Differentiations & Conversions (Organic) — NEET Notes

Distinguishing tests — the diagnostic toolkit

A distinguishing test is a one-shot chemical assay that exploits the unique reactivity of a functional group. NEET expects you to know reagent, observation, and use for every test on the canonical list. Six tests dominate the question pool: Tollens, Fehling, Lucas, Hinsberg, Iodoform and Carbylamine. They appear by themselves, in matching-type questions, and embedded inside multi-step conversion problems where the test confirms the product.

Tollens, Fehling and Schiff — the aldehyde set

The three classical "aldehyde detectors" run on a single chemical idea — an aldehyde is easily oxidised, a ketone is not. Tollens' reagent is ammoniacal silver nitrate, [Ag(NH₃)₂]OH. The aldehyde reduces Ag⁺ to metallic silver, which deposits on the clean glass wall as a mirror. The ketone leaves the reagent unchanged. Aromatic aldehydes like benzaldehyde also give a positive Tollens — it is the most general aldehyde diagnostic.

Fehling's solution is a deep-blue complex of Cu²⁺ with sodium-potassium tartrate in alkaline medium. Warmed with an aldehyde, the Cu²⁺ is reduced to Cu⁺ and precipitates as red Cu₂O. The trap is selectivity: Fehling works for aliphatic aldehydes only; aromatic aldehydes give a Fehling-negative.

Schiff's reagent is fuchsin dye decolourised by SO₂. Aldehydes restore the magenta colour by sequestering the SO₂; ketones do not.

Iodoform test — the methyl ketone fingerprint

The iodoform test is one of NEET's most over-tested diagnostics. The reagent is iodine in alkali — written as I₂/NaOH or equivalently NaOI. The substrate is iodinated three times on a methyl group adjacent to a carbonyl, then cleaved by hydroxide to release the yellow precipitate of iodoform, CHI₃. The smell is unmistakable: antiseptic, like an old hospital.

The structural rule is sharp. A compound is iodoform-positive if it contains either:

• a CH₃-CO- group (a methyl ketone or acetaldehyde itself), or
• a CH₃-CH(OH)- group (an alcohol that can be oxidised to a methyl ketone in situ; ethanol counts because oxidation gives acetaldehyde).

So acetaldehyde, acetone, butan-2-one, acetophenone, ethanol and isopropanol are all iodoform-positive. Methanol, formaldehyde, propanal, propan-1-ol and benzaldehyde are all negative. The test cleaves the carbon chain: the product is iodoform plus the carboxylate of one less carbon.

Lucas test — classifying alcohols by SN1 reactivity

The Lucas test transforms an alcohol into an alkyl chloride in the presence of concentrated HCl and anhydrous ZnCl₂. ZnCl₂ is a Lewis acid: it activates the hydroxyl by coordinating to the oxygen, making the C-O bond easier to break. The reaction proceeds through a carbocation (SN1) for 2 degrees and 3 degrees alcohols, and through SN2 only for 1 degrees alcohols. The alkyl chloride is insoluble in the aqueous Lucas reagent and appears as a separate cloudy layer — hence the term turbidity.

Because carbocation stability is 3 degrees > 2 degrees > 1 degrees, the time taken to develop turbidity is the diagnostic clock.

Hinsberg test — sorting 1 degrees, 2 degrees, 3 degrees amines

Hinsberg's reagent is benzenesulphonyl chloride, C₆H₅SO₂Cl, reacted with the unknown amine in aqueous KOH. The principle is the same in all three cases — the N-H of the amine attacks SO₂Cl, displacing Cl^- to form a sulphonamide. What differs is whether the sulphonamide has any N-H left, and whether that N-H is acidic enough to deprotonate in alkali.

A 1 degrees amine (R-NH₂) forms R-NH-SO₂Ph. The remaining N-H is strongly acidic — sulphonyl is electron-withdrawing — and dissolves in KOH as the sodium salt. Observation: clear solution. Acidification later restores the solid sulphonamide.

A 2 degrees amine (R₂NH) forms R₂N-SO₂Ph. No N-H remains, so there is nothing for KOH to deprotonate. Observation: insoluble in alkali — the sulphonamide stays as a solid.

A 3 degrees amine (R₃N) has no N-H at all and cannot displace chloride. Observation: no reaction; the tertiary amine separates as an oily layer that can be removed and recovered unchanged.

Carbylamine test & the azo dye test

The carbylamine reaction — also called the isocyanide test — converts a primary amine into an isocyanide (R-N≡C) when heated with chloroform and alcoholic KOH. The product is one of the most foul-smelling compounds in organic chemistry; even microgram quantities are detectable by nose. The mechanism runs through a dichlorocarbene intermediate generated by KOH acting on CHCl₃; the carbene inserts into the N-H bond and the dichloride eliminates to leave the N=C linkage. Only primary amines (aliphatic or aromatic) have the right N-H pattern; secondary and tertiary amines give no carbylamine.

