JEE Mains Chemistry · Aldehydes, Ketones and Carboxylic Acids

Preparation of Aldehydes and Ketones

Aldehydes and ketones are made by stopping an oxidation or a reduction at the carbonyl level, by named reactions such as Rosenmund, Stephen, Etard and Gattermann–Koch, and from alkynes, gem-dihalides and alkenes.

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

Why this matters

Twenty-five PYQs, one numerical, one from 2026, the largest page in the chapter. Eleven name a reaction or its reagents; eight ask how far a reagent takes an alcohol, an ester, a nitrile or an acid; six make a carbonyl compound from a hydrocarbon or one of its halides.

Concept 1 of 3: Named routes to aldehydes and ketones

Each named method starts from a different compound and uses a reagent that stops at the carbonyl group. Learn each one as a set of three: the starting compound, the reagent and the name.

Definition

Rosenmund reduction: an acyl chloride is hydrogenated over palladium on barium sulphate. The catalyst is partly poisoned (with sulphur or quinoline), so the aldehyde is not reduced further.
Stephen reduction: $\mathrm{SnCl_2}$ and HCl reduce a nitrile to an imine salt; hydrolysis then gives the aldehyde.
Etard reaction: chromyl chloride in $\mathrm{CS_2}$ oxidises the methyl group of toluene to a chromium complex, and water hydrolyses it to benzaldehyde.
Chromic oxide in acetic anhydride (273–283 K) traps the aldehyde as benzylidene diacetate, $\mathrm{C_6H_5CH(OCOCH_3)_2}$ , so it is not oxidised further; acid hydrolysis releases it.
Gattermann–Koch reaction: benzene, CO and HCl with anhydrous $\mathrm{AlCl_3}$ and CuCl give benzaldehyde.
Friedel–Crafts acylation gives aryl ketones and stops after one acyl group, because the ketone deactivates the ring.

Name	Starting compound	Reagents	Product
Rosenmund reduction	Acyl chloride $\mathrm{RCOCl}$	$\mathrm{H_2}$ , Pd– $\mathrm{BaSO_4}$ (poisoned)	Aldehyde $\mathrm{RCHO}$
Stephen reduction	Nitrile $\mathrm{RC{\equiv}N}$	$\mathrm{SnCl_2}$ , HCl, then $\mathrm{H_3O^+}$	Aldehyde $\mathrm{RCHO}$ , through the imine $\mathrm{RCH{=}NH}$
Etard reaction	Toluene	$\mathrm{CrO_2Cl_2}$ in $\mathrm{CS_2}$ , then $\mathrm{H_3O^+}$	Benzaldehyde, through $\mathrm{C_6H_5CH(OCrOHCl_2)_2}$
Chromic oxide oxidation	Toluene	$\mathrm{CrO_3}$ in $\mathrm{(CH_3CO)_2O}$ , 273–283 K, then $\mathrm{H_3O^+}$	Benzaldehyde, through benzylidene diacetate
Gattermann–Koch reaction	Benzene	CO, HCl, anhydrous $\mathrm{AlCl_3}$ and CuCl	Benzaldehyde
Friedel–Crafts acylation	Benzene	$\mathrm{RCOCl}$ , anhydrous $\mathrm{AlCl_3}$	Aryl ketone $\mathrm{C_6H_5COR}$ ; with $\mathrm{C_6H_5COCl}$ , benzophenone
Dialkylcadmium route	Acyl chloride $\mathrm{RCOCl}$	$\mathrm{R'_2Cd}$	Ketone $\mathrm{RCOR'}$

Rosenmund starts from an acyl chloride, Stephen from a nitrile, Etard from toluene and Gattermann–Koch from benzene.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 21 Jan 2026 Shift 2 · Q32Moderate

Example 1 · Aldehydes, Ketones and Carboxylic Acids · Preparation of Aldehydes and Ketones

Match List-I with List-II.

	List-I Reagents		List-II Reaction Name (Involving aldehydes)
(A)	$H_{2},\ Pd - {BaSO}_{4}$	(I)	Etard Reaction
(B)	${SnCl}_{2},\ HCl$	(II)	Rosenmund Reduction
(C)	${CrO}_{2}{Cl}_{2},\ {CS}_{2}$	(III)	Gattermann-Koch Reaction
(D)	$CO,\ HCl$ , anhyd. ${AlCl}_{3}$	(IV)	Stephen Reaction

Choose the correct answer from the options given below:

The Stephen reduction needs the water step

\mathrm{SnCl_2}

and HCl stop at the imine salt. Only hydrolysis with

\mathrm{H_3O^+}

turns it into the aldehyde, so a scheme without that step does not give RCHO yet.

Etard and Gattermann–Koch start from different rings

Etard oxidises a methyl group already on the ring (toluene). Gattermann–Koch adds a new CHO group to benzene itself. A match list often swaps these two.

