Chemistry रसायनशास्त्र

Quick summary

Coordination compounds contain a central metal atom or ion bonded to a fixed number of ions or molecules called ligands. This topic covers Werner's theory, important terms, IUPAC naming, isomerism, the effective atomic number (EAN) rule, and valence bond theory (VBT) with its limitations.

Key points

• Example: [Co(NH₃)₆]Cl₃. The part in square brackets is the coordination sphere; Cl⁻ ions outside are counter ions.
• Werner's theory (1893): metals show two valencies. Primary valency = oxidation state, ionisable, satisfied by negative ions. Secondary valency = coordination number, non-ionisable, satisfied by ligands, and directed in space (gives geometry).
• Ligands: monodentate (Cl⁻, NH₃, H₂O, CN⁻), bidentate (ethylenediamine 'en', oxalate C₂O₄²⁻), polydentate (EDTA⁴⁻ is hexadentate).
• Ambidentate ligands can bind through two different atoms: NO₂⁻ (via N, nitro) or ONO⁻ (via O, nitrito); SCN⁻ / NCS⁻.
• Chelate: ring formed when a polydentate ligand binds the same metal. Chelates are extra stable.
• Coordination number (CN): number of donor atoms bonded to the metal. CN 4 → tetrahedral or square planar; CN 6 → octahedral.
• Isomerism: structural (ionisation, hydrate, linkage, coordination) and stereo (geometrical cis–trans, optical).
• [Co(NH₃)₅Br]SO₄ and [Co(NH₃)₅SO₄]Br are ionisation isomers.
• EAN rule (Sidgwick): EAN = Z − oxidation state + 2 × CN. Many stable complexes have EAN equal to the next noble gas number.
• VBT: metal uses hybrid orbitals: sp³ (tetrahedral), dsp² (square planar), d²sp³ (inner orbital octahedral, low spin), sp³d² (outer orbital octahedral, high spin).
• Strong field ligands (CN⁻, CO) cause pairing of electrons; weak field ligands (F⁻, H₂O) do not.
• Limitations of VBT: does not explain colour, gives no quantitative idea of magnetic data beyond spin, does not separate strong and weak ligands clearly.

Important terms

• Ligand (लिगंड) – ion or molecule that donates an electron pair to the metal.
• Denticity – number of donor atoms in one ligand.
• Chelation – ring formation by a polydentate ligand.
• Inner orbital complex – uses (n − 1)d orbitals (d²sp³).
• Outer orbital complex – uses nd orbitals (sp³d²).

Formulas

• Oxidation state x: sum of charges = charge on complex. In [Fe(CN)₆]⁴⁻: x + 6(−1) = −4 → x = +2
• EAN = Z − (oxidation state) + 2 × (coordination number)
• Naming rules: cation first; ligands in alphabetical order before metal; prefixes di, tri, tetra (bis, tris for complex names); anionic ligands end in -o (chlorido, cyanido); anionic complex ends in -ate (ferrate, cuprate); oxidation state in Roman numerals.

5 important questions

Q1. Write IUPAC names of K₄[Fe(CN)₆] and [Co(NH₃)₅Cl]Cl₂.

K₄[Fe(CN)₆] – potassium hexacyanidoferrate(II). [Co(NH₃)₅Cl]Cl₂ – pentaamminechloridocobalt(III) chloride.

Q2. What are ambidentate ligands? Give two examples.

Ligands that have two different donor atoms but bind through only one at a time. NO₂⁻ binds through N (nitro) or O (nitrito); SCN⁻ binds through S (thiocyanato) or N (isothiocyanato). They give linkage isomers.

Q3. Calculate the EAN of Fe in K₄[Fe(CN)₆] (Z of Fe = 26).

• Oxidation state of Fe: x + 6(−1) = −4 → x = +2
• Coordination number = 6
• EAN = 26 − 2 + 2 × 6 = 36
• 36 = atomic number of krypton, so the complex is stable.

Q4. Explain the structure of [Ni(CN)₄]²⁻ using VBT.

• Ni²⁺ = 3d⁸ (two unpaired electrons in free ion).
• CN⁻ is a strong field ligand; the two unpaired 3d electrons pair up, freeing one 3d orbital.
• One 3d, one 4s and two 4p orbitals mix → dsp² hybridisation.
• Four CN⁻ donate electron pairs → square planar shape.
• No unpaired electrons → diamagnetic.

Q5. State the main postulates of Werner's theory.

• Metal shows primary (ionisable) and secondary (non-ionisable) valencies.
• Primary valency = oxidation state; satisfied by anions.
• Secondary valency = coordination number; satisfied by ligands (neutral or negative).
• Secondary valencies are directed in space and decide geometry.
• Example: CoCl₃·6NH₃ gives 3 Cl⁻ with AgNO₃ → [Co(NH₃)₆]Cl₃; CoCl₃·5NH₃ gives 2 Cl⁻ → [Co(NH₃)₅Cl]Cl₂.

Exam tip

IUPAC naming and VBT structures are the most frequent questions. For VBT, show the box diagram of the free ion and of the complex.

Quick summary

Redox reactions are reactions in which oxidation and reduction happen together. This topic covers the electronic concept of oxidation and reduction, oxidation number rules, oxidising and reducing agents, balancing redox equations by the oxidation number and ion-electron (half-reaction) methods, equivalent weight, and redox titrations.

