CBSE · Class 12 · Chemistry · Chapter 9

Amines — Formula Sheet

Board Formulas

19 formulas

1.Reduction of Nitro Compounds
$C_{6} H_{5} NO_{2} Fe + HCl (or Sn + HCl, or H_{2} / Pd) C_{6} H_{5} NH_{2}$
Iron scrap + HCl is preferred: the FeCl₂ formed hydrolyses and releases HCl, so only a little acid is needed to start the reaction. In acid the amine is present as its salt; NaOH sets free the amine.
2.Ammonolysis of Alkyl Halides
$R-X + NH_{3} \to R-NH_{3}^{+} X^{-} NaOH R-NH_{2} + NaX + H_{2} O$
Gives a mixture of 1°, 2° and 3° amines and a quaternary ammonium salt, because the amine formed is itself a nucleophile. A large excess of NH₃ favours the 1° amine. Halide reactivity: RI > RBr > RCl.
3.Reduction of Nitriles and Amides
$R-C \equiv N H_{2} / Ni or LiAlH_{4} R-CH_{2} -NH_{2}; R-CONH_{2} (i) LiAlH_{4} (ii) H_{2} O R-CH_{2} -NH_{2}$
R–X → R–CN → R–CH₂NH₂ gives an amine with one carbon more than the alkyl halide (ascent of the series). Amide reduction keeps every carbon.
4.Gabriel Phthalimide Synthesis★
$C_{6} H_{4} (CO)_{2} NH KOH (ethanolic) C_{6} H_{4} (CO)_{2} N^{-} K^{+} R-X C_{6} H_{4} (CO)_{2} N-R NaOH(aq) R-NH_{2}$
Gives pure 1° aliphatic amines only. Aniline and other aromatic 1° amines cannot be made, because aryl halides do not undergo nucleophilic substitution with the phthalimide anion. By-product: sodium phthalate.
5.Hoffmann Bromamide Degradation★
$R-CONH_{2} + Br_{2} + 4 NaOH \to R-NH_{2} + Na_{2} CO_{3} + 2 NaBr + 2 H_{2} O$
The amine has ONE CARBON FEWER than the amide — the carbonyl carbon leaves as carbonate while R moves from C to N. Works for aryl amides too: benzamide → aniline.
6.Base Dissociation Constant
$K_{b} = \frac{[ R-NH _{3}^{+} ] [ OH ^{-} ]}{[ R-NH _{2} ]}, p K_{b} = - lo g K_{b}$
$K_{b}$ : Base dissociation constant of the amine in water · $p K_{b}$ : Negative logarithm of K_b (smaller value = stronger base)
Larger K_b (smaller pK_b) ⇒ stronger base. pK_b of NH₃ = 4.75; simple alkylamines 3.0–4.2 (stronger than NH₃); aniline 9.38 (much weaker).
7.Basicity of Aliphatic Amines in Water★
$(C_{2} H_{5})_{2} NH > (C_{2} H_{5})_{3} N > C_{2} H_{5} NH_{2} > NH_{3}; (CH_{3})_{2} NH > CH_{3} NH_{2} > (CH_{3})_{3} N > NH_{3}$
In the gas phase only the +I effect matters: 3° > 2° > 1° > NH₃. In water, solvation of the ammonium ion and steric hindrance also count, which changes the order.
8.Aromatic Amines are Weaker Bases
$Basicity: CH_{3} NH_{2} > NH_{3} > C_{6} H_{5} NH_{2} (p K_{b} = 3.38, 4.75, 9.38)$
In aniline the N lone pair is delocalised into the ring by resonance, so it is less available to accept a proton. Electron-releasing groups (–CH₃, –OCH₃) on the ring increase basicity; electron-withdrawing groups (–NO₂) decrease it.
9.Carbylamine (Isocyanide) Test★
$R-NH_{2} + CHCl_{3} + 3 KOH Δ R-NC + 3 KCl + 3 H_{2} O$
Only 1° amines (aliphatic AND aromatic) give the foul-smelling isocyanide. 2° and 3° amines do not. Ethanolic KOH is used.
10.Nitrous Acid with 1° Aliphatic Amines
$R-NH_{2} + HNO_{2} NaNO_{2} + HCl [R-N_{2}^{+} Cl^{-}] H_{2} O R-OH + N_{2} ↑ + HCl$
