Class 12 Chemistry Ch 9 Amines Formula Sheet PDF Download

Ans. You can download the Amines Class 12 Chemistry Formula Sheet PDF directly from this Collegedunia page. Both the Normal and HD versions are available and free of cost.

Ans. Yes. This page reflects the current 2026-27 syllabus for Class 12 Chemistry. Amines is fully retained in the new edition with no formula cuts; every reaction in Sections 9.1 to 9.10 of the NCERT remains examinable.

Ans. The Formula Sheet PDF runs approximately 9 pages and covers the master reaction table, symbol glossary, pK_b reference, quick-fact MCQ cards, and four common numerical pattern templates.

Ans. Both produce primary amines but differ in the starting material, the carbon count change, and the substrate scope. Hoffmann bromamide degradation starts from an amide (R-CONH₂), uses Br₂ with 4 equivalents of NaOH, and the product amine has one carbon fewer than the amide because the R group migrates from carbonyl C to N (alkyl and aryl amides both work). Gabriel phthalimide synthesis starts from phthalimide, alkylates with R-X (S_N2), then hydrolyses to give a pure 1° amine with the same carbon count as the alkyl halide; it works only for aliphatic amines because aryl halides do not undergo S_N2 on the phthalimide anion.

Ans. In aniline (C₆H₅-NH₂), the lone pair on nitrogen is delocalised into the aromatic ring through five resonance structures, making it much less available to accept a proton. Once protonated, the anilinium ion (C₆H₅-NH₃⁺) has only two Kekule resonance structures, so the cation is far less stabilised than the neutral aniline. Methylamine has no such delocalisation - the methyl group is a +I donor that increases electron density on N, so the lone pair is fully available and the methylammonium cation is stabilised by hyperconjugation. Numerically, pK_b (aniline) = 9.38 versus pK_b (methylamine) = 3.38, a roughly 10⁶-fold difference in K_b.

Ans. In an aryl diazonium salt (Ar-N₂⁺), the positive charge on the terminal nitrogen is delocalised into the aromatic ring by resonance, spreading the charge over two nitrogen atoms and the ortho/para ring carbons. This delocalisation stabilises the cation enough for the salt to exist in aqueous solution at 273-278 K. In an alkyl diazonium salt (R-N₂⁺), there is no aromatic ring to accept the charge; the localised positive nitrogen is immediately attacked by water or any nucleophile, releasing N₂ gas and giving R-OH (or other substitution / elimination products). That is why HNO₂ on a 1° aliphatic amine gives R-OH plus a measurable volume of N₂, while the same reagent on a 1° aromatic amine gives an isolable diazonium chloride.

Ans. Friedel-Crafts reactions need a Lewis acid catalyst, typically AlCl₃. The nitrogen lone pair of aniline is more basic than a typical aromatic electron pair, so AlCl₃ coordinates strongly to N rather than activating the alkyl halide. This generates an Ar-N⁺H₂-AlCl₃^- complex in which the N now bears a formal positive charge. The aryl ring connected to this strongly electron-withdrawing group is heavily deactivated, so any electrophile (the R⁺ or RCO⁺) cannot attack. The standard workaround is to protect the amine as acetanilide (C₆H₅-NHCOCH₃), perform the Friedel-Crafts on the acetanilide, then hydrolyse back to the substituted aniline.

Ans. The Balz-Schiemann reaction is the only Class 12 route to aryl fluoride from an aryl diazonium salt. The diazonium chloride (Ar-N₂⁺Cl^-) is first treated with HBF₄ to precipitate aryl diazonium fluoroborate (Ar-N₂⁺BF₄^-), which is then heated dry to release N₂ and BF₃, leaving Ar-F. The formula table lists it as Schiemann under NCERT Section 9.9 with reagent "HBF₄, dry heat" and the common-use cue "Only route to Ar-F".

Ans. Sandmeyer uses cuprous halide (CuCl, CuBr, or CuCN) dissolved in the corresponding halogen acid and gives high yields of aryl halide or aryl nitrile. Gattermann uses copper powder (Cu metal) in HCl or HBr and gives lower-yield aryl halides. Both convert Ar-N₂⁺ to Ar-X (X = Cl, Br), but the catalyst tells them apart in an MCQ: Cu(I) salt = Sandmeyer, Cu metal + HX = Gattermann. The formula table shows both rows side by side under NCERT Section 9.9.

Ans. In the gas phase, basicity follows the pure inductive (+I) effect: 3° > 2° > 1° > NH₃. In aqueous solution, the order flips because the protonated amine (R_nNH_(4-n)⁺) needs N-H bonds to hydrogen-bond with water; fewer N-H bonds means less solvation, which destabilises the conjugate acid. The aqueous order is therefore 2° > 1° > 3° > NH₃ for the methylamine series, and 2° > 3° > 1° > NH₃ for bulkier ethyl groups. Aromatic amines (aniline) sit far below NH₃ because the lone pair delocalises into the ring.

Ans. The Hinsberg test uses benzenesulphonyl chloride (C₆H₅SO₂Cl) in aqueous KOH. A primary amine (R-NH₂) reacts to give an N-monoalkyl sulphonamide (PhSO₂NHR) whose remaining N-H is acidic enough to be deprotonated by NaOH, so the product is soluble in alkali. A secondary amine (R₂NH) gives an N,N-dialkyl sulphonamide (PhSO₂NR₂) with no N-H, so it is insoluble in NaOH. A tertiary amine (R₃N) has no N-H to start with and does not react at all - it remains free amine, which on acidification dissolves as the ammonium salt. The three outcomes - NaOH-soluble product, NaOH-insoluble product, no reaction - cleanly separate all three classes in one test.