Class 12 Chemistry · NCERT chapter 13 · 6 topics · free to read
Classification & Nomenclature
Amines are derivatives of ammonia where one, two, or three H atoms are replaced by alkyl/aryl groups, giving primary (1°), secondary (2°), and tertiary (3°) amines. IUPAC names use the suffix '-amine' on the parent chain. Classification determines basicity, reactions, and identification tests.
R—NH_2 (1°), R_2NH (2°), R_3N (3°)
What are Amines?
Amines are organic derivatives of ammonia (NH₃) where one or more hydrogen atoms are replaced by alkyl or aryl groups. They are the most important class of organic nitrogen compounds and serve as building blocks for drugs, dyes, and neurotransmitters.
Start with NH_3 . Replace one H with an organic group R — you get RNH_2 , a primary amine. Replace two — R_2NH , a secondary amine. Replace all three — R_3N , a tertiary amine. The nitrogen still has a lone pair in every case, which is why amines act as bases and nucleophiles.
Primary (1°) amine: one H replaced → RNH_2 (e.g., CH_3NH_2 , methylamine)
All three classes have a lone pair on nitrogen — the source of basicity and nucleophilicity
If a 4th group bonds to N (forming R_4N^+ ), it's a quaternary ammonium salt — not an amine (no lone pair)
Aliphatic vs Aromatic Amines
Amines are further classified as aliphatic (alkyl groups on N) or aromatic (at least one aryl group directly on N). Arylamines like aniline behave very differently from alkylamines due to resonance between the lone pair and the benzene ring.
If all groups on N are alkyl → aliphatic amine (e.g., ethylamine C_2H_5NH_2 ). If at least one group is aryl (directly bonded benzene ring) → aromatic amine (e.g., aniline C_6H_5NH_2 ). This distinction is critical because the benzene ring's resonance withdraws the nitrogen's lone pair, making arylamines much weaker bases than alkylamines.
Only alkyl groups on N
At least one aryl group on N
N lone pair freely available
N lone pair delocalized into ring
Stronger bases ( pK_b 3–4 )
Weaker bases ( pK_b 9.4 for aniline)
E.g., CH_3NH_2 , (C_2H_5)_2NH
E.g., C_6H_5NH_2 (aniline), C_6H_5NHCH_3
IUPAC Naming
In IUPAC nomenclature, amines are named by replacing '-e' of the parent alkane with '-amine'. For secondary and tertiary amines, the largest alkyl chain is the parent and other groups are named with the prefix 'N-'. Aromatic amines use 'benzenamine' (though 'aniline' is the accepted IUPAC name).
The IUPAC system treats -NH_2 as a suffix (unlike halogens which are prefixes). The steps:
Step 2: Replace the '-e' of the alkane name with '-amine' (ethane → ethanamine)
Step 3: Number the chain to give the lowest locant to -NH_2
Step 4: For 2° and 3° amines, choose the largest group as parent; other groups get N- prefix (e.g., N -methylethanamine)
Step 5: If -NH_2 is a substituent (not the principal group), use prefix 'amino-'
Quick recall
Give the IUPAC name of CH_3CH_2NHCH_3 .
N -methylethanamine. The longer chain is ethane (2C) → parent = ethanamine. The methyl on N gets the N- prefix.
Common Naming System
In common naming, amines are named by listing all alkyl/aryl groups on nitrogen in alphabetical order, followed by the suffix '-amine' (written as one word).
Common names are simpler: just list the groups on N alphabetically and add '-amine'. For identical groups, use di-/tri- prefixes.
CH_3NH_2 → methylamine (one methyl group on N)
(CH_3)_2NH → dimethylamine (two methyl groups on N)
CH_3NHC_2H_5 → ethylmethylamine (alphabetical: ethyl before methyl)
C_6H_5CH_2NH_2 → benzylamine (not 'phenylmethylamine' in common naming)
NEET questions on nomenclature are typically 1-mark direct questions. Know both IUPAC and common names for all amines up to 4 carbons — they appear almost every year.