Aliphatic diazonium salts are unstable even in the cold; N₂ is given off quantitatively (used to estimate amino acids and proteins). Aromatic 1° amines instead give diazonium salts that are stable for a short time in solution at 273–278 K.
11.Acylation of Amines
$C_{2} H_{5} NH_{2} + CH_{3} COCl pyridine C_{2} H_{5} NHCOCH_{3} + HCl$
1° and 2° amines (which have N–H) react with acid chlorides, anhydrides or esters; 3° amines do not. Pyridine, a stronger base, removes HCl and pushes the reaction forward. With C₆H₅COCl it is called benzoylation.
12.Hinsberg Test (benzenesulphonyl chloride)★
$C_{6} H_{5} SO_{2} Cl + H-NHR \to C_{6} H_{5} SO_{2} NHR + HCl$
1° amine → N-alkylbenzenesulphonamide, whose N–H is acidic, so it is SOLUBLE in alkali. 2° amine → N,N-dialkylbenzenesulphonamide with no N–H, INSOLUBLE in alkali. 3° amine → no reaction.
13.Bromination of Aniline
$C_{6} H_{5} NH_{2} + 3 Br_{2} H_{2} O 2,4,6-Br_{3} C_{6} H_{2} NH_{2} ↓ + 3 HBr$
–NH₂ activates the ring so strongly that bromine water gives a white precipitate of 2,4,6-tribromoaniline at room temperature. For a monobromo product, acetylate first, brominate, then hydrolyse.
14.Controlled Nitration through Acetanilide
$C_{6} H_{5} NH_{2} (CH_{3} CO)_{2} O C_{6} H_{5} NHCOCH_{3} HNO_{3} / H_{2} SO_{4} p-O_{2} NC_{6} H_{4} NHCOCH_{3} H^{+} or OH^{-} p-O_{2} NC_{6} H_{4} NH_{2}$
Direct nitration gives tarry oxidation products and a large amount of m-nitroaniline, because in strong acid aniline becomes the m-directing anilinium ion. Protecting –NH₂ as –NHCOCH₃ makes p-nitroaniline the major product.
15.Diazotisation★
$C_{6} H_{5} NH_{2} + NaNO_{2} + 2 HCl 273 - 278 K C_{6} H_{5} N_{2}^{+} Cl^{-} + NaCl + 2 H_{2} O$
Keep the mixture ice-cold (273–278 K). The salt decomposes when warmed, so it is used immediately rather than stored. Arenediazonium ions are stabilised by resonance with the ring; aliphatic ones are not.
16.Sandmeyer and Gattermann Reactions
$ArN_{2}^{+} X^{-} CuCl/HCl, CuBr/HBr or CuCN/KCN ArCl, ArBr or ArCN + N_{2}$
Sandmeyer uses Cu(I) salts. Gattermann uses copper powder with the halogen acid: ArN₂⁺X⁻ + Cu/HX → ArX + N₂ + CuX. Sandmeyer usually gives the better yield.
17.Replacement by Iodide and Fluoride
$ArN_{2}^{+} Cl^{-} + KI \to ArI + KCl + N_{2}; ArN_{2}^{+} Cl^{-} HBF_{4} ArN_{2}^{+} BF_{4}^{-} Δ ArF + BF_{3} + N_{2}$
Iodine is hard to introduce into benzene directly, so KI on the diazonium salt is the standard route to iodobenzene. Aryl fluorides come from heating the diazonium fluoroborate.
18.Replacement by H and by OH
$ArN_{2}^{+} Cl^{-} + H_{3} PO_{2} + H_{2} O \to ArH + N_{2} + H_{3} PO_{3} + HCl; ArN_{2}^{+} Cl^{-} + H_{2} O \to ArOH + N_{2} + HCl$
Hypophosphorous acid (or ethanol, which is oxidised to ethanal) replaces –N₂⁺ by H — useful to remove an –NH₂ group that was only there to direct substitution. Warming with water gives phenol.
19.Coupling Reactions (azo dyes)
$C_{6} H_{5} N_{2}^{+} Cl^{-} + C_{6} H_{5} OH OH^{-} C_{6} H_{5} -N=N-C_{6} H_{4} -OH (p) + Cl^{-} + H_{2} O$
The diazonium ion is a weak electrophile that attacks the para position of phenol (mildly alkaline), giving orange p-hydroxyazobenzene. With aniline (mildly acidic) it gives yellow p-aminoazobenzene. Here the N₂ is retained.