Preparation of aldehydes and ketones

How are aldehydes and ketones prepared?

In plain words: an aldehyde sits halfway between an alcohol and an acid, so the trick in most aldehyde routes is to stop at the right point, with a mild reducing agent coming down from an acid derivative or a controlled oxidant going up from a hydrocarbon. Ketones do not oxidise easily, so they are simpler to stop at.

Methods for both

1. Oxidation of alcohols. Controlled oxidation of primary alcohols gives aldehydes (for example, with PCC, or by removing the aldehyde as it forms); stronger aqueous oxidation can continue to the acid. Secondary alcohols give ketones (NCERT page 231, referring to Unit 7).
2. Dehydrogenation of alcohols. Alcohol vapours passed over heavy metal catalysts (Ag or Cu) give aldehydes from primary and ketones from secondary alcohols; the method suits volatile alcohols and is used in industry (NCERT page 231).
3. Ozonolysis of alkenes. Ozonolysis followed by zinc dust and water gives aldehydes, ketones or both, depending on the alkene’s substitution (NCERT pages 231–232).
4. Hydration of alkynes. Water adds to ethyne in the presence of H₂SO₄ and HgSO₄ to give acetaldehyde; under these acid/mercury(II) conditions, other alkynes give ketones (NCERT page 232).

Preparation of aldehydes

Rosenmund reduction and Stephen reaction

• Rosenmund reduction. An acyl chloride is hydrogenated over palladium on barium sulphate: C₆H₅COCl + H₂ → C₆H₅CHO + HCl (Pd–BaSO₄), benzoyl chloride to benzaldehyde (NCERT page 232).
• Stephen reaction. Nitriles “are reduced to corresponding imine with stannous chloride in the presence of hydrochloric acid”, which hydrolysis turns into the aldehyde (NCERT page 232). The reaction sequence is RCN → RCH=NH → RCHO, with SnCl₂/HCl for reduction and water for hydrolysis. In the acid the imine is present protonated, as an iminium salt, so the middle formula is a shorthand, not a balanced reagent equation (reasoned); name it the imine, as NCERT does, in an exam.
• DIBAL-H. Diisobutylaluminium hydride, AlH(i-Bu)₂, selectively reduces nitriles to imines, which water hydrolyses to aldehydes; the isolated C=C bond is retained in the NCERT example, CH₃CH=CHCH₂CH₂CN → CH₃CH=CHCH₂CH₂CHO. Esters can also give aldehydes with a controlled amount of DIBAL-H at low temperature (commonly about −78 °C), under dry conditions, followed by aqueous work-up. Excess reagent or warmer conditions can give further reduction to alcohols. NCERT illustrates: CH₃(CH₂)₉COOC₂H₅ → CH₃(CH₂)₉CHO (NCERT page 232).

Etard and Gattermann–Koch reactions

Strong oxidising agents take toluene all the way to benzoic acid. The first three methods below modify its methyl group; Gattermann–Koch instead introduces a new formyl carbon into an aromatic ring (NCERT page 232):

• Etard reaction. Chromyl chloride, CrO₂Cl₂, in CS₂ converts the methyl group of toluene into a chromium complex, C₆H₅CH(OCrOHCl₂)₂, which H₃O⁺ hydrolyses to benzaldehyde (NCERT pages 232–233).
• Chromic oxide. CrO₃ in acetic anhydride at 273–283 K gives benzylidene diacetate, C₆H₅CH(OCOCH₃)₂, which aqueous acid hydrolyses to benzaldehyde (NCERT page 233).
• Side-chain chlorination, then hydrolysis. Cl₂ in light turns toluene into benzal chloride, C₆H₅CHCl₂, which water at 373 K hydrolyses to benzaldehyde; this is a commercial method (NCERT page 233). Light initiates radical side-chain chlorination; the extent of chlorination must be controlled to favour CHCl₂ over CH₂Cl or CCl₃. The absence of iron alone does not specify the reaction conditions. Compare the chlorination of toluene (reasoned).
• Gattermann–Koch reaction. Benzene or a derivative with carbon monoxide and hydrogen chloride, in the presence of anhydrous AlCl₃ with CuCl, gives benzaldehyde or a substituted benzaldehyde (NCERT page 233, whose catalyst wording is the slip below).
• NCERT slip (page 233): the text says “anhydrous aluminium chloride or cuprous chloride”, as if either would do, while its scheme labels the arrow “Anhyd. AlCl₃/CuCl”. Beyond the NCERT text: AlCl₃ is the Lewis acid, and when CO is not used at high pressure, traces of a copper(I) chloride co-catalyst are often necessary, so the usual system combines the two chlorides; the method does not apply to phenols and phenol ethers (Wikipedia, “Gattermann reaction”). In an exam, write CO and HCl with anhydrous AlCl₃/CuCl, as NCERT’s scheme does.
• Spelling note: NCERT writes “Gatterman-Koch” here (page 233), and on page 275 (Unit 9, Amines) it prints both “Gatterman reaction” and “Gattermann reaction”, for a different reaction that carries the same name. This page uses Gattermann, the spelling NCERT also prints. Expect either spelling in a question.

Preparation of ketones

1. From acyl chlorides. A dialkylcadmium, made from cadmium chloride and a Grignard reagent, converts an acyl chloride into a ketone (NCERT page 233): 2RMgX + CdCl₂ → R₂Cd + 2Mg(X)Cl, then 2R′COCl + R₂Cd → 2R′COR + CdCl₂.
2. From nitriles. A Grignard reagent adds to the nitrile, and hydrolysis of the product gives a ketone: CH₃CH₂CN + C₆H₅MgBr, in dry ether, then a separate H₃O⁺ work-up, gives C₂H₅COC₆H₅, propiophenone (NCERT page 234).
3. Friedel–Crafts acylation. Benzene or a substituted benzene with an acid chloride and anhydrous AlCl₃ gives the ketone, C₆H₅COR or C₆H₅COAr (NCERT page 234), the acylation of suitable aromatic hydrocarbons.

