रसायनशास्त्र Chemistry

थोडक्यात सारांश

Nuclear chemistry deals with changes in the nucleus of an atom. This topic covers nuclear particles and forces, mass defect and binding energy, the n/p ratio and stability, radioactive decay (α, β, γ), the rate of decay and half-life, artificial transmutation, nuclear fission and fusion, and applications such as radiocarbon dating and radioisotopes in medicine and agriculture.

महत्त्वाचे मुद्दे

• The nucleus contains protons and neutrons (nucleons). Mass number A = protons + neutrons; atomic number Z = protons.
• Isotopes: same Z, different A (¹²C, ¹⁴C). Isobars: same A, different Z (⁴⁰Ar, ⁴⁰Ca). Isotones: same number of neutrons.
• Nucleons are held by the very strong, short-range nuclear force.
• Mass defect (Δm): the mass of a nucleus is less than the total mass of its separate nucleons. This missing mass appears as binding energy (E = Δmc²).
• Binding energy per nucleon is highest near iron (Fe-56, about 8.8 MeV), so medium nuclei are the most stable. Heavy nuclei gain stability by fission, light nuclei by fusion.
• n/p ratio: stable light nuclei have n/p ≈ 1; heavier stable nuclei need more neutrons (up to about 1.5). Nuclei above the stability belt emit β⁻; those below emit positrons or capture electrons.
• α decay: loses ⁴₂He → A decreases by 4, Z by 2. β⁻ decay: a neutron changes to a proton → Z increases by 1, A unchanged. γ emission: energy only, no change in A or Z.
• Group displacement law (Soddy and Fajans): α emission moves the element two places left in the periodic table; β emission moves it one place right.
• Radioactive decay is a first-order process: rate does not depend on temperature or pressure.
• Nuclear fission: a heavy nucleus (U-235) splits into two medium nuclei when hit by a slow neutron, releasing about 200 MeV and 2–3 neutrons; this can start a chain reaction (used in reactors and atom bombs).
• Nuclear fusion: light nuclei (isotopes of hydrogen) combine to form a heavier nucleus; it powers the Sun and needs very high temperature.
• Applications: C-14 dating of old organic material, Co-60 in cancer therapy, I-131 for thyroid study and treatment, P-32 as a tracer in agriculture, irradiation of food.

महत्त्वाच्या संज्ञा

• Radioactivity (किरणोत्सर्ग) – spontaneous emission of radiation by unstable nuclei, discovered by Henri Becquerel (1896).
• Half-life (t½) – time for half of the radioactive nuclei to decay.
• Decay constant (λ) – fraction of nuclei decaying per unit time.
• Chain reaction – a self-sustaining series of fissions caused by released neutrons.
• Critical mass – minimum mass of fissile material needed to sustain a chain reaction.

सूत्रे

• E = Δmc²; 1 amu (u) ≡ 931.5 MeV
• Δm = [Z mₚ + (A − Z) mₙ] − M(nucleus)
• N = N₀ e^(−λt); λ = (2.303/t) log (N₀/N)
• t½ = 0.693/λ; average life τ = 1/λ
• After n half-lives: N = N₀ (½)ⁿ
• C-14 dating: t = (2.303/λ) log (N₀/N), t½ of C-14 ≈ 5730 years

५ महत्त्वाचे प्रश्न

Q1. Define mass defect and binding energy.

Mass defect is the difference between the total mass of free protons and neutrons and the actual mass of the nucleus. Binding energy is the energy equivalent of this mass defect (E = Δmc²); it is the energy needed to break the nucleus into its nucleons.

Q2. ²³⁸₉₂U emits one α and two β particles. Find the product.

α: A = 238 − 4 = 234, Z = 92 − 2 = 90. Two β⁻: Z = 90 + 2 = 92, A stays 234. Product is ²³⁴₉₂U, an isotope of uranium.

Q3. Calculate the binding energy of ⁴₂He. (mₚ = 1.00728 u, mₙ = 1.00867 u, mass of He nucleus = 4.00151 u)

• Mass of nucleons = 2 × 1.00728 + 2 × 1.00867 = 2.01456 + 2.01734 = 4.03190 u
• Δm = 4.03190 − 4.00151 = 0.03039 u
• BE = 0.03039 × 931.5 = 28.31 MeV
• BE per nucleon = 28.31/4 ≈ 7.08 MeV

Q4. A wooden sample has 25% of the C-14 activity of fresh wood. Find its age (t½ = 5730 years).

• N/N₀ = 0.25 = (½)², so 2 half-lives have passed.
• Age = 2 × 5730 = 11,460 years
• Check: λ = 0.693/5730 = 1.209 × 10⁻⁴ per year; t = (2.303/λ) log 4 = (2.303 × 0.602)/1.209 × 10⁻⁴ ≈ 11,460 years ✓

Q5. Distinguish between nuclear fission and nuclear fusion.

• Fission: heavy nucleus splits; fusion: light nuclei join.
• Fission starts with slow neutrons at ordinary temperature; fusion needs millions of degrees.
• Fission gives radioactive waste; fusion products are mostly not radioactive.
• Fusion gives more energy per unit mass.
• Fission is used in nuclear power stations; fusion powers the Sun and stars (controlled fusion is still under research).

परीक्षेसाठी टीप

For decay problems, balance both A and Z on each side. For half-life questions, first check if the fraction is a simple power of ½ — it saves time.

