Chemistry रसायनशास्त्र

Quick summary

Stereochemistry studies the three-dimensional arrangement of atoms in molecules. Isomers with the same structure but different spatial arrangement are called stereoisomers. This topic covers optical isomerism, chirality, R/S configuration, geometrical (cis–trans, E/Z) isomerism and conformations of ethane and butane.

Key points

• Stereoisomers have the same molecular formula and same bonding sequence but differ in spatial arrangement.
• Optical activity: the ability to rotate plane-polarised light. Measured with a polarimeter. Dextrorotatory (+, d) rotates to the right; laevorotatory (−, l) to the left.
• A carbon bonded to four different groups is a chiral centre (asymmetric carbon). Example: lactic acid, CH₃CH(OH)COOH.
• Enantiomers are non-superimposable mirror images. They have the same physical properties except the direction of rotation of light.
• Diastereomers are stereoisomers that are not mirror images. They have different physical properties.
• A meso compound has chiral centres but also a plane of symmetry, so it is optically inactive. Example: meso-tartaric acid.
• A racemic mixture is a 50:50 mix of enantiomers; it shows no rotation (external compensation).
• Number of optical isomers = 2ⁿ (n = chiral centres), when the molecule is not symmetric. Tartaric acid has only 3 (d, l, meso).
• CIP rules (Cahn–Ingold–Prelog): give priority by higher atomic number. Keep the lowest priority group away; if 1 → 2 → 3 is clockwise, it is R; anticlockwise is S.
• Geometrical isomerism occurs due to restricted rotation around C=C or in rings. Cis: same groups on the same side; trans: opposite sides. E/Z uses CIP priorities.
• Conformations arise by rotation about a C–C single bond. Ethane: staggered (most stable) and eclipsed (least stable).
• n-Butane: anti (most stable) > gauche > partially eclipsed > fully eclipsed (least stable).

Important terms

• Chirality (हस्तता) – property of a molecule of being non-superimposable on its mirror image.
• Specific rotation – rotation by a solution of 1 g/mL concentration in a 1 dm tube.
• Resolution – separation of a racemic mixture into its enantiomers.
• Fischer projection – 2D drawing; horizontal lines point towards the viewer, vertical lines away.
• Newman projection – view along a C–C bond, front carbon as a point, back carbon as a circle.

Formulas

• Specific rotation: [α] = α/(l × c), α = observed rotation (degrees), l = tube length (dm), c = concentration (g/mL)
• Number of stereoisomers = 2ⁿ (for unsymmetrical molecules)
• Enantiomeric excess: ee (%) = (observed [α] / [α] of pure enantiomer) × 100
• Priority examples: −Br > −Cl > −OH > −NH₂ > −COOH > −CHO > −CH₃ > −H

5 important questions

Q1. What is a meso compound? Why is it optically inactive?

A meso compound has two or more chiral centres but has an internal plane of symmetry. The rotation by one half is cancelled by the other half (internal compensation). Example: meso-tartaric acid.

Q2. Distinguish between enantiomers and diastereomers.

Enantiomers are non-superimposable mirror images with identical physical properties except the sign of rotation. Diastereomers are not mirror images and differ in melting point, boiling point and solubility. Cis and trans-2-butene are diastereomers.

Q3. A solution of 0.2 g/mL of a sugar in a 2 dm tube shows rotation +13.3°. Find its specific rotation.

• [α] = α/(l × c)
• = 13.3 / (2 × 0.2)
• = 13.3 / 0.4
• = +33.25°

Q4. Assign R or S to (+)-lactic acid drawn with H pointing back, and explain the steps.

• Groups on chiral C: −OH, −COOH, −CH₃, −H.
• Priorities: −OH (O, 8) = 1; −COOH (C with O,O,O) = 2; −CH₃ = 3; −H = 4.
• Keep H at the back. Trace 1 → 2 → 3.
• Clockwise → R; anticlockwise → S.
• Note: R/S and +/− are not related; the sign is found only by experiment. (+)-Lactic acid is S.

Q5. Discuss the conformations of n-butane with an energy diagram.

• Rotate about C2–C3 bond; view in Newman projection.
• Anti (CH₃ groups 180° apart): lowest energy, most stable.
• Gauche (60°): slight steric strain.
• Partially eclipsed (120°) and fully eclipsed (0°): torsional and steric strain; fully eclipsed is least stable.
• Energy curve shows minima at anti and gauche, maxima at eclipsed forms.

Exam tip

Practice R/S assignment on 4–5 molecules and draw Newman projections neatly. A labelled energy diagram of butane fetches full marks.

Quick summary

Hydrocarbons are compounds of carbon and hydrogen only. This topic covers alkanes, alkenes, alkynes and aromatic hydrocarbons (arenes): their preparation, important reactions and mechanisms, Markovnikov's rule, aromaticity and electrophilic substitution in benzene.

