General Methods of Preparation of Aliphatic Aldehydes & Ketones

ChemistryAldehydes And KetonesFor NEET aspirants

The general methods of preparation of aliphatic aldehydes and ketones start from functional groups you already know: alcohols, acyl chlorides, esters, nitriles, alkenes, alkynes and gem-dihalides. The central idea is control. Oxidation or reduction must stop exactly at the carbonyl level, not one step short or one step beyond. This page covers every NCERT method for the preparation of aldehydes and ketones, together with the reagent choices (PCC, DIBAL-H, poisoned palladium, dialkylcadmium) that JEE Main, JEE Advanced and NEET questions test again and again.

Key reactions: quick reference

1° alcohol, PCC
$RC H_{2} OH PCC CH_{2} Cl_{2} RCHO$
2° alcohol, dichromate
$RCH (OH) R^{'} K_{2} Cr_{2} O_{7}, H^{+} RCO R^{'}$
Dehydrogenation
$RC H_{2} OH Cu, 573 K RCHO + H_{2}$
Rosenmund reduction
$RCOCl + H_{2} Pd/BaSO_{4} RCHO + HCl$
Stephen reduction
$RC \equiv N (i) SnCl_{2}, HCl (ii) H_{3} O^{+} RCHO$
DIBAL-H
$RCOO R^{'} or RC \equiv N (i) DIBAL-H, - 78^{\circ} C (ii) H_{2} O RCHO$
Nitrile + Grignard
$RC \equiv N (i) R’MgBr (ii) H_{3} O^{+} RCO R^{'}$
Dialkylcadmium
$2 R^{'} COCl + R_{2} Cd \to 2 R^{'} COR + CdC l_{2}$
Ozonolysis
$R_{2} C = CH R^{'} (i) O_{3} (ii) Zn, H_{2} O R_{2} CO + R^{'} CHO$
Hydration of ethyne
$HC \equiv CH + H_{2} O HgSO_{4}, H_{2} SO_{4} 333 K C H_{3} CHO$

1. The Big Picture

An aldehyde or ketone sits at a middle oxidation level: above an alcohol, below a carboxylic acid. Every method on this page belongs to one of four families.

Controlled oxidation of alcohols, or their dehydrogenation over copper.
Controlled reduction of acid derivatives (acyl chlorides, esters, nitriles), stopping before the alcohol.
Carbon-carbon bond formation with organometallic reagents (Grignard, dialkylcadmium) to make ketones.
Addition or cleavage of multiple bonds: hydration of alkynes, ozonolysis of alkenes, hydrolysis of gem-dihalides.

Figure 1: Eight starting points for the preparation of aldehydes and ketones: alcohols, acyl chlorides, esters, nitriles, alkenes, alkynes, gem-dihalides and calcium salts of carboxylic acids.

2. From Alcohols

Oxidation of primary alcohols to aldehydes

A primary alcohol is oxidised first to an aldehyde and then to a carboxylic acid:

RC H_{2} OH [O] RCHO [O] RCOOH

The aldehyde is only a halfway stage, and ordinary oxidants such as acidified K₂Cr₂O₇ or KMnO₄ carry it on to the acid. To stop at the aldehyde, use pyridinium chlorochromate (PCC), C₅H₅NH⁺CrO₃Cl⁻, in dichloromethane:

RC H_{2} OH C_{5} H_{5} NH^{+} CrO_{3} Cl^{-} (PCC) CH_{2} Cl_{2} RCHO

PCC works in the absence of water. Chromium(VI) oxidises an aldehyde to an acid through its hydrate, RCH(OH)₂, and without water that hydrate cannot form, so oxidation stops. With acidified dichromate, a low-boiling aldehyde such as ethanal can also be saved by distilling it out as soon as it forms.

Oxidation of secondary alcohols to ketones

Secondary alcohols are oxidised to ketones by chromic acid (H₂CrO₄), acidified K₂Cr₂O₇ or PCC:

RCH (OH) R^{'} H_{2} CrO_{4} RCO R^{'}

The reaction usually stops at the ketone, because oxidising it further would require breaking a C-C bond. Tertiary alcohols have no hydrogen on the carbinol carbon and are not oxidised under these conditions.

Figure 2: Oxidation of alcohols: PCC stops a primary alcohol at the aldehyde, while excess acidified K₂Cr₂O₇ continues to the acid; a secondary alcohol stops at the ketone.