थोडक्यात सारांश

Bioinorganic chemistry studies the role of metal ions and other inorganic elements in living systems. This topic covers essential and trace elements, the sodium–potassium pump, metalloproteins such as haemoglobin and myoglobin, the role of Mg, Zn, Co, Cu and Mo in biology, toxicity of heavy metals and chelation therapy, and metal-based drugs like cisplatin.

महत्त्वाचे मुद्दे

• Bulk (macro) elements: C, H, O, N, P, S, Na, K, Mg, Ca, Cl. Trace elements: Fe, Zn, Cu, Mn, Co, Mo, Se, I, Cr. Some elements are toxic at any level (Hg, Pb, Cd, As).
• Na⁺/K⁺ pump: an enzyme (Na⁺/K⁺-ATPase) in the cell membrane uses one ATP to push 3 Na⁺ out and bring 2 K⁺ in. It keeps high K⁺ inside and high Na⁺ outside, needed for nerve impulses.
• Ca²⁺: bones and teeth (hydroxyapatite), muscle contraction, blood clotting.
• Haemoglobin (Hb): four subunits, each with a haem group — Fe(II) held by four N atoms of a porphyrin ring and a fifth N from a histidine. O₂ binds at the sixth position. Iron stays Fe(II) while carrying O₂.
• Cooperative binding: when one subunit takes O₂, the others take it more easily, giving an S-shaped (sigmoidal) curve. Myoglobin (one haem, in muscle) gives a hyperbolic curve and stores O₂.
• CO poisoning: CO binds to haem iron about 200–250 times more strongly than O₂, blocking oxygen transport.
• Chlorophyll: Mg²⁺ in a porphyrin-type (chlorin) ring; absorbs red and blue light in photosynthesis.
• Vitamin B₁₂ (cobalamin): contains cobalt in a corrin ring; needed to form red blood cells.
• Zinc enzymes: carbonic anhydrase (CO₂ + H₂O ⇌ HCO₃⁻ + H⁺), carboxypeptidase.
• Cytochromes (Fe), ferredoxins (Fe–S clusters) and copper proteins (plastocyanin) carry electrons. Nitrogenase (Fe–Mo) fixes N₂ into NH₃ in root-nodule bacteria.
• Toxicity: Pb (anaemia, brain damage), Hg (Minamata disease, nervous system), Cd (itai-itai disease, bones and kidneys), As (skin lesions, cancer). Excess Fe or Cu is also harmful.
• Chelation therapy: a chelating agent binds the toxic metal into a soluble complex that is excreted. EDTA (as Ca-EDTA) for lead, BAL (British anti-Lewisite) for Hg and As, desferrioxamine for iron overload, D-penicillamine for copper (Wilson's disease).
• Cisplatin, cis-[Pt(NH₃)₂Cl₂], is an anticancer drug that binds to DNA; the trans isomer is inactive.

महत्त्वाच्या संज्ञा

• Metalloprotein – protein containing a metal ion as an essential part.
• Porphyrin – large flat ring with four pyrrole units that binds metals through four N atoms.
• Chelation (कीलेशन) – binding of a metal ion by a ligand through two or more donor atoms.

सूत्रे

• Na⁺/K⁺ pump: 3Na⁺(in) + 2K⁺(out) + ATP → 3Na⁺(out) + 2K⁺(in) + ADP + Pi

५ महत्त्वाचे प्रश्न

Q1. Why is CO poisonous?

CO binds to the Fe(II) of haemoglobin much more strongly than O₂, forming carboxyhaemoglobin. Hb can no longer carry oxygen, so tissues suffer oxygen shortage.

Q2. Name the metal in chlorophyll and vitamin B₁₂.

Chlorophyll contains magnesium (Mg²⁺). Vitamin B₁₂ contains cobalt.

Q3. Explain the structure and function of haemoglobin.

Four globin chains, haem structure, Fe(II) coordination (four porphyrin N + histidine + O₂), oxy- and deoxy-Hb, cooperative effect and sigmoid curve, comparison with myoglobin, why Fe is not oxidised to Fe(III) inside the protein pocket.

Q4. Describe the sodium–potassium pump.

Concentration difference across the membrane, role of ATP, stoichiometry (3 Na⁺ out, 2 K⁺ in), importance in nerve conduction and osmotic balance; diagram.

Q5. Write a note on metal toxicity and chelation therapy.

Toxic effects of Pb, Hg, Cd and As with diseases; principle of chelation; agents (Ca-EDTA, BAL, desferrioxamine, D-penicillamine) matched to metals; why Ca-EDTA and not Na₂-EDTA is used (to avoid removing body calcium).

परीक्षेसाठी टीप

A table of metal → biomolecule → function covers most short questions; learn haemoglobin in detail for the long answer.

थोडक्यात सारांश

Carbohydrates are optically active polyhydroxy aldehydes or ketones, or compounds that give these on hydrolysis. They are the main energy source of living things. This topic covers classification, the open-chain and cyclic structures of glucose and fructose, mutarotation, important reactions, interconversion (chain lengthening and shortening), and an introduction to disaccharides and polysaccharides.