Key points

• Oxidation: loss of electrons, or increase in oxidation number. Reduction: gain of electrons, or decrease in oxidation number. Memory aid: OIL RIG (Oxidation Is Loss, Reduction Is Gain).
• Oxidising agent gains electrons and itself gets reduced (KMnO₄, K₂Cr₂O₇, H₂O₂, Cl₂). Reducing agent loses electrons and gets oxidised (H₂, Zn, Fe²⁺, H₂S, oxalic acid).
• Oxidation number rules: element in free state = 0; monoatomic ion = its charge; O = −2 (but −1 in peroxides, +2 in OF₂); H = +1 (but −1 in metal hydrides like NaH); F = −1 always; sum in a neutral molecule = 0; sum in an ion = its charge.
• Disproportionation: the same element is both oxidised and reduced. Example: 2H₂O₂ → 2H₂O + O₂; Cl₂ + 2NaOH → NaCl + NaClO + H₂O.
• Balancing (oxidation number method): find changes in oxidation numbers, equalise increase and decrease, then balance O with H₂O and H with H⁺ (acid) or OH⁻ (base).
• Ion-electron method: split into oxidation and reduction half-reactions, balance each for atoms and charge (using electrons), multiply to equalise electrons, and add.
• Equivalent weight of an oxidant or reductant = molar mass ÷ number of electrons gained or lost per formula unit.
• KMnO₄ in acidic medium gains 5 electrons (MnO₄⁻ → Mn²⁺); in neutral/weakly basic medium gains 3 (→ MnO₂).
• K₂Cr₂O₇ in acid gains 6 electrons per formula unit (Cr₂O₇²⁻ → 2Cr³⁺).
• Redox titrations: KMnO₄ vs oxalic acid or Mohr's salt (KMnO₄ is a self-indicator: first permanent pink colour is the end point). K₂Cr₂O₇ vs Fe²⁺ uses diphenylamine as indicator. Iodometry uses starch.

Important terms

• Oxidation number (ऑक्सिडीकरण अंक) – charge an atom would have if all bonds were ionic.
• Half-reaction – oxidation or reduction part of a redox reaction shown separately.
• Self-indicator – titrant whose own colour shows the end point (KMnO₄).
• Normality – gram equivalents of solute per litre of solution.
• Disproportionation – simultaneous oxidation and reduction of one element.

Formulas

• Equivalent weight = Molar mass / n (n = electrons transferred per formula unit)
• Normality N = Molarity × n
• Titration: N₁V₁ = N₂V₂
• Acidic permanganate: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O
• Dichromate: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O
• Oxalate: C₂O₄²⁻ → 2CO₂ + 2e⁻
• Fe²⁺ → Fe³⁺ + e⁻

5 important questions

Q1. Find the oxidation number of Cr in K₂Cr₂O₇ and of S in H₂SO₄.

K₂Cr₂O₇: 2(+1) + 2x + 7(−2) = 0 → 2x = 12 → x = +6. H₂SO₄: 2(+1) + x + 4(−2) = 0 → x = +6.

Q2. What is a disproportionation reaction? Give an example.

A redox reaction in which the same element in one substance is oxidised and reduced at the same time. Example: 2H₂O₂ → 2H₂O + O₂; oxygen goes from −1 to −2 (in H₂O) and to 0 (in O₂).

Q3. Balance by the ion-electron method: MnO₄⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺ (acidic medium).

• Reduction: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O
• Oxidation: Fe²⁺ → Fe³⁺ + e⁻ (multiply by 5)
• 5Fe²⁺ → 5Fe³⁺ + 5e⁻
• Add: MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O
• Check charge: left 17+, right 17+ ✓

Q4. Find the equivalent weight of KMnO₄ in acidic medium and the normality of a 0.02 M solution. (KMnO₄ = 158 g/mol)

• Mn changes from +7 to +2 → n = 5
• Equivalent weight = 158/5 = 31.6 g
• Normality = Molarity × n = 0.02 × 5 = 0.1 N

Q5. 20 mL of 0.1 N oxalic acid needs 25 mL of KMnO₄ for complete titration. Find the normality and strength of KMnO₄ in g/L.

• N₁V₁ = N₂V₂ → 0.1 × 20 = N₂ × 25
• N₂ = 2/25 = 0.08 N
• Strength = N × Eq. wt = 0.08 × 31.6
• = 2.528 g/L
• End point: colourless to permanent light pink (KMnO₄ acts as self-indicator; titration done hot, about 60 °C, in dilute H₂SO₄).

Exam tip

Always check both atoms and total charge at the end of balancing. In titration problems write N₁V₁ = N₂V₂ with labels for each side.

Quick summary

Titrimetric (volumetric) analysis finds the amount of a substance by measuring the volume of a standard solution that reacts completely with it. This topic covers basic terms, primary and secondary standards, the four types of titrations (acid–base, redox, precipitation, complexometric), theory of indicators, titration curves, and calculations.