Solved Example 1

Give the main organic product: (a) butan-1-ol with PCC in CH₂Cl₂; (b) butan-1-ol heated with excess acidified K₂Cr₂O₇; (c) pentan-2-ol with acidified K₂Cr₂O₇.

Solution:

(a) PCC works without water and stops at the aldehyde. Butanal, CH₃CH₂CH₂CHO.

(b) Excess aqueous Cr(VI) with heating oxidises the aldehyde further. Butanoic acid, CH₃CH₂CH₂COOH.

(c) A secondary alcohol stops at the ketone, because further oxidation would break a C-C bond. Pentan-2-one, CH₃COCH₂CH₂CH₃.

Dehydrogenation of alcohols

When alcohol vapour is passed over heated copper at 573 K, hydrogen is removed instead of being oxidised away. This method is used industrially.

RC H_{2} OH Cu, 573 K RCHO + H_{2} RCH (OH) R^{'} Cu, 573 K RCO R^{'} + H_{2}

A tertiary alcohol has no hydrogen on the carbinol carbon to lose, so under the same conditions it is dehydrated to an alkene.

Figure 3: Dehydrogenation of alcohols over copper at 573 K: 1° alcohols give aldehydes, 2° alcohols give ketones, and 3° alcohols are dehydrated to alkenes.

Solved Example 2

The vapours of three isomeric C₄H₁₀O alcohols are each passed over copper at 573 K: butan-1-ol, butan-2-ol and 2-methylpropan-2-ol. Name the product in each case.

Solution:

Butan-1-ol is a 1° alcohol, so it loses H₂ to give butanal.

Butan-2-ol is a 2° alcohol, so it loses H₂ to give butan-2-one.

2-Methylpropan-2-ol is a 3° alcohol with no hydrogen on the carbinol carbon, so it cannot be dehydrogenated. It loses water instead, giving 2-methylpropene.

3. From Acid Derivatives by Partial Reduction

In principle, reducing a carboxylic acid should give an aldehyde. In practice the usual reagent, LiAlH₄, reduces the acid all the way to a primary alcohol. LiAlH₄ is a powerful reducing agent and aldehydes are reduced very easily, so any aldehyde that forms is immediately reduced further:

RCOOH LiAlH_{4} [RCHO] LiAlH_{4} RC H_{2} OH

The solution has two parts. First, start from an acid derivative that is more easily reduced than the acid itself: an acyl chloride, an ester or a nitrile. Second, use an aluminium hydride that is less reactive than LiAlH₄. The two standard reagents, lithium tri-tert-butoxyaluminium hydride and diisobutylaluminium hydride (DIBAL-H), each carry only one hydride. Both are sterically hindered, so they transfer hydride with more difficulty.

Figure 4: Hydride reagents compared: LiAlH₄ over-reduces, while bulky LiAlH(OCMe₃)₃ and DIBAL-H stop at the aldehyde.

Figure 5: Partial reduction to aldehydes: acyl chlorides with LiAlH(OCMe₃)₃, and esters and nitriles with DIBAL-H, at low temperature.

Acyl chlorides with LiAlH(OCMe₃)₃

Acyl chlorides are reduced to aldehydes by lithium tri-tert-butoxyaluminium hydride, LiAlH[OC(CH₃)₃]₃, in ether at $- 78^{\circ} C$ . The carboxylic acid is first converted into the acyl chloride with thionyl chloride:

RCOOH SOCl_{2} RCOCl (i) LiAlH(OCMe_{3})_{3}, Et_{2} O, - 78^{\circ} C (ii) H_{2} O RCHO

For example, 3-methoxy-4-methylbenzoyl chloride gives 3-methoxy-4-methylbenzaldehyde under these conditions, so the method works for aromatic acid chlorides too.

Hydride adds. The single Al-H hydride attacks the carbonyl carbon, and the $π$ electrons move onto oxygen.
Chloride leaves. The alkoxide re-forms the carbonyl bond and expels Cl⁻, a good leaving group. Lithium chloride is the by-product.
Reduction stops. At $- 78^{\circ} C$ the bulky reagent reacts with the aldehyde only slowly, so it survives.

Figure 6: Mechanism of acyl chloride reduction by lithium tri-tert-butoxyaluminium hydride: hydride addition, then loss of chloride.

Rosenmund reduction

Rosenmund reduction: an acyl chloride is hydrogenated over palladium supported on barium sulphate, with the catalyst partially poisoned by sulphur or quinoline. The C-Cl bond is reduced to C-H, giving the aldehyde.