महत्त्वाचे मुद्दे

• General formula of many carbohydrates: Cₓ(H₂O)ᵧ (hence "hydrates of carbon"), but this is not a strict rule.
• Classification: monosaccharides (cannot be hydrolysed further: glucose, fructose), oligosaccharides (2–10 units: sucrose, maltose, lactose), polysaccharides (many units: starch, cellulose, glycogen).
• Aldoses have a −CHO group (glucose is an aldohexose); ketoses have a C=O group (fructose is a ketohexose).
• Reducing sugars reduce Tollens' and Fehling's reagents (all monosaccharides, maltose, lactose). Sucrose is non-reducing because the reducing groups of glucose and fructose are joined in the glycosidic bond.
• Evidence for the open-chain structure of glucose: forms oxime with NH₂OH (C=O present); gives n-hexane with HI/red P (six-carbon straight chain); forms a penta-acetate (five −OH groups); oxidised by Br₂ water to gluconic acid (−CHO present); oxidised by HNO₃ to saccharic (glucaric) acid (a −CH₂OH at the end).
• Objections to the open chain: glucose does not give Schiff's test or a bisulphite addition product; its penta-acetate does not react with NH₂OH; it exists as two forms (α and β); it shows mutarotation.
• Cyclic (pyranose) structure: the C-5 −OH adds to the C-1 aldehyde, forming a six-membered ring (like pyran) and a new chiral centre at C-1 (anomeric carbon). This gives α-D-glucose and β-D-glucose (anomers).
• Mutarotation: specific rotation of a fresh glucose solution changes slowly to an equilibrium value of +52.7° (α form +112°, β form +18.7°), because α and β forms interconvert through the open chain.
• Glucose and fructose form the same osazone with excess phenylhydrazine, because only C-1 and C-2 react, and the rest of the molecule is identical.
• Kiliani–Fischer synthesis lengthens an aldose by one carbon (via cyanohydrin). Ruff degradation shortens it by one carbon.
• Polysaccharides: starch (amylose + amylopectin, α-glucose units), cellulose (β-glucose units, β-1,4 links; humans cannot digest it), glycogen (animal starch, highly branched).

महत्त्वाच्या संज्ञा

• Monosaccharide (एकशर्करा) – simplest sugar, not hydrolysable.
• Anomers – cyclic sugars differing only at the anomeric carbon (α and β).
• Epimers – sugars differing in configuration at only one carbon other than C-1 (glucose and mannose at C-2).
• Glycosidic linkage – C–O–C bond joining two sugar units.
• Osazone – crystalline derivative formed by sugars with excess phenylhydrazine.
• D/L configuration – based on the −OH on the chiral carbon farthest from C=O; on the right in Fischer projection = D.

सूत्रे

• Glucose: C₆H₁₂O₆, CH₂OH(CHOH)₄CHO
• Fructose: C₆H₁₂O₆, CH₂OH(CHOH)₃COCH₂OH
• Sucrose: C₁₂H₂₂O₁₁ → glucose + fructose (hydrolysis; invert sugar)
• Oxidation: CHO → COOH (Br₂ water) gives gluconic acid
• With HI/P: glucose → n-hexane (CH₃(CH₂)₄CH₃)
• Number of stereoisomers of an aldohexose (4 chiral C in open chain) = 2⁴ = 16

५ महत्त्वाचे प्रश्न

Q1. What is mutarotation?

It is the change in specific rotation of a freshly prepared solution of a sugar until it reaches a constant value. Glucose changes from +112° (α) or +18.7° (β) to +52.7°, because α and β forms interconvert through the open-chain form and reach equilibrium.

Q2. Why is sucrose a non-reducing sugar?

In sucrose, C-1 of glucose and C-2 of fructose — the two reducing (carbonyl) carbons — are joined in the glycosidic bond. No free aldehyde or ketone group can form, so sucrose does not reduce Fehling's or Tollens' reagent.

Q3. Give evidence for the open-chain structure of glucose.

• Molecular formula C₆H₁₂O₆; with HI/P gives n-hexane → straight 6-carbon chain.
• With NH₂OH forms oxime; with HCN forms cyanohydrin → carbonyl group present.
• Br₂ water oxidises it to gluconic acid (same carbon number) → aldehyde group.
• Acetylation gives penta-acetate → five −OH groups.
• HNO₃ gives saccharic acid → a primary alcohol group (−CH₂OH) also present.

Q4. Why do glucose and fructose give the same osazone?

• Osazone formation involves only C-1 and C-2.
• In glucose, C-1 (CHO) reacts first, then C-2 is oxidised and reacts.
• In fructose, C-2 (C=O) reacts first, then C-1 is oxidised and reacts.
• C-3 to C-6 are identical in both, so the products are the same.
• Mannose also gives the same osazone (C-2 epimer).

Q5. Explain the Kiliani–Fischer synthesis (arabinose → glucose and mannose).

• Aldopentose + HCN → cyanohydrin (new chiral C formed; two epimers).
• Hydrolysis → aldonic acids.
• Heating → lactones.
• Reduction with Na/Hg → aldohexoses with one more carbon.
• Arabinose gives glucose and mannose (epimers at C-2).

परीक्षेसाठी टीप

Learn glucose evidence as a list of reaction → conclusion. Draw the Haworth projection of α- and β-D-glucopyranose neatly; it is a frequent diagram question.

थोडक्यात सारांश

Amino acids are organic compounds containing both an amino group (−NH₂) and a carboxyl group (−COOH). α-Amino acids join through peptide bonds to form peptides and proteins. This topic covers classification of amino acids, zwitterions and the isoelectric point, synthesis, the peptide bond, structure of proteins (primary to quaternary), denaturation, and tests for proteins.