Preparation of Amines
Amines are prepared by reduction of nitro compounds (Sn/HCl or catalytic hydrogenation), ammonolysis of alkyl halides (gives mixture of 1°/2°/3°), Gabriel phthalimide synthesis (pure 1° amines only), and Hofmann bromamide degradation (amine with one carbon less than the amide). Each method has distinct selectivity.
Nitro groups (–NO₂) can be reduced to amino groups (–NH₂) using Sn/HCl, Fe/HCl, or catalytic hydrogenation (H₂/Pd). This is the most important method for preparing aromatic amines like aniline.
Nitrobenzene → aniline is the single most important amine preparation in NCERT. The nitro group is reduced step by step: -NO_2 arrow -NO arrow -NHOH arrow -NH_2 .
C_6H_5NO_2 Sn/HCl C_6H_5NH_2
Sn/HCl (or Fe/HCl): most common laboratory method for aniline from nitrobenzene
Catalytic hydrogenation ( H_2 /Pd or Pt or Ni): clean reduction, no side products, works for both aliphatic and aromatic nitro compounds
Acidic medium (HCl) initially gives anilinium salt C_6H_5NH_3^+Cl^- ; treat with NaOH to liberate free aniline
For aliphatic nitro compounds: R—NO_2 H_2/Pd R—NH_2 (LiAlH _4 also works)
Ammonolysis of Alkyl Halides
Alkyl halides react with ammonia via nucleophilic substitution to give amines. The problem: the product (1° amine) is itself a nucleophile and reacts further, giving a mixture of 1°, 2°, 3° amines and quaternary ammonium salt.
Ammonia ( NH_3 ) is a nucleophile. It attacks the electrophilic carbon of an alkyl halide ( R—X ) via S_N2 . But the product RNH_2 is also a nucleophile — so it attacks another molecule of R—X to give R_2NH , which attacks again to give R_3N , and finally R_4N^+X^- .
R—X NH_3 RNH_2 R—X R_2NH R—X R_3N R—X R_4N^+X^-
Ammonolysis = cleavage of C—X bond by NH_3 (nucleophilic substitution)
Product is always a mixture — 1°, 2°, 3° amines + quaternary salt
Using excess NH₃ favors the 1° amine (by probability — most NH₃ molecules react first)
Works best with 1° alkyl halides (SN2 mechanism)
Aryl halides ( Ar—X ) do NOT undergo ammonolysis under normal conditions (C—X bond too strong)
Gabriel Phthalimide Synthesis
Phthalimide is treated with KOH to form potassium phthalimide, which reacts with alkyl halides. Hydrolysis of the N-alkylphthalimide gives a pure primary amine. The method fails for aryl halides (no SN2 on sp² carbon).
The Gabriel synthesis solves the mixture problem of ammonolysis. The nitrogen in phthalimide is flanked by two C=O groups, making the N—H acidic. Treatment with KOH gives the potassium salt, which is a good nucleophile but can only undergo one alkylation (no further substitution possible because both N—H bonds are gone).
Step 3: Hydrolysis with NaOH_ (aq) or hydrazinolysis ( NH_2NH_2 ) → pure 1° amine + phthalic acid/phthalhydrazide
Only gives primary amines — impossible to get 2° or 3° (N has no more H to replace)
Fails for aryl halides — Ar—X doesn't undergo SN2
Quick recall
Can you prepare aniline by Gabriel phthalimide synthesis?
No. Gabriel synthesis requires an SN2 step with R—X. Aryl halides (like C_6H_5Cl ) don't undergo SN2 because the C—Cl bond has partial double-bond character and the sp^2 carbon is sterically shielded. Use nitrobenzene reduction instead.
Hofmann Bromamide Degradation
A primary amide (RCONH₂) reacts with Br₂ and NaOH to give a primary amine with one carbon fewer than the parent amide. This is a degradation reaction — the carbonyl carbon is lost as Na₂CO₃.
The Hofmann bromamide reaction is unique: it removes one carbon from the chain. An amide with n carbons gives an amine with n-1 carbons. The carbonyl carbon is lost as carbonate.
Nitriles (R—C≡N) are reduced to primary amines by LiAlH₄ or catalytic hydrogenation (H₂/Ni). Amides (RCONH₂) are reduced to amines by LiAlH₄. Both give primary amines with the same carbon count as the starting material.