RCOCl + H_{2} Pd/BaSO_{4} S or quinoline RCHO + HCl

The partial deactivation of the catalyst is essential. A fully active palladium catalyst would also reduce the aldehyde to a primary alcohol. Formaldehyde cannot be made by this method because formyl chloride (HCOCl) is unstable.

Figure 7: Rosenmund reduction: H₂ over Pd/BaSO₄ poisoned with sulphur or quinoline reduces an acyl chloride to an aldehyde without over-reduction.

Esters with DIBAL-H

Esters are reduced to aldehydes by DIBAL-H in hexane at $- 78^{\circ} C$ , followed by water. The amount of reagent must be carefully controlled (one equivalent), and the low temperature is essential to avoid over-reduction.

RCOO R^{'} (i) DIBAL-H, hexane, - 78^{\circ} C (ii) H_{2} O RCHO + R^{'} OH

Coordination. Aluminium in DIBAL-H is electron-deficient (a Lewis acid) and binds the carbonyl oxygen.
Hydride transfer. The hydride moves from aluminium to the carbonyl carbon, giving a tetrahedral aluminium intermediate.
Work-up. This intermediate is relatively stable at $- 78^{\circ} C$ and does not expel R'O⁻ while cold. Hydrolysis at the end liberates the aldehyde and R'OH.

Figure 8: Mechanism of DIBAL-H reduction of an ester. At

- 78^{\circ} C

the hydride adds and the Al-bound intermediate holds; at work-up, water removes Al (and destroys leftover DIBAL-H), and the freed

O^{-}

pushes out

R^{'} O^{-}

to give the aldehyde.

Nitriles with DIBAL-H or SnCl₂/HCl (Stephen reduction)

DIBAL-H adds one hydride to the carbon of the $C \equiv N$ group, giving an N-aluminium imine that is stable until hydrolysis:

RC \equiv N DIBAL-H, hexane RCH = NAl (B u^{i})_{2} H_{2} O RCHO

The Stephen reduction achieves the same result with stannous chloride and hydrogen chloride. The nitrile is reduced to an aldimine, isolated as its hydrochloride, which is hydrolysed to the aldehyde:

RC \equiv N SnCl_{2}, HCl RCH = NH \cdot HCl H_{3} O^{+} RCHO + N H_{4} Cl

Figure 9: Nitriles to aldehydes. DIBAL-H adds one hydride (Route 1) and Stephen reduction with

SnC l_{2} /HCl

gives the imine salt directly (Route 2); both meet at the iminium ion, which water hydrolyses to the aldehyde and

N H_{4}^{+}

Solved Example 3

Suggest reagents for each conversion and explain why LiAlH₄ is not suitable: (a) ethyl butanoate to butanal; (b) butanoyl chloride to butanal; (c) pentanenitrile to pentanal.

Solution:

(a) DIBAL-H in hexane at $- 78^{\circ} C$ , then H₂O. The tetrahedral intermediate survives while cold and gives the aldehyde on work-up.

(b) H₂ over poisoned Pd/BaSO₄ (Rosenmund), or LiAlH(OCMe₃)₃ at $- 78^{\circ} C$ .

LiAlH₄ is unsuitable because it reduces the aldehyde as soon as it forms. It would give butan-1-ol in (a) and (b), and pentan-1-amine in (c).

Solved Example 4

(a) Why is the palladium catalyst poisoned in Rosenmund reduction? (b) Why can methanal not be prepared by this method?

Solution:

(a) Fully active palladium would also hydrogenate the aldehyde to a primary alcohol. Barium sulphate lowers the activity, and sulphur or quinoline blocks the most active sites, so only the more reactive C-Cl bond of the acyl chloride is reduced. Poisoning stops the reaction at the aldehyde.

(b) Methanal would need formyl chloride, HCOCl, as the starting material. Formyl chloride is unstable and decomposes to CO and HCl, so it cannot be used.

4. Ketones Using Organometallic Reagents

Nitriles with Grignard reagents

A Grignard reagent adds once to the carbon of a nitrile. The product is the magnesium salt of an imine, which acid hydrolysis converts into a ketone. The imine carbon cannot be attacked a second time, so the reaction does not go on to an alcohol.

