Login

Concepts

Read through concept notes with worked explanations and diagrams, chapter by chapter.

Concept 03 / 04Amines

Chemical Reactions of Amines: Alkylation, Acylation, Carbylamine, Nitrous Acid, Hinsberg Test and Ring Substitution

Reactions of amines as nucleophiles and bases: alkylation to quaternary salts, acylation and benzoylation, carbylamine reaction, reaction with nitrous acid, reaction with benzenesulfonyl chloride (Hinsberg test), and electrophilic substitution in aniline with control by acetylation (bromination, nitration, sulfonation) and the failure of Friedel-Crafts reactions.

Figure 1
The lone pair on nitrogen makes amines nucleophiles and bases. Alkylation. Amines react with alkyl halides (SN2), progressively forming secondary and tertiary amines and finally quaternary ammonium salts. Acylation. Primary and secondary amines react with acid chlorides, anhydrides or esters (nucleophilic substitution) to give amides: C2H5NH2+CH3COCl→C2H5NHCOCH3+HCl\mathrm{C_{2}H_{5}NH_{2} + CH_{3}COCl \rightarrow C_{2}H_{5}NHCOCH_{3} + HCl}; pyridine is added to remove HCl. Benzoylation with benzoyl chloride: CH3NH2+C6H5COCl→CH3NHCOC6H5+HCl\mathrm{CH_{3}NH_{2} + C_{6}H_{5}COCl \rightarrow CH_{3}NHCOC_{6}H_{5} + HCl}. Tertiary amines have no N-H and are not acylated. Carbylamine reaction (isocyanide test). Only primary amines (aliphatic and aromatic), heated with chloroform and ethanolic KOH, give isocyanides (carbylamines) with an extremely unpleasant smell: RNH2+CHCl3+3KOH→RNC+3KCl+3H2O\mathrm{RNH_{2} + CHCl_{3} + 3KOH \rightarrow RNC + 3KCl + 3H_{2}O}. Reaction with nitrous acid (NaNO2\mathrm{NaNO_{2}} + HCl, generated in situ). Primary aliphatic amines give unstable diazonium salts that decompose at once, releasing nitrogen quantitatively and giving alcohols: RNH2+HNO2→[RN2+Cl−]→ROH+N2+HCl\mathrm{RNH_{2} + HNO_{2} \rightarrow [RN_{2}^{+}Cl^{-}] \rightarrow ROH + N_{2} + HCl} (the N2\mathrm{N_{2}} evolved is used to estimate amino acids and proteins). Primary aromatic amines at 273273 to 278 K278\,\mathrm{K} give stable benzenediazonium salts (diazotisation): C6H5NH2+NaNO2+2HCl→C6H5N2+Cl−+NaCl+2H2O\mathrm{C_{6}H_{5}NH_{2} + NaNO_{2} + 2HCl \rightarrow C_{6}H_{5}N_{2}^{+}Cl^{-} + NaCl + 2H_{2}O}. Secondary amines give yellow oily N-nitrosamines; tertiary aliphatic amines form soluble salts. Hinsberg test (see figure). Benzenesulfonyl chloride (C6H5SO2Cl\mathrm{C_{6}H_{5}SO_{2}Cl}, Hinsberg's reagent) distinguishes the three classes: primary amines give N-alkylbenzenesulfonamides; the remaining N-H is strongly acidic (flanked by the sulfonyl group), so the product dissolves in alkali; secondary amines give N,N-dialkylbenzenesulfonamides, with no acidic H, insoluble in alkali; tertiary amines do not react. (p-Toluenesulfonyl chloride is often used instead.) Electrophilic substitution in aniline. The −NH2\mathrm{-NH_{2}} group is a very strong activating ortho/para director. Bromination: with bromine water at room temperature, aniline gives a white precipitate of 2,4,6-tribromoaniline at once. To get a monosubstituted product, the activating effect is moderated by acetylation: aniline → acetanilide (with acetic anhydride); acetanilide + Br2\mathrm{Br_{2}} in acetic acid → p-bromoacetanilide (major); hydrolysis → p-bromoaniline. The acetyl group's carbonyl pulls the N lone pair into competing resonance, making −NHCOCH3\mathrm{-NHCOCH_{3}} only moderately activating (and bulky, favouring para). Nitration: direct nitration gives tarry oxidation products and, because anilinium ions form in acid (meta-directing), about 51%51\% para, 47%47\% meta and 2%2\% ortho nitroaniline. Protecting as acetanilide gives mainly p-nitroacetanilide, hydrolysed to p-nitroaniline. Sulfonation: aniline with concentrated H2SO4\mathrm{H_{2}SO_{4}} forms anilinium hydrogensulfate, which on heating at 453453 to 473 K473\,\mathrm{K} gives p-aminobenzenesulfonic acid (sulfanilic acid), which exists as a zwitterion. Aniline does not undergo Friedel-Crafts reactions: it forms a salt with the Lewis acid AlCl3\mathrm{AlCl_{3}}, placing a positive charge on nitrogen that strongly deactivates the ring. Key Takeaways: - Acylation needs N-H (1∘1^{\circ}, 2∘2^{\circ}); carbylamine test is specific to 1∘1^{\circ} amines. - Nitrous acid: 1∘1^{\circ} aliphatic → alcohol + N2\mathrm{N_{2}}; 1∘1^{\circ} aromatic → diazonium salt at 00 to 5∘C5^{\circ}\mathrm{C}. - Hinsberg: 1∘1^{\circ} product alkali-soluble, 2∘2^{\circ} insoluble, 3∘3^{\circ} no reaction. - Aniline + bromine water → 2,4,6-tribromoaniline; acetylation controls substitution. - Pitfall: significant meta product in direct nitration of aniline comes from the anilinium ion, not from the free amine. Brain-Teaser 1: How would you distinguish ethanamine, N-ethylethanamine and N,N-diethylethanamine? Solution: Hinsberg test: ethanamine gives a sulfonamide soluble in KOH; N-ethylethanamine gives a sulfonamide insoluble in KOH; triethylamine does not react. Alternatively, only ethanamine gives the carbylamine smell. Brain-Teaser 2: Outline the preparation of p-nitroaniline from aniline. Solution: Acetylate aniline with acetic anhydride to acetanilide; nitrate with HNO3/H2SO4\mathrm{HNO_{3}/H_{2}SO_{4}} to get mainly p-nitroacetanilide (the acetyl group moderates activation and prevents oxidation and anilinium formation); hydrolyse the amide with acid or base to p-nitroaniline. Brain-Teaser 3: 0.59 g0.59\,\mathrm{g} of a primary aliphatic amine liberates 224 mL224\,\mathrm{mL} of N2\mathrm{N_{2}} (at 273 K273\,\mathrm{K}, 1 atm1\,\mathrm{atm}) with nitrous acid. Find its molar mass and a possible structure. Solution: 224 mL=0.01 mol224\,\mathrm{mL} = 0.01\,\mathrm{mol} N2\mathrm{N_{2}}, so 0.01 mol0.01\,\mathrm{mol} amine. M=0.590.01=59 g mol−1M = \frac{0.59}{0.01} = 59\,\mathrm{g\,mol^{-1}}: C3H7NH2\mathrm{C_{3}H_{7}NH_{2}} (propan-1-amine or propan-2-amine).