महत्त्वाचे मुद्दे

• In α-amino acids, the −NH₂ is on the carbon next to −COOH. General formula: R−CH(NH₂)−COOH.
• About 20 standard amino acids occur in proteins. All except glycine (R = H) are optically active; natural ones mainly have the L-configuration.
• Classification by R group: neutral (glycine, alanine), acidic (aspartic acid, glutamic acid), basic (lysine, arginine, histidine), aromatic (phenylalanine, tyrosine), sulphur-containing (cysteine, methionine).
• Essential amino acids cannot be made by the human body and must come from food (e.g. valine, leucine, lysine, tryptophan). Non-essential ones are made by the body (glycine, alanine).
• Zwitterion: in water an amino acid exists as a dipolar ion, H₃N⁺−CHR−COO⁻. This explains its high melting point, solubility in water and low solubility in organic solvents.
• Amino acids are amphoteric: they react with both acids and bases.
• Isoelectric point (pI): the pH at which the amino acid has no net charge and does not move in an electric field.
• Synthesis: Strecker synthesis (aldehyde + NH₃ + HCN, then hydrolysis); Gabriel phthalimide synthesis; amination of α-halo acids.
• Peptide bond: −CO−NH− formed between the −COOH of one amino acid and the −NH₂ of another, with loss of water. The peptide bond is planar (partial double bond character).
• Protein structure: primary (sequence of amino acids), secondary (α-helix and β-pleated sheet, held by hydrogen bonds), tertiary (overall 3D folding: H-bonds, disulphide bridges, ionic and hydrophobic interactions), quaternary (arrangement of two or more polypeptide chains, e.g. haemoglobin).
• Denaturation: loss of biological activity when heat, pH change or chemicals disturb the secondary/tertiary structure; primary structure stays intact (boiling an egg).
• Tests: Biuret test (violet colour with alkaline CuSO₄; for peptide bonds), Ninhydrin test (purple/blue colour with amino acids), Xanthoproteic test (yellow with conc. HNO₃, aromatic rings).

महत्त्वाच्या संज्ञा

• Zwitterion (द्विध्रुवी आयन) – molecule with both positive and negative charges but neutral overall.
• Isoelectric point (pI) – pH at which net charge is zero.
• Peptide – chain of amino acids joined by peptide bonds (dipeptide, tripeptide, polypeptide).
• Denaturation – destruction of protein's 3D structure and activity.
• N-terminal / C-terminal – end with free −NH₂ / free −COOH of a peptide chain.

सूत्रे

• Glycine: H₂N−CH₂−COOH; Alanine: H₂N−CH(CH₃)−COOH
• Dipeptide formation: H₂N−CH₂−COOH + H₂N−CH(CH₃)−COOH → H₂N−CH₂−CO−NH−CH(CH₃)−COOH + H₂O (Gly-Ala)
• Isoelectric point for a simple neutral amino acid: pI = (pKa₁ + pKa₂)/2
• Number of possible peptides from n different amino acids (each used once) = n!

५ महत्त्वाचे प्रश्न

Q1. What is a zwitterion? Why do amino acids have high melting points?

A zwitterion is a dipolar ion with both + and − charges, like H₃N⁺CH₂COO⁻. Because amino acids exist as zwitterions in the solid, strong ionic attractions hold them together, giving high melting points (they often decompose before melting).

Q2. Define isoelectric point. Calculate pI of glycine (pKa₁ = 2.34, pKa₂ = 9.60).

Isoelectric point is the pH at which an amino acid carries no net charge. pI = (2.34 + 9.60)/2 = 11.94/2 = 5.97.

Q3. How many tripeptides can be formed from glycine, alanine and valine, each used once? Name them.

• Number = 3! = 3 × 2 × 1 = 6
• Gly-Ala-Val, Gly-Val-Ala, Ala-Gly-Val, Ala-Val-Gly, Val-Gly-Ala, Val-Ala-Gly
• Order matters because the chain has an N-terminal and a C-terminal end.

Q4. Explain the Strecker synthesis of alanine.

• Start with acetaldehyde, CH₃CHO.
• React with NH₃ and HCN → α-aminonitrile, CH₃CH(NH₂)CN.
• Acid hydrolysis converts −CN to −COOH.
• Product: alanine, CH₃CH(NH₂)COOH (racemic mixture, since the method is not stereoselective).

Q5. Describe the four levels of protein structure.

• Primary: order of amino acids linked by peptide bonds (decides all other levels).
• Secondary: α-helix (H-bonds within the chain, right-handed coil) and β-pleated sheet (H-bonds between neighbouring chains, e.g. silk).
• Tertiary: further folding into globular or fibrous shapes; held by disulphide bridges, H-bonds, ionic and hydrophobic forces.
• Quaternary: assembly of two or more polypeptide subunits (haemoglobin has four).
• Denaturation destroys secondary to quaternary levels.

परीक्षेसाठी टीप

Questions on zwitterion, isoelectric point and protein structure are common. Draw an α-helix and a β-sheet in simple sketches with H-bonds marked.

थोडक्यात सारांश

Nucleic acids (DNA and RNA) are biopolymers that store and transfer genetic information. Their monomers are nucleotides. This topic covers the components of nucleic acids, nucleosides and nucleotides, the Watson–Crick double helix of DNA, Chargaff's rules, types of RNA, replication, and an outline of protein synthesis and the genetic code.