RC \equiv N + R^{'} MgBr ether R R^{'} C = NMgBr H_{3} O^{+} RCO R^{'}

NCERT's example uses an aromatic Grignard reagent: propanenitrile and phenylmagnesium bromide give propiophenone (1-phenylpropan-1-one).

C H_{3} C H_{2} CN + C_{6} H_{5} MgBr (i) ether (ii) H_{3} O^{+} C_{6} H_{5} COC H_{2} C H_{3}

Acyl chlorides with dialkylcadmium

A dialkylcadmium is prepared from a Grignard reagent and cadmium chloride. It reacts with an acyl chloride to give a ketone:

2 RMgBr + CdC l_{2} \to R_{2} Cd + 2 MgBrCl

2 R^{'} COCl + R_{2} Cd \to 2 R^{'} COR + CdC l_{2}

A Grignard reagent cannot be used here because it attacks the ketone as fast as it forms, giving a tertiary alcohol. Dialkylcadmium is much less reactive, so it attacks the reactive acyl chloride but leaves the ketone alone.

Figure 10: Ketones from nitriles with Grignard reagents, and from acyl chlorides with dialkylcadmium. The

C = N - MgBr

salt and the mild

R_{2} Cd

both stop at one addition, so the product is a ketone, not a

3^{\circ}

alcohol.

Solved Example 5

Identify the missing compounds in each sequence.

(a) C H_{3} CH = C H_{2} NBS A Mg, ether B CH_{3} CN C H_{3} O^{+} D

(b) HC \equiv N + C_{3} H_{5} MgBr (cyclopropyl) \to A H_{3} O^{+} B

Solution:

(a) NBS brominates the allylic position: A = 3-bromopropene, CH₂=CHCH₂Br. With magnesium it gives B = allylmagnesium bromide, CH₂=CHCH₂MgBr.

B adds to the nitrile carbon of CH₃CN: C = the imine salt CH₃C(=NMgBr)CH₂CH=CH₂. Hydrolysis gives the ketone D = pent-4-en-2-one, CH₃COCH₂CH=CH₂.

(b) By the same addition-hydrolysis pattern, A = the imine salt (C₃H₅)CH=NMgBr and B = cyclopropanecarbaldehyde, C₃H₅CHO. This is the expected exam answer. In a real flask, the acidic hydrogen of HCN would destroy the Grignard reagent. Aldehydes are made from Grignard reagents in practice with triethyl orthoformate.

Solved Example 6

Complete the sequences.

(a) C H_{3} C \equiv N + C_{6} H_{11} MgBr (cyclohexyl) \to A H_{3} O^{+} B

(b) cyclohexene NBS A Mg, ether B (i) CH_{3} CN (ii) H_{3} O^{+} C

Solution:

(a) The Grignard reagent adds to the nitrile carbon: A = the imine salt CH₃C(=NMgBr)C₆H₁₁. Hydrolysis gives B = cyclohexyl methyl ketone (1-cyclohexylethanone).

(b) NBS brominates the allylic carbon: A = 3-bromocyclohexene. With magnesium it gives B = cyclohex-2-enylmagnesium bromide. Addition to acetonitrile and hydrolysis give C = 1-(cyclohex-2-en-1-yl)ethanone.

Solved Example 7

Butan-2-one is to be made from acetyl chloride. A student uses ethylmagnesium bromide and gets 3-methylpentan-3-ol instead. Explain, and give a better reagent.

Solution:

Ethylmagnesium bromide does convert acetyl chloride into butan-2-one, but a Grignard reagent attacks ketones readily. A second molecule adds to the ketone at once, and hydrolysis gives the tertiary alcohol 3-methylpentan-3-ol.

The better reagent is diethylcadmium, made from C₂H₅MgBr and CdCl₂. It is reactive enough for the acyl chloride but not for the ketone:

2 C H_{3} COCl + (C_{2} H_{5})_{2} Cd \to 2 C H_{3} CO C_{2} H_{5} + CdC l_{2}

5. From Alkenes: Ozonolysis

Ozone adds across a C=C bond to form an ozonide. Treating the ozonide with zinc dust and water (reductive work-up) cleaves it into two carbonyl compounds. Zinc destroys the hydrogen peroxide formed, which would otherwise oxidise any aldehyde to an acid.

R_{2} C = CH R^{'} O_{3} ozonide Zn, H_{2} O R_{2} C = O + R^{'} CHO

To predict the products, erase the double bond and attach an oxygen to each of its two carbons. A carbon that carried a hydrogen becomes an aldehyde group; a carbon carrying two alkyl groups becomes a ketone.