Nitrile reduction is a chain-preserving method — the product amine has the same number of carbons as the nitrile (unlike Hofmann degradation which loses one). LiAlH_4 is the reagent of choice in the lab; catalytic H_2 /Ni works industrially.
R—C N LiAlH_4 or H_2/Ni R—CH_2—NH_2
Nitrile → 1° amine: R—CN 4[H] R—CH_2NH_2 (carbon count stays same; one extra C comes from CN)
LiAlH _4 reduces both nitriles and amides but NOT nitro groups efficiently (use Sn/HCl for that)
Catalytic hydrogenation ( H_2 /Pd-C or Ni) is milder and works for nitriles
Quick recall
How would you convert ethanenitrile ( CH_3CN ) to ethanamine ( CH_3CH_2NH_2 )?
Reduce with LiAlH_4 (in dry ether) followed by hydrolysis, OR use H_2 /Ni. Both give CH_3CH_2NH_2 — a 1° amine with 2 carbons (same as the nitrile).
Physical Properties
Amines have lower boiling points than alcohols but higher than alkanes of similar molecular mass due to N—H hydrogen bonding (weaker than O—H). Lower amines are water-soluble due to H-bonding with water. Amines have characteristic fishy smell. 3° amines have no N—H bonds, so no intermolecular H-bonding among themselves.
Primary and secondary amines can form intermolecular N—H···N hydrogen bonds. Since N is less electronegative than O, these H-bonds are weaker than O—H···O bonds in alcohols, resulting in lower boiling points.
Nitrogen ( EN = 3.0 ) is less electronegative than oxygen ( EN = 3.5 ). So the N—H bond is less polar than O—H , and the hydrogen bonds between amine molecules are weaker than those between alcohol molecules.
1° amines ( RNH_2 ): two N—H bonds → can form H-bonds as both donor and acceptor
2° amines ( R_2NH ): one N—H bond → weaker H-bonding than 1° amines
3° amines ( R_3N ): no N—H bond → cannot donate H-bonds (can only accept via lone pair)
H-bond strength: O—H > N—H > N—H (no H-bond for 3° amines among themselves)
Boiling Points
For similar molecular mass: alcohols > amines > alkanes. Among amines, 1° > 2° > 3° for the same molecular formula because of decreasing ability to form intermolecular H-bonds.
Boiling point depends on intermolecular forces. More H-bonding → higher bp. Since O—H bonds are stronger than N—H bonds, alcohols boil higher than amines of similar molecular mass. Alkanes, with only London forces, boil lowest.
C_2H_5OH (ethanol, MW 46)
78°C
C_2H_5NH_2 (ethylamine, MW 45)
17°C
CH_3OCH_3 (dimethyl ether, MW 46)
−25°C
C_3H_8 (propane, MW 44)
−42°C
Among isomeric amines (same molecular formula): 1° > 2° > 3° boiling point
Reason: 1° has 2 N—H bonds (more H-bonding), 2° has 1 N—H, 3° has 0 N—H
As chain length increases, boiling point increases (more London forces)
Aromatic amines have higher bp than aliphatic amines of similar MW (stronger London forces from the ring)
Solubility in Water
Lower amines (up to ~5 carbons) are soluble in water because they can form H-bonds with water. Solubility decreases as the hydrophobic alkyl chain grows. Aniline is sparingly soluble in water but dissolves in dilute HCl.
Amines form N—H and O—H hydrogen bonds with water molecules. As the hydrocarbon portion grows, the hydrophobic effect dominates and solubility drops.
Methylamine, dimethylamine, trimethylamine — all are gases at room temperature and highly soluble in water
Amines up to ~5 carbon atoms are appreciably soluble in water
Higher amines (≥6 C) are practically insoluble in water
Aniline is only slightly soluble in water ( 3.6 g/100 mL at 20°C) — the hydrophobic benzene ring dominates
Aniline dissolves readily in dilute HCl → forms water-soluble anilinium chloride: C_6H_5NH_2 + HCl arrow C_6H_5NH_3^+Cl^-
Quick recall
Why does aniline dissolve in dilute HCl but not in water?
