Clemmensen and Wolff-Kishner Reduction: C=O to CH2, Mechanism and Exam Traps

ChemistryClass 12, Aldehydes, Ketones and Carboxylic AcidsFor JEE and NEET aspirants

The Clemmensen and Wolff-Kishner reductions both turn the C=O group of an aldehyde or ketone into a $C H_{2}$ group, so a simple aldehyde or ketone becomes an alkane. Clemmensen reduction uses zinc amalgam and concentrated hydrochloric acid; Wolff-Kishner reduction first makes the hydrazone with hydrazine and then heats it with KOH or NaOH in ethylene glycol. One works in acid and the other in base, so the rest of the molecule decides which to use. Both are in the Class 12 NCERT chapter on aldehydes, ketones and carboxylic acids, and JEE Main, JEE Advanced and NEET test them in reaction sequences.

At a glance

Clemmensen:
$C H_{3} COC H_{3} Zn-Hg conc. HCl C H_{3} C H_{2} C H_{3}$
Wolff-Kishner, step 1:
$C H_{3} COC H_{3} + N H_{2} N H_{2} (C H_{3})_{2} C = NN H_{2} + H_{2} O$
Wolff-Kishner, step 2:
$(C H_{3})_{2} C = NN H_{2} KOH, ethylene glycol Δ (453-473 K) C H_{3} C H_{2} C H_{3} + N_{2} ↑$
Change: ketone C=O becomes $C H_{2}$ ; aldehyde $CHO$ becomes $C H_{3}$ .
Choose: acid-sensitive molecule, use Wolff-Kishner; base-sensitive molecule, use Clemmensen.
Left alone: the benzene ring and $COOH$ by both ( $COOH$ ends as its salt in Wolff-Kishner); esters by Clemmensen.

1. The two reductions side by side

	Clemmensen	Wolff-Kishner
Reagent	Zinc amalgam (Zn-Hg) + conc. HCl, heated	$N H_{2} N H_{2}$ , then KOH or NaOH in ethylene glycol
Medium	Strongly acidic	Strongly basic, about 453-473 K
By-product	$H_{2} O$ (Zn goes to $ZnC l_{2}$ )	$N_{2}$ gas and $H_{2} O$
Use it for	Molecules stable to hot acid, such as aryl alkyl ketones from Friedel-Crafts acylation	Molecules that acid would damage, such as acetals or 3° alcohols
Avoid it for	Acid-sensitive groups	Base-sensitive groups: esters hydrolyse, alkyl halides substitute or eliminate

Ordinary reducing agents stop earlier: $NaB H_{4}$ , $LiAl H_{4}$ and catalytic hydrogenation turn an aldehyde or ketone into an alcohol, not into a $C H_{2}$ group. In the syllabus these two are the standard ways to remove the oxygen completely; HI with red phosphorus, and the thioacetal route with Raney nickel, also do it.

2. Mechanism

2.1 Wolff-Kishner

Hydrazine first condenses with the carbonyl group, like any ammonia derivative, giving the hydrazone $R_{2} C = NN H_{2}$ and water. Hot alkali removes the two N–H protons one at a time, the carbon picks up two protons from the solvent, and nitrogen leaves as $N_{2}$ gas.

Figure 1: Wolff-Kishner reduction after the hydrazone has formed. Twice, hydroxide removes an N–H proton and the carbon picks up a proton from water; the escape of

N_{2}

gas makes the change one-way.

Hydrazone. $N H_{2} N H_{2}$ adds to C=O and loses water: $R_{2} C = O$ becomes $R_{2} C = N - N H_{2}$ .
First proton shift. $O H^{-}$ removes an N–H proton; the electrons form N=N and the C=N electrons move onto carbon, which takes $H^{+}$ from water: $R_{2} CH - N = NH$ .
Loss of nitrogen. $O H^{-}$ removes the second N–H proton; $N_{2}$ leaves and the carbanion $R_{2} C H^{-}$ takes $H^{+}$ from water: $R_{2} C H_{2}$ .

2.2 Clemmensen

The Clemmensen reduction happens on the surface of the zinc, and its exact path is not settled. One fact matters for exams: the alcohol is not an intermediate, because alcohols are not reduced to hydrocarbons under these conditions. So a ketone goes to $C H_{2}$ directly, while an alcohol already in the molecule is not reduced (although hot HCl can still replace a tertiary OH by Cl).

Exam Trick

"Clemmensen in acid, Kishner in kali (alkali)." Look at the rest of the molecule first: if hot HCl would damage it, choose Wolff-Kishner; if hot KOH would damage it, choose Clemmensen.