R-NH₂ + CHCl₃ + 3 KOH  →  R-N≡C + 3 KCl + 3 H₂O

The azo dye test is the analogous diagnostic for primary aromatic amines. Treat aniline (or any ArNH₂) with NaNO₂ + HCl at 0–5 degrees C; the diazonium salt ArN₂⁺ forms. Couple this with an alkaline solution of β-naphthol or N,N-dimethylaniline, and an intensely coloured azo dye (typically orange-red to scarlet) precipitates. Aliphatic primary amines instead form an unstable diazonium that decomposes to alcohol + N₂ — no dye. The azo test therefore separates aromatic 1 degrees amines from everything else.

Bromine water, Beilstein and Lassaigne — the supporting tests

Bromine water is a reddish-brown solution of Br₂. When shaken with an alkene or alkyne, the colour disappears as the bromine adds across the multiple bond to form a colourless vicinal dibromide. The test distinguishes saturated alkanes (no reaction) from unsaturated hydrocarbons (instant decolourisation). Phenols and anilines, being activated aromatics, also decolourise bromine water — and additionally precipitate 2,4,6-tribromophenol or 2,4,6-tribromoaniline. NEET often pairs this with a "white precipitate" clue.

The Beilstein test screens for halogens. A clean copper wire is dipped in the unknown and held in a Bunsen flame; if halogen is present, the flame turns bluish-green from volatile copper halide. The Beilstein test is fast but not specific — some non-halogen compounds give false positives.

The sodium fusion test (Lassaigne's) is the definitive screen for N, S, and halogens. The sample is fused with sodium metal, converting any nitrogen to NaCN, sulphur to Na₂S and halogen to NaX. The aqueous extract then reveals:

• Nitrogen: Prussian-blue precipitate after adding FeSO₄ + acidified FeCl₃.
• Sulphur: violet colouration with sodium nitroprusside, or black PbS with lead acetate.
• Halogen: white AgCl, pale yellow AgBr or yellow AgI with AgNO₃/HNO₃.

Wurtz & Wurtz-Fittig — coupling halides into hydrocarbons

The Wurtz reaction couples two alkyl halides in the presence of sodium metal in dry ether to give a symmetrical alkane of double the carbon count. The mechanism is radical: Na donates an electron to R-X, forming R• and Na⁺X^-; two R• radicals couple to give R-R. The reaction is the classical route to even-carbon symmetrical alkanes; ethane from CH₃I, butane from C₂H₅I, and so on. The major NEET trap is that Wurtz with two different alkyl halides gives a mess of three products and is therefore not useful for unsymmetrical alkanes.

The Wurtz-Fittig reaction couples an aryl halide with an alkyl halide in the presence of sodium, again in dry ether. The product is an alkylbenzene — toluene from bromobenzene + methyl iodide, ethylbenzene from bromobenzene + ethyl iodide. The aryl-alkyl coupling beats the two symmetric couplings (which still occur as side reactions) when one excess reagent dominates the radical pool.

Sandmeyer — diazonium to aryl halide

The Sandmeyer reaction converts an aryl diazonium salt (ArN₂⁺) into an aryl halide using a copper(I) catalyst. The diazonium is generated from a primary aromatic amine using NaNO₂/HCl at 0–5 degrees C. The cuprous salt then donates its halide to the aryl group while N₂ is expelled.

• ArN₂⁺ + CuCl → Ar-Cl + N₂
• ArN₂⁺ + CuBr → Ar-Br + N₂
• ArN₂⁺ + CuCN → Ar-CN + N₂

The reaction is irreplaceable because direct halogenation of benzene with Cl₂/FeCl₃ cannot select a specific ring position when other directing groups are present, whereas the -NH₂ group can be installed in advance and then swapped for Cl, Br or CN via the diazonium. Aryl fluorides are made instead by the Balz-Schiemann variant (ArN₂⁺BF₄^- on heating), while aryl iodides come from a simple KI exchange.

Reimer-Tiemann & Kolbe — formylation and carboxylation of phenol

The Reimer-Tiemann reaction introduces a -CHO group at the ortho position of phenol. Phenol is treated with CHCl₃ + aqueous NaOH at 340 K; the alkali generates dichlorocarbene (:CCl₂), which attacks the ortho-position of the phenoxide ion. Hydrolysis of the resulting gem-dichloride gives salicylaldehyde (2-hydroxybenzaldehyde) as the major product. A small amount of para isomer also forms but ortho dominates because of intramolecular hydrogen bonding stabilising the transition state.

The Kolbe reaction (Kolbe-Schmitt) introduces a -COOH group at the ortho position of phenol. Sodium phenoxide is heated with CO₂ at 400 K and 4–7 atm. The phenoxide carbon attacks CO₂, giving (after acidification) salicylic acid (2-hydroxybenzoic acid). Salicylic acid is the precursor of aspirin (acetylsalicylic acid) — esterify the -OH with acetic anhydride and the synthesis is complete.

Friedel-Crafts alkylation & acylation

Friedel-Crafts reactions add a carbon substituent to an aromatic ring in the presence of a Lewis acid catalyst — most commonly anhydrous AlCl₃.