Key points

• Titration: slow addition of a solution of known concentration (titrant, in burette) to the solution being analysed (titrand / analyte, in conical flask) until the reaction is complete.
• Equivalence point: exact amount of titrant needed has been added (theoretical). End point: point at which the indicator changes colour (practical). A good indicator makes them almost equal.
• Primary standard: pure, stable, non-hygroscopic, high molar mass, easily available — e.g. oxalic acid (H₂C₂O₄·2H₂O), anhydrous Na₂CO₃, K₂Cr₂O₇, borax. A secondary standard (NaOH, KMnO₄, HCl) must be standardised against a primary standard.
• Acid–base titrations: indicators are weak organic acids or bases whose two forms have different colours (Ostwald's theory). Phenolphthalein (pH 8.3–10, colourless → pink) and methyl orange (pH 3.1–4.4, red → yellow).
• Choice of indicator: strong acid–strong base: either; weak acid–strong base: phenolphthalein; strong acid–weak base: methyl orange; weak acid–weak base: no sharp end point.
• Redox titrations: KMnO₄ (self-indicator, in dilute H₂SO₄, colourless → pale pink), K₂Cr₂O₇ (diphenylamine indicator), iodometry and iodimetry (starch indicator, blue colour).
• Precipitation titrations: argentometric; Mohr's method (K₂CrO₄ indicator, red-brown Ag₂CrO₄ at the end), Volhard's method (Fe³⁺ indicator, back titration with KSCN), Fajans' method (adsorption indicators like fluorescein).
• Complexometric titrations: EDTA (hexadentate) forms 1:1 complexes with metal ions; Eriochrome Black T indicator (wine red → blue) at pH 10 buffer; used to find hardness of water.
• Titration curves: plot of pH against volume of titrant; the steep jump near equivalence decides which indicator can be used.

Important terms

• Standard solution (प्रमाणित द्रावण) – solution whose concentration is exactly known.
• Normality (N) – gram equivalents of solute per litre.
• Back titration – excess reagent is added and the unused part is titrated.

Formulas

• Normality equation: N₁V₁ = N₂V₂
• Molarity form: (M₁V₁)/n₁ = (M₂V₂)/n₂ (n = stoichiometric coefficients)
• Strength (g L⁻¹) = Normality × equivalent weight = Molarity × molar mass
• Equivalent weight of KMnO₄ in acidic medium = M/5 = 158/5 = 31.6
• Equivalent weight of oxalic acid dihydrate = 126/2 = 63

5 important questions

Q1. Distinguish between end point and equivalence point.

Equivalence point is when chemically equivalent amounts of titrant and analyte have reacted. End point is when the indicator visibly changes colour. The difference between them is the titration error.

Q2. Why is KMnO₄ called a self-indicator?

MnO₄⁻ is deep purple and its reduced form Mn²⁺ is almost colourless. The first extra drop after the end point leaves the solution pale pink, so no separate indicator is needed.

Q3. 25 mL of NaOH solution required 20 mL of 0.1 N HCl for neutralisation. Find the normality and strength of NaOH.

• N₁V₁ = N₂V₂ → N(NaOH) × 25 = 0.1 × 20
• N(NaOH) = 2/25 = 0.08 N
• Strength = 0.08 × 40 = 3.2 g L⁻¹

Q4. Explain Ostwald's theory of acid–base indicators.

Phenolphthalein as a weak acid HIn ⇌ H⁺ + In⁻ (colourless / pink). Le Chatelier: in acid HIn dominates, in base In⁻. Methyl orange as a weak base. Indicator range pH = pKIn ± 1. Limitation: does not explain the colour itself (quinonoid theory does).

Q5. Explain complexometric titration with EDTA and its use to find hardness of water.

Structure and hexadentate nature of EDTA, 1:1 complex, pH 10 ammonia buffer, EBT indicator colour change, calculation of total hardness as ppm of CaCO₃.

Exam tip

Always write the balanced reaction before using N₁V₁ = N₂V₂, and check whether the question gives normality or molarity.

Quick summary

Chromatography is a group of methods for separating the components of a mixture. Separation happens because the components are distributed differently between a stationary phase and a mobile phase that moves over it. This topic covers the principle and classification, paper, thin-layer and column chromatography, Rf value, and an introduction to gas chromatography and HPLC.

Key points

• Russian botanist M. S. Tswett (1906) first separated plant pigments on a column of chalk (CaCO₃); the name means "colour writing".
• Stationary phase: a solid or a liquid held on a solid support. Mobile phase: a liquid or gas that moves through it.
• Adsorption chromatography: components are held on the surface of a solid (column chromatography with silica gel or alumina, TLC).
• Partition chromatography: components divide between two liquids, one held stationary on a support (paper chromatography — water held in cellulose fibres is the stationary phase).
• Other types: ion-exchange, gel (size-exclusion), affinity, gas chromatography (GC), HPLC.
• Paper chromatography: spot near the bottom of Whatman filter paper, dip in solvent (ascending, descending or radial). Colourless spots are made visible with a locating reagent: ninhydrin for amino acids (purple), iodine vapour, UV light.
• TLC: glass plate coated with a thin layer of silica gel G (gypsum binder). Faster and sharper than paper; used to check purity and follow reactions.
• Column chromatography: adsorbent packed in a glass column; mixture added at top; solvents of increasing polarity (eluent) wash out components one by one (elution). Less strongly adsorbed component comes out first.
• Rf value is constant for a substance under fixed conditions (same solvent, paper, temperature), so it helps identify it. It always lies between 0 and 1.
• Gas chromatography: mobile phase is an inert carrier gas (He, N₂); used for volatile substances; detected by flame ionisation or thermal conductivity detectors. Result is shown by retention time.
• HPLC: liquid mobile phase pumped at high pressure through a column of very fine particles; fast, sensitive; used for drugs and food analysis.