Figure 11: Reductive ozonolysis of 2-methylbut-2-ene gives ethanal and propanone, with a shortcut for predicting ozonolysis products.

Solved Example 8

(a) Give the products of reductive ozonolysis of pent-2-ene. (b) Which alkene of formula C₆H₁₂ gives only propanone on ozonolysis with Zn/H₂O?

Solution:

(a) CH₃CH=CHCH₂CH₃: cut the double bond and add O to each carbon. The products are ethanal (CH₃CHO) and propanal (CH₃CH₂CHO).

(b) Only propanone means both double-bond carbons carry two methyl groups, and the molecule is symmetrical: (CH₃)₂C=C(CH₃)₂, 2,3-dimethylbut-2-ene.

Solved Example 9

An organic compound C₆H₁₀ (A) on successive reduction gives C₆H₁₂ (B) and C₆H₁₄ (C). Ozonolysis of A followed by hydrolysis gives two aldehydes, C₂H₄O (D) and C₂H₂O₂ (E). B on oxidation with alkaline KMnO₄ followed by acidification gives an acid C₃H₆O₂ (F). Identify A to F. JEE 2001 pattern

Solution:

A takes up two molecules of H₂ in steps, so it is an open-chain diene. B oxidises to a single C₃ acid, propanoic acid, so B is symmetrical: CH₃CH₂CH=CHCH₂CH₃.

Ozonolysis of A gives ethanal and glyoxal, so the diene is conjugated: CH₃CH=CH-CH=CHCH₃. Its 1,4-addition of H₂ shifts the double bond to the centre, giving B.

A = hexa-2,4-diene; B = hex-3-ene; C = hexane; D = ethanal (CH₃CHO); E = glyoxal (OHC-CHO); F = propanoic acid (CH₃CH₂COOH).

Check: ozonolysis of B gives only propanal, as expected for a symmetrical alkene.

6. From Alkynes

Hydration with mercuric sulphate

Water adds to alkynes only with the catalytic help of both acid and mercury(II) ions: dilute H₂SO₄ with HgSO₄ at about 333 K. Ethyne gives ethanal; every other alkyne gives a ketone, because water adds by Markovnikov's rule.

HC \equiv CH + H_{2} O HgSO_{4}, dil. H_{2} SO_{4} 333 K C H_{3} CHO

C H_{3} C \equiv CH + H_{2} O HgSO_{4}, dil. H_{2} SO_{4} C H_{3} COC H_{3}

$π$ -complex. Hg²⁺ forms a complex with the triple bond, making it open to attack by water.
Water attacks. Loss of H⁺ gives a mercury-substituted enol.
H⁺ replaces Hg²⁺. This gives the enol and regenerates the catalyst.
Tautomerism. The enol rearranges to the far more stable keto form.

Figure 12: Mechanism of mercury(II)-catalysed hydration of ethyne:

π

-complex, mercury-substituted enol, enol, and keto-enol tautomerism to ethanal.

Hydroboration-oxidation of terminal alkynes

Hydroboration is faster with alkynes than with alkenes. A bulky dialkylborane (R₂BH) adds once, with boron going to the terminal carbon. Treating the vinylborane with alkaline hydrogen peroxide gives an enol with the -OH on the end carbon, which tautomerises to an aldehyde.

RC \equiv CH R_{2}^{'} BH RCH = CHB R_{2}^{'} H_{2} O_{2}, OH^{-} [RCH = CHOH] \to RC H_{2} CHO

Hydration and hydroboration-oxidation therefore put the oxygen on opposite ends of a terminal alkyne.

Figure 13: Hydration of but-1-yne gives butan-2-one (Markovnikov), while hydroboration-oxidation gives butanal (anti-Markovnikov).

Solved Example 10

Propyne is treated separately with (a) HgSO₄/dil. H₂SO₄ and (b) a bulky R₂BH, then H₂O₂/NaOH. Give the products and explain the difference.

Solution:

(a) Markovnikov hydration puts -OH on C-2. The enol CH₃C(OH)=CH₂ tautomerises to propanone, CH₃COCH₃.

(b) Boron adds to the less hindered terminal carbon, and oxidation replaces it with -OH on C-1. The enol CH₃CH=CHOH tautomerises to propanal, CH₃CH₂CHO.

The two methods place the oxygen on opposite ends of the triple bond.