Key points

• Alkanes (CₙH₂ₙ₊₂) are saturated, sp³, and mostly give substitution reactions.
• Preparation of alkanes: Wurtz reaction, Kolbe's electrolysis, decarboxylation of sodium salts with soda lime, hydrogenation of alkenes.
• Halogenation of methane is a free radical chain reaction: initiation, propagation, termination.
• Alkenes (CₙH₂ₙ) have a C=C (one σ + one π bond), sp². They give electrophilic addition reactions.
• Markovnikov's rule: in addition of HX to an unsymmetrical alkene, H goes to the carbon that already has more hydrogens (because the more stable carbocation forms).
• Anti-Markovnikov (peroxide/Kharasch effect): HBr in presence of peroxide adds the opposite way by a free radical mechanism. Works only with HBr.
• Ozonolysis of alkenes gives aldehydes/ketones and helps locate the double bond.
• Alkynes (CₙH₂ₙ₋₂) have C≡C, sp, linear. Terminal alkynes are weakly acidic (sp carbon has 50% s-character) and form acetylides with sodium or ammoniacal AgNO₃.
• Aromaticity (Hückel's rule): a cyclic, planar, fully conjugated molecule with (4n + 2) π electrons is aromatic. Benzene has 6 π electrons (n = 1).
• Benzene gives electrophilic substitution (not addition), keeping its stable aromatic ring: nitration, halogenation, sulphonation, Friedel–Crafts alkylation and acylation.
• Mechanism: formation of electrophile → attack on ring to form a σ-complex (arenium ion) → loss of H⁺ to restore aromaticity.
• Directive influence: −OH, −NH₂, −CH₃, −Cl are ortho/para directors; −NO₂, −CHO, −COOH, −CN are meta directors. Halogens are o/p directing but deactivating.

Important terms

• Saturated hydrocarbon (संतृप्त) – contains only single C–C bonds.
• Electrophilic addition – addition started by attack of an electrophile on a π bond.
• Aromatic compound – cyclic planar conjugated compound obeying Hückel's rule.
• Arenium ion (σ-complex) – carbocation intermediate in aromatic substitution.
• Ozonolysis – cleavage of C=C by ozone followed by Zn/H₂O.

Formulas

Key reactions:

• Wurtz: 2CH₃Br + 2Na → CH₃–CH₃ + 2NaBr (dry ether)
• Decarboxylation: CH₃COONa + NaOH → CH₄ + Na₂CO₃ (CaO, heat)
• Markovnikov: CH₃CH=CH₂ + HBr → CH₃CHBrCH₃
• Peroxide effect: CH₃CH=CH₂ + HBr → CH₃CH₂CH₂Br (peroxide)
• Nitration: C₆H₆ + HNO₃ → C₆H₅NO₂ + H₂O (conc. H₂SO₄); electrophile NO₂⁺
• Friedel–Crafts alkylation: C₆H₆ + CH₃Cl → C₆H₅CH₃ + HCl (anhydrous AlCl₃)
• Hückel's rule: π electrons = 4n + 2 (n = 0, 1, 2 …)

5 important questions

Q1. State Markovnikov's rule with an example.

When an unsymmetrical reagent like HX adds to an unsymmetrical alkene, the negative part (X) goes to the carbon having fewer hydrogen atoms. Propene + HBr gives 2-bromopropane as the main product.

Q2. Why are terminal alkynes acidic?

The carbon of ≡C–H is sp hybridised with 50% s-character. It holds electrons closer and is more electronegative. So the C–H bond is polarised and H⁺ can be removed by strong bases like NaNH₂.

Q3. Explain the mechanism of nitration of benzene.

• Step 1: HNO₃ + 2H₂SO₄ → NO₂⁺ + H₃O⁺ + 2HSO₄⁻ (nitronium ion formed).
• Step 2: NO₂⁺ attacks the π cloud of benzene → arenium ion (σ-complex), stabilised by resonance. This is the slow step.
• Step 3: HSO₄⁻ removes H⁺ → nitrobenzene; aromaticity is restored.
• Overall it is electrophilic substitution.

Q4. Explain the free radical mechanism of chlorination of methane.

• Initiation: Cl₂ → 2Cl• (light).
• Propagation: Cl• + CH₄ → CH₃• + HCl; CH₃• + Cl₂ → CH₃Cl + Cl•.
• Termination: Cl• + Cl• → Cl₂; CH₃• + Cl• → CH₃Cl; CH₃• + CH₃• → C₂H₆.
• Further substitution gives CH₂Cl₂, CHCl₃ and CCl₄.

Q5. State Hückel's rule. Which of these are aromatic: benzene, cyclobutadiene, cyclopentadienyl anion?

• Rule: cyclic, planar, fully conjugated, with (4n + 2) π electrons.
• Benzene: 6 π electrons (n = 1) → aromatic.
• Cyclobutadiene: 4 π electrons (4n type) → antiaromatic, not aromatic.
• Cyclopentadienyl anion: 6 π electrons, planar, conjugated → aromatic.

Exam tip

Learn three mechanisms well: free radical halogenation, electrophilic addition of HBr (with peroxide effect), and nitration of benzene. Write each step with arrows and intermediates.

Quick summary

Alkyl halides (haloalkanes, R–X) and aryl halides (haloarenes, Ar–X) are compounds in which a halogen is bonded to an sp³ carbon or to a benzene ring. This topic covers their classification, preparation, nucleophilic substitution (SN1 and SN2) and elimination (E1 and E2) mechanisms, and why aryl halides are much less reactive.