Important terms

• Eluent – solvent used to wash components through a column.
• Retention factor, Rf (धारण गुणांक) – ratio of distance moved by the substance to distance moved by the solvent front.
• Retention time – time taken by a component to pass through a GC/HPLC column.

Formulas

• Rf = distance travelled by the solute from the origin / distance travelled by the solvent front from the origin

5 important questions

Q1. In a paper chromatogram the solvent front moved 7.5 cm and a spot moved 3.0 cm. Find Rf.

Rf = 3.0/7.5 = 0.40.

Q2. Why is Rf always less than 1?

A solute can at most move as far as the solvent; it is slowed down by the stationary phase, so its distance is smaller than that of the solvent front.

Q3. Explain the principle and procedure of thin-layer chromatography.

Preparation and activation of the plate, spotting, development in a closed jar, drying, locating spots (iodine chamber/UV), calculating Rf. Applications: purity check, monitoring reactions, identifying drugs and dyes. Advantages over paper chromatography.

Q4. Describe column chromatography with a labelled diagram.

Packing the column, loading the sample, elution with solvents in order of increasing polarity (hexane → benzene → chloroform → ethyl acetate → methanol), collecting fractions, separation of chlorophyll pigments as an example.

Q5. Write a note on gas chromatography.

Carrier gas, injection port, column in oven, detector and recorder; principle of partition; retention time; uses (essential oils, alcohol in blood, pesticide residues). Limitation: only volatile, thermally stable compounds.

Exam tip

A neat diagram of a TLC plate showing origin, spots and solvent front, with the Rf formula, scores full marks in short notes.

Quick summary

Amines are organic derivatives of ammonia in which one or more hydrogen atoms are replaced by alkyl or aryl groups. This topic covers classification, nomenclature, methods of preparation, basic character, chemical reactions, distinction between 1°, 2° and 3° amines, and diazonium salts.

Key points

• Classification: primary (RNH₂), secondary (R₂NH), tertiary (R₃N). Quaternary ammonium salts: R₄N⁺X⁻.
• Nitrogen in amines is sp³ with one lone pair; the shape is pyramidal.
• Preparation: reduction of nitro compounds (Sn/HCl or H₂/Ni), reduction of nitriles and amides (LiAlH₄), Gabriel phthalimide synthesis (gives pure 1° aliphatic amines), Hofmann bromamide degradation (amide → amine with one carbon less).
• Basicity: the lone pair on N accepts a proton. Aliphatic amines are stronger bases than NH₃ (+I effect of alkyl groups). Aromatic amines like aniline are weaker bases than NH₃ because the lone pair is delocalised into the benzene ring.
• In water, the order for methylamines is (CH₃)₂NH > CH₃NH₂ > (CH₃)₃N > NH₃, due to a balance of +I effect, solvation (H-bonding) and steric hindrance.
• Hinsberg's test (benzenesulphonyl chloride) distinguishes 1°, 2° and 3° amines.
• Carbylamine test: only 1° amines (aliphatic and aromatic) give a foul-smelling isocyanide with CHCl₃ and alcoholic KOH.
• Reaction with nitrous acid (HNO₂): 1° aliphatic amines give alcohol and N₂ gas; 1° aromatic amines at 0–5 °C give diazonium salts; 2° amines give yellow oily N-nitrosoamines.
• Diazonium salts (C₆H₅N₂⁺Cl⁻) are very useful: Sandmeyer reaction (CuCl/HCl → chlorobenzene; CuCN → benzonitrile), replacement by −OH (warm water), −I (KI), −H (H₃PO₂), and coupling with phenol or aniline to form azo dyes.
• The −NH₂ group in aniline is strongly ortho/para directing; aniline with bromine water gives a white precipitate of 2,4,6-tribromoaniline.

Important terms

• Amine (अमाइन) – compound with −NH₂, −NH− or −N< group attached to carbon.
• Diazotisation – conversion of a 1° aromatic amine into a diazonium salt with NaNO₂ + HCl at 0–5 °C.
• Coupling reaction – diazonium salt joins with phenol/aniline to give coloured azo compounds.
• pKb – measure of base strength; lower pKb means stronger base.
• Acetylation – replacing H of −NH₂ by CH₃CO−, used to protect the amino group.

Formulas

Key reactions:

• Hofmann bromamide: RCONH₂ + Br₂ + 4NaOH → RNH₂ + Na₂CO₃ + 2NaBr + 2H₂O
• Carbylamine: RNH₂ + CHCl₃ + 3KOH → RNC + 3KCl + 3H₂O
• Diazotisation: C₆H₅NH₂ + NaNO₂ + 2HCl → C₆H₅N₂⁺Cl⁻ + NaCl + 2H₂O (0–5 °C)
• Sandmeyer: C₆H₅N₂⁺Cl⁻ + CuCl → C₆H₅Cl + N₂ (CuCl/HCl)
• Coupling: C₆H₅N₂⁺Cl⁻ + C₆H₅OH → p-hydroxyazobenzene (orange dye) + HCl (alkaline medium)

5 important questions

Q1. Why is aniline a weaker base than methylamine?