महत्त्वाचे मुद्दे

• Nucleic acids were first isolated by Friedrich Miescher (1869) from cell nuclei.
• Components: a pentose sugar, a nitrogenous base and phosphoric acid.
• Sugars: DNA has 2-deoxy-D-ribose (no –OH at C-2′); RNA has D-ribose.
• Bases: purines — adenine (A) and guanine (G); pyrimidines — cytosine (C), thymine (T, DNA only) and uracil (U, RNA only).
• Nucleoside = base + sugar (joined by an N-glycosidic bond at C-1′). Nucleotide = nucleoside + phosphate (ester at C-5′).
• Nucleotides join by 3′,5′-phosphodiester bonds to form a polynucleotide chain with a 5′ end and a 3′ end.
• Watson–Crick model (1953), based on X-ray data of Rosalind Franklin and Maurice Wilkins: two antiparallel strands coiled into a right-handed double helix; sugar–phosphate backbone outside, bases inside.
• Base pairing: A pairs with T by two H-bonds; G pairs with C by three H-bonds. So GC-rich DNA is more stable (higher melting temperature).
• Dimensions (B-DNA): diameter about 2 nm, one turn about 3.4 nm, about 10 base pairs per turn, 0.34 nm between base pairs.
• Chargaff's rules: in DNA, A = T and G = C, so purines = pyrimidines; the ratio (A + T)/(G + C) varies from species to species.
• Replication is semi-conservative (Meselson–Stahl, 1958): each new DNA has one old and one new strand.
• RNA is usually single-stranded. mRNA carries the code from DNA, tRNA brings amino acids (cloverleaf shape, anticodon), rRNA forms ribosomes.
• Central dogma: DNA → (transcription) → RNA → (translation) → protein.
• Genetic code: triplet code of 64 codons; 61 code for 20 amino acids, 3 are stop codons (UAA, UAG, UGA); AUG is the start codon (methionine). The code is nearly universal and degenerate.

महत्त्वाच्या संज्ञा

• Nucleotide (न्यूक्लिओटाइड) – base + sugar + phosphate; monomer of nucleic acids.
• Codon – three-base sequence on mRNA coding for one amino acid.
• Denaturation of DNA – separation of strands by heat (melting).

सूत्रे

• Example: if a DNA has 30 % A, then T = 30 %, G + C = 40 %, so G = C = 20 %.

५ महत्त्वाचे प्रश्न

Q1. Distinguish between a nucleoside and a nucleotide.

A nucleoside is a base joined to a pentose sugar (adenosine). A nucleotide is a nucleoside with a phosphate group on the sugar (adenosine monophosphate, AMP).

Q2. A DNA sample contains 18 % cytosine. Find the percentage of the other bases.

G = C = 18 %, so G + C = 36 %. A + T = 64 %, and A = T = 32 %.

Q3. Describe the double helical structure of DNA.

Antiparallel strands, right-handed helix, backbone and base positions, complementary base pairing with number of H-bonds, major and minor grooves, dimensions; draw a labelled diagram and a base-pair sketch.

Q4. Explain the replication of DNA.

Unwinding by helicase, replication fork, primer, DNA polymerase adding nucleotides in the 5′ → 3′ direction, leading and lagging strands (Okazaki fragments, joined by ligase), semi-conservative nature with the Meselson–Stahl experiment.

Q5. Write a note on the types of RNA and their role in protein synthesis.

mRNA (codons), tRNA (anticodon, amino-acid arm, cloverleaf), rRNA (ribosome); transcription in the nucleus and translation on ribosomes; start and stop codons; peptide bond formation.

परीक्षेसाठी टीप

Chargaff's rule numericals are quick marks: A = T, G = C, and all four add to 100 %.

थोडक्यात सारांश

Polymers are very large molecules (macromolecules) made by joining many small repeating units called monomers. This topic covers classification of polymers, addition and condensation polymerisation with mechanisms, molecular mass averages, natural and synthetic rubber, and important polymers like polythene, PVC, nylon-6,6, terylene and Bakelite, along with biodegradable polymers.

महत्त्वाचे मुद्दे

• Monomer: small molecule that joins to form a polymer (ethene → polythene). The process is polymerisation.
• Classification by source: natural (cellulose, starch, proteins, natural rubber), semi-synthetic (cellulose acetate, rayon), synthetic (nylon, PVC).
• By structure: linear (HDPE), branched (LDPE), cross-linked/network (Bakelite, melamine).
• By monomers: homopolymer (one type of monomer: polythene) and copolymer (two or more types: Buna-S, nylon-6,6).
• By forces/properties: elastomers (weak forces, rubber), fibres (strong H-bonds, nylon, terylene), thermoplastics (soften on heating and can be remoulded: PVC, polythene), thermosetting (become hard permanently on heating: Bakelite).
• Addition (chain-growth) polymerisation: monomers with double bonds add without losing any molecule. Free-radical mechanism: initiation (by peroxide), propagation, termination.
• Condensation (step-growth) polymerisation: monomers with two functional groups join with loss of small molecules like H₂O or HCl (nylon-6,6, terylene).
• Ziegler–Natta catalyst (TiCl₄ + Al(C₂H₅)₃) gives linear, high-density polythene and stereoregular polymers.
• Natural rubber is cis-1,4-polyisoprene. Vulcanisation (heating with sulphur) forms S cross-links, making rubber harder, more elastic and less sticky.
• Synthetic rubbers: neoprene (from chloroprene), Buna-S (butadiene + styrene), Buna-N (butadiene + acrylonitrile, oil-resistant).
• Biodegradable polymers: PHBV and nylon-2-nylon-6 are broken down by microbes; they help reduce plastic pollution.
• Glass transition temperature (Tg): below Tg a polymer is hard and glassy; above it, soft and rubbery.

महत्त्वाच्या संज्ञा

• Polymer (बहुवारिक) – large molecule of repeating units.
• Degree of polymerisation – number of repeating units in a polymer chain.
• Thermoplastic – polymer that softens on heating repeatedly.
• Thermosetting polymer – polymer that sets hard permanently on heating.
• Vulcanisation – heating rubber with sulphur to form cross-links.
• Polydispersity index (PDI) – M̄w/M̄n; equals 1 for a polymer with all chains of the same length.