Key points

• Classification: 1°, 2° and 3° alkyl halides depending on how many carbons are attached to the C bearing X. Also allylic, benzylic, vinylic and aryl halides.
• Preparation of R–X: from alcohols (HX/ZnCl₂, PCl₅, PCl₃, SOCl₂ – SOCl₂ is best because by-products SO₂ and HCl are gases), from alkenes (HX addition), Finkelstein reaction (R–Cl + NaI in acetone → R–I), Swarts reaction (R–Br + AgF → R–F).
• Preparation of Ar–X: electrophilic halogenation of benzene (Cl₂/FeCl₃) and the Sandmeyer reaction (ArN₂⁺Cl⁻ + CuCl → ArCl).
• SN2: one step, bimolecular, rate = k[RX][Nu⁻]. Backside attack → Walden inversion of configuration. Reactivity: CH₃X > 1° > 2° > 3°. Favoured by polar aprotic solvents.
• SN1: two steps through a carbocation, rate = k[RX]. Gives racemisation (with some inversion). Reactivity: 3° > 2° > 1° > CH₃X. Favoured by polar protic solvents.
• E2: one step, anti-periplanar H and X, strong base (alc. KOH). E1: through carbocation.
• Saytzeff (Zaitsev) rule: in elimination, the more substituted alkene is the main product (2-bromobutane → but-2-ene mainly).
• Substitution vs elimination: aqueous KOH gives alcohol (substitution); alcoholic KOH gives alkene (elimination).
• Aryl halides are less reactive towards nucleophiles because (1) the C–X bond gets partial double-bond character by resonance, (2) the C is sp² and holds the bond more tightly, (3) the phenyl cation is unstable, (4) electron-rich ring repels the nucleophile.
• Electron-withdrawing groups (–NO₂) at ortho/para positions make aryl halides react (2,4-dinitrochlorobenzene hydrolyses easily). Benzyne (elimination–addition) mechanism occurs with NaNH₂.
• Wurtz reaction: 2R–X + 2Na → R–R; Wurtz–Fittig: Ar–X + R–X + 2Na → Ar–R.

Important terms

• Nucleophile (केंद्रकस्नेही) – electron-rich species that attacks a positive centre.
• Leaving group – group that departs with the bond pair (I⁻ > Br⁻ > Cl⁻ > F⁻).
• Walden inversion – inversion of configuration in SN2.

Formulas

5 important questions

Q1. Why is SOCl₂ preferred for converting alcohols to alkyl chlorides?

R–OH + SOCl₂ → R–Cl + SO₂↑ + HCl↑. Both by-products are gases and escape, so pure alkyl chloride is left.

Q2. Why does tert-butyl bromide undergo SN1 faster than methyl bromide?

It forms a stable 3° carbocation (stabilised by +I and hyperconjugation of nine C–H bonds). Methyl cation is very unstable. Bulky groups also block SN2 at the tertiary carbon.

Q3. Explain the SN2 mechanism with stereochemistry.

• Example: OH⁻ + CH₃Br → CH₃OH + Br⁻.
• Nucleophile attacks from the side opposite to Br; five groups in the transition state (C in trigonal bipyramid form).
• Second-order kinetics; inversion of configuration (umbrella flip), e.g. (R)-2-bromobutane → (S)-butan-2-ol.
• Effect of substrate, nucleophile, leaving group and solvent.

Q4. Explain the E2 mechanism and Saytzeff rule.

One-step removal of β-H by base and loss of X⁻ in anti arrangement. Example: 2-bromobutane + alc. KOH → but-2-ene (major, about 80 %) + but-1-ene (minor). More substituted alkene is more stable due to hyperconjugation.

Q5. Why are aryl halides less reactive than alkyl halides towards nucleophilic substitution?

Draw resonance structures of chlorobenzene showing C=Cl⁺ character, explain sp² carbon, instability of phenyl cation and repulsion by π electrons. Show how –NO₂ groups at o/p positions activate the ring (Meisenheimer complex).

Exam tip

A comparison table of SN1 and SN2 is a guaranteed 3–4 mark answer. Always draw the transition state for SN2.

Quick summary

Alcohols (R–OH) and phenols (Ar–OH) contain a hydroxyl group; ethers (R–O–R′) have an oxygen between two carbon groups. This topic covers classification, preparation, acidity, distinction tests and key reactions such as oxidation, pinacol–pinacolone rearrangement, Reimer–Tiemann and Kolbe reactions, and cleavage of ethers by HI.

Key points

• Alcohols: monohydric (ethanol), dihydric (ethylene glycol), trihydric (glycerol); also 1°, 2°, 3°.
• Preparation: hydration of alkenes, hydrolysis of alkyl halides, reduction of aldehydes/ketones (H₂/Ni, NaBH₄, LiAlH₄), Grignard reagent with HCHO (1°), other aldehydes (2°) and ketones (3°).
• Lucas test (conc. HCl + anhydrous ZnCl₂): 3° alcohols give turbidity at once, 2° in about 5 minutes, 1° not at room temperature.
• Oxidation: 1° → aldehyde → carboxylic acid; 2° → ketone; 3° resist mild oxidation.
• Glycols: cleaved by HIO₄ or Pb(OAc)₄ into two carbonyl compounds. Pinacol–pinacolone rearrangement: pinacol + acid → pinacolone through a carbocation and 1,2-methyl shift.
• Phenols are more acidic than alcohols: the phenoxide ion is stabilised by resonance; alkoxide is not. Phenol (pKa about 10) reacts with NaOH; ethanol (pKa about 16) does not.
• Electron-withdrawing groups (–NO₂) increase phenol acidity; electron-donating groups (–CH₃) decrease it. Picric acid is strongly acidic.
• Reactions of phenol: bromine water gives white precipitate of 2,4,6-tribromophenol; Reimer–Tiemann (CHCl₃ + NaOH → salicylaldehyde, via dichlorocarbene); Kolbe's reaction (sodium phenoxide + CO₂, 125 °C, pressure → salicylic acid); FeCl₃ gives violet colour.
• Ethers: prepared by Williamson synthesis (R–ONa + R′–X → R–O–R′, SN2; use 1° halide). Fairly unreactive; cleaved by hot conc. HI.
• Anisole + HI → phenol + CH₃I (never iodobenzene, because the Ar–O bond does not break).
• Epoxides (three-membered cyclic ethers) open easily with acid or base because of ring strain.

Important terms

• Phenoxide ion – C₆H₅O⁻, resonance-stabilised conjugate base of phenol.
• Dichlorocarbene (:CCl₂) – electrophile in Reimer–Tiemann reaction.
• Hydroxyl group (हायड्रॉक्सिल गट) – –OH functional group.