In aniline the lone pair on N is delocalised into the benzene ring by resonance, so it is less available to accept H⁺. In methylamine the CH₃ group pushes electrons (+I) to N, increasing availability. So methylamine is the stronger base.

Q2. What is the carbylamine test? Which amines give it?

When a primary amine is warmed with chloroform and alcoholic KOH, a foul-smelling isocyanide (carbylamine) is formed. Only primary amines give it; secondary and tertiary amines do not.

Q3. Explain Hinsberg's test to distinguish 1°, 2° and 3° amines.

• Reagent: benzenesulphonyl chloride (C₆H₅SO₂Cl) with aqueous KOH.
• 1° amine forms N-alkylbenzenesulphonamide which has an acidic H on N; it dissolves in alkali.
• 2° amine forms N,N-dialkylbenzenesulphonamide with no H on N; insoluble in alkali.
• 3° amine does not react (no H on N).

Q4. Explain Hofmann bromamide degradation with mechanism outline.

• Amide + Br₂ + NaOH → primary amine with one carbon less.
• Example: CH₃CONH₂ → CH₃NH₂.
• Steps: N-bromoamide forms; base removes H; the alkyl group migrates from C to N as Br⁻ leaves (isocyanate forms); hydrolysis gives amine and CO₂ (as carbonate).
• Used to step down a carbon chain.

Q5. What are diazonium salts? Give their preparation and any three reactions.

• Formula ArN₂⁺X⁻, e.g. benzenediazonium chloride.
• Prepared by diazotisation of aniline with NaNO₂ and HCl at 0–5 °C (unstable above this).
• With CuCl/HCl → chlorobenzene (Sandmeyer).
• With KI → iodobenzene.
• With warm water → phenol.
• With phenol (alkaline) → orange azo dye (coupling).

Exam tip

Name reactions — Gabriel, Hofmann, Sandmeyer, carbylamine, coupling — are favourite short questions. Write each with a balanced equation and the condition.

Quick summary

Heterocyclic compounds are cyclic compounds in which the ring contains at least one atom other than carbon (a heteroatom) such as N, O or S. This topic covers classification and nomenclature, and the structure, aromaticity, preparation and reactions of the five-membered rings furan, pyrrole and thiophene and the six-membered ring pyridine.

Key points

• Common heteroatoms: nitrogen, oxygen and sulphur.
• Five-membered: furan (O), pyrrole (N–H), thiophene (S). Six-membered: pyridine (N). Fused: indole, quinoline, isoquinoline.
• Numbering starts at the heteroatom (position 1). In five-membered rings, positions 2 and 5 are called α, and 3 and 4 are β.
• Aromaticity of pyrrole, furan, thiophene: four carbon atoms give 4 π electrons, and the heteroatom gives one lone pair (2 electrons) to the ring. Total 6 π electrons → aromatic by Hückel's rule.
• Order of aromaticity: benzene > thiophene > pyrrole > furan (O is most electronegative and holds its lone pair most tightly).
• These five-membered rings are π-excessive: they undergo electrophilic substitution more easily than benzene, mainly at the 2-position (α), because the intermediate formed by attack at C-2 is more resonance-stabilised.
• Reactivity in electrophilic substitution: pyrrole > furan > thiophene > benzene.
• Pyrrole is a very weak base, because its N lone pair is part of the aromatic sextet. Its N–H is weakly acidic.
• Pyridine: N is sp² and its lone pair is in an sp² orbital in the plane of the ring, not part of the π sextet. So pyridine is basic (weaker than aliphatic amines, stronger than pyrrole).
• Pyridine is π-deficient: electrophilic substitution is difficult and occurs at the 3-position (β) under harsh conditions; nucleophilic substitution occurs at 2 and 4 (Chichibabin reaction with NaNH₂ gives 2-aminopyridine).
• Paal–Knorr synthesis: 1,4-dicarbonyl compounds give furan (with acid/P₂O₅), pyrrole (with NH₃) or thiophene (with P₂S₅).
• Industrial note: furan from furfural (obtained from pentosans in agricultural waste like corn cobs and bran).

Important terms

• Heteroatom – atom other than carbon in a ring (N, O, S).
• Heterocycle (विषमचक्रीय संयुग) – ring compound containing a heteroatom.
• π-excessive ring – ring with more π electron density per atom than benzene (pyrrole).
• π-deficient ring – ring with less π electron density than benzene (pyridine).
• Chichibabin reaction – amination of pyridine at C-2 with sodamide.

Formulas

Key reactions:

• Paal–Knorr: hexane-2,5-dione + NH₃ → 2,5-dimethylpyrrole + 2H₂O
• Pyridine + NaNH₂ → 2-aminopyridine (Chichibabin)
• Pyrrole + CHCl₃ + KOH → pyrrole-2-carbaldehyde (Reimer–Tiemann); 3-chloropyridine also forms by ring expansion
• Basicity order: piperidine > pyridine > aniline > pyrrole

5 important questions

Q1. Why is pyrrole aromatic?

Pyrrole is cyclic and planar. Each carbon gives one p electron (4 total) and nitrogen gives its lone pair (2), making 6 π electrons in a continuous ring of p orbitals. 6 = 4n + 2 with n = 1, so it is aromatic.

