}\text{NaOH} \longrightarrow \text{CHI}_{3} + \text{CH}_{3}\text{COONa} + _{3}\text{NaI} + _{3}\text{H}_{2}\text{O}$

Ethanal is the only aldehyde that gives it. Ethanol is the only primary alcohol that gives it. Propanone, butanone, pentan-2-one, propan-2-ol and acetophenone also give it.

A test with 2,4-dinitrophenylhydrazine is positive for every aldehyde and ketone. So it separates this family from alcohols and acids, and the tests above divide the family.

Set methanal and $0.5\ \text{g}$. It gives four silver atoms for each molecule, so more than twice as much silver as ethanal does from the same mass, even though methanal is lighter. Then try benzaldehyde.

Practise with new numbers

Try it in Python

The program uses the stoichiometry of the Tollens test. For a given mass of each aldehyde it finds the silver that forms.

Common mistakes

• Expecting Fehling's test from benzaldehyde. Only aliphatic aldehydes give it.
• Giving ketones the Tollens' test. Ketones are not oxidised by these mild reagents.
• Writing the iodoform test for every aldehyde. Only ethanal gives it.
• Using Clemmensen on an acid-sensitive compound. Use Wolff-Kishner.
• Forgetting that NaBH$_4$ and LiAlH$_4$ add a hydride, not hydrogen gas. The acid step gives the O-H.

Remember these

• Aldehyde to primary alcohol, ketone to secondary alcohol: NaBH$_4$, LiAlH$_4$, or H$_2$ and a metal.
• Clemmensen: Zn-Hg and conc. HCl. Wolff-Kishner: NH$_2$NH$_2$, then KOH in ethylene glycol.
• Tollens: silver mirror, aldehydes. Fehling: red Cu$_2$O, aliphatic aldehydes. Iodoform: yellow CHI$_3$, methyl ketones, ethanal, ethanol.
• Ketones are oxidised only by strong oxidants, and the carbon chain breaks.

Check yourself

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

1. Which test separates benzaldehyde from ethanal: Tollens' or Fehling's?

2. Which of pentan-2-one and pentan-3-one gives the iodoform test?

3. What does Wolff-Kishner reduction turn acetophenone into?

4. Why is a ketone harder to oxidise than an aldehyde?

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 sections on reduction, oxidation and the tests.