3. Where they are used

3.1 Straight-chain alkylbenzenes

Friedel-Crafts alkylation with 1-chloropropane gives mostly isopropylbenzene, because the n-propyl electrophile rearranges by a 1,2-hydride shift to the 2° (isopropyl) carbocation. To make n-propylbenzene, acylate first (the acylium ion does not rearrange) and then reduce the C=O:

C_{6} H_{6} + C H_{3} C H_{2} COCl anhyd. AlC l_{3} C_{6} H_{5} COC H_{2} C H_{3} + HCl

C_{6} H_{5} COC H_{2} C H_{3} Zn-Hg conc. HCl C_{6} H_{5} C H_{2} C H_{2} C H_{3}

3.2 Alkanes and cycloalkanes

A simple aliphatic aldehyde or ketone gives the alkane with the same carbon skeleton: ethanal gives ethane, propanone gives propane and cyclohexanone gives cyclohexane.

JEE Advanced

Keto esters. With Zn-Hg and HCl the ketone of a keto ester is reduced and the ester survives. Under Wolff-Kishner conditions the hot alkali also hydrolyses the ester, so the product is the carboxylate salt (the acid after acidification).

Huang-Minlon modification. In practice the carbonyl compound, hydrazine hydrate and KOH are heated together in diethylene glycol in one pot; water is distilled off so the temperature can rise to about 470 K.

Hydrazones are not the end. Hydrazine, phenylhydrazine, 2,4-dinitrophenylhydrazine (2,4-DNP) and semicarbazide all give C=N derivatives of a carbonyl group. Forming any of these is not a reduction; only the hydrazone, heated with strong base, loses $N_{2}$ and becomes $C H_{2}$ .

4. Worked examples

Solved Example 1

Propanone and ethanal are separately treated with Zn-Hg and concentrated HCl. Name the products.

Solution:

Both are reduced, C=O to $C H_{2}$ : propanone gives propane, $C H_{3} C H_{2} C H_{3}$ , and ethanal gives ethane, $C H_{3} C H_{3}$ (its $CHO$ becomes $C H_{3}$ ).

Solved Example 2

The C=O group of 5-hydroxy-5-methylhexan-2-one,

C H_{3} COC H_{2} C H_{2} C (C H_{3})_{2} OH

, must become

C H_{2}

while its OH group stays. Which reduction should be used, and what is the product?

Solution:

Wolff-Kishner. Hot concentrated HCl would turn the tertiary alcohol into a chloride or an alkene. In base the OH is safe and only the C=O is reduced: the product is 2-methylhexan-2-ol, $C H_{3} C H_{2} C H_{2} C H_{2} C (C H_{3})_{2} OH$ .

Solved Example 3

Ethyl 4-oxopentanoate,

C H_{3} COC H_{2} C H_{2} COO C_{2} H_{5}

, is to be converted into ethyl pentanoate. Which reagent is better?

Solution:

Zn-Hg and HCl (Clemmensen). It reduces the ketone to $C H_{2}$ and leaves the ester: $C H_{3} C H_{2} C H_{2} C H_{2} COO C_{2} H_{5}$ . Under Wolff-Kishner conditions hot KOH would hydrolyse the ester, and hydrazine would also attack it (here the ester and the ketone would close into a cyclic hydrazide).

5. Exam traps

Watch out

The C=O becomes a $C H_{2}$ (methylene) group, not CH–OH. $NaB H_{4}$ and $LiAl H_{4}$ stop at the alcohol.
An aldehyde's $CHO$ becomes $C H_{3}$ : ethanal gives ethane, not methane.
Acid-sensitive molecule: Wolff-Kishner. Base-sensitive molecule (for example an ester): Clemmensen.
Zn-Hg and HCl also reduce an aromatic $N O_{2}$ group to $N H_{2}$ (metal and acid), so a nitro aryl ketone gives the amino alkylbenzene.
Neither reduces the benzene ring or a $COOH$ group; Clemmensen also leaves an ester.
Hydrazine alone stops at the hydrazone; the reduction needs strong base and heat.
2,4-Dinitrophenylhydrazine and semicarbazide make derivatives for tests; they do not reduce anything.
Friedel-Crafts acylation followed by Clemmensen reduction gives a straight-chain alkylbenzene without rearrangement.

Past-paper questions on Clemmensen and Wolff-Kishner reduction

Our past papers: JEE Main 2025-2026, JEE Advanced 2022-2026, NEET 2018-2019, 2022-2026. We found 8 questions in them that test this:

JEE Main 4 (2025-2026)
JEE Advanced 3 (2023-2024, 2026)
NEET 1 (2023)

Each question was read and checked by hand (list checked 2026-10-14). A question that shows its structures only as pictures can be missed by our search. Each opens in a new tab, with the question to try and our step-by-step solution.

Learn it with the chapter

Related named reactions

Friedel-Crafts reactionAn alkyl or acyl group is put on a benzene ring with anhydrous $AlC l_{3}$ ; rearrangement, polyalkylation and the rings that do not react.

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