7. Hydrolysis of gem-Dihalides

A gem-dihalide has both halogens on the same carbon. Boiling it with aqueous alkali replaces both halogens by -OH groups. The resulting gem-diol is unstable and loses water to give a carbonyl compound. A terminal gem-dihalide (RCHX₂) gives an aldehyde; a non-terminal one (RCX₂R') gives a ketone.

C H_{3} C H_{2} CHC l_{2} aq. KOH [C H_{3} C H_{2} CH (OH)_{2}] - H_{2} O C H_{3} C H_{2} CHO

Figure 14: Hydrolysis of gem-dihalides through unstable gem-diols: a terminal gem-dihalide gives an aldehyde and a non-terminal one gives a ketone.

Solved Example 11

Give the product when 1,1-dichloropropane is boiled with aqueous alkali.

Solution:

Both chlorines are on the same terminal carbon. Aqueous KOH replaces them with -OH groups, giving the unstable gem-diol CH₃CH₂CH(OH)₂, which loses water.

C H_{3} C H_{2} CHC l_{2} + 2 KOH \to [C H_{3} C H_{2} CH (OH)_{2}] + 2 KCl \to C H_{3} C H_{2} CHO + H_{2} O

Product: propanal.

8. Dry Distillation of Calcium Salts

Heating the calcium salt of a fatty acid gives a ketone and calcium carbonate:

(C H_{3} COO)_{2} Ca Δ C H_{3} COC H_{3} + CaC O_{3}

Aldehydes are obtained by heating the calcium salt of an acid together with calcium formate, to about 675 K:

(C H_{3} COO)_{2} Ca + (HCOO)_{2} Ca 675 K 2 C H_{3} CHO + 2 CaC O_{3}

Note: when two different calcium salts are heated together, the carboxylate groups pair up in every possible way. The mixture of calcium acetate and calcium formate therefore also gives some methanal (from two formates) and propanone (from two acetates). This makes the method less useful for pure products.

Solved Example 12

What is formed on dry distillation of (a) calcium propanoate alone; (b) a mixture of calcium acetate and calcium formate?

Solution:

(a) Two propanoate groups join, losing CaCO₃:

(C H_{3} C H_{2} COO)_{2} Ca Δ C H_{3} C H_{2} COC H_{2} C H_{3} + CaC O_{3}

Product: pentan-3-one.

(b) An acetate paired with a formate gives ethanal, the intended product. Two acetates give propanone and two formates give methanal, so these form as by-products.

9. Choosing the Right Method

Starting material	Reagent	Product	Watch for
1° alcohol	PCC, CH₂Cl₂	Aldehyde	Aqueous Cr(VI) goes on to the acid
2° alcohol	K₂Cr₂O₇/H⁺, PCC, or Cu at 573 K	Ketone	3° alcohols do not give carbonyls
Acyl chloride	H₂, Pd/BaSO₄ (poisoned) or LiAlH(OCMe₃)₃	Aldehyde	Not for HCHO; active Pd over-reduces
Acyl chloride	R₂Cd	Ketone	RMgX would give a 3° alcohol
Ester	DIBAL-H, $- 78^{\circ} C$ , then H₂O	Aldehyde	Warm conditions over-reduce
Nitrile	SnCl₂/HCl or DIBAL-H, then H₃O⁺	Aldehyde	Hydrolysis step is essential
Nitrile	R'MgX, then H₃O⁺	Ketone	Adds one new C-C bond
Alkene	O₃, then Zn/H₂O	Aldehydes and/or ketones	Without Zn, aldehydes become acids
Terminal alkyne	HgSO₄/H₂SO₄	Methyl ketone (ethanal from ethyne)	Markovnikov
Terminal alkyne	R₂BH, then H₂O₂/OH⁻	Aldehyde	Anti-Markovnikov
gem-Dihalide	aq. KOH	Aldehyde or ketone	Terminal vs non-terminal

Practice Questions

These questions come from the practice sets for this topic. Try each one before opening the answer.