Formulas

5 important questions

Q1. Why is phenol acidic but ethanol is almost neutral?

After losing H⁺, phenoxide ion spreads its negative charge over the ring by resonance and becomes stable. Ethoxide has no resonance and +I effect of ethyl increases its charge, so ethanol hardly ionises.

Q2. How will you distinguish between 1°, 2° and 3° alcohols?

Use the Lucas test (turbidity speed) or Victor Meyer's test (red, blue, colourless). Oxidation products also differ: aldehyde, ketone, no reaction.

Q3. Explain the pinacol–pinacolone rearrangement with mechanism.

• Protonation of one –OH, loss of water → 3° carbocation.
• 1,2-shift of a methyl group to the cationic carbon → more stable oxocarbenium ion.
• Loss of H⁺ → pinacolone (3,3-dimethylbutan-2-one).

Q4. Explain the Reimer–Tiemann reaction with mechanism.

CHCl₃ + OH⁻ → :CCl₂ (dichlorocarbene). Phenoxide attacks :CCl₂ at the ortho position; the –CHCl₂ group is hydrolysed to –CHO. Product after acidification: salicylaldehyde (o-hydroxybenzaldehyde), with a little para isomer.

Q5. Describe the Williamson synthesis and the action of HI on ethers.

Show the SN2 step; explain why tert-butyl halide gives alkene instead. Give HI cleavage of diethyl ether (cold: C₂H₅I + C₂H₅OH; hot excess: 2C₂H₅I) and of anisole.

Exam tip

Name reactions (Reimer–Tiemann, Kolbe, Williamson, pinacol) carry easy marks — write the equation, reagents and conditions for each.

Quick summary

Aldehydes (R–CHO) and ketones (R–CO–R′) contain the carbonyl group (C=O). The polar C=O makes the carbon electrophilic, so the typical reaction is nucleophilic addition. This topic covers structure, preparation, nucleophilic addition, aldol, Cannizzaro, benzoin, Perkin, Knoevenagel, Wittig and Mannich reactions, and oxidation and reduction tests.

Key points

• Carbonyl carbon is sp² hybridised, trigonal planar (about 120°). C=O is polar: C is δ+, O is δ−.
• Preparation: oxidation of alcohols (1° → aldehyde with PCC, 2° → ketone), Rosenmund reduction (RCOCl + H₂ / Pd–BaSO₄ → RCHO), from nitriles with Grignard reagent (R–C≡N + R′MgX → ketone after hydrolysis), Friedel–Crafts acylation (aromatic ketones), ozonolysis of alkenes.
• Reactivity order: HCHO > other aldehydes > ketones (alkyl groups push electrons by +I and block the nucleophile).
• Nucleophilic addition: HCN (cyanohydrin), NaHSO₃ (crystalline adduct, used for purification), Grignard reagents (alcohols), ammonia derivatives (addition–elimination giving oximes, hydrazones, 2,4-DNP derivatives).
• Aldol condensation: aldehydes/ketones with α-H, dilute NaOH → β-hydroxy carbonyl compound, which loses water on heating to give an α,β-unsaturated compound.
• Cannizzaro reaction: aldehydes with no α-H (HCHO, C₆H₅CHO) + conc. NaOH → one molecule oxidised to acid salt, one reduced to alcohol (disproportionation).
• Benzoin condensation: 2 C₆H₅CHO with KCN (alcoholic) → benzoin.
• Perkin reaction: benzaldehyde + acetic anhydride + sodium acetate → cinnamic acid.
• Knoevenagel: aldehyde + active methylene compound (malonic ester) with weak base (pyridine/piperidine).
• Wittig reaction: carbonyl + phosphorus ylide (Ph₃P=CH₂) → alkene; fixes C=C at the exact position.
• Mannich reaction: ketone with α-H + HCHO + secondary amine → β-amino ketone (Mannich base).
• Tests: Tollens' (silver mirror) and Fehling's (red Cu₂O) are given by aliphatic aldehydes, not by ketones. Benzaldehyde gives Tollens' but not Fehling's. Iodoform test for CH₃CO– group.
• Reductions: Clemmensen (Zn–Hg / conc. HCl) and Wolff–Kishner (NH₂NH₂ / KOH, heat) convert C=O to CH₂.

Important terms

• α-hydrogen – H on the carbon next to C=O; it is acidic due to resonance-stabilised enolate.
• Carbonyl group (कार्बोनिल गट) – C=O group.
• Ylide – neutral molecule with adjacent opposite charges (Ph₃P⁺–CH₂⁻).

Formulas

5 important questions

Q1. Why are aldehydes more reactive than ketones towards nucleophilic addition?

Ketones have two alkyl groups that reduce the positive charge on carbon by +I effect and cause more steric hindrance. Aldehydes have only one, so their carbonyl carbon is more electrophilic and easier to reach.

Q2. Give a test to distinguish acetaldehyde from acetone.

Tollens' reagent: acetaldehyde gives a silver mirror, acetone does not. (Fehling's solution: red precipitate with acetaldehyde only.)

Q3. Explain the mechanism of aldol condensation of acetaldehyde.

• OH⁻ removes α-H → enolate ion (resonance stabilised).
• Enolate carbon attacks carbonyl C of second CH₃CHO → alkoxide.
• Alkoxide takes H⁺ from water → 3-hydroxybutanal (aldol).
• On heating: loss of H₂O → but-2-enal (crotonaldehyde).

Q4. Explain the Cannizzaro reaction with mechanism.

• OH⁻ adds to C=O of HCHO/C₆H₅CHO → tetrahedral anion.
• Hydride ion (H⁻) transfer to a second aldehyde molecule (rate-determining).
• Products: carboxylate + alcohol (2C₆H₅CHO → C₆H₅COONa + C₆H₅CH₂OH).
• Crossed Cannizzaro with HCHO: HCHO is always oxidised.

