Named Reactions — Concepts, Formulas & Examples

Named reactions are classic reactions identified with their discoverers. CBSE Class 12 and NEET test about ten to fifteen of them across organic chemistry. Expect two questions a year on named reactions.

Core Concepts

Wurtz reaction

$2\text{R-X} + 2\text{Na} \xrightarrow{\text{dry ether}} \text{R-R} + 2\text{NaX}$ . Two alkyl halides couple via sodium in dry ether to form a longer alkane. Used for symmetric alkanes.

Limitation: If two different alkyl halides are used, three products form (R-R, R’-R’, and R-R’), making it useless for asymmetric alkane synthesis. For this reason, Wurtz is only practical for symmetric alkanes like ethane from CH $_3$ Cl.

Wurtz-Fittig

Mixed version. Aryl halide + alkyl halide + Na gives alkyl-substituted aromatic. Useful for side-chain attachment to benzene.

\text{C}_6\text{H}_5\text{Br} + \text{CH}_3\text{Br} + 2\text{Na} \xrightarrow{\text{dry ether}} \text{C}_6\text{H}_5\text{CH}_3 + 2\text{NaBr}

The aryl-alkyl coupling product is the major product because aryl-aryl coupling (Fittig reaction) and alkyl-alkyl coupling (Wurtz) are slower under these conditions.

Kolbe electrolysis

Electrolysis of sodium carboxylates gives alkanes by decarboxylation. $2\text{RCOONa} + 2\text{H}_2\text{O} \to \text{R-R} + 2\text{CO}_2 + \text{H}_2 + 2\text{NaOH}$ .

At the anode: $\text{RCOO}^- \rightarrow \text{RCOO}\cdot + e^- \rightarrow \text{R}\cdot + \text{CO}_2$ . Two R radicals couple to form R-R.

This reaction produces an alkane with an even number of carbons. If you electrolyse sodium acetate (CH $_3$ COONa), you get ethane (C $_2$ H $_6$ ), not methane.

Friedel-Crafts

Alkylation or acylation of benzene using AlCl $_3$ catalyst. Alkylation — benzene + RCl → alkylbenzene. Acylation — benzene + RCOCl → aryl ketone. Does not work with strongly deactivated rings.

Alkylation:

\text{C}_6\text{H}_6 + \text{RCl} \xrightarrow{\text{AlCl}_3} \text{C}_6\text{H}_5\text{R} + \text{HCl}

Acylation:

\text{C}_6\text{H}_6 + \text{RCOCl} \xrightarrow{\text{AlCl}_3} \text{C}_6\text{H}_5\text{COR} + \text{HCl}

Key limitations:

Does NOT work on rings deactivated by strong electron-withdrawing groups (NO $_2$ , COOH, CHO)
Alkylation suffers from polyalkylation (product is more reactive than starting material)
Acylation does NOT suffer from polyalkylation (the acyl group deactivates the ring)
Vinyl and aryl halides do not react (they cannot form stable carbocations)

When a question asks “prepare an alkylbenzene without polyalkylation,” the answer is usually Friedel-Crafts acylation followed by Clemmensen reduction. Acylation gives a single product; then reduce C=O to CH $_2$ using Zn-Hg/HCl.

Cannizzaro reaction

Aldehydes without alpha hydrogens (e.g. HCHO, benzaldehyde) disproportionate in concentrated alkali — one is reduced to alcohol, the other oxidised to carboxylate. $2\text{HCHO} + \text{NaOH} \to \text{HCOONa} + \text{CH}_3\text{OH}$ .

Requirement: The aldehyde must lack $\alpha$ -hydrogens. If it has $\alpha$ -H, it undergoes aldol condensation instead.

Examples that undergo Cannizzaro: HCHO, C $_6$ H $_5$ CHO (benzaldehyde), (CH $_3$ ) $_3$ CCHO (pivaldehyde), furfural.

Examples that undergo aldol instead: CH $_3$ CHO (acetaldehyde), CH $_3$ CH $_2$ CHO (propanal), CH $_3$ COCH $_3$ (acetone).

Crossed Cannizzaro: When HCHO is mixed with another aldehyde (say benzaldehyde) in NaOH, HCHO is preferentially oxidised (to HCOONa) and benzaldehyde is preferentially reduced (to C $_6$ H $_5$ CH $_2$ OH). HCHO is a better reducing agent.

Aldol condensation

Aldehydes or ketones with alpha hydrogens self-condense in dilute alkali to give beta-hydroxy carbonyl compounds, which may dehydrate to alpha-beta unsaturated products.

2\text{CH}_3\text{CHO} \xrightarrow{\text{dil. NaOH}} \text{CH}_3\text{CH(OH)CH}_2\text{CHO}

The product is a $\beta$ -hydroxy aldehyde (aldol = aldehyde + alcohol). On heating, it loses water to form $\alpha$ , $\beta$ -unsaturated aldehyde (crotonaldehyde): $\text{CH}_3\text{CH=CHCHO}$ .