Q2. Why is pyridine more basic than pyrrole?

In pyridine the N lone pair lies in an sp² orbital outside the π system, so it is free to accept H⁺. In pyrrole the lone pair is part of the aromatic sextet; using it would destroy aromaticity. So pyridine is much more basic.

Q3. Why does electrophilic substitution in pyrrole occur at the 2-position?

• Attack at C-2 gives a carbocation intermediate with three resonance structures (charge spread over C-3, C-5 and N).
• Attack at C-3 gives an intermediate with only two resonance structures.
• More delocalisation means a more stable intermediate and lower activation energy.
• So 2-substitution is preferred (e.g. nitration, sulphonation, acylation).

Q4. Compare the reactivity of pyrrole, furan, thiophene and benzene towards electrophiles.

• All three five-membered rings are π-excessive (6 electrons over 5 atoms).
• Pyrrole: N is least electronegative of N and O; donates lone pair best → most reactive.
• Furan: O holds lone pair more tightly → less reactive than pyrrole.
• Thiophene: S is large; 3p–2p overlap is weaker, but it is most aromatic → least reactive of the three.
• Benzene: least reactive. Order: pyrrole > furan > thiophene > benzene.

Q5. Describe the Paal–Knorr synthesis of furan, pyrrole and thiophene.

• Starting material: a 1,4-dicarbonyl compound (e.g. hexane-2,5-dione).
• Heat with acid or P₂O₅ (dehydration) → 2,5-dimethylfuran.
• Heat with NH₃ or a primary amine → 2,5-dimethylpyrrole.
• Heat with P₂S₅ → 2,5-dimethylthiophene.
• One method gives all three rings — a common exam point.

Exam tip

Draw the orbital picture of pyrrole and pyridine side by side; it answers both aromaticity and basicity questions in one diagram.

Quick summary

Organometallic compounds contain at least one direct bond between a carbon atom and a metal (or metalloid like B, Si). This topic covers classification, the Grignard reagent (preparation and many reactions), organolithium and organozinc compounds, metal carbonyls, ferrocene, and the 18-electron rule, with industrial uses.

Key points

• A compound is organometallic only if a metal–carbon bond is present. Sodium ethoxide (C₂H₅ONa) is not organometallic (Na is bonded to O).
• Classification by bond type: ionic (Na, K compounds), σ-bonded (Grignard reagent, R₂Zn, (C₂H₅)₄Pb), π-bonded (ferrocene, Zeise's salt), and metal carbonyls.
• Grignard reagent (RMgX), discovered by Victor Grignard (Nobel Prize 1912). Prepared from an alkyl halide and Mg turnings in dry ether.
• The C–Mg bond is polar; carbon carries a partial negative charge. So R acts as a carbanion-like nucleophile and a strong base.
• Grignard reagent must be kept away from water, alcohols and acids, because it reacts to give alkanes.
• With carbonyl compounds: HCHO → 1° alcohol; other aldehydes → 2° alcohol; ketones → 3° alcohol; CO₂ → carboxylic acid (after hydrolysis).
• Organolithium (RLi) are more reactive than Grignard reagents; organozinc (R₂Zn, Frankland) are less reactive. Organozinc is used in the Reformatsky reaction.
• Metal carbonyls: CO bonds to metals by σ-donation from C plus π-back-bonding from metal d orbitals into CO π* orbitals. Examples: Ni(CO)₄ (tetrahedral), Fe(CO)₅ (trigonal bipyramidal), Cr(CO)₆ (octahedral).
• 18-electron rule: stable organometallic complexes often have 18 valence electrons around the metal (like a noble gas). Ni(CO)₄: 10 + 4 × 2 = 18.
• Ferrocene Fe(C₅H₅)₂ is a 'sandwich' compound: Fe²⁺ between two cyclopentadienyl rings; 18 electrons; aromatic-like reactions (Friedel–Crafts acylation).
• Uses: Mond process (Ni(CO)₄ to purify nickel), Wilkinson's catalyst [RhCl(PPh₃)₃] for hydrogenation, Ziegler–Natta catalyst (TiCl₄ + Al(C₂H₅)₃) for polythene, Grignard reagents in synthesis.

Important terms

• Organometallic compound (कार्बधातू संयुग) – compound with a direct metal–carbon bond.
• Hapticity (η) – number of ligand atoms bonded to the metal (ferrocene rings are η⁵).
• Back-bonding – metal gives d electrons back to empty ligand π* orbitals.
• Sandwich compound – metal between two flat aromatic rings.
• Dry ether – water-free ether; it stabilises RMgX by coordination.

Formulas

• Preparation: CH₃I + Mg → CH₃MgI (dry ether)
• With water: CH₃MgI + H₂O → CH₄ + Mg(OH)I
• With HCHO: HCHO + CH₃MgI → CH₃CH₂OMgI → (H₂O/H⁺) CH₃CH₂OH (1° alcohol)
• With acetone: (CH₃)₂CO + CH₃MgI → (CH₃)₃COMgI → (CH₃)₃COH (2-methylpropan-2-ol, 3°)
• With CO₂: CO₂ + CH₃MgI → CH₃COOMgI → CH₃COOH
• 18-electron count = metal group number electrons + 2 per CO (or as per ligand) ± charge

5 important questions

Q1. What are organometallic compounds? Is CH₃COONa organometallic?