Q5. Write notes on the Wittig and Benzoin reactions.

Equations, reagents, mechanism outline (betaine / oxaphosphetane for Wittig; cyanide as catalyst and umpolung idea for benzoin) and uses.

Exam tip

Make a one-page chart of all named carbonyl reactions with reagent, condition and product; questions often ask "complete the reaction".

Quick summary

Carboxylic acids contain the –COOH group. Their derivatives are acid chlorides, anhydrides, esters and amides, in which the –OH is replaced by another group. This topic covers acidity and the effect of substituents, preparation, important reactions (HVZ, decarboxylation, reduction), and the relative reactivity of acid derivatives towards nucleophilic acyl substitution.

Key points

• Acidity: RCOOH ⇌ RCOO⁻ + H⁺. The carboxylate ion is stabilised by resonance (two equal C–O bonds), so carboxylic acids are much more acidic than alcohols and phenols.
• Effect of substituents: electron-withdrawing groups increase acidity, electron-donating groups decrease it. Cl₃C–COOH > Cl₂CH–COOH > ClCH₂–COOH > CH₃COOH. HCOOH is stronger than CH₃COOH.
• The nearer the –Cl is to –COOH, the stronger the acid (2-chlorobutanoic > 3-chloro > 4-chloro).
• Carboxylic acids form hydrogen-bonded dimers, giving high boiling points.
• Preparation: oxidation of 1° alcohols/aldehydes, hydrolysis of nitriles (R–CN + 2H₂O → RCOOH + NH₃), Grignard reagent + CO₂ then H₃O⁺, oxidation of alkylbenzenes (toluene → benzoic acid with KMnO₄).
• Hell–Volhard–Zelinsky (HVZ): RCH₂COOH + Br₂ / red P → RCHBrCOOH (α-halogenation).
• Decarboxylation: sodium salt + soda lime (NaOH + CaO), heat → alkane with one C less. Kolbe electrolysis gives R–R.
• Reduction: LiAlH₄ reduces –COOH to –CH₂OH.
• Derivatives reactivity (towards nucleophiles): acid chloride > anhydride > ester > amide. Better leaving group (Cl⁻) means higher reactivity; amide is least reactive due to strong resonance of N lone pair.
• Esterification (Fischer): RCOOH + R′OH ⇌ RCOOR′ + H₂O with conc. H₂SO₄; reversible.
• Hofmann bromamide: RCONH₂ + Br₂ + 4KOH → RNH₂ (one C less).
• Unsaturated acids: acrylic acid (CH₂=CH–COOH) and cinnamic acid; dicarboxylic acids: oxalic, malonic (loses CO₂ on heating), succinic (forms anhydride).

Important terms

• Carboxylate ion (कार्बोक्सिलेट आयन) – RCOO⁻, resonance-stabilised.
• Acyl group – R–CO–.
• Nucleophilic acyl substitution – addition of a nucleophile to C=O followed by loss of the leaving group.

Formulas

• pKa values (approx.): HCOOH 3.75, CH₃COOH 4.76, ClCH₂COOH 2.86, C₆H₅COOH 4.20.

5 important questions

Q1. Why is chloroacetic acid stronger than acetic acid?

Cl has a −I effect. It pulls electron density away from the –COO⁻ group and spreads the negative charge, making the anion more stable. So ClCH₂COOH loses H⁺ more easily.

Q2. What is the HVZ reaction?

α-Halogenation of a carboxylic acid with Cl₂ or Br₂ in the presence of red phosphorus. Example: CH₃COOH + Cl₂ (red P) → ClCH₂COOH + HCl.

Q3. Explain the effect of substituents on the acidity of carboxylic acids with examples.

• Inductive effect: −I groups (Cl, NO₂, CN) raise acidity, +I groups (alkyl) lower it.
• Number and position of substituents (mono-, di-, trichloroacetic acid; α vs β vs γ).
• Aromatic acids: p-nitrobenzoic acid > benzoic acid > p-methylbenzoic acid. Ortho effect: almost all ortho-substituted benzoic acids are stronger than benzoic acid.

Q4. Explain the mechanism of acid-catalysed esterification.

Protonation of C=O → attack by R′OH → tetrahedral intermediate → proton transfer → loss of H₂O → loss of H⁺ → ester. Isotope labelling (¹⁸O in alcohol ends up in the ester) shows the C–OH bond of the acid breaks.

Q5. Why is an acid chloride more reactive than an amide?

Compare leaving-group ability (Cl⁻ weak base, good leaving group; NH₂⁻ strong base, poor leaving group) and resonance donation (strong from N, weak from Cl) to the carbonyl carbon. Show resonance structures.

Exam tip

For acidity questions always mention the stability of the anion and the inductive/resonance effect by name.

Quick summary

This topic explains the behaviour of gases. It covers the gas laws and ideal gas equation, the postulates of the kinetic theory of gases, molecular speeds, deviations of real gases from ideal behaviour, the van der Waals equation and critical phenomena.