Reimer-Tiemann

Phenol + CHCl $_3$ + NaOH → salicylaldehyde. Formylation of phenol at the ortho position.

\text{C}_6\text{H}_5\text{OH} + \text{CHCl}_3 + 3\text{NaOH} \rightarrow \text{o-HOC}_6\text{H}_4\text{CHO} + 3\text{NaCl} + 2\text{H}_2\text{O}

The reactive intermediate is dichlorocarbene (:CCl $_2$ ), an electrophile. If CCl $_4$ is used instead of CHCl $_3$ , the product is salicylic acid (Kolbe-Schmitt pathway).

Hoffmann bromamide

Primary amide + Br $_2$ + NaOH → primary amine with one fewer carbon. Useful for amine preparation.

\text{RCONH}_2 + \text{Br}_2 + 4\text{NaOH} \rightarrow \text{RNH}_2 + \text{Na}_2\text{CO}_3 + 2\text{NaBr} + 2\text{H}_2\text{O}

The product amine has one carbon less than the starting amide. If you start with ethanamide (CH $_3$ CONH $_2$ ), you get methylamine (CH $_3$ NH $_2$ ).

Additional important named reactions

Clemmensen reduction: Ketone/aldehyde + Zn-Hg/conc. HCl → alkane (C=O reduced to CH $_2$ ). Works in acidic medium.

Wolff-Kishner reduction: Ketone/aldehyde + N $_2$ H $_4$ /KOH/ethylene glycol (high T) → alkane. Works in basic medium. Use this when acid-sensitive groups are present.

Hell-Volhard-Zelinsky (HVZ): Carboxylic acid + Br $_2$ /P → $\alpha$ -bromoacid. Specifically brominates the $\alpha$ -carbon of carboxylic acids.

Sandmeyer reaction: Diazonium salt + CuX → aryl halide. ArN $_2^+$ Cl $^-$ + CuCl → ArCl + N $_2$ . Also works with CuBr and CuCN.

Gattermann reaction: Like Sandmeyer but uses Cu powder instead of CuX salts, and gives the same products.

Kolbe reaction (Kolbe-Schmitt): Sodium phenoxide + CO $_2$ (high pressure) → sodium salicylate. This is how aspirin’s precursor is made.

Summary Table

Reaction	Substrate	Reagent	Product	Key condition
Wurtz	R-X	Na/dry ether	R-R (alkane)	Symmetric only
Friedel-Crafts alkylation	ArH + RCl	AlCl $_3$	ArR	No deactivated rings
Friedel-Crafts acylation	ArH + RCOCl	AlCl $_3$	ArCOR	No polyacylation
Cannizzaro	RCHO (no $\alpha$ -H)	Conc. NaOH	ROH + RCOONa	Must lack $\alpha$ -H
Aldol	RCHO (has $\alpha$ -H)	Dil. NaOH	$\beta$ -hydroxy carbonyl	Must have $\alpha$ -H
Hoffmann	RCONH $_2$	Br $_2$ /NaOH	RNH $_2$	Loses one carbon
Clemmensen	RCOR’	Zn-Hg/HCl	RCH $_2$ R’	Acidic medium
Wolff-Kishner	RCOR’	N $_2$ H $_4$ /KOH	RCH $_2$ R’	Basic medium
Sandmeyer	ArN $_2^+$	CuX	ArX	From diazonium salt

Worked Examples

Two benzaldehyde molecules in conc NaOH give benzyl alcohol and sodium benzoate. Must be without alpha H.

Nitrobenzene cannot undergo F-C because the NO $_2$ group is a strong deactivator, and the Lewis acid (AlCl $_3$ ) coordinates with the NO $_2$ lone pair, poisoning the catalyst.

To reduce a ketone to a CH $_2$ group: if the molecule is acid-stable, use Clemmensen (Zn-Hg/HCl). If it has acid-sensitive groups (like an acetal or an acid-labile protecting group), use Wolff-Kishner (N $_2$ H $_4$ /KOH). Both give the same product but under opposite pH conditions.

Toluene → side chain oxidation with KMnO $_4$ /H $^+$ → benzoic acid. This is not a named reaction but often confused with one. The named reaction route would be: toluene → benzaldehyde (by Etard reaction with CrO $_2$ Cl $_2$ ) → benzoic acid (by further oxidation).

Common Mistakes

Confusing Wurtz and Wurtz-Fittig. Wurtz is all alkyl; Wurtz-Fittig is alkyl + aryl.

Writing that Cannizzaro works with acetaldehyde. It does not — acetaldehyde has alpha hydrogens and goes aldol instead.

Saying Friedel-Crafts works with any aromatic. It fails with deactivated rings.