Aniline is a base. Dilute HCl protonates it to form anilinium chloride ( C_6H_5NH_3^+Cl^- ), which is an ionic compound and dissolves readily in water. Pure aniline's hydrophobic benzene ring makes it sparingly soluble.
Basicity of Amines
Amine basicity depends on the availability of the nitrogen lone pair. Alkyl groups (+I effect) increase basicity while aryl groups (resonance) decrease it. The basicity order in aqueous solution is 2° > 1° > 3° (for methyl series) due to the combined effect of induction, solvation, and steric factors. Aniline is a very weak base because the lone pair is delocalized into the benzene ring.
R_2NH > RNH_2 > R_3N > NH_3 > C_6H_5NH_2 (in water, for CH_3 series)
What Makes Amines Basic?
Amines are basic because the lone pair on nitrogen can accept a proton (Brønsted base) or donate to a Lewis acid. Anything that makes the lone pair more available increases basicity; anything that delocalizes or shields it decreases basicity.
The lone pair on nitrogen is the sole source of basicity. When an amine accepts a proton from water: R—NH_2 + H_2O leftharpoons R—NH_3^+ + OH^- The equilibrium constant for this is K_b . Higher K_b (or lower pK_b ) = stronger base.
Electron-donating groups (alkyl, +I effect) → push electron density onto N → lone pair more available → stronger base
Electron-withdrawing groups ( -NO_2 , -CN , −I effect) → pull electron density away from N → weaker base
Resonance (aryl group on N) → lone pair delocalized into ring → much weaker base
Steric hindrance (bulky groups around N) → hard for H^+ to approach → reduces basicity
Solvation of the conjugate acid ( R—NH_3^+ ) stabilizes it → increases basicity in solution
Inductive Effect on Basicity
Alkyl groups have a +I (electron-donating) effect that increases electron density on nitrogen, making the lone pair more available for protonation. More alkyl groups → stronger +I effect → expected to increase basicity.
Alkyl groups are electron-donating (+I effect). Attaching more alkyl groups to nitrogen should progressively increase electron density on N and make the amine a stronger base. If only inductive effect mattered, the basicity order would be: 3° > 2° > 1° > NH_3 .
In gas phase, 3° is the strongest base — exactly what +I effect predicts
But in aqueous solution, the order changes — this is where the anomaly lies
The gas phase vs solution discrepancy is one of the most asked topics in NEET. Understanding this single anomaly can fetch you 4-8 marks over your career.
Basicity Order in Solution
In aqueous solution, the basicity order for methylamines is: (CH₃)₂NH > CH₃NH₂ > (CH₃)₃N > NH₃. The 2° amine is the strongest base (not 3°) because of the combined interplay of +I effect, solvation of the conjugate acid, and steric hindrance.
Here's the surprise: in water, dimethylamine (2°) is the strongest base, not trimethylamine (3°). Three competing effects determine the final order:
+I effect: favors 3° > 2° > 1° > NH₃ (more alkyl groups = more electron donation)
Solvation effect: the conjugate acid ( R_nNH_ 4-n ^+ ) must be stabilized by H-bonding with water. More N—H bonds on the cation → better solvation. Favors 1° > 2° > 3°
Steric effect: bulky alkyl groups hinder proton approach and hinder solvation of cation. Hurts 3° the most
Net result for CH_3 series in water: (CH_3)_2NH > CH_3NH_2 > (CH_3)_3N > NH_3
In aniline, the nitrogen lone pair is delocalized into the benzene ring by resonance, making it less available for protonation. This is why aniline (pKb = 9.4) is ~10⁶ times weaker a base than methylamine (pKb = 3.4).