Key points

• Boyle's law: at constant temperature, PV = constant.
• Charles' law: at constant pressure, V/T = constant.
• Avogadro's law: equal volumes of gases at same T and P contain equal numbers of molecules.
• Ideal gas equation: PV = nRT. R = 8.314 J K⁻¹ mol⁻¹ = 0.0821 L atm K⁻¹ mol⁻¹.
• Kinetic theory postulates: gas molecules are tiny points with negligible volume; they move randomly in straight lines; collisions are perfectly elastic; there are no attractive forces between molecules; average kinetic energy is proportional to absolute temperature.
• Kinetic gas equation: PV = ⅓ mN c², where c is the root mean square speed.
• Three molecular speeds: most probable (α), average (ū) and root mean square (c). Ratio α : ū : c = 1 : 1.128 : 1.224.
• Maxwell distribution: at higher temperature the curve becomes flatter and its peak moves to higher speed.
• Real gases deviate from ideal behaviour at high pressure and low temperature. Compressibility factor Z = PV/nRT. For an ideal gas Z = 1.
• The ideal postulates fail because real molecules have a finite volume and they attract each other.
• van der Waals equation corrects both: (P + an²/V²)(V − nb) = nRT. 'a' measures attraction; 'b' measures the volume of molecules.
• Critical temperature (Tc): the temperature above which a gas cannot be liquefied however high the pressure.
• Boyle temperature: temperature at which a real gas behaves ideally over a range of pressure; T_B = a/Rb.

Important terms

• Ideal gas (आदर्श वायू) – gas that obeys PV = nRT at all T and P.
• Compressibility factor (Z) – PV/nRT; shows deviation from ideal behaviour.
• Critical pressure (Pc) – pressure needed to liquefy a gas at its critical temperature.
• Critical volume (Vc) – volume of one mole of gas at Tc and Pc.
• Mean free path – average distance a molecule travels between two collisions.

Formulas

• PV = nRT
• Kinetic gas equation: PV = ⅓ mNc²
• c (rms) = √(3RT/M); ū = √(8RT/πM); α = √(2RT/M) (M in kg/mol, R = 8.314)
• Average kinetic energy per mole = (3/2)RT
• van der Waals: (P + an²/V²)(V − nb) = nRT
• Critical constants: Tc = 8a/27Rb, Pc = a/27b², Vc = 3b
• Boyle temperature: T_B = a/Rb
• Z = PV/nRT

5 important questions

Q1. Why do real gases deviate from ideal behaviour?

Two postulates of kinetic theory are wrong for real gases. Real molecules occupy some volume, and they attract each other. These effects become large at high pressure and low temperature.

Q2. Define critical temperature. Give the relation with van der Waals constants.

Critical temperature is the temperature above which a gas cannot be liquefied by pressure alone. Tc = 8a/27Rb.

Q3. Calculate the rms speed of O₂ molecules at 27 °C.

• T = 300 K, M = 32 g/mol = 0.032 kg/mol
• c = √(3RT/M) = √(3 × 8.314 × 300 / 0.032)
• = √(7482.6 / 0.032) = √233831
• c ≈ 483.6 m/s

Q4. 2 mol of a gas occupy 10 L at 300 K. Find the pressure (ideal gas). (R = 0.0821 L atm K⁻¹ mol⁻¹)

• P = nRT/V
• = 2 × 0.0821 × 300 / 10
• = 49.26 / 10
• P = 4.93 atm

Q5. Derive the van der Waals equation (outline).

• Volume correction: molecules have own volume; free volume = V − nb (b = excluded volume per mole, four times the actual volume of molecules).
• Pressure correction: molecules near the wall are pulled inward, so observed P is less than ideal pressure.
• Correction is proportional to (n/V)², written an²/V²; ideal P = P + an²/V².
• Put both in PV = nRT → (P + an²/V²)(V − nb) = nRT.
• At low P and high T the corrections become small, and it reduces to PV = nRT.

Exam tip

Always convert temperature to kelvin and molar mass to kg/mol before using speed formulas. The van der Waals derivation is a favourite long question.

Quick summary

Liquids have a definite volume but no definite shape. Their molecules are close together and held by intermolecular forces, but they can still move. This topic covers vapour pressure, surface tension, viscosity, their measurement, the effect of temperature, and liquid crystals.

Key points

• Intermolecular forces in liquids: London (dispersion) forces, dipole–dipole forces and hydrogen bonding.
• Vapour pressure is the pressure of vapour in equilibrium with the liquid at a given temperature. It increases with temperature.
• A liquid boils when its vapour pressure equals the outside pressure. Water boils at 100 °C at 1 atm, and at a lower temperature on mountains.
• Surface tension (पृष्ठताण): the force acting per unit length on the surface, at right angles to any line on it. Surface molecules are pulled inward, so a liquid tries to reduce its surface area. That is why drops are spherical.
• Surface tension decreases with rise in temperature and becomes zero at the critical temperature.
• Surface tension is measured by the stalagmometer (drop count or drop weight method) or the capillary rise method.
• Viscosity: internal resistance to flow of a liquid, due to friction between layers moving at different speeds.
• Viscosity of a liquid decreases with rise in temperature (for gases it increases).
• Viscosity is measured with an Ostwald viscometer by comparing flow times of the liquid and water.
• Parachor is an additive and constitutive property linking surface tension, molar mass and densities; it helps in finding structures.
• Liquid crystals are a state between solid and liquid. They flow like liquids but have some order like crystals. Types: smectic, nematic and cholesteric. They are used in LCD displays and in temperature indicators.
• Liquids with strong hydrogen bonds (water, glycerol) have high surface tension, high viscosity and high boiling points.

Important terms

• Vapour pressure – pressure of vapour in equilibrium with its liquid at a fixed temperature.
• Surface tension (γ) – force per unit length on the surface; SI unit N m⁻¹.
• Surface energy – work needed to increase surface area by one unit; J m⁻², equal in value to surface tension.
• Coefficient of viscosity (η) – force per unit area needed to keep unit velocity gradient between layers. SI unit Pa s (N s m⁻²); CGS unit poise.
• Fluidity – reciprocal of viscosity, 1/η.