Confusing Clemmensen and Wolff-Kishner. Both reduce C=O to CH $_2$ , but Clemmensen is acidic (Zn-Hg/HCl) and Wolff-Kishner is basic (N $_2$ H $_4$ /KOH). Using the wrong one can destroy acid- or base-sensitive groups.

Forgetting that Hoffmann bromamide loses one carbon. If you start with a C $_3$ amide, you get a C $_2$ amine. Students often draw a product with the same carbon count.

Exam Weightage and Revision

This topic is a repeat performer in board papers and entrance exams. NEET typically asks one to two questions on the core mechanisms, CBSE boards give three to six marks, and state PMT papers often include a diagram-based long answer. The PYQs cluster around a small set of facts — lock those and you clear the topic.

JEE Main 2024 tested Friedel-Crafts acylation vs alkylation. NEET 2023 asked about the Hoffmann bromamide product. CBSE 2024 boards had a named reaction identification question. Two questions per year across exams is the norm.

When a question gives a scenario, identify the core mechanism first, then match it to the concepts above. Most wrong answers come from reading the scenario too quickly.

Make a two-column list — reaction name and one-line description. Memorise ten. NEET picks one question from this list every year.

Practice Questions

Q1. What product forms when formaldehyde undergoes Cannizzaro reaction?

$2\text{HCHO} + \text{NaOH} \rightarrow \text{CH}_3\text{OH} + \text{HCOONa}$ . One formaldehyde is reduced to methanol (CH $_3$ OH), and the other is oxidised to sodium formate (HCOONa). This works because HCHO has no $\alpha$ -hydrogens.

Q2. How would you prepare toluene from benzene using a named reaction?

Friedel-Crafts alkylation: $\text{C}_6\text{H}_6 + \text{CH}_3\text{Cl} \xrightarrow{\text{AlCl}_3} \text{C}_6\text{H}_5\text{CH}_3 + \text{HCl}$ . Alternatively, Wurtz-Fittig: $\text{C}_6\text{H}_5\text{Br} + \text{CH}_3\text{Br} + 2\text{Na} \rightarrow \text{C}_6\text{H}_5\text{CH}_3 + 2\text{NaBr}$ .

Q3. Why does acetaldehyde undergo aldol condensation but not Cannizzaro?

Acetaldehyde (CH $_3$ CHO) has $\alpha$ -hydrogens (the H atoms on CH $_3$ adjacent to the carbonyl). In dilute NaOH, the base removes an $\alpha$ -H to form an enolate ion, which attacks another aldehyde molecule — this is aldol condensation. Cannizzaro requires no $\alpha$ -H because the aldehyde cannot form an enolate.

Q4. What is the product of Hoffmann bromamide reaction on benzamide (C $_6$ H $_5$ CONH $_2$ )?

Aniline (C $_6$ H $_5$ NH $_2$ ). The amide loses one carbon (the C=O carbon) and gains an NH $_2$ group directly on the benzene ring. Benzamide (7C) → aniline (6C).

Q5. Name the reaction: $\text{C}_6\text{H}_5\text{OH} + \text{CHCl}_3 + \text{NaOH} \rightarrow \text{salicylaldehyde}$

Reimer-Tiemann reaction. The intermediate dichlorocarbene (:CCl $_2$ ) acts as an electrophile and attacks the ortho position of phenol. Hydrolysis of the intermediate gives salicylaldehyde (2-hydroxybenzaldehyde).

FAQs

How do I know whether an aldehyde undergoes Cannizzaro or aldol? Check for $\alpha$ -hydrogens. If the aldehyde has H atoms on the carbon adjacent to the C=O (like acetaldehyde, propanal), it undergoes aldol in dilute NaOH. If it has no $\alpha$ -H (like HCHO, benzaldehyde, pivaldehyde), it undergoes Cannizzaro in concentrated NaOH. The concentration of NaOH also matters — dilute for aldol, concentrated for Cannizzaro.

Why is Friedel-Crafts acylation preferred over alkylation? Acylation does not suffer from polysubstitution (the acyl group is deactivating), does not undergo rearrangement (the acylium ion is resonance-stabilised), and gives a single clean product. The ketone product can then be reduced to an alkyl group by Clemmensen or Wolff-Kishner if needed.

What is the difference between Sandmeyer and Gattermann reactions? Both convert diazonium salts to aryl halides. Sandmeyer uses CuX (CuCl, CuBr, CuCN) as catalyst. Gattermann uses Cu powder instead. The products are the same. Gattermann is sometimes considered simpler but less selective.

Can the Wurtz reaction be used for asymmetric alkanes? Technically yes, but practically no. Using two different alkyl halides (say CH $_3$ Cl and C $_2$ H $_5$ Cl) gives three products: ethane, butane, and propane. The desired cross-coupling product (propane) is only 1/3 of the mixture, making purification difficult and yield low.

Named reactions are organic chemistry’s greatest hits. Each one is a tool you add to your synthesis toolbox.