Compounds containing at least one direct metal–carbon bond, like CH₃MgBr and ferrocene. CH₃COONa is not organometallic, because Na is bonded to oxygen, not carbon.

Q2. Why must a Grignard reagent be prepared in dry ether?

RMgX reacts instantly with water to form an alkane, so moisture destroys it. Dry ether also coordinates with Mg through its oxygen lone pairs and stabilises the reagent.

Q3. How will you prepare ethanol, propan-2-ol and 2-methylpropan-2-ol from CH₃MgI?

• Ethanol (1°): CH₃MgI + HCHO, then hydrolysis.
• Propan-2-ol (2°): CH₃MgI + CH₃CHO, then hydrolysis → CH₃CH(OH)CH₃.
• 2-methylpropan-2-ol (3°): CH₃MgI + CH₃COCH₃, then hydrolysis.
• In each case the R group adds to the carbonyl carbon, then H₂O/H⁺ gives the alcohol.

Q4. Apply the 18-electron rule to Fe(CO)₅ and Cr(CO)₆.

• Fe is in Group 8 → 8 valence electrons; 5 CO × 2 = 10; total = 18 ✓
• Cr is in Group 6 → 6 electrons; 6 CO × 2 = 12; total = 18 ✓
• Both are stable, diamagnetic carbonyls.
• Shapes: Fe(CO)₅ trigonal bipyramidal; Cr(CO)₆ octahedral.

Q5. Describe the structure and bonding of ferrocene.

• Formula Fe(η⁵-C₅H₅)₂; orange crystalline solid.
• Fe²⁺ lies between two parallel cyclopentadienyl (C₅H₅⁻) rings: sandwich structure.
• Each C₅H₅⁻ ring has 6 π electrons (aromatic).
• Electron count: Fe²⁺ (6) + 2 × 6 = 18.
• All Fe–C distances equal; rings are aromatic and undergo Friedel–Crafts acylation.

Exam tip

Grignard reactions are the most-asked part. Make a table of 'reactant + RMgX → product' for HCHO, RCHO, R₂CO, CO₂ and H₂O and learn it.

Quick summary

In a molecular rearrangement, an atom or group moves from one atom to another inside the same molecule, giving a product with a new carbon skeleton or a new functional group position. This topic covers the classification of rearrangements and the mechanisms of the pinacol–pinacolone, Beckmann, Hofmann, Curtius, benzilic acid, Claisen and Fries rearrangements.

Key points

• Classification by the atom receiving the group: electron-deficient carbon (pinacol, Wagner–Meerwein, benzilic acid), electron-deficient nitrogen (Beckmann, Hofmann, Curtius, Lossen), electron-deficient oxygen (Baeyer–Villiger), and aromatic/sigmatropic types (Claisen, Fries).
• Migratory aptitude (rough order): aryl > H ≈ alkyl; among alkyls 3° > 2° > 1° > methyl. Groups move with their electron pair (1,2-shift).
• Pinacol–pinacolone: 1,2-diol + H⁺ → carbocation → 1,2-methyl shift → ketone (pinacolone).
• Wagner–Meerwein: carbocation rearranges to a more stable carbocation by an alkyl or hydride shift (seen in SN1/E1 reactions).
• Beckmann rearrangement: ketoxime + acid (H₂SO₄, PCl₅) → N-substituted amide. The group anti to the –OH migrates. Cyclohexanone oxime → caprolactam (monomer of nylon-6).
• Hofmann rearrangement (bromamide): RCONH₂ + Br₂ + NaOH → RNH₂ (one carbon less) through an N-bromoamide and an isocyanate intermediate.
• Curtius rearrangement: acyl azide (RCON₃) on heating → isocyanate (R–N=C=O) + N₂ → amine on hydrolysis.
• Benzilic acid rearrangement: benzil + KOH → potassium benzilate → benzilic acid (1,2-aryl shift to carbonyl carbon).
• Claisen rearrangement: allyl phenyl ether heated (about 200 °C) → o-allylphenol; a concerted [3,3]-sigmatropic shift.
• Fries rearrangement: phenyl ester + AlCl₃ → o- and p-hydroxy ketones. Low temperature favours para, high temperature favours ortho.
• Baeyer–Villiger oxidation: ketone + peracid → ester (O inserted next to the carbonyl).

Important terms

• Migrating group – the group that moves; it carries its bonding electrons.
• Isocyanate – R–N=C=O, intermediate in Hofmann and Curtius reactions.
• Rearrangement (पुनर्रचना) – change in the skeleton or functional group position within a molecule.

Formulas

5 important questions

Q1. What happens when acetophenone oxime is treated with conc. H₂SO₄?

It undergoes Beckmann rearrangement. The phenyl group anti to –OH migrates to nitrogen, giving acetanilide (CH₃CONHC₆H₅).

Q2. Convert acetamide to methylamine.

Heat acetamide with Br₂ and NaOH (Hofmann bromamide reaction): CH₃CONH₂ + Br₂ + 4NaOH → CH₃NH₂ + Na₂CO₃ + 2NaBr + 2H₂O.

Q3. Explain the mechanism of the Hofmann rearrangement.