Drill 10 more on named routes to aldehydes and ketones

Concept 2 of 3: Reagents that stop at the aldehyde or ketone

Many reagents pass through the aldehyde level, but most do not stop there. Oxidants in water go on to the acid; strong hydrides go on to the alcohol. The question is always which reagent halts at the carbonyl group.

Definition

PCC (pyridinium chlorochromate) in dichloromethane oxidises a 1° alcohol to the aldehyde and stops, because no water is present.
Chromic acid ( $\mathrm{CrO_3}$ – $\mathrm{H_2SO_4}$ , the Jones reagent), acidified dichromate and hot $\mathrm{KMnO_4}$ take a 1° alcohol on to the carboxylic acid. In water the aldehyde forms a hydrate, and the hydrate is oxidised again.
A 2° alcohol stops at the ketone with any of these oxidants; a 3° alcohol is not oxidised.
DIBAL-H at low temperature adds only one hydride to an ester or a nitrile, and work-up gives the aldehyde. $\mathrm{LiAlH_4}$ takes an ester down to two alcohols; $\mathrm{NaBH_4}$ does not reduce an ester at all.
Hydroboration–oxidation of a terminal alkene gives the 1° alcohol, and PCC then gives the aldehyde, with oxygen on the end carbon.

Reagent	Acts on	Stops at
PCC in $\mathrm{CH_2Cl_2}$	1° alcohol $\mathrm{RCH_2OH}$	Aldehyde $\mathrm{RCHO}$
$\mathrm{CrO_3}$ – $\mathrm{H_2SO_4}$ (Jones) or $\mathrm{K_2Cr_2O_7}$ – $\mathrm{H_2SO_4}$	1° alcohol; 2° alcohol	Carboxylic acid $\mathrm{RCOOH}$ ; ketone
Hot $\mathrm{KMnO_4}$	1° alcohol or aldehyde	Carboxylic acid
Cu at 573 K	1° or 2° alcohol vapour	Aldehyde or ketone (dehydrogenation)
DIBAL-H at low temperature, then $\mathrm{H_2O}$	Ester $\mathrm{RCOOR'}$ or nitrile $\mathrm{RCN}$	Aldehyde $\mathrm{RCHO}$
$\mathrm{LiAlH_4}$ , then $\mathrm{H_3O^+}$	Ester $\mathrm{RCOOR'}$	Two alcohols, $\mathrm{RCH_2OH}$ and $\mathrm{R'OH}$
Dilute $\mathrm{H_2SO_4}$ , water	Ester $\mathrm{RCOOR'}$	Acid $\mathrm{RCOOH}$ and alcohol $\mathrm{R'OH}$ (hydrolysis)
$\mathrm{BH_3}$ ; $\mathrm{H_2O_2}$ , $\mathrm{OH^-}$ ; then PCC	Terminal alkene $\mathrm{RCH{=}CH_2}$	Aldehyde $\mathrm{RCH_2CHO}$
MnO at about 573 K	Benzoic acid vapour	Benzaldehyde, in one step

PCC and DIBAL-H are the two reagents built to stop at the aldehyde.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2021 · Paper 19 · Q37Moderate

Example 2 · Aldehydes, Ketones and Carboxylic Acids · Preparation of Aldehydes and Ketones

Match List-I with List-II: List-I (Chemical Reaction) List-II (Reagent used) (a)

CH_{3}COOCH_{2}CH_{3}\rightarrow CH_{3}CH_{2}OH

(i)

CH_{3}MgBr/H_{3}O^{+}

(1 equivalent) (b)

CH_{3}COOCH_{3}\rightarrow CH_{3}CHO

(ii)

H_{2}SO_{4}/H_{2}O

(c)

CH_{3}C \equiv N \rightarrow CH_{3}CHO

(iii) DIBAL-H/

H_{2}O

(d)

CH_{3}C \equiv N \rightarrow CH_{3}COCH_{3}

(iv)

SnCl_{2},HCl/H_{2}O

Choose the most appropriate match:

PCC stops at the aldehyde; the Jones reagent does not

Both are chromium(VI) reagents. PCC is used without water and stops at RCHO. The Jones reagent is aqueous and takes a 1° alcohol on to RCOOH.

Hydroboration puts the oxygen on the end carbon

Acid-catalysed hydration or

\mathrm{HgSO_4}

hydration follows Markovnikov's rule and leads to a ketone. Only hydroboration–oxidation puts OH on the terminal carbon, so only that route leads to the aldehyde.

Drill 7 more on reagents that stop at the aldehyde or ketone

Concept 3 of 3: Carbonyl compounds from alkynes, gem-dihalides and alkenes

A hydrocarbon becomes a carbonyl compound when water adds across a triple bond, when two halogens on one carbon are hydrolysed, or when a double bond is cut. Where the oxygen lands decides aldehyde or ketone.

