Oxidation, Reduction & Distinguishing Tests

Oxidation of Aldehydes and Ketones

Aldehydes are easily oxidised to carboxylic acids — even by mild oxidising agents such as Tollens' reagent and Fehling's solution (and of course by strong oxidants like KMnO$_4$, K$_2$Cr$_2$O$_7$).

Ketones resist oxidation. They are oxidised only by strong oxidants under vigorous conditions, which break C-C bonds on either side of the carbonyl to give a mixture of carboxylic acids with fewer carbons.

This big difference — aldehydes oxidise easily, ketones do not — is the basis of the tests that distinguish aldehydes from ketones (Tollens and Fehling).

Key Point: aldehydes oxidise easily (even mild reagents); ketones resist oxidation (need strong oxidants, C-C cleavage). This distinguishes them.

Reduction of Aldehydes and Ketones

To alcohols (reduce C=O to CH-OH): use NaBH$_4$ or LiAlH$_4$ (or H$_2$/Ni). Aldehydes → primary alcohols, ketones → secondary alcohols.

To hydrocarbons (reduce C=O all the way to CH$_2$): two named reactions remove the oxygen completely:

• Clemmensen reduction: Zn-Hg (zinc amalgam) + concentrated HCl → CH$_2$. Used when the molecule is acid-stable.
• Wolff-Kishner reduction: NH$_2$-NH$_2$ (hydrazine), then KOH/heat (in ethylene glycol) → CH$_2$. Used when the molecule is base-stable.

Both convert >C=O into >CH$_2$ (e.g. acetophenone → ethylbenzene).

Key Point: NaBH$_4$/LiAlH$_4$ reduce C=O to alcohol; Clemmensen (Zn-Hg/HCl) and Wolff-Kishner (NH$_2$NH$_2$/KOH) reduce C=O all the way to CH$_2$.

Distinguishing Tests

Tollens' test (silver mirror): an aldehyde reduces Tollens' reagent [ammoniacal AgNO$_3$, Ag(NH$_3$)$_2$⁺] to a shiny silver mirror (metallic Ag). Ketones give no reaction.

Fehling's test: an aliphatic aldehyde reduces Fehling's solution (blue Cu$^{2+}$) to a red-brown precipitate of Cu$_2$O. Ketones and aromatic aldehydes (like benzaldehyde) do not respond.

2,4-DNP (Brady's) test: both aldehydes and ketones give an orange-yellow precipitate — confirms a carbonyl group (but does not distinguish aldehyde from ketone).

Iodoform test: compounds with a CH$_3$-CO- group (methyl ketones, ethanal) or a CH$_3$-CH(OH)- group give a yellow precipitate of iodoform (CHI$_3$) with I$_2$ / NaOH. Used to detect a methyl ketone or ethanol/acetaldehyde.

Key Point: Tollens (silver mirror) and Fehling (red Cu$_2$O) → aldehydes only; 2,4-DNP → any carbonyl; iodoform → CH$_3$CO- or CH$_3$CH(OH)- groups.

Solved Examples

Example 1: Distinguish aldehyde from ketone

How would you distinguish propanal from propanone?

Solution: Use Tollens' reagent: propanal gives a silver mirror; propanone gives no reaction. (Fehling's solution also works — propanal gives red Cu$_2$O, propanone does not.)

Example 2: Reduce to alcohol

What product forms when propanone is reduced with NaBH$_4$?

Solution: The C=O is reduced to CH-OH, giving the secondary alcohol propan-2-ol.

Example 3: Clemmensen reduction

What is formed when acetophenone undergoes Clemmensen reduction?

Solution: The C=O is reduced to CH$_2$, giving ethylbenzene (C$_6$H$_5$CH$_2$CH$_3$) (reagents: Zn-Hg / conc. HCl).

Example 4: Wolff-Kishner

Give the product and reagents for the Wolff-Kishner reduction of cyclohexanone.

Solution: Cyclohexanone → cyclohexane using hydrazine (NH$_2$NH$_2$) then KOH/heat (the C=O becomes CH$_2$).

Example 5: Iodoform-positive?

Which of ethanal, propanal and propanone give a positive iodoform test?

Solution: Ethanal (CH$_3$CHO) and propanone (CH$_3$COCH$_3$) have the CH$_3$CO- group → positive (yellow CHI$_3$). Propanal (CH$_3$CH$_2$CHO) has no CH$_3$CO- → negative.

Example 6: Fehling vs benzaldehyde

Does benzaldehyde respond to Fehling's test?

Solution: No. Fehling's solution oxidises only aliphatic aldehydes; aromatic aldehydes such as benzaldehyde do not give the red Cu$_2$O precipitate (though benzaldehyde does give a Tollens' silver mirror).

Example 7: 2,4-DNP scope

Two unknown liquids both give an orange precipitate with 2,4-DNP. What does this tell you?

Solution: Both contain a carbonyl group (aldehyde or ketone). The 2,4-DNP test confirms a carbonyl but cannot distinguish an aldehyde from a ketone — a further test (Tollens/Fehling) is needed.

Example 8: Choosing the reduction

You must convert a ketone to an alkane but the molecule contains an acid-sensitive group. Which reduction?

Solution: Use the Wolff-Kishner reduction (NH$_2$NH$_2$, then KOH/heat) — it works in basic conditions, avoiding the strong acid of Clemmensen.

Example 9: Identify by tests

An unknown gives a silver mirror with Tollens and a yellow precipitate with iodoform. What is it likely to be?

Solution: Silver mirror ⇒ an aldehyde; positive iodoform ⇒ a CH$_3$CO-/CH$_3$CH(OH)- unit. The aldehyde with a CH$_3$-CO/CH-OH pattern that fits both is ethanal (CH$_3$CHO) (it is an aldehyde and has the CH$_3$CO- arrangement).

Example 10: Oxidation of a ketone

Why do ketones require strong oxidising agents, and what kind of products form?

Solution: Ketones have no H on the carbonyl carbon, so oxidation must break a C-C bond. Only strong oxidants under vigorous conditions do this, giving a mixture of smaller carboxylic acids.