Class 12 Chemistry Chapter 9 Amines Handwritten Notes PDF

Q. Are the Amines Class 12 handwritten notes aligned with the 2026-27 syllabus?

Yes. The scanned notebook covers every topic retained in the current 2026-27 NCERT print, including amine classification, IUPAC nomenclature, the five major preparation routes (nitro reduction, nitrile reduction, amide reduction, Gabriel phthalimide synthesis, Hofmann bromamide degradation), physical properties, basicity in gas and aqueous phases, distinction tests (carbylamine, Hinsberg, nitrous acid), and the full diazonium chemistry block. Sub-topics dropped from the rationalised edition (some industrial routes and a few specialised reactions) have been removed from the notebook to match the current syllabus exactly.

Q. How many name reactions does the notebook cover for Chapter 9?

The scan maps seven core name reactions: Gabriel phthalimide synthesis, Hofmann bromamide degradation, carbylamine reaction, Hinsberg test, Sandmeyer reaction, Gattermann reaction, and azo coupling. Each reaction has its mechanism arrows drawn in two-colour ink so the electron flow direction is unmistakable, and the diazonium replacement map on page 16 collects Cl, Br, I, CN, F, and OH in a single foldable panel.

Q. Why is the basicity order of methylamines different in gas phase and water?

In the gas phase only the intrinsic +I effect matters, so basicity follows 3^∘ > 2^∘ > 1^∘ > NH₃. In water, the conjugate acid (R_nNH_(4-n)⁺) is stabilised by hydrogen bonding with solvent, and fewer N-H bonds means fewer hydrogen bonds. Trimethylammonium has only one N-H so it is solvated poorly, while dimethylammonium with two N-H bonds is solvated best. The combined effect reverses the order to 2^∘ > 1^∘ > 3^∘ > NH₃. The notebook draws both ladders side by side on page 8.

Q. What is the expected CBSE Board weightage for Amines in 2026?

CBSE typically allocates 4 to 6 marks to this chapter, usually as one 2-marker on a distinction test or basicity comparison and one 3-marker on diazonium conversion or a name reaction mechanism. Occasionally the 5-mark long-answer question pulls entirely from this chapter when it combines a preparation route with a diazonium synthesis. JEE Main and NEET each carry one to two additional questions per shift on top of this.

Q. How do you distinguish between primary, secondary, and tertiary amines?

The Hinsberg test is the cleanest method. The amine is reacted with benzenesulfonyl chloride (C₆H₅SO₂Cl) followed by KOH. Primary amines give a sulfonamide with an acidic N-H, which dissolves in KOH; secondary amines give a sulfonamide with no N-H, insoluble in KOH; tertiary amines do not react with the sulfonyl chloride at all. The carbylamine reaction (CHCl₃ + KOH) is a complementary test that gives an offensive isocyanide only with primary amines. The notebook gives a colour-coded decision tree on page 13.

Q. What is the difference between Sandmeyer, Gattermann, and Balz-Schiemann reactions for diazonium salts?

All three convert an aryl diazonium salt (Ar-N₂⁺) to a substituted arene, but the reagents and products differ. The Sandmeyer reaction uses cuprous halide (CuCl, CuBr, or CuCN) dissolved in the corresponding halogen acid and gives high-yield Ar-Cl, Ar-Br, or Ar-CN. The Gattermann reaction uses copper powder (Cu metal) with HCl or HBr, gives the same Ar-X products but at lower yield. The Balz-Schiemann reaction is the only Class 12 route to aryl fluoride: the diazonium chloride is first converted to the aryl diazonium fluoroborate with HBF₄, then dry-heated to release N₂ and BF₃ leaving Ar-F. In MCQs, the catalyst is the key: Cu(I) salt = Sandmeyer, Cu metal + HX = Gattermann, HBF₄ + heat = Schiemann.

Q. Why is azo coupling done in mild base with phenol and mild acid with aniline, and what dyes are produced?

Both couplings need the aryl diazonium electrophile (Ar-N₂⁺) at 273-278 K. With phenol, mild NaOH converts PhOH to the much more nucleophilic phenoxide ion (PhO^-), which couples with benzenediazonium chloride at the para position to give p-hydroxyazobenzene, an orange dye. With aniline, mild HCl is used so that the aniline N is not fully protonated; in mild acid medium it remains a free amine and couples at the para position to give p-aminoazobenzene, a yellow dye. If the medium were strong acid for the aniline coupling, the substrate would protonate fully to anilinium ion and the coupling would fail.

Q. Why does Friedel-Crafts alkylation fail on aniline, and how is this fixed in synthesis?

The Friedel-Crafts reaction needs a Lewis acid (typically AlCl₃) to generate the electrophile. Aniline carries a basic nitrogen lone pair that coordinates strongly to AlCl₃, forming an Ar-NH₂-AlCl₃ complex in which N now bears a formal positive charge. This deactivates the ring toward electrophilic attack. The standard fix is to acetylate the amine first with acetic anhydride to give acetanilide (Ar-NHCOCH₃), perform the Friedel-Crafts on the acetanilide where the amide is only weakly activating, and then hydrolyse the acetyl group at the end to recover the substituted aniline. The same acetylation strategy is used for clean para-bromination and para-nitration of aniline.

Q. Are these handwritten notes enough to prepare for JEE Main and NEET 2026 questions from this chapter?

The notebook gives complete conceptual coverage of every NCERT topic plus the highest-yield mechanisms, but JEE Main and NEET demand timed practice on previous-year questions. Pair the notes with the Collegedunia NCERT Exemplar Solutions for this chapter and the last five years of JEE Main and NEET PYQs to convert this chapter into 6 to 10 marks across both exams.