Friedel–Crafts acylation is unsuitable for strongly deactivated rings; amino groups can bind AlCl₃ and prevent the usual reaction. Grignard steps require dry apparatus and no unprotected acidic O–H or N–H groups.

• NCERT slip (page 234): NCERT labels propiophenone “1-Phenylpropanone”, without the locant of the carbonyl group. The name is ambiguous: 1-phenylpropan-2-one, C₆H₅CH₂COCH₃, is a different ketone. NCERT itself writes 1-phenylpropan-1-one on page 229; use that name in an exam. NCERT uses the same name again in Exercise 8.7 (vi) (page 256); either reading is a ketone with α-hydrogens, so the expected answer there, aldol condensation, is the same (reasoned).

Solved examples

Common mistakes to avoid

Quick revision summary

• Both: oxidation or dehydrogenation (Ag or Cu) of 1° and 2° alcohols; ozonolysis of alkenes; H₂SO₄/HgSO₄ hydration of alkynes (only ethyne gives an aldehyde under these conditions).
• Aldehydes: Rosenmund (RCOCl, H₂, Pd–BaSO₄); Stephen (RCN, SnCl₂/HCl, then H₃O⁺); DIBAL-H for nitriles and esters; Etard (CrO₂Cl₂); CrO₃ in acetic anhydride; benzal chloride hydrolysis; Gattermann–Koch (CO, HCl, AlCl₃ with CuCl).
• Ketones: R₂Cd with acyl chlorides; Grignard reagents with nitriles, then hydrolysis; Friedel–Crafts acylation.

Questions students ask

What is the role of barium sulphate in the Rosenmund reduction?

BaSO₄ supports the palladium catalyst: NCERT’s Rosenmund reduction uses palladium on barium sulphate (page 232). Catalyst activity and reaction conditions are chosen to favour the aldehyde over further hydrogenation; additional deactivating agents can be used, but are not universal. Do not confuse this with Lindlar’s catalyst, “palladised charcoal partially deactivated with poisons like sulphur compounds or quinoline”, used to stop alkyne hydrogenation at a cis alkene (NCERT Class 11 Chemistry Part II, page 309); beyond the NCERT text, the conventional Lindlar catalyst is Pd on CaCO₃, modified with lead compounds and commonly quinoline (Wikipedia, “Lindlar catalyst”).

Why can formaldehyde not be prepared by the Rosenmund reduction?

The required starting material would be formyl chloride, HCOCl, which is unstable under ordinary preparation and storage conditions. Rosenmund reduction is therefore not a standard route to formaldehyde. Gattermann–Koch introduces –CHO onto an aromatic ring and makes aromatic aldehydes; it does not make formaldehyde. Methanol oxidation or dehydrogenation provides the familiar route to formaldehyde.

Why does the Stephen reaction give an aldehyde and not an amine?

The Stephen reaction gives an aldehyde because SnCl₂ and HCl reduce the nitrile only as far as the imine, RCH=NH, which water then hydrolyses to RCHO (NCERT page 232). Beyond the NCERT text, the imine is held as its salt, [RCH=NH₂]⁺Cl⁻, which separates as a tin chloride compound, and its hydrolysis also gives ammonium chloride (Wikipedia, “Stephen aldehyde synthesis”). Stronger reducing agents carry a nitrile past the imine to a primary amine (reduction of nitriles), so the reducing agent decides the product; DIBAL-H also stops at the imine (NCERT page 232).

Why is carbon disulphide used in the Etard reaction?

CS₂ is the solvent in NCERT’s Etard scheme; it is not the oxidising reagent. CrO₂Cl₂ converts the benzylic methyl group into a chromium-containing intermediate, and hydrolytic work-up releases benzaldehyde (NCERT pages 232–233). The key distinction from strong aqueous oxidation is this controlled intermediate and work-up. This reaction scheme is for understanding the conversion; solvent and chromium handling require an approved laboratory procedure.

Where are these named reactions taught, and how should I use them for exam preparation?

These reactions appear in NCERT Class 12 Chemistry Part II, Unit 8, Section 8.2, pages 231–234. They are useful preparation methods to study alongside the unit’s general organic chemistry. A syllabus heading about general methods of preparation does not establish that every named method will be tested; check the official syllabus for your examination year instead of treating this page as a guarantee of question coverage.

Practice questions

Try each question first, then open the answer to check your working.

1. Q1. Why can toluene not be turned into benzaldehyde with acidified KMnO₄, and what is used instead?

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   Strong oxidising agents oxidise the methyl group right through to –COOH, giving benzoic acid. Chromyl chloride (Etard) or CrO₃ in acetic anhydride converts it to an intermediate that resists further oxidation and is hydrolysed to benzaldehyde (NCERT pages 232–233).

2. Q2. What does but-2-yne, CH₃C≡CCH₃, give on hydration with dilute H₂SO₄ and HgSO₄?

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   Butan-2-one, CH₃COCH₂CH₃: under these conditions only ethyne gives an aldehyde; other ordinary alkynes give ketones (NCERT page 232). This symmetrical alkyne can give only one ketone, the same one that Example 3 makes from a nitrile (reasoned).