Exercises
Q 8.1: Hybridisation states of each carbon in CH₂=C=O, CH₃CH=CH₂, (CH₃)₂C=O, CH₂=CHCN, C₆H₅NO₂
CH₂=C=O (ketene): CH₂ carbon = sp² (double bond, trigonal planar); C=O carbon = sp (two double bonds, linear)
CH₃CH=CH₂ (propene): CH₃ carbon = sp³; CH= carbon = sp²; =CH₂ carbon = sp²
(CH₃)₂C=O (acetone): Both CH₃ carbons = sp³; C=O carbon = sp²
CH₂=CHCN (acrylonitrile): CH₂= carbon = sp²; CH= carbon = sp²; CN carbon = sp (triple bond)
C₆H₅NO₂ (nitrobenzene): All ring carbons = sp²
Q 8.2: Indicate σ and π bonds
CH₃CH₂CH₂Cl (1-chloropropane): All C–C and C–H bonds are σ bonds; C–Cl is a σ bond. Total: all bonds σ, no π bonds.
(CH₃)₂C=CH₂ (isobutylene): All C–C single bonds and C–H bonds are σ; the C=C has 1 σ + 1 π bond.
CH₂=C=CH₂ (allene): All single bonds σ; each C=C: 1σ + 1π. Total: σ bonds plus 2 π bonds in the two C=C double bonds.
CH₃NO₂ (nitromethane): C–H: σ bonds; C–N: σ bond; N=O: 1σ + 1π; N–O⁻: σ bond.
HCONHCH₃ (N-methylformamide): C=O: 1σ + 1π; all other bonds σ.
CH₃CH=CHCHO (but-2-enal): C=C: 1σ + 1π; C=O: 1σ + 1π; all other bonds σ.
Q 8.3: Bond-line formulas
Isopropyl alcohol (2-propanol): A V-shape with OH at the middle carbon: /\OH where angles represent CH₃ groups at each end and OH at vertex.
2,2-Dimethylbutanal: Four-carbon chain, two methyl groups at C-2, CHO at C-1.
Heptanone (heptan-2-one): Seven-carbon chain with C=O at C-2: CH₃CO(CH₂)₄CH₃
Q 8.4: IUPAC names of compounds
(a) A compound with structure CH₃CH(OH)CH₂CH₃: Butan-2-ol
(b) CH₃CH₂COCH₂CH₃: Pentan-3-one
(c) CH₃CH=CHCH₂CH₃: Pent-2-ene
(d) CH₃CH₂CH₂NH₂: Propan-1-amine
(e) C₆H₅OH: Phenol
(f) ClCH₂CH₂OH: 2-Chloroethan-1-ol
Q 8.5: Correct IUPAC names
(a) 2,2-Dimethylpentane (not 2,2-dimethylpentane); principal chain has 5 carbons → 2,2-dimethylpentane is correct if both methyls are on C-2.
(b) 2,4,7-Trimethyloctane: count to give lowest locants; check if numbering from other end gives lower locants → use the name that gives lowest set.
(c) 2-Chloro-4-methylpentane (substituents get lowest locants from appropriate end).
(d) But-3-yn-1-ol: triple bond at C-3, OH at C-1 (chain numbered to give OH lower number) — but actually, functional group gets lowest locant; correct IUPAC: But-3-yn-1-ol if OH at one end. The given names should be evaluated by IUPAC 2013 rules. Answer: select the name with lowest set of locants.
Q 8.6: First five members of homologous series
(a) HCOOH (formic acid series — carboxylic acids):
HCOOH (methanoic acid), CH₃COOH (ethanoic acid), CH₃CH₂COOH (propanoic acid), CH₃(CH₂)₂COOH (butanoic acid), CH₃(CH₂)₃COOH (pentanoic acid)
(b) CH₃COCH₃ (ketone series):
CH₃COCH₃ (propan-2-one), CH₃COC₂H₅ (butan-2-one), CH₃CO(CH₂)₂CH₃ (pentan-2-one), C₂H₅COC₂H₅ (pentan-3-one), CH₃CO(CH₂)₃CH₃ (hexan-2-one)
(c) H–CH=CH₂ (alkene — H₂C=CH₂ ethene):
C₂H₄ (ethene), C₃H₆ (propene), C₄H₈ (but-1-ene), C₅H₁₀ (pent-1-ene), C₆H₁₂ (hex-1-ene)
Q 8.7: Structural formulas and functional groups
(a) 2,2,4-Trimethylpentane (isooctane): CH₃C(CH₃)₂CH₂CH(CH₃)CH₃ — no functional group (alkane)
(b) 2-Hydroxy-1,2,3-propanetricarboxylic acid (citric acid): HOOCCH₂C(OH)(COOH)CH₂COOH — functional groups: –OH (hydroxyl), –COOH (carboxyl) ×3
(c) Hexanedial: OHCCH₂CH₂CH₂CH₂CHO — functional group: –CHO (aldehyde) ×2
Q 8.8: Identify functional groups
(a) A compound containing –