Reduction and oxidation reactions

Every carbonyl compound sits at a fixed rung on an oxidation-level ladder, and the whole of this topic is about moving up or down that ladder with the right reagent. Push an aldehyde or ketone down one rung and you get an alcohol; push it down three rungs and you strip the oxygen out entirely, landing on an alkane; push an aldehyde up one rung and you get a carboxylic acid. Ketones refuse to go up at all under ordinary conditions, and that single refusal is the basis of the classic aldehyde-versus-ketone test. Three named methods compete for the alkane route — Clemmensen, Wolff–Kishner, and red phosphorus with HI — and picking the right one for a given molecule is the examiner's favourite twist. Layered on top is the iodoform reaction, a reagent that both oxidises and cleaves, giving a bright yellow precipitate that fingerprints one specific structural fragment. This article organises all five transformations around the oxidation-level ladder so the reagent choices stop looking like a memorised list and start looking like consequences of one idea.

The core insight

Carbonyl compounds occupy a middle oxidation level between alcohols and carboxylic acids, so every named reaction here is a reagent choice for moving one, two, or three rungs up or down that ladder while respecting whatever else is on the molecule.

The core idea

Read the ladder first. Carbon oxidation state rises from alkane (lowest) through alcohol, to aldehyde/ketone, to carboxylic acid (highest). Reduction moves down the ladder, oxidation moves up. Every reagent in this article is answering the question "how many rungs, in which direction, and can this substrate tolerate the conditions."

One rung down — carbonyl to alcohol. Hydride reducing agents deliver a hydride ion to the electrophilic carbonyl carbon; aqueous or alcoholic work-up then protonates the resulting alkoxide.

\text{RCHO} \xrightarrow{\text{NaBH}_4 \text{ or LiAlH}_4 \text{ or H}_2/\text{Ni}} \text{RCH}_2\text{OH}

\text{RCOR}' \xrightarrow{\text{NaBH}_4 \text{ or LiAlH}_4 \text{ or H}_2/\text{Ni}} \text{RCH(OH)R}'

where: R and R' are hydrocarbon groups (or H for the parent aldehyde); an aldehyde gives a primary alcohol, a ketone gives a secondary alcohol, since the carbonyl carbon already carries one or two carbon substituents respectively.

Three rungs down — carbonyl to alkane. This is a complete deoxygenation, $\text{C=O} \rightarrow \text{CH}_2$ , and three named routes exist because no single reagent set survives every kind of substrate.

\text{R–CO–R}' \xrightarrow{Zn(Hg)/\text{conc. HCl}} \text{R–CH}_2\text{–R}' \quad \text{(Clemmensen)}

\text{R–CO–R}' \xrightarrow{NH_2NH_2,\ KOH/\text{ethylene glycol},\ \Delta} \text{R–CH}_2\text{–R}' + N_2 + H_2O \quad \text{(Wolff–Kishner)}

\text{R–CO–R}' \xrightarrow{\text{Red P} + HI,\ \Delta} \text{R–CH}_2\text{–R}' \quad \text{(Red P + HI)}

where: Clemmensen runs in strong acid so it fails on acid-labile groups; Wolff–Kishner runs in strong base at high temperature via a hydrazone intermediate that loses $N_2$ , so it fails on base-labile groups; red phosphorus with HI is a strongly reducing, direct route reserved for substrates that tolerate neither of the other two extremes.

One rung up (aldehydes only) — carbonyl to acid. Mild oxidants add an oxygen to the aldehyde's remaining C–H bond; ketones have no such hydrogen on the carbonyl carbon, so they resist under the same conditions.

\text{RCHO} \xrightarrow{\text{Tollens' reagent, Fehling's solution, or mild }K_2Cr_2O_7} \text{RCOOH}

where: Tollens' reagent is $\text{Ag(NH}_3)_2^+$ in ammoniacal solution, reduced to a silver mirror; Fehling's solution is a $\text{Cu}^{2+}$ –tartrate complex, reduced to a brick-red $\text{Cu}_2\text{O}$ precipitate. Both are simultaneously oxidants for aldehydes and distinguishing tests against ketones, which give no visible change.

A combined test — the iodoform reaction. Specific to the $\text{CH}_3\text{CO–}$ fragment (methyl ketones, acetaldehyde, and alcohols oxidisable to these), it both oxidises and cleaves the carbon chain, ending in a yellow precipitate.

\text{CH}_3\text{COR} + 3I_2 + 4NaOH \rightarrow CHI_3\downarrow + RCOONa + 3NaI + 3H_2O

where: the three methyl hydrogens are progressively iodinated and then cleaved off as $\text{CHI}_3$ (iodoform, yellow, characteristic smell) by base, leaving the rest of the chain as a carboxylate salt.

Figure

The carbon oxidation ladder from alkane through alcohol and carbonyl to carboxylic acid with reduction arrows pointing down and the single oxidation arrow pointing up only from aldehyde

Why it works

The choice between Clemmensen, Wolff–Kishner, and red phosphorus with HI is not arbitrary — it follows from what functional groups the rest of the molecule can survive.

Step 1. Clemmensen reduction runs in concentrated hydrochloric acid on a zinc–mercury amalgam surface. The acidified metal surface delivers hydrogen in a stepwise fashion to the activated carbonyl carbon, ultimately replacing $\text{C=O}$ with $\text{CH}_2$ and releasing water. Because the medium is strongly acidic, any acid-labile group elsewhere in the molecule (for example a group that hydrolyses or rearranges under strong acid) would be destroyed, so Clemmensen is reserved for acid-stable substrates.