In aniline ( C_6H_5NH_2 ), the lone pair on nitrogen is in a p -orbital that overlaps with the -system of the benzene ring. This delocalization means the lone pair is shared between nitrogen and the ring — it's not fully available to grab a proton.
pK_b(aniline) = 9.4 vs pK_b(CH_3NH_2) = 3.4
Aniline's lone pair is delocalized into the ring → 5 resonance structures (lone pair on N, and on ortho and para C atoms)
The C—N bond in aniline has partial double bond character (shorter than C—N in alkylamines)
Aniline is ~ 10^6 times weaker base than CH_3NH_2 in water
Electron-withdrawing groups (e.g., -NO_2 ) on the ring further decrease basicity
Electron-donating groups (e.g., -CH_3 , -OCH_3 ) on the ring increase basicity slightly
Quick recall
Arrange in order of increasing basicity: aniline, p -nitroaniline, p -toluidine, cyclohexylamine.
p -nitroaniline < aniline < p -toluidine < cyclohexylamine. Cyclohexylamine is an aliphatic amine (no resonance with ring) → strongest. p -Nitroaniline has the strongest EWG → weakest.
Chemical Reactions of Amines
Amines undergo acylation (Schotten-Baumann), reaction with nitrous acid (HNO₂ — the key distinguishing test for 1°/2°/3° amines), carbylamine test (isocyanide test for 1° amines), Hinsberg test (benzenesulfonyl chloride to distinguish all three classes), and electrophilic aromatic substitution (aniline is strongly activating, ortho/para directing).
R—NH_2 + CHCl_3 + 3KOH arrow R—NC + 3KCl + 3H_2O
Acylation
Primary and secondary amines react with acid chlorides (RCOCl), anhydrides ((RCO)₂O), or esters to replace an N—H hydrogen with an acyl group (–COR), forming substituted amides. Tertiary amines have no N—H, so they don't undergo acylation.
Acylation replaces an N—H hydrogen with an acyl group ( —COR ). The reaction is a nucleophilic substitution at the acyl carbon. Since 3° amines have no N—H, they cannot be acylated.
The reaction is called Schotten-Baumann reaction when done in aqueous NaOH
Acetylation of aniline: C_6H_5NH_2 + (CH_3CO)_2O arrow C_6H_5NHCOCH_3 (acetanilide)
Reaction with Nitrous Acid (HNO₂)
HNO₂ (generated in situ from NaNO₂ + HCl at 0–5°C) reacts differently with 1°, 2°, and 3° amines. Primary aliphatic → unstable diazonium (→ alcohol + N₂). Primary aromatic → stable diazonium salt. Secondary → N-nitrosoamine (yellow oily liquid). Tertiary aliphatic → no clear reaction. Tertiary aromatic → C-nitroso compound (electrophilic substitution on ring).
Nitrous acid ( HNO_2 ) is unstable and made in situ: NaNO_2 + HCl arrow HNO_2 + NaCl . It reacts with the lone pair on N, and the product depends entirely on the amine class.
1° aliphatic amine + HNO_2 → unstable diazonium salt → decomposes to alcohol + N_2 gas (brisk effervescence)
1° aromatic amine + HNO_2 at 0–5°C → stable diazonium salt ArN_2^+Cl^- (crucial intermediate — see next concept)
2° amine (both aliphatic & aromatic) + HNO_2 → N -nitrosoamine ( R_2N—N=O ) — yellow oily substance, insoluble in water
3° aliphatic amine + HNO_2 → forms a soluble salt (no useful reaction)
3° aromatic amine + HNO_2 → C -nitroso compound (electrophilic substitution at para position of ring)
1° aliphatic
Alcohol + N₂↑ (diazonium decomposes)
1° aromatic (0–5°C)
Stable diazonium salt ArN₂⁺Cl⁻
2° (any)
N-nitrosoamine (yellow oil)
3° aliphatic
Ammonium salt (no useful product)
3° aromatic
p-Nitroso compound (ring substitution)
Carbylamine (Isocyanide) Test
Primary amines (and only 1°) react with chloroform and alcoholic KOH to form isocyanides (carbylamines), which have an extremely offensive smell. This is a definitive test for primary amines — 2° and 3° amines do not respond.
Heat a primary amine with chloroform ( CHCl_3 ) and alcoholic KOH . If you detect a foul, nauseating smell (isocyanide/carbylamine), the amine is primary.