• OH⁻ removes N–H; Br₂ gives N-bromoamide.
• Second N–H removed → bromamide anion.
• R migrates from C to N as Br⁻ leaves → isocyanate.
• Isocyanate + water → carbamic acid → amine + CO₂ (as carbonate).
• R migrates with retention of configuration.

Q4. Explain the Beckmann rearrangement with mechanism and stereochemistry.

Protonation of oxime –OH, loss of water with simultaneous anti migration, nitrilium ion, attack by water, tautomerisation to amide. Show how syn and anti oximes give different amides. Mention caprolactam for nylon-6.

Q5. Write notes on the benzilic acid and Fries rearrangements.

Equations, mechanism outline (OH⁻ attack and 1,2-phenyl shift for benzilic; acylium ion and AlCl₃ complex for Fries), and the effect of temperature on the Fries ortho/para ratio.

Exam tip

Remember "anti group migrates" for Beckmann and "one carbon less" for Hofmann; these two lines are the most tested facts.

Quick summary

Photochemistry studies chemical reactions that are started by the absorption of light (UV or visible). This topic covers the difference between thermal and photochemical reactions, the laws of photochemistry, quantum yield and the reasons for high and low values, photosensitisation, fluorescence and phosphorescence (Jablonski diagram), and chemiluminescence.

Key points

• Thermal reactions take place in the dark and depend strongly on temperature. Photochemical reactions need light; their rate depends on light intensity and is only slightly affected by temperature. ΔG can even be positive for a photochemical reaction.
• Grotthuss–Draper law: only light that is absorbed can cause a photochemical change.
• Stark–Einstein law of photochemical equivalence: each molecule taking part in the primary process absorbs one quantum of light (one photon).
• Einstein: energy of one mole of photons, E = Nₐhc/λ. For λ = 400 nm, E ≈ 299 kJ mol⁻¹.
• Quantum yield (φ) = number of molecules reacted / number of photons absorbed.
• High quantum yield (φ > 1): chain reactions. H₂ + Cl₂ → 2HCl has φ about 10⁴–10⁶. Decomposition of HI has φ = 2.
• Low quantum yield (φ < 1): deactivation of excited molecules by collision, fluorescence, recombination of fragments (e.g. H₂ + Br₂ has a low φ at room temperature).
• Photosensitisation: a substance absorbs light and passes the energy to another molecule that cannot absorb it directly. Example: Hg vapour sensitises the dissociation of H₂; chlorophyll in photosynthesis.
• Fluorescence: emission of light from the excited singlet state (S₁ → S₀); stops almost at once when light is removed (about 10⁻⁸ s). Fluorite, fluorescein.
• Phosphorescence: emission from the triplet state (T₁ → S₀) after intersystem crossing; continues after the light is removed (ZnS).
• Jablonski diagram shows absorption, vibrational relaxation, internal conversion, intersystem crossing, fluorescence and phosphorescence.
• Chemiluminescence: light given out by a chemical reaction (oxidation of luminol; firefly light is bioluminescence).

Important terms

• Photon – one quantum of light energy, E = hν.
• Quantum yield (क्वांटम उत्पन्न) – molecules reacted per photon absorbed.
• Intersystem crossing – non-radiative change from a singlet to a triplet state.

Formulas

• E (one photon) = hν = hc/λ (h = 6.626 × 10⁻³⁴ J s, c = 3 × 10⁸ m s⁻¹)
• One einstein = Nₐhc/λ = 0.1196/λ J mol⁻¹ (λ in m)
• φ = moles reacted / einsteins absorbed

5 important questions

Q1. Define quantum yield. Why is it very high for H₂ + Cl₂?

φ = molecules reacted per photon absorbed. One photon splits Cl₂ into two Cl atoms, which start a chain (Cl + H₂ → HCl + H; H + Cl₂ → HCl + Cl) that repeats thousands of times, so φ is about 10⁴–10⁶.

Q2. Distinguish between fluorescence and phosphorescence.

Fluorescence comes from the singlet excited state and stops immediately when light is cut off. Phosphorescence comes from the triplet state after intersystem crossing and continues for some time after the light is removed.

Q3. Calculate the energy of one einstein of light of wavelength 400 nm.

• E = Nₐhc/λ = (6.022 × 10²³ × 6.626 × 10⁻³⁴ × 3 × 10⁸)/(400 × 10⁻⁹)
• = (1.197 × 10⁻¹)/(4 × 10⁻⁷) = 2.99 × 10⁵ J mol⁻¹ ≈ 299 kJ mol⁻¹

Q4. State and explain the laws of photochemistry.

Grotthuss–Draper law with an example; Stark–Einstein law; primary and secondary processes; why many reactions do not show φ = 1 (secondary reactions). Give the HI example: HI + hν → H + I; H + HI → H₂ + I; I + I → I₂, so φ = 2.

Q5. Explain the Jablonski diagram.

Draw S₀, S₁, S₂ and T₁ levels with vibrational sub-levels. Mark absorption, vibrational relaxation, internal conversion, fluorescence, intersystem crossing and phosphorescence with straight (radiative) and wavy (non-radiative) arrows.

Exam tip

Learn the H₂–Cl₂ chain and HI decomposition steps; they are used to explain both high quantum yield and quantum yield = 2.