Definition

Alkyne hydration ( $\mathrm{HgSO_4}$ , dilute $\mathrm{H_2SO_4}$ ) goes through an enol that tautomerises. Addition follows Markovnikov's rule, so ethyne gives ethanal and every other alkyne gives a ketone.
Gem-dihalides hydrolyse to a carbonyl compound: two halogens on an end carbon ( $\mathrm{RCHX_2}$ ) give an aldehyde; two on a middle carbon ( $\mathrm{RCX_2R'}$ ) give a ketone.
Side-chain chlorination of toluene gives benzal chloride, $\mathrm{C_6H_5CHCl_2}$ , which hydrolyses to benzaldehyde.
Ozonolysis cuts a C=C; each carbon of the double bond becomes a C=O.
Hydroformylation (the oxo process) adds H and CHO across a C=C, giving an aldehyde one carbon longer, mostly the straight chain.

Starting compound	Reagents	Product
Ethyne $\mathrm{HC{\equiv}CH}$	$\mathrm{H_2O}$ , $\mathrm{HgSO_4}$ , dilute $\mathrm{H_2SO_4}$	Ethanal $\mathrm{CH_3CHO}$
Terminal alkyne $\mathrm{RC{\equiv}CH}$	$\mathrm{H_2O}$ , $\mathrm{HgSO_4}$ , dilute $\mathrm{H_2SO_4}$	Methyl ketone $\mathrm{RCOCH_3}$
Terminal gem-dihalide $\mathrm{RCHCl_2}$	Aqueous KOH (hydrolysis)	Aldehyde $\mathrm{RCHO}$
Internal gem-dihalide $\mathrm{RCCl_2R'}$	Aqueous KOH (hydrolysis)	Ketone $\mathrm{RCOR'}$
Toluene	$\mathrm{Cl_2}$ and light, then water at 373 K	Benzaldehyde, through $\mathrm{C_6H_5CHCl_2}$
Alkene	$\mathrm{O_3}$ , then Zn and water	Aldehydes or ketones, one from each end of the C=C
Alkene $\mathrm{RCH{=}CH_2}$	CO and $\mathrm{H_2}$ , cobalt or rhodium catalyst	Aldehyde $\mathrm{RCH_2CH_2CHO}$ , one carbon longer
Methane	$\mathrm{O_2}$ over a molybdenum oxide catalyst, heat	Methanal $\mathrm{HCHO}$

Only ethyne gives an aldehyde on hydration; an end-carbon gem-dihalide gives an aldehyde on hydrolysis.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2021 · Paper 25 · Q31Moderate

Example 3 · Aldehydes, Ketones and Carboxylic Acids · Preparation of Aldehydes and Ketones

\underset{\left( C_{4}H_{8}Cl_{2} \right)}{A}\overset{\text{~Hydrolysis~}}{\rightarrow}\underset{\left( C_{4}H_{8}O \right)}{B}

B

reacts with Hydroxyl amine but does not give Tollen's test. Identify A and B.

Alkyne hydration gives an aldehyde only from ethyne

Every other alkyne gives a ketone, because the OH goes to the more substituted carbon. Propanal, for example, cannot be made this way.

The position of the two halogens decides the product

Both halogens must sit on one carbon. On an end carbon they give an aldehyde, on a middle carbon a ketone. A 1,2-dihalide is not a gem-dihalide and does not give a carbonyl compound this way.

Drill 5 more on carbonyl compounds from alkynes, gem-dihalides and alkenes

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 (3)

Named routes to aldehydes and ketones7 rows

Name	Starting compound	Reagents	Product
Rosenmund reduction	Acyl chloride $\mathrm{RCOCl}$	$\mathrm{H_2}$ , Pd– $\mathrm{BaSO_4}$ (poisoned)	Aldehyde $\mathrm{RCHO}$
Stephen reduction	Nitrile $\mathrm{RC{\equiv}N}$	$\mathrm{SnCl_2}$ , HCl, then $\mathrm{H_3O^+}$	Aldehyde $\mathrm{RCHO}$ , through the imine $\mathrm{RCH{=}NH}$
Etard reaction	Toluene	$\mathrm{CrO_2Cl_2}$ in $\mathrm{CS_2}$ , then $\mathrm{H_3O^+}$	Benzaldehyde, through $\mathrm{C_6H_5CH(OCrOHCl_2)_2}$
Chromic oxide oxidation	Toluene	$\mathrm{CrO_3}$ in $\mathrm{(CH_3CO)_2O}$ , 273–283 K, then $\mathrm{H_3O^+}$	Benzaldehyde, through benzylidene diacetate
Gattermann–Koch reaction	Benzene	CO, HCl, anhydrous $\mathrm{AlCl_3}$ and CuCl	Benzaldehyde
Friedel–Crafts acylation	Benzene	$\mathrm{RCOCl}$ , anhydrous $\mathrm{AlCl_3}$	Aryl ketone $\mathrm{C_6H_5COR}$ ; with $\mathrm{C_6H_5COCl}$ , benzophenone
Dialkylcadmium route	Acyl chloride $\mathrm{RCOCl}$	$\mathrm{R'_2Cd}$	Ketone $\mathrm{RCOR'}$

Rosenmund starts from an acyl chloride, Stephen from a nitrile, Etard from toluene and Gattermann–Koch from benzene.