R—NH_2 + CHCl_3 + 3KOH R—NC + 3KCl + 3H_2O
Only 1° amines give this test — both aliphatic and aromatic
Product: isocyanide ( R—NC ) — has a horrible, penetrating odor
2° and 3° amines do NOT give carbylamine test (no two N—H bonds available)
The test is also used to detect CHCl_3 — warm with aniline + KOH; foul smell = chloroform present
Mechanism involves dichlorocarbene ( :CCl_2 ) as intermediate, generated from CHCl_3 + KOH
Hinsberg Test
Benzenesulfonyl chloride (C₆H₅SO₂Cl, Hinsberg's reagent) reacts with 1° amines to give a sulfonamide soluble in NaOH (acidic N—H), with 2° amines to give a sulfonamide insoluble in NaOH (no N—H), and gives no reaction with 3° amines.
The Hinsberg test is the single test that distinguishes all three classes of amines in one shot. The key lies in whether the sulfonamide product has an acidic N—H or not.
1° amine + C_6H_5SO_2Cl → N -alkylsulfonamide ( RSO_2NHR' ) → dissolves in NaOH (N—H is acidic, forms soluble sodium salt)
2° amine + C_6H_5SO_2Cl → N,N -dialkylsulfonamide ( RSO_2NR'_2 ) → insoluble in NaOH (no N—H, not acidic)
3° amine → no reaction with C_6H_5SO_2Cl (no N—H to substitute)
The sulfonamide from 1° amine has one N—H → weakly acidic ( pK_a 10 ) due to electron-withdrawing -SO_2- group
1° amine
Precipitate forms → dissolves in NaOH → reappears on acidification
2° amine
Precipitate forms → does NOT dissolve in NaOH
3° amine
No precipitate (no reaction)
Electrophilic Substitution of Aniline
The –NH₂ group is a strong activating group and ortho/para director in electrophilic aromatic substitution. Aniline undergoes bromination, nitration, and sulphonation, but special care is needed because –NH₂ is so activating that polysubstitution occurs easily.
The -NH_2 group donates electron density into the benzene ring through resonance (lone pair → ring). This makes the ring highly electron-rich, especially at ortho and para positions, causing very fast electrophilic substitution.
Bromination: Aniline + Br_2 (aq) → 2,4,6-tribromoanilne (white precipitate) — no Lewis acid catalyst needed!
Mono-bromination: First acetylate aniline → brominate → hydrolyze: gives p -bromoaniline selectively
Nitration: Direct nitration ( HNO_3/H_2SO_4 ) oxidizes -NH_2 → must protect by acetylation first, then nitrate, then hydrolyze
Sulphonation: Aniline + conc. H_2SO_4 at 180°C → sulfanilic acid ( p -aminobenzenesulfonic acid) — used in making sulfa drugs
-NH_2 is one of the strongest activating groups — even stronger than -OH and -OR
Quick recall
Why does aniline give 2,4,6-tribromoanilne directly with Br_2 /water, without any catalyst?
The -NH_2 group is so strongly activating (lone pair donation into ring) that the ring is electron-rich enough to react with Br_2 directly — no Lewis acid ( FeBr_3 ) needed. All three ortho/para positions get brominated simultaneously because of the extreme activation.
Diazonium Salts
Aromatic diazonium salts (ArN₂⁺X⁻) are prepared by treating 1° aromatic amines with NaNO₂ + HCl at 0–5°C (diazotization). They are versatile synthetic intermediates: Sandmeyer reaction replaces N₂⁺ with Cl/Br/CN, Schiemann gives ArF, and coupling with phenols/amines produces azo dyes. The N₂⁺ group is the best leaving group in organic chemistry.
Treatment of a primary aromatic amine with nitrous acid (NaNO₂ + HCl) at 0–5°C converts the –NH₂ group into a –N₂⁺ (diazonium) group. The low temperature is critical — above 5°C, the diazonium salt decomposes to phenol + N₂.
Diazotization converts -NH_2 into -N_2^+ (diazo group). The reaction must be done at 0–5°C in an ice bath because diazonium salts are thermally unstable and decompose above ~5°C.