सूत्रे

• Number-average molecular mass: M̄n = ΣNᵢMᵢ / ΣNᵢ
• Weight-average molecular mass: M̄w = ΣNᵢMᵢ² / ΣNᵢMᵢ
• PDI = M̄w / M̄n (≥ 1)
• Degree of polymerisation = M̄n / mass of repeating unit

५ महत्त्वाचे प्रश्न

Q1. Distinguish between thermoplastic and thermosetting polymers.

Thermoplastics are linear or slightly branched and soften on heating, so they can be remoulded many times (PVC, polythene). Thermosetting polymers form a 3D cross-linked network on heating and become permanently hard; they cannot be remoulded (Bakelite).

Q2. What is vulcanisation of rubber? Why is it done?

Vulcanisation is heating natural rubber with sulphur (about 3–5%) at 373–415 K. Sulphur forms cross-links between chains. This makes rubber stronger, more elastic, less sticky and usable over a wider temperature range (tyres).

Q3. A polymer sample has 100 molecules of molar mass 10,000 g/mol and 200 molecules of molar mass 20,000 g/mol. Calculate M̄n, M̄w and PDI.

• ΣNM = 100 × 10⁴ + 200 × 2 × 10⁴ = 1 × 10⁶ + 4 × 10⁶ = 5 × 10⁶; ΣN = 300
• M̄n = 5 × 10⁶/300 = 16,667 g/mol
• ΣNM² = 100 × 10⁸ + 200 × 4 × 10⁸ = 9 × 10¹⁰
• M̄w = 9 × 10¹⁰/5 × 10⁶ = 18,000 g/mol
• PDI = 18,000/16,667 ≈ 1.08

Q4. Explain the free-radical mechanism of polymerisation of ethene.

• Initiation: benzoyl peroxide breaks into radicals R•; R• + CH₂=CH₂ → R−CH₂−CH₂•.
• Propagation: R−CH₂−CH₂• + CH₂=CH₂ → R−CH₂−CH₂−CH₂−CH₂•, and so on; chain grows.
• Termination: two growing radicals combine (or disproportionate) and the chain stops.
• Product: polythene, −(CH₂−CH₂)ₙ−.
• High-pressure free-radical method gives branched LDPE.

Q5. Write the preparation of nylon-6,6 and Bakelite.

• Nylon-6,6: n H₂N(CH₂)₆NH₂ + n HOOC(CH₂)₄COOH → −[NH(CH₂)₆NH−CO(CH₂)₄CO]ₙ− + 2n H₂O; heated under pressure; "6,6" = six carbons in each monomer.
• Bakelite: phenol + formaldehyde (acid or base catalyst) → o- and p-hydroxymethylphenols → Novolac (linear) → heating with more HCHO gives cross-linked Bakelite.
• Both are condensation polymers.

परीक्षेसाठी टीप

Classification questions are common: always give one example for each class. Learn monomers of six key polymers from the table.

थोडक्यात सारांश

Dyes are coloured organic compounds that can fix themselves to a fabric and give it a lasting colour. This topic covers the qualities of a good dye, Witt's theory of colour (chromophores and auxochromes), classification of dyes by structure and by method of application, and the synthesis and uses of methyl orange, Congo red, malachite green, phenolphthalein, fluorescein, alizarin and indigo.

महत्त्वाचे मुद्दे

• A dye must: have a suitable colour, attach firmly to the fibre, and be fast (resist washing, light and sweat). Not every coloured substance is a dye.
• W. H. Perkin made the first synthetic dye, mauveine, in 1856.
• Witt's theory (1876): colour comes from chromophores — unsaturated groups such as –N=N– (azo), C=O, –NO₂, C=C, quinonoid rings. The compound containing a chromophore is a chromogen.
• Auxochromes (–OH, –NH₂, –NHR, –NR₂) do not give colour by themselves but deepen it and help the dye bind to the fibre by salt formation.
• Modern view: colour arises from absorption of visible light by an extended conjugated system; longer conjugation shifts absorption to longer wavelength (bathochromic shift).
• Classification by structure: nitro and nitroso, azo (methyl orange, Congo red), triphenylmethane (malachite green), phthaleins and xanthenes (phenolphthalein, fluorescein), anthraquinone (alizarin), indigoid (indigo).
• acid dyes (sulphonic acid groups; wool, silk) – e.g. orange II
• basic dyes (amino groups; wool, silk, acrylic) – malachite green
• direct dyes (applied straight to cotton from water) – Congo red
• mordant dyes (need a metal ion, Al³⁺/Cr³⁺, as fixing agent) – alizarin
• vat dyes (reduced to a soluble leuco form, applied, then oxidised in air) – indigo
• disperse dyes (polyester), reactive dyes (form covalent bonds with cotton)
• Classification by application:
• Methyl orange: diazotised sulphanilic acid coupled with N,N-dimethylaniline. Used as an acid–base indicator (red below pH 3.1, yellow above pH 4.4).
• Phenolphthalein: phthalic anhydride + 2 phenol with conc. H₂SO₄. Colourless in acid, pink in base (pH 8.3–10).
• Fluorescein: phthalic anhydride + resorcinol with ZnCl₂; strong green fluorescence in alkaline solution; used in tracing water leaks.
• Malachite green: benzaldehyde + 2 N,N-dimethylaniline (conc. H₂SO₄ or ZnCl₂) → leuco base, oxidised (PbO₂) → dye.
• Alizarin: anthraquinone → sulphonation → fusion with NaOH (and KClO₃) → alizarin (1,2-dihydroxyanthraquinone).