Compound (A), C₅H₁₂O, when refluxed with K₂Cr₂O₇ and dil. H₂SO₄ was converted to (B), C₅H₁₀O. Compound (B) formed a derivative with 2,4-dinitrophenylhydrazine but did not give a positive iodoform test. What are (A) and (B)?
Show answer
(B) reacts with 2,4-DNP, so it is a carbonyl compound. It survives refluxing dichromate, so it is a ketone, and it is not a methyl ketone (no iodoform). The only such C₅ ketone is (B) = pentan-3-one, so (A) = pentan-3-ol.
An organic compound (A) contains C = 60.4%, H = 13.33% and O = 26.27%. When vaporised, 0.06 g of (A) displaces 22.4 mL of air at STP. (A) reacts readily with Na and on mild oxidation gives (B), which forms a crystalline semicarbazone and responds to the iodoform test. Give the structure of (A).
Show answer
Molar mass = 0.06 × 22400 / 22.4 = 60 g mol⁻¹. The percentages fit C₃H₈O. Reaction with Na shows an -OH group. Oxidation gives a methyl ketone (semicarbazone plus iodoform), which must be propanone. (A) = propan-2-ol, CH₃CH(OH)CH₃; (B) = propanone.
Two isomeric gem-dihalides C₅H₁₀Br₂ on hydrolysis give two isomeric compounds C₅H₁₀O. Both form oximes and do not react with Tollens’ reagent. One gives the iodoform reaction and the other does not. On strong oxidation with K₂Cr₂O₇/dil. H₂SO₄, both give a mixture of propanoic acid and acetic acid. Identify the dihalides and carbonyl compounds.
Show answer
Both products are ketones (oximes, but no Tollens’ test). 2,2-Dibromopentane gives pentan-2-one, which gives iodoform. 3,3-Dibromopentane gives pentan-3-one, which does not. Strong oxidation cleaves both next to the carbonyl group, giving acetic and propanoic acids.
Compound (A), C₅H₁₁Br, forms a Grignard reagent that gives n-pentane with dilute acid. With aqueous KOH, (A) gives (B), C₅H₁₂O, which gives a positive Lucas test. (B) with K₂Cr₂O₇/dil. H₂SO₄ gives (C), which gives a negative iodoform test. Deduce (A), (B) and (C).
Show answer
n-Pentane from the Grignard reagent means an unbranched chain. A positive Lucas test (turbidity within minutes) points to a 2° alcohol. The ketone is not a methyl ketone, so the -OH is on C-3. (A) = 3-bromopentane, (B) = pentan-3-ol, (C) = pentan-3-one.
Carry out the conversion of but-1-yne into pentan-2-one in not more than three steps. JEE 1999
Show answer
Step 1: NaNH₂ in liquid NH₃ gives sodium but-1-ynide, $C H_{3} C H_{2} C \equiv CNa$ .
Step 2: CH₃I gives pent-2-yne, $C H_{3} C H_{2} C \equiv CC H_{3}$ .
Step 3: HgSO₄/dil. H₂SO₄ hydration gives pentan-2-one.
Pent-2-yne is unsymmetrical, so hydration also gives some pentan-3-one alongside pentan-2-one.

Common Mistakes to Avoid

Watch out

Using excess acidified K₂Cr₂O₇ to make an aldehyde. The aldehyde goes on to the carboxylic acid. Use PCC, or distil the aldehyde out as it forms.
Using LiAlH₄ on acyl chlorides, esters or nitriles to make aldehydes. It over-reduces to alcohols (or amines from nitriles).
Forgetting the low temperature for DIBAL-H with esters. The tetrahedral intermediate is only stable when cold.
Writing Rosenmund reduction for methanal, or using unpoisoned palladium.
Using a Grignard reagent with an acyl chloride to make a ketone. The ketone reacts further to a tertiary alcohol; use dialkylcadmium.
Doing ozonolysis without zinc. Oxidative work-up turns aldehyde fragments into carboxylic acids.
Expecting an aldehyde from hydration of a terminal alkyne. Only ethyne gives an aldehyde; other alkynes give methyl ketones. Use hydroboration-oxidation for the aldehyde.
Expecting a carbonyl compound from a tertiary alcohol. It resists mild oxidation and dehydrates to an alkene over hot copper.

Frequently Asked Questions

What are the general methods of preparation of aldehydes and ketones?

Aldehydes and ketones are prepared by oxidation or dehydrogenation of alcohols, partial reduction of acyl chlorides, esters and nitriles (Rosenmund, Stephen, DIBAL-H), reaction of nitriles with Grignard reagents or acyl chlorides with dialkylcadmium, ozonolysis of alkenes, hydration of alkynes, hydrolysis of gem-dihalides, and dry distillation of calcium salts.

Why does PCC stop the oxidation of a primary alcohol at the aldehyde?