HNO_2 is generated in situ: NaNO_2 + HCl arrow HNO_2 + NaCl
Temperature must be 0–5°C — higher temperatures decompose the product
Only primary aromatic amines give stable diazonium salts (aliphatic ones decompose instantly)
The diazonium group ( -N_2^+ ) is an excellent leaving group — N_2 is a very stable molecule
Benzenediazonium chloride is a white crystalline solid, stable in cold aqueous solution
Sandmeyer Reaction
The diazonium group is replaced by Cl⁻, Br⁻, or CN⁻ when treated with the corresponding cuprous salt (CuCl, CuBr, or CuCN) in the presence of the corresponding acid. This is called the Sandmeyer reaction.
In the Sandmeyer reaction, cuprous salts ( Cu_2Cl_2 , Cu_2Br_2 , or CuCN ) catalyze the replacement of -N_2^+ with the corresponding nucleophile. This allows you to introduce -Cl , -Br , or -CN onto the benzene ring — groups that are difficult to introduce directly.
ArN_2^+Cl^- + CuCl/HCl arrow ArCl + N_2
ArN_2^+Cl^- + CuBr/HBr arrow ArBr + N_2
ArN_2^+Cl^- + CuCN/HCN arrow ArCN + N_2
All Sandmeyer reactions release N_2 gas
The Cu(I) salt acts as a catalyst for the radical mechanism
Other Diazonium Replacements
Beyond Sandmeyer, diazonium salts can be converted to ArF (Balz-Schiemann), ArI (KI treatment), ArOH (boiling with water), ArH (H₃PO₂ reduction), and ArNO₂ (HBF₄/NaNO₂). Each reaction replaces the N₂⁺ group with a different nucleophile.
The diazonium salt is a universal precursor to almost any substituted benzene. Different reagents replace -N_2^+ with different groups:
Balz-Schiemann reaction: ArN_2^+Cl^- + HBF_4 arrow ArN_2^+BF_4^- ArF + BF_3 + N_2 — the ONLY good way to put F on a benzene ring
Replacement by I: ArN_2^+Cl^- + KI arrow ArI + KCl + N_2 — no Cu catalyst needed (I⁻ is nucleophilic enough)
Replacement by H: ArN_2^+Cl^- + H_3PO_2 + H_2O arrow ArH + N_2 + H_3PO_3 — removes the amino group entirely
Replacement by -NO_2 : ArN_2^+BF_4^- + NaNO_2 Cu ArNO_2 — useful for putting -NO_2 at unusual positions
Quick recall
How do you convert aniline to fluorobenzene?
Aniline NaNO_2/HCl, 0–5°C C_6H_5N_2^+Cl^- HBF_4 C_6H_5N_2^+BF_4^- C_6H_5F + BF_3 + N_2 . This is the Balz-Schiemann reaction.
Coupling Reaction
Diazonium salts undergo electrophilic substitution with activated aromatic compounds (phenols, amines) to form azo compounds (Ar—N=N—Ar'). These brightly colored compounds are the basis of the azo dye industry. Coupling occurs at the para position of the activated ring.
The diazonium ion ( ArN_2^+ ) is a weak electrophile. It can't attack benzene (too electron-poor), but it CAN attack activated rings — phenols (in weakly alkaline medium) and amines (in weakly acidic medium). The product has an -N=N- bridge between two aromatic rings.
Coupling with phenol: done in weakly alkaline medium (NaOH converts ArOH to ArO⁻, a stronger activator)
Coupling with aniline: done in weakly acidic medium (strong acid would protonate -NH_2 to -NH_3^+ , deactivating the ring)
Product: azo compound ( Ar—N=N—Ar' ) — the -N=N- group is called the azo group
Coupling occurs preferentially at the para position of the activated ring
Azo compounds are brightly colored (yellow, orange, red) due to extended conjugation through the -N=N- bridge
Quick recall
Why must coupling with phenol be done in weakly alkaline (not strongly alkaline) medium?
Weakly alkaline medium converts phenol to phenoxide ( ArO^- ), which is a stronger activator than phenol. But strongly alkaline medium would convert ArN_2^+ to a diazohydroxide ( ArN=N—OH ), which is not electrophilic and doesn't couple. So you need just enough base to activate the phenol, but not so much that you destroy the diazonium ion.