महत्त्वाच्या संज्ञा

• Chromophore (रंगदायी गट) – group responsible for colour.
• Auxochrome – group that intensifies colour and fixes the dye.
• Mordant – metal salt that binds a dye to the fibre.
• Leuco compound – colourless reduced form of a dye.

सूत्रे

५ महत्त्वाचे प्रश्न

Q1. What are chromophores and auxochromes? Give two examples of each.

Chromophores are unsaturated groups that cause colour, e.g. –N=N–, –NO₂. Auxochromes are groups that deepen colour and help binding, e.g. –OH, –NH₂.

Q2. Why does phenolphthalein turn pink in alkaline solution?

In base it loses protons and the lactone ring opens to give a quinonoid dianion with an extended conjugated system, which absorbs visible light and appears pink. In acid it stays in the colourless lactone (benzenoid) form.

Q3. Give the synthesis and uses of methyl orange.

• Sulphanilic acid + Na₂CO₃ + NaNO₂ + HCl at 0–5 °C → diazonium salt.
• Coupling with N,N-dimethylaniline at the para position → helianthin (red, acid form).
• Treatment with NaOH → sodium salt, methyl orange.
• Uses: indicator for strong acid–weak base titrations; dyeing wool.

Q4. Classify dyes on the basis of application with one example each.

Write acid, basic, direct, mordant, vat, disperse and reactive dyes with the fibre each suits and how it is fixed.

Q5. Write the synthesis of fluorescein and alizarin.

Fluorescein: phthalic anhydride + 2 resorcinol, ZnCl₂, about 200 °C. Alizarin: anthraquinone + fuming H₂SO₄ → anthraquinone-2-sulphonic acid; fusion with NaOH and KClO₃; acidification. State a use of each.

परीक्षेसाठी टीप

Learn each dye as a set: reactants, condition, structural class, application class and one use. That set answers both short and long questions.

थोडक्यात सारांश

Drugs are chemicals that are used to prevent, diagnose or cure disease. This topic covers basic terms, the classification of drugs by therapeutic action, and the synthesis and uses of important drugs: aspirin, paracetamol, sulphanilamide, isoniazid, chloroquine and an introduction to antibiotics, with a note on drug safety.

महत्त्वाचे मुद्दे

• Chemotherapy: treatment of disease with chemicals that harm the disease-causing organism more than the patient. The term was introduced by Paul Ehrlich, who made salvarsan (an arsenic drug for syphilis, 1910).
• Pharmacophore: the part of a drug molecule responsible for its biological action.
• Therapeutic index = LD₅₀/ED₅₀; a larger value means a safer drug.
• Analgesics relieve pain: non-narcotic (aspirin, paracetamol, ibuprofen) and narcotic (morphine, codeine — addictive).
• Antipyretics lower fever (paracetamol, aspirin). Anti-inflammatory: aspirin, ibuprofen.
• Aspirin (acetylsalicylic acid): salicylic acid + acetic anhydride (conc. H₂SO₄ or H₃PO₄ catalyst). Also prevents blood clotting; can irritate the stomach.
• Paracetamol (p-acetamidophenol): p-aminophenol + acetic anhydride. p-Aminophenol is made from phenol (nitration → p-nitrophenol → reduction) or from nitrobenzene by electrolytic reduction.
• Sulpha drugs (antibacterial): sulphanilamide from acetanilide → chlorosulphonation (ClSO₃H) → p-acetamidobenzenesulphonyl chloride → NH₃ → hydrolysis. They block bacterial folic acid synthesis by imitating p-aminobenzoic acid (PABA).
• Antibiotics: substances made by micro-organisms that kill or stop other microbes. Penicillin was discovered by Alexander Fleming (1928). Broad-spectrum: chloramphenicol, tetracycline; narrow-spectrum: penicillin G.
• Antitubercular: isoniazid (INH) — 4-methylpyridine (γ-picoline) → oxidation → isonicotinic acid → ethyl ester → hydrazine hydrate → isonicotinic acid hydrazide.
• Antimalarials: quinine (natural), chloroquine (synthetic 4-aminoquinoline).
• Others: antacids (Mg(OH)₂, ranitidine), antihistamines (cetirizine), tranquilisers (diazepam), antiseptics (Dettol, 0.2 % phenol) and disinfectants (1 % phenol), local anaesthetics (lidocaine).

महत्त्वाच्या संज्ञा

• Drug (औषध) – substance used to treat or prevent disease.
• LD₅₀ – dose that kills 50 % of test animals.
• Broad-spectrum antibiotic – effective against many types of bacteria (Gram-positive and Gram-negative).

सूत्रे

• Aspirin synthesis: C₆H₄(OH)COOH + (CH₃CO)₂O → C₆H₄(OCOCH₃)COOH + CH₃COOH

५ महत्त्वाचे प्रश्न

Q1. What is the difference between an antiseptic and a disinfectant?

Antiseptics kill or stop microbes on living tissue (Dettol, 0.2 % phenol, iodine tincture). Disinfectants are used on non-living objects like floors and toilets (1 % phenol, chlorine). The same substance can be either, depending on concentration.

Q2. How do sulpha drugs act?

Sulphanilamide is structurally like p-aminobenzoic acid (PABA). Bacteria take it up instead of PABA, so they cannot make folic acid and stop multiplying. Humans get folic acid from food, so they are not affected.

Q3. Give the synthesis and uses of paracetamol.

Phenol → (dil. HNO₃) p-nitrophenol → (Sn/HCl or H₂/Ni) p-aminophenol → (acetic anhydride) paracetamol. Uses: antipyretic and analgesic. Overdose damages the liver.