Formulas

• Newton's law of viscous flow: F = η A (dv/dx)
• Ostwald viscometer: η₁/η₂ = (d₁ t₁)/(d₂ t₂), d = density, t = flow time
• Stalagmometer (drop number): γ₁/γ₂ = (n₂ d₁)/(n₁ d₂), n = number of drops of a fixed volume
• Drop weight method: γ₁/γ₂ = m₁/m₂ (m = mass of one drop)
• Capillary rise: γ = ½ h r d g (contact angle zero)
• Parachor: [P] = M γ^(1/4) / (D − d), M = molar mass, D = density of liquid, d = density of vapour
• 1 poise = 0.1 Pa s

5 important questions

Q1. Why are liquid drops spherical?

Surface tension makes a liquid reduce its surface area to the minimum. For a given volume, a sphere has the smallest surface area. So free drops take a spherical shape.

Q2. What is the effect of temperature on viscosity of a liquid?

As temperature rises, molecules get more kinetic energy and intermolecular attraction becomes less effective. Layers slide more easily, so viscosity decreases.

Q3. In an Ostwald viscometer, water takes 120 s and a liquid takes 90 s to flow. Density of liquid = 0.8 g/cm³, water = 1.0 g/cm³, η(water) = 1.0 cP. Find η of the liquid.

• η₁/η₂ = (d₁ t₁)/(d₂ t₂)
• η₁ = 1.0 × (0.8 × 90)/(1.0 × 120)
• = 72/120
• η₁ = 0.60 cP

Q4. In a stalagmometer, a fixed volume gives 25 drops of water and 40 drops of a liquid of density 0.9 g/cm³. γ(water) = 72 dyne/cm. Find γ of the liquid.

• γ₁/γ₂ = (n₂ d₁)/(n₁ d₂), liquid = 1, water = 2
• γ₁ = 72 × (25 × 0.9)/(40 × 1.0)
• = 72 × 22.5/40 = 72 × 0.5625
• γ₁ = 40.5 dyne/cm

Q5. Describe the determination of viscosity by Ostwald's viscometer.

• Clean, dry viscometer fixed vertically in a constant-temperature bath.
• Fill a fixed volume of water, suck it above the upper mark, note time t₂ for flow between two marks.
• Repeat with the liquid to get t₁.
• Find densities d₁ and d₂.
• Calculate η₁ = η₂ (d₁ t₁)/(d₂ t₂). Mention the diagram: bulb, two marks, capillary, lower reservoir.

Exam tip

Both viscometer and stalagmometer numericals use simple ratios. Write the formula, then substitute carefully; keep units of density the same for both liquids.

Quick summary

Solids have a definite shape and volume because their particles are held in fixed positions. This topic covers crystalline and amorphous solids, laws of crystallography, unit cells and crystal systems, Miller indices, Bragg's law of X-ray diffraction, packing in cubic lattices, calculation of density, and defects in crystals.

Key points

• Crystalline solids: regular 3D arrangement (long-range order), sharp melting point, anisotropic (NaCl, quartz). Amorphous solids: short-range order only, melt over a range, isotropic (glass, rubber).
• Laws of crystallography: law of constancy of interfacial angles, law of rational indices, law of symmetry.
• Symmetry elements of a cube: 9 planes, 13 axes and 1 centre of symmetry — 23 elements in all.
• Unit cell: smallest repeating unit of a crystal lattice, described by edge lengths a, b, c and angles α, β, γ.
• Seven crystal systems (cubic, tetragonal, orthorhombic, monoclinic, triclinic, hexagonal, rhombohedral) and 14 Bravais lattices.
• Atoms per unit cell (Z): simple cubic 1, body-centred cubic (bcc) 2, face-centred cubic (fcc) 4.
• Packing efficiency: simple cubic 52.4 %, bcc 68 %, fcc/hcp 74 %.
• Miller indices (h k l): take intercepts in units of a, b, c → take reciprocals → clear fractions to smallest whole numbers.
• Bragg's law: X-rays reflected from parallel planes interfere constructively when nλ = 2d sinθ.
• For cubic crystals, the ratio of interplanar distances d₁₀₀ : d₁₁₀ : d₁₁₁ helps decide sc, bcc or fcc.
• NaCl structure: fcc, coordination number 6:6, 4 NaCl units per cell. CsCl: bcc-type, 8:8, 1 unit per cell.
• Defects: Schottky (equal cation and anion vacancies, density falls; NaCl, KCl) and Frenkel (ion moves to an interstitial site, density unchanged; AgCl, ZnS). AgBr shows both.

Important terms

• Crystal lattice (स्फटिक जालक) – regular 3D arrangement of points representing particles.
• Coordination number – number of nearest neighbours of a particle.
• Interplanar spacing (d) – distance between parallel lattice planes.

Formulas

• Density ρ = Z × M / (a³ × Nₐ) (a in cm, ρ in g cm⁻³, Nₐ = 6.022 × 10²³ mol⁻¹)
• Bragg's law: nλ = 2d sinθ (n = order, λ = wavelength, θ = glancing angle)
• Cubic lattice: d(hkl) = a / √(h² + k² + l²)
• Radius relations: sc a = 2r; bcc √3 a = 4r; fcc √2 a = 4r

5 important questions

Q1. Find the Miller indices of a plane with intercepts 2a, 3b and c.

Intercepts: 2, 3, 1 → reciprocals: ½, ⅓, 1 → multiply by 6: 3, 2, 6. Miller indices = (3 2 6).

Q2. Distinguish between Schottky and Frenkel defects.

Schottky: pair of cation and anion missing, density decreases, found in ionic solids with similar-sized ions (NaCl). Frenkel: smaller ion leaves its site and sits in an interstitial position, density unchanged, found where ions differ greatly in size (AgCl, ZnS).

Q3. NaCl has an fcc unit cell with edge length 5.64 Å. Calculate its density (M = 58.5 g mol⁻¹).

