JEE Mains Chemistry · Aldehydes, Ketones and Carboxylic Acids

Reductions: Clemmensen, Wolff-Kishner and Hydrides

Clemmensen (in acid) and Wolff–Kishner (in base) turn C=O into CH₂; LiAlH₄, NaBH₄ and DIBAL-H reduce each group to a different level, or leave it alone.

All Aldehydes, Ketones and Carboxylic Acids notes JEE Mains Chemistry strategy

Why this matters

Twenty-one PYQs, none numerical, two from 2026. Thirteen turn a C=O into CH₂ by Clemmensen or Wolff–Kishner and ask which other groups survive; eight ask how far LiAlH₄, NaBH₄ or DIBAL-H reduces each group in a molecule.

Concept 1 of 2: Clemmensen and Wolff–Kishner reductions

Both methods remove the oxygen of an aldehyde or ketone completely and leave

\mathrm{CH_2}

. They differ in the medium: Clemmensen is strongly acidic, Wolff–Kishner strongly basic and hot. Choose the one that the rest of the molecule can survive.

Definition

Clemmensen: zinc amalgam and concentrated HCl. $\mathrm{{>}C{=}O \to {>}CH_2}$ .
Wolff–Kishner: hydrazine forms the hydrazone, and heating with KOH in ethylene glycol (about 470 K) drives off $\mathrm{N_2}$ , leaving $\mathrm{CH_2}$ .
Neither method stops at the alcohol, and neither touches an isolated C=C or a COOH group.
A molecule that cannot stand acid (a 3° or benzylic alcohol, which dehydrates) needs Wolff–Kishner. A molecule that cannot stand hot base (a C–Cl bond, which is substituted or eliminated) needs Clemmensen.
Friedel–Crafts acylation followed by Clemmensen puts a straight alkyl chain on a ring. Direct alkylation with a 1° halide would rearrange.

Feature of the substrate	Clemmensen: Zn-Hg, conc. HCl	Wolff–Kishner: NH₂NH₂, KOH, glycol, heat
Aldehyde or ketone C=O	Reduced to $\mathrm{CH_2}$	Reduced to $\mathrm{CH_2}$ , with loss of $\mathrm{N_2}$
Medium	Strongly acidic, aqueous	Strongly basic, about 470 K
Isolated C=C	Unchanged	Unchanged
COOH group	Unchanged	Unchanged (present as the carboxylate until acidified)
3° or benzylic OH	Dehydrated; avoid this method	Unchanged; use this method
C–Cl bond in the chain	Survives the acid; use this method	Substituted or eliminated by hot base; avoid this method
Amide $\mathrm{CONH_2}$	Hydrolysed to COOH by the hot acid	Hydrolysed to the carboxylate by the hot base

Same result on the C=O; the rest of the molecule decides the method.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 22 Jan 2026 Shift 1 · Q38Moderate

Example 1 · Aldehydes, Ketones and Carboxylic Acids · Reductions: Clemmensen, Wolff-Kishner and Hydrides

Match the LIST-I with LIST-II

	List-I Reagents		List-II Name of Reaction involving carbonyl compound
(A)	${NH}_{2} - {NH}_{2},\ KOH$	(I)	Tollen's Test
(B)	$Ag\left( {NH}_{3} \right)_{2}OH$	(II)	Clemmensen Reduction
(C)	Aq. ${CuSO}_{4}$ , sodium potassium tartrate, KOH	(III)	Wolff-Kishner Reduction
(D)	$Zn - Hg,\ HCl$	(IV)	Fehling's Test

Choose the correct answer from the options given below

Neither method stops at the alcohol

Clemmensen and Wolff–Kishner give

\mathrm{CH_2}

, not CHOH. An option showing the alcohol is the product of a hydride such as

\mathrm{NaBH_4}

, not of these reagents.

Choose the method by what else is in the molecule

A statement that a molecule 'can be reduced' by one of these methods is false if the medium destroys another group: acid dehydrates a 3° alcohol, hot base removes a chlorine.

Drill 12 more on clemmensen and wolff–kishner reductions

Concept 2 of 2: How far LiAlH₄, NaBH₄ and DIBAL-H reduce each group

The three hydride reagents differ in strength.

\mathrm{LiAlH_4}

reduces almost every polar multiple bond.

\mathrm{NaBH_4}

reduces only aldehydes and ketones. DIBAL-H at low temperature adds a single hydride, so esters and nitriles stop at the aldehyde.

Definition

$\mathrm{LiAlH_4}$ reduces aldehydes and ketones to alcohols, acids and esters to 1° alcohols, amides and nitriles to amines. It does not reduce an isolated C=C.
$\mathrm{NaBH_4}$ reduces aldehydes and ketones only. An ester, an acid, an amide or a lactam in the same molecule survives, so $\mathrm{NaBH_4}$ can reduce the ketone of a keto-ester alone.
DIBAL-H (diisobutylaluminium hydride) at about −78 °C adds one hydride to an ester, a lactone or a nitrile. The intermediate breaks down to the aldehyde only on work-up.
A lactone (cyclic ester) with DIBAL-H gives a hydroxy aldehyde, which may close to a cyclic hemiacetal (lactol).