Reagents that stop at the aldehyde or ketone9 rows

Reagent	Acts on	Stops at
PCC in $\mathrm{CH_2Cl_2}$	1° alcohol $\mathrm{RCH_2OH}$	Aldehyde $\mathrm{RCHO}$
$\mathrm{CrO_3}$ – $\mathrm{H_2SO_4}$ (Jones) or $\mathrm{K_2Cr_2O_7}$ – $\mathrm{H_2SO_4}$	1° alcohol; 2° alcohol	Carboxylic acid $\mathrm{RCOOH}$ ; ketone
Hot $\mathrm{KMnO_4}$	1° alcohol or aldehyde	Carboxylic acid
Cu at 573 K	1° or 2° alcohol vapour	Aldehyde or ketone (dehydrogenation)
DIBAL-H at low temperature, then $\mathrm{H_2O}$	Ester $\mathrm{RCOOR'}$ or nitrile $\mathrm{RCN}$	Aldehyde $\mathrm{RCHO}$
$\mathrm{LiAlH_4}$ , then $\mathrm{H_3O^+}$	Ester $\mathrm{RCOOR'}$	Two alcohols, $\mathrm{RCH_2OH}$ and $\mathrm{R'OH}$
Dilute $\mathrm{H_2SO_4}$ , water	Ester $\mathrm{RCOOR'}$	Acid $\mathrm{RCOOH}$ and alcohol $\mathrm{R'OH}$ (hydrolysis)
$\mathrm{BH_3}$ ; $\mathrm{H_2O_2}$ , $\mathrm{OH^-}$ ; then PCC	Terminal alkene $\mathrm{RCH{=}CH_2}$	Aldehyde $\mathrm{RCH_2CHO}$
MnO at about 573 K	Benzoic acid vapour	Benzaldehyde, in one step

PCC and DIBAL-H are the two reagents built to stop at the aldehyde.

Carbonyl compounds from alkynes, gem-dihalides and alkenes8 rows

Starting compound	Reagents	Product
Ethyne $\mathrm{HC{\equiv}CH}$	$\mathrm{H_2O}$ , $\mathrm{HgSO_4}$ , dilute $\mathrm{H_2SO_4}$	Ethanal $\mathrm{CH_3CHO}$
Terminal alkyne $\mathrm{RC{\equiv}CH}$	$\mathrm{H_2O}$ , $\mathrm{HgSO_4}$ , dilute $\mathrm{H_2SO_4}$	Methyl ketone $\mathrm{RCOCH_3}$
Terminal gem-dihalide $\mathrm{RCHCl_2}$	Aqueous KOH (hydrolysis)	Aldehyde $\mathrm{RCHO}$
Internal gem-dihalide $\mathrm{RCCl_2R'}$	Aqueous KOH (hydrolysis)	Ketone $\mathrm{RCOR'}$
Toluene	$\mathrm{Cl_2}$ and light, then water at 373 K	Benzaldehyde, through $\mathrm{C_6H_5CHCl_2}$
Alkene	$\mathrm{O_3}$ , then Zn and water	Aldehydes or ketones, one from each end of the C=C
Alkene $\mathrm{RCH{=}CH_2}$	CO and $\mathrm{H_2}$ , cobalt or rhodium catalyst	Aldehyde $\mathrm{RCH_2CH_2CHO}$ , one carbon longer
Methane	$\mathrm{O_2}$ over a molybdenum oxide catalyst, heat	Methanal $\mathrm{HCHO}$

Only ethyne gives an aldehyde on hydration; an end-carbon gem-dihalide gives an aldehyde on hydrolysis.

Watch out for (6)

The Stephen reduction needs the water step
→ Named routes to aldehydes and ketones
Etard and Gattermann–Koch start from different rings
→ Named routes to aldehydes and ketones
PCC stops at the aldehyde; the Jones reagent does not
→ Reagents that stop at the aldehyde or ketone
Hydroboration puts the oxygen on the end carbon
→ Reagents that stop at the aldehyde or ketone
Alkyne hydration gives an aldehyde only from ethyne
→ Carbonyl compounds from alkynes, gem-dihalides and alkenes
The position of the two halogens decides the product
→ Carbonyl compounds from alkynes, gem-dihalides and alkenes

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