Friedel-Crafts alkylation uses an alkyl halide. The Lewis acid abstracts the halide to give a carbocation that then attacks the ring:

C₆H₆ + R-Cl  ^AlCl₃→  C₆H₅-R + HCl

Alkylation has two famous limitations. First, the alkyl group it installs is activating, so the product is more nucleophilic than the starting material and is alkylated again — giving over-alkylation. Second, rearrangement of the carbocation produces unexpected products: for example, n-propyl chloride gives largely iso-propylbenzene because the primary carbocation rearranges to the more stable secondary one.

Friedel-Crafts acylation avoids both problems. It uses an acid chloride (R-COCl) or anhydride [(R-CO)₂O] to install a -COR group. The acylium ion R-C⁺=O is resonance-stabilised and does not rearrange; the resulting ketone is deactivated, so monoacylation is clean and selective.

Aldol & crossed aldol — α-carbon couplings

An aldol reaction couples two carbonyl molecules through their α-carbons. The first molecule loses an α-H to base, becoming a nucleophilic enolate; the enolate attacks the carbonyl carbon of the second molecule, giving a β-hydroxy aldehyde or ketone — the "aldol." Heating, or strong acid, dehydrates the aldol to an α,β-unsaturated carbonyl. Classical aldol on acetaldehyde gives 3-hydroxybutanal; on heating, this loses water to give but-2-enal (crotonaldehyde).

The aldol is only useful between two identical carbonyls (self-condensation) — otherwise four products result from random mixing. The crossed aldol overcomes this when one of the two partners has no α-hydrogen. Then only the partner with α-H can enolise; the partner without α-H acts purely as the electrophile. Benzaldehyde + acetaldehyde, formaldehyde + acetone, and benzaldehyde + acetone are all clean crossed aldols.

Cannizzaro & crossed Cannizzaro — disproportionation without α-H

The Cannizzaro reaction is the disproportionation of an aldehyde lacking α-hydrogens in the presence of concentrated alkali. One molecule of the aldehyde is oxidised to the carboxylate; another is reduced to the primary alcohol. The hydride is transferred directly between the two molecules through an OH^--addition intermediate. Aldehydes with α-H instead choose the aldol pathway, so the two reactions are mutually exclusive on the same substrate.

Classical examples: 2 HCHO + conc. NaOH → HCOONa + CH₃OH. 2 C₆H₅CHO + conc. KOH → C₆H₅COOK + C₆H₅CH₂OH. Trimethylacetaldehyde (pivaldehyde) and furfural also undergo Cannizzaro.

The crossed Cannizzaro uses formaldehyde (HCHO) as a sacrificial reductant against a second non-enolisable aldehyde. Formaldehyde is the more easily oxidised partner, so it is preferentially oxidised to formate while the other aldehyde is reduced to its alcohol. This is the standard way to make benzyl alcohol from benzaldehyde — HCHO + PhCHO → HCOONa + PhCH₂OH.

HVZ & Hofmann bromamide — installing N at α and at C-1

The Hell-Volhard-Zelinsky (HVZ) reaction halogenates the α-carbon of a carboxylic acid bearing an α-H. The acid is treated with Cl₂ or Br₂ in the presence of a catalytic quantity of red phosphorus. The α-halo acid (R-CHX-COOH) is the platform from which α-hydroxy acids (hydrolysis), α-amino acids (ammonolysis) and α,β-unsaturated acids (elimination) are all reached. HVZ does not work on formic acid (no α-C) or on acids without an α-H such as benzoic acid or pivalic acid.

The Hofmann bromamide degradation converts a primary amide (R-CONH₂) to a primary amine (R-NH₂) with the loss of one carbon. Treat the amide with bromine and aqueous (or alcoholic) KOH. The mechanism passes through an N-bromoamide, a nitrene-like intermediate and an isocyanate that loses CO₂ on hydrolysis to leave R-NH₂. Hofmann bromamide is invaluable when a pure 1 degrees amine is needed — direct ammonolysis of alkyl halides gives mixtures, but Hofmann gives a clean primary amine.

Functional group interconversion — the roadmap

The endgame of organic chemistry on NEET is the functional-group ladder. Every NEET conversion question reduces to walking up or down this ladder: RH ↔ RX ↔ ROH ↔ RCHO ↔ RCOOH ↔ RCONH₂ ↔ RNH₂. Memorise both directions of every arrow. Knowing the ladder lets you answer "convert ethane to ethylamine in five steps" without thinking — you just walk left to right.

Reverse arrows matter equally. ROH → RX with SOCl₂, PCl₃, PCl₅ or HX. RCOOH → ROH with LiAlH₄. RNH₂ → ROH with HNO₂ at 0 degrees C (aliphatic) or via diazonium (aromatic). RCN → RNH₂ with LiAlH₄ or H₂/Ni; RCN → RCOOH by acidic hydrolysis. Internalising both directions of every link is what separates rote-memorisers from rank-shifters.