• Z = 4; a = 5.64 × 10⁻⁸ cm; a³ = 1.794 × 10⁻²² cm³
• ρ = (4 × 58.5) / (1.794 × 10⁻²² × 6.022 × 10²³) = 234 / 108.0
• ρ ≈ 2.17 g cm⁻³ (experimental value 2.16 g cm⁻³)

Q4. Derive Bragg's equation.

• Two parallel planes distance d apart; X-rays strike at angle θ.
• The ray reflected from the lower plane travels an extra path = 2d sinθ.
• Constructive interference when the path difference = whole number of wavelengths: nλ = 2d sinθ.
• Example: λ = 1.54 Å, first-order reflection at θ = 30° → d = 1.54/(2 × 0.5) = 1.54 Å.

Q5. Explain the seven crystal systems with their axial lengths and angles.

Make a table: cubic (a = b = c; all 90°), tetragonal (a = b ≠ c; all 90°), orthorhombic (a ≠ b ≠ c; all 90°), monoclinic (a ≠ b ≠ c; α = γ = 90°, β ≠ 90°), triclinic (all different), hexagonal (a = b ≠ c; α = β = 90°, γ = 120°), rhombohedral (a = b = c; α = β = γ ≠ 90°). Give one example each.

Exam tip

Learn Z, coordination number and packing for sc, bcc and fcc as one table. Density and Bragg's law numericals are favourites.

Quick summary

Thermodynamics studies energy changes in physical and chemical processes. This topic covers system and surroundings, state functions, the first law, internal energy and enthalpy, heat capacities, work in reversible and irreversible processes, Hess's law, and an introduction to entropy, the second law and Gibbs free energy.

Key points

• System: the part studied. Surroundings: everything else. Systems are open, closed or isolated.
• State functions depend only on initial and final states (U, H, S, G, P, V, T). Path functions depend on the path (q, w).
• Processes: isothermal (T constant), adiabatic (q = 0), isobaric (P constant), isochoric (V constant), reversible and irreversible.
• First law: energy can neither be created nor destroyed. ΔU = q + w (IUPAC sign: work done on the system is positive).
• Enthalpy: H = U + PV. At constant pressure, ΔH = q_p. At constant volume, ΔU = q_v.
• ΔH = ΔU + Δn(g)RT for reactions involving gases.
• Cp − Cv = R for one mole of an ideal gas. γ = Cp/Cv.
• Reversible isothermal expansion of an ideal gas gives maximum work. ΔU = 0, so q = −w.
• Hess's law: the total enthalpy change of a reaction is the same whether it occurs in one step or several steps.
• Kirchhoff's equation gives the effect of temperature on enthalpy of reaction.
• Second law: heat does not flow by itself from a colder to a hotter body; the entropy of the universe increases in a spontaneous process.
• Gibbs free energy: G = H − TS. At constant T and P, ΔG < 0 means spontaneous, ΔG = 0 means equilibrium, ΔG > 0 non-spontaneous.
• Third law: entropy of a perfect crystal at 0 K is zero.

Important terms

• Internal energy (अंतर्गत ऊर्जा, U) – total energy stored in a system.
• Enthalpy (H) – heat content at constant pressure.
• Entropy (S) – measure of randomness or disorder.
• Heat capacity – heat needed to raise the temperature by 1 K.
• Extensive property – depends on amount (mass, volume, U, H). Intensive property – does not (T, P, density).

Formulas

• ΔU = q + w
• Work against constant external pressure: w = −P_ext ΔV
• Reversible isothermal work: w = −2.303 nRT log(V₂/V₁) = −2.303 nRT log(P₁/P₂)
• ΔH = ΔU + ΔnRT
• Cp − Cv = R (per mole)
• Kirchhoff: ΔH₂ − ΔH₁ = ΔCp (T₂ − T₁)
• ΔS = q_rev/T; ΔG = ΔH − TΔS
• ΔG° = −2.303 RT log K

5 important questions

Q1. State the first law of thermodynamics and write its mathematical form.

Energy can neither be created nor destroyed, only changed from one form to another. The energy of an isolated system is constant. ΔU = q + w.

Q2. What are state functions? Give two examples.

Properties whose value depends only on the present state of the system, not on how it was reached. Examples: internal energy and enthalpy. Heat and work are not state functions.

Q3. Calculate the work done when 2 mol of an ideal gas expands reversibly and isothermally at 300 K from 10 L to 20 L.

• w = −2.303 nRT log(V₂/V₁)
• = −2.303 × 2 × 8.314 × 300 × log 2
• = −2.303 × 4988.4 × 0.3010
• = −3458 J ≈ −3.46 kJ (work done by the gas)

Q4. For N₂(g) + 3H₂(g) → 2NH₃(g), ΔH = −92.4 kJ at 298 K. Find ΔU.

• Δn(g) = 2 − (1 + 3) = −2
• ΔnRT = −2 × 8.314 × 298 = −4955 J = −4.96 kJ
• ΔU = ΔH − ΔnRT = −92.4 − (−4.96)
• ΔU = −87.44 kJ

Q5. Explain the criteria of spontaneity in terms of ΔG.

• G = H − TS; ΔG = ΔH − TΔS at constant T and P.
• ΔH negative, ΔS positive → ΔG always negative, spontaneous at all T.
• ΔH positive, ΔS negative → never spontaneous.
• ΔH negative, ΔS negative → spontaneous at low T.
• ΔH positive, ΔS positive → spontaneous at high T (e.g. melting of ice above 0 °C).

Exam tip

Use the IUPAC sign convention throughout and say so in the answer. Write R = 8.314 J K⁻¹ mol⁻¹ and convert J to kJ at the end.