Group	LiAlH₄, then H₃O⁺	NaBH₄	DIBAL-H at low temperature, then H₂O
Aldehyde $\mathrm{RCHO}$	$\mathrm{RCH_2OH}$	$\mathrm{RCH_2OH}$	$\mathrm{RCH_2OH}$
Ketone $\mathrm{RCOR'}$	$\mathrm{RCH(OH)R'}$	$\mathrm{RCH(OH)R'}$	$\mathrm{RCH(OH)R'}$
Ester $\mathrm{RCOOR'}$	$\mathrm{RCH_2OH + R'OH}$	No reaction	$\mathrm{RCHO + R'OH}$
Lactone (cyclic ester)	Diol	No reaction	Hydroxy aldehyde (or its lactol)
Nitrile $\mathrm{RC{\equiv}N}$	$\mathrm{RCH_2NH_2}$	No reaction	$\mathrm{RCHO}$
Isolated C=C	Unchanged	Unchanged	Unchanged

NaBH₄ is the selective one; DIBAL-H is the one that stops at the aldehyde.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 1 February 2024 · Q48Moderate

Example 2 · Aldehydes, Ketones and Carboxylic Acids · Reductions: Clemmensen, Wolff-Kishner and Hydrides

Match List - I with List - II.

	List - I (Reactions)		List - II (Reagents)
(A)	$CH_{3}(CH_{2})_{5}COOC_{2}H_{5} \rightarrow CH_{3}(CH_{2})_{5}CHO$	(I)	$CH_{3}MgBr,\ H_{2}O$
(B)	$C_{6}H_{5}COC_{6}H_{5} \rightarrow C_{6}H_{5}CH_{2}C_{6}H_{5}$	(II)	$Zn(Hg)$ and conc. $HCl$
(C)	$C_{6}H_{5}CHO \rightarrow C_{6}H_{5}CH(OH)CH_{3}$	(III)	$NaBH_{4},\ H^{+}$
(D)	$CH_{3}COCH_{2}COOC_{2}H_{5} \rightarrow CH_{3}CH(OH)CH_{2}COOC_{2}H_{5}$	(IV)	DIBAL-H, $H_{2}O$

Choose the correct answer from options given below:

NaBH₄ leaves esters, acids and amides alone

In a molecule with a ketone and an ester,

\mathrm{NaBH_4}

reduces the ketone only.

\mathrm{LiAlH_4}

would reduce both.

DIBAL-H must be cold

DIBAL-H stops at the aldehyde only at low temperature with one equivalent. Warm, or in excess, it reduces the ester on to the alcohol.

Drill 7 more on how far lialh₄, nabh₄ and dibal-h reduce each group

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Reference tables (2)

Clemmensen and Wolff–Kishner reductions7 rows

Feature of the substrate	Clemmensen: Zn-Hg, conc. HCl	Wolff–Kishner: NH₂NH₂, KOH, glycol, heat
Aldehyde or ketone C=O	Reduced to $\mathrm{CH_2}$	Reduced to $\mathrm{CH_2}$ , with loss of $\mathrm{N_2}$
Medium	Strongly acidic, aqueous	Strongly basic, about 470 K
Isolated C=C	Unchanged	Unchanged
COOH group	Unchanged	Unchanged (present as the carboxylate until acidified)
3° or benzylic OH	Dehydrated; avoid this method	Unchanged; use this method
C–Cl bond in the chain	Survives the acid; use this method	Substituted or eliminated by hot base; avoid this method
Amide $\mathrm{CONH_2}$	Hydrolysed to COOH by the hot acid	Hydrolysed to the carboxylate by the hot base

Same result on the C=O; the rest of the molecule decides the method.

How far LiAlH₄, NaBH₄ and DIBAL-H reduce each group6 rows

Group	LiAlH₄, then H₃O⁺	NaBH₄	DIBAL-H at low temperature, then H₂O
Aldehyde $\mathrm{RCHO}$	$\mathrm{RCH_2OH}$	$\mathrm{RCH_2OH}$	$\mathrm{RCH_2OH}$
Ketone $\mathrm{RCOR'}$	$\mathrm{RCH(OH)R'}$	$\mathrm{RCH(OH)R'}$	$\mathrm{RCH(OH)R'}$
Ester $\mathrm{RCOOR'}$	$\mathrm{RCH_2OH + R'OH}$	No reaction	$\mathrm{RCHO + R'OH}$
Lactone (cyclic ester)	Diol	No reaction	Hydroxy aldehyde (or its lactol)
Nitrile $\mathrm{RC{\equiv}N}$	$\mathrm{RCH_2NH_2}$	No reaction	$\mathrm{RCHO}$
Isolated C=C	Unchanged	Unchanged	Unchanged

NaBH₄ is the selective one; DIBAL-H is the one that stops at the aldehyde.

Watch out for (4)

Neither method stops at the alcohol
→ Clemmensen and Wolff–Kishner reductions
Choose the method by what else is in the molecule
→ Clemmensen and Wolff–Kishner reductions
NaBH₄ leaves esters, acids and amides alone
→ How far LiAlH₄, NaBH₄ and DIBAL-H reduce each group
DIBAL-H must be cold
→ How far LiAlH₄, NaBH₄ and DIBAL-H reduce each group

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