Step 2. Wolff–Kishner reduction takes the opposite route to the same destination. The carbonyl first condenses with hydrazine to form a hydrazone; strong base (KOH) at high temperature in a high-boiling solvent like ethylene glycol then decomposes the hydrazone, expelling nitrogen gas and leaving $\text{CH}_2$ in place of $\text{C=O}$ . Since the medium here is strongly basic rather than acidic, this route is chosen precisely when the substrate carries acid-sensitive groups that Clemmensen would destroy — the two methods are complementary, not interchangeable.

Step 3. Red phosphorus with HI is a third, more forceful option: a direct, strongly reducing combination that achieves the same net deoxygenation without the specific acid or base sensitivity of the other two, though it is less commonly the first choice on the syllabus.

Step 4. The oxidation side follows the same substrate logic in reverse. An aldehyde carbon still carries one hydrogen bonded directly to the carbonyl carbon; mild oxidants can pull that hydrogen off along with adding oxygen, giving the acid. A ketone carbon has no such hydrogen — both remaining bonds go to carbon — so mild oxidants like Tollens' and Fehling's simply have nothing to abstract, and no reaction is observed. This structural difference, not reagent strength, is why the aldehyde/ketone oxidation test works as a distinguishing test at all.

Worked example

Problem: Given cyclohexanone, describe how you would convert it to (a) cyclohexanol and (b) cyclohexane, choosing the appropriate reagents, and (c) explain why cyclohexanone itself gives a negative Tollens' test.

Step 1. For (a), only one rung of reduction is needed, $\text{C=O} \rightarrow \text{C–OH}$ , and cyclohexanone carries no other sensitive functional group, so any hydride reagent works cleanly:

\text{cyclohexanone} \xrightarrow{NaBH_4,\ \text{MeOH}} \text{cyclohexanol}

Sodium borohydride selectively reduces the ketone carbonyl to the secondary alcohol.

Step 2. For (b), a full deoxygenation to the alkane level is needed. Cyclohexanone has no acid- or base-sensitive groups elsewhere, so either Clemmensen or Wolff–Kishner works; taking Clemmensen:

\text{cyclohexanone} \xrightarrow{Zn(Hg)/\text{conc. HCl},\ \Delta} \text{cyclohexane}

The amalgam surface delivers hydrogen stepwise under the strongly acidic medium, removing the oxygen entirely as water.

Step 3. For (c), cyclohexanone is a ketone: its carbonyl carbon is bonded to two ring carbons and has no hydrogen attached directly to it. Tollens' reagent oxidises by abstracting exactly that kind of carbonyl-adjacent hydrogen, which simply is not present, so cyclohexanone gives no silver mirror — a negative Tollens' test, consistent with it being a ketone, not an aldehyde.

Competitive stretch

If instead the substrate were 4-hydroxycyclohexanone (a ketone bearing an acid-labile hydroxyl elsewhere in the ring), converting it to the fully deoxygenated hydrocarbon would call for Wolff–Kishner, not Clemmensen — the strong HCl of Clemmensen risks side reactions on the hydroxyl-bearing carbon, whereas the basic hydrazine/KOH route leaves it untouched.

Common traps

Using Clemmensen (strongly acidic) on an acid-sensitive substrate, or Wolff–Kishner (strongly basic, high temperature) on a base-sensitive one — the correct method depends on what else is in the molecule, not on personal preference.
Believing the iodoform test is positive for any methyl-substituted ketone — it is specific to the $\text{CH}_3\text{CO–}$ fragment (methyl ketones and acetaldehyde only, plus alcohols oxidisable to these), not to methyl groups located elsewhere on the chain.
Forgetting formaldehyde is the exception among aldehydes: Tollens'/Fehling's positive (it is an aldehyde) but iodoform negative, because it has no methyl group attached to the carbonyl carbon.
Assuming ketones oxidise as readily as aldehydes under mild oxidants — under Tollens' or Fehling's conditions ketones give no reaction at all, which is exactly the basis of the standard distinguishing test.

Ask the tutor: unsure which reduction method — Clemmensen, Wolff–Kishner, or red phosphorus with HI — fits a molecule carrying other sensitive groups? Ask the AI tutor to walk through the choice with you.

Why does the same iodoform reaction both oxidise and cleave the molecule?

The alkaline iodine first replaces all three hydrogens of the methyl group with iodine through repeated substitution, forming a $\text{CI}_3$ group next to the carbonyl. That trihalomethyl ketone is then attacked by hydroxide at the carbonyl carbon and cleaves, expelling the stable $\text{CHI}_3$ (iodoform) anion and leaving the rest of the chain as a carboxylate salt. Both steps happen in the same pot, which is why one set of conditions gives a substitution and a cleavage together.

If I have a ketone with an acid-sensitive ester group elsewhere, can I still fully deoxygenate it to the alkane?

Yes, but not with Clemmensen — concentrated HCl would attack the ester. Choose Wolff–Kishner instead: it forms the hydrazone with hydrazine and then decomposes it under basic, non-aqueous-acid conditions (KOH, ethylene glycol, heat), leaving an acid-sensitive ester elsewhere on the molecule untouched. If the molecule instead carried a base-sensitive group, you would flip the choice back to Clemmensen or fall back on red phosphorus with HI.

Why doesn't the iodoform test work on ethanol directly, only after it is oxidised?

It does work on ethanol, but only because ethanol is first oxidised in situ by the same alkaline iodine solution to acetaldehyde, which does carry the required $\text{CH}_3\text{CO–}$ fragment; the iodoform chemistry then proceeds exactly as it would on acetaldehyde itself. Alcohols that cannot be oxidised to a methyl ketone or to acetaldehyde — for instance a primary alcohol other than ethanol, which oxidises to an aldehyde without a methyl group directly on the carbonyl carbon — give a negative test even though they are alcohols, because the necessary fragment is never generated.