Q4. Describe the synthesis of sulphanilamide from acetanilide.

• Acetanilide + ClSO₃H → p-acetamidobenzenesulphonyl chloride (the –NHCOCH₃ group protects –NH₂ and directs para).
• + NH₃ → p-acetamidobenzenesulphonamide.
• Acid hydrolysis removes the acetyl group → sulphanilamide.

Q5. Write notes on antibiotics and antitubercular drugs.

Definition, Fleming and penicillin, broad and narrow spectrum with examples, bactericidal vs bacteriostatic; synthesis of isoniazid in steps; why TB drugs are given in combination (to prevent resistance).

परीक्षेसाठी टीप

For each drug write: class, synthesis in arrows, and one use. Names of discoverers (Ehrlich, Fleming) are common one-mark questions.

थोडक्यात सारांश

Green chemistry is the design of chemical products and processes that reduce or eliminate the use and production of hazardous substances. It aims to prevent pollution at the source instead of cleaning it later. This topic covers the meaning and need of green chemistry, the twelve principles given by Paul Anastas and John Warner (1998), atom economy, green solvents and reagents, alternative energy sources like microwaves and ultrasound, and examples of green syntheses.

महत्त्वाचे मुद्दे

• Green chemistry is also called sustainable chemistry. Its aim: less waste, safer chemicals, less energy, renewable raw materials.
• Need: traditional processes use toxic solvents and reagents, produce large amounts of waste, use much energy, and cause pollution and health hazards.
• Twelve principles (Anastas and Warner): 1. Prevent waste. 2. Maximise atom economy. 3. Less hazardous chemical syntheses. 4. Design safer chemicals. 5. Safer solvents and auxiliaries. 6. Design for energy efficiency. 7. Use renewable feedstocks. 8. Reduce derivatives (avoid unnecessary protecting groups). 9. Use catalysts rather than stoichiometric reagents. 10. Design for degradation. 11. Real-time analysis for pollution prevention. 12. Inherently safer chemistry for accident prevention.
• Atom economy (Barry Trost) measures how many atoms of the reactants end up in the desired product. Addition and rearrangement reactions have 100% atom economy; substitution and elimination reactions are lower.
• Green solvents: water, supercritical CO₂ (used for decaffeination of coffee), ionic liquids, and solvent-free reactions.
• Alternative energy: microwave-assisted reactions (faster heating, shorter times), ultrasound (sonochemistry), photochemical reactions.
• Catalysis: catalysts are used in small amounts and can be reused; biocatalysts (enzymes) work in water at mild conditions.
• Examples: the Boots–Hoechst–Celanese (BHC) route to ibuprofen uses 3 catalytic steps instead of the older 6-step Boots process and gives much less waste; dry cleaning with liquid CO₂ instead of perchloroethylene; making adipic acid from glucose using enzymes instead of benzene; hydrogen peroxide as a clean oxidant (gives only water).
• Green chemistry is linked to the idea of the E-factor (Roger Sheldon): kg of waste per kg of product. Lower is better.

महत्त्वाच्या संज्ञा

• Green chemistry (हरित रसायनशास्त्र) – chemistry that avoids pollution at the source.
• Atom economy – percentage of reactant mass that ends up in the desired product.
• E-factor – mass of waste ÷ mass of product.
• Supercritical fluid – substance above its critical temperature and pressure with properties of both gas and liquid.
• Renewable feedstock – raw material from biomass that can be replaced naturally.

सूत्रे

• % Atom economy = (Molar mass of desired product / Sum of molar masses of all reactants) × 100
• % Yield = (actual yield / theoretical yield) × 100
• E-factor = total mass of waste / mass of product

५ महत्त्वाचे प्रश्न

Q1. Define green chemistry. Who gave the twelve principles?

Green chemistry is the design of chemical products and processes that reduce or eliminate hazardous substances. The twelve principles were given by Paul Anastas and John Warner in 1998.

Q2. Why is supercritical CO₂ called a green solvent?

It is non-toxic, non-flammable, cheap and easily available. After use it simply turns into gas and can be recycled, leaving no solvent residue. It is used for extracting caffeine from coffee and in dry cleaning.

Q3. Calculate the atom economy for the preparation of ethene from ethanol: C₂H₅OH → C₂H₄ + H₂O.

• Molar masses: C₂H₅OH = 46 g/mol; C₂H₄ = 28 g/mol
• % Atom economy = (28/46) × 100
• = 60.9%
• The rest (water) is a by-product, so elimination reactions have lower atom economy.

Q4. Calculate the atom economy for the addition reaction CH₂=CH₂ + Br₂ → CH₂Br−CH₂Br.

• Reactants: C₂H₄ = 28, Br₂ = 159.8; total = 187.8
• Product C₂H₄Br₂ = 187.8
• % Atom economy = (187.8/187.8) × 100 = 100%
• All atoms of the reactants are in the product; addition reactions are ideal from the atom-economy view.

Q5. Explain any five principles of green chemistry with examples.

• Prevention of waste: better to avoid waste than treat it (BHC ibuprofen process).
• Atom economy: prefer addition and rearrangement reactions.
• Safer solvents: water, supercritical CO₂ instead of benzene or CCl₄.
• Catalysis: catalysts and enzymes instead of excess reagents.
• Energy efficiency: microwave reactions at room pressure, shorter time.
• Design for degradation: biodegradable polymers like PHBV.

परीक्षेसाठी टीप

Atom economy numericals are easy marks. Write the balanced equation, molar masses of all reactants and the desired product, then use the formula.