PCC (pyridinium chlorochromate) is used in dry dichloromethane. Chromium(VI) oxidises an aldehyde to a carboxylic acid through the aldehyde hydrate, RCH(OH)₂, which forms only when water is present. Without water the hydrate cannot form, so oxidation stops cleanly at the aldehyde.

What is Rosenmund reduction and why is the catalyst poisoned?

Rosenmund reduction converts an acyl chloride into an aldehyde using hydrogen over palladium on barium sulphate. The catalyst is partially poisoned with sulphur or quinoline so it reduces only the C-Cl bond. An active catalyst would reduce the aldehyde further to a primary alcohol. It cannot make methanal.

What is the Stephen reaction?

In the Stephen reaction, a nitrile is reduced by stannous chloride and hydrogen chloride to an aldimine, which separates as its hydrochloride. Hydrolysis of this imine salt with water gives the aldehyde. For example, ethanenitrile gives ethanal. It is an NCERT method for aldehydes from nitriles.

How does DIBAL-H convert an ester into an aldehyde?

DIBAL-H (diisobutylaluminium hydride) has one hydride and a Lewis acidic aluminium. It binds the ester carbonyl oxygen and delivers one hydride, forming a tetrahedral intermediate that is stable at −78 °C. When water is added at the end, this intermediate breaks down to the aldehyde and an alcohol.

Why is dialkylcadmium used instead of a Grignard reagent to make ketones from acyl chlorides?

A Grignard reagent attacks the ketone as soon as it forms, giving a tertiary alcohol. Dialkylcadmium, R₂Cd, is far less reactive: it reacts with the highly reactive acyl chloride but not with the ketone product. The ketone therefore survives and can be isolated.

Which preparation methods of aldehydes and ketones are most important for JEE Main and JEE Advanced?

JEE questions often match reagents to named reductions (Rosenmund, Stephen, DIBAL-H) and ask for products of ozonolysis or alkyne hydration. JEE Advanced adds reagent selectivity: PCC versus dichromate, R₂Cd versus RMgX, and hydration versus hydroboration of terminal alkynes, often inside multi-step sequences.

Which methods of preparing aldehydes and ketones should NEET students know?

NEET follows NCERT, which covers oxidation and dehydrogenation of alcohols, ozonolysis of alkenes, hydration of alkynes, Rosenmund reduction, Stephen reaction, DIBAL-H reduction of nitriles and esters, and ketones from acyl chlorides with dialkylcadmium or from nitriles with Grignard reagents. Learn the exact reagents and conditions.

Previous year questions on General Methods of Preparation of Aliphatic Aldehydes & Ketones

2 questions from past papers, each with a step-by-step solution.

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1. The Big Picture

2. From Alcohols

Oxidation of primary alcohols to aldehydes

Oxidation of secondary alcohols to ketones

Dehydrogenation of alcohols

3. From Acid Derivatives by Partial Reduction

Acyl chlorides with LiAlH(OCMe3)3

Rosenmund reduction

Esters with DIBAL-H

Nitriles with DIBAL-H or SnCl2/HCl (Stephen reduction)

4. Ketones Using Organometallic Reagents

Nitriles with Grignard reagents

Acyl chlorides with dialkylcadmium

5. From Alkenes: Ozonolysis

6. From Alkynes

Hydration with mercuric sulphate

Hydroboration-oxidation of terminal alkynes

7. Hydrolysis of gem-Dihalides

8. Dry Distillation of Calcium Salts

9. Choosing the Right Method

Practice Questions

Common Mistakes to Avoid

Frequently Asked Questions

What are the general methods of preparation of aldehydes and ketones?

Why does PCC stop the oxidation of a primary alcohol at the aldehyde?

What is Rosenmund reduction and why is the catalyst poisoned?

What is the Stephen reaction?

How does DIBAL-H convert an ester into an aldehyde?

Why is dialkylcadmium used instead of a Grignard reagent to make ketones from acyl chlorides?

Which preparation methods of aldehydes and ketones are most important for JEE Main and JEE Advanced?

Which methods of preparing aldehydes and ketones should NEET students know?

Previous year questions on General Methods of Preparation of Aliphatic Aldehydes & Ketones

Ready to master Aldehydes And Ketones?

Acyl chlorides with LiAlH(OCMe₃)₃

Nitriles with DIBAL-H or SnCl₂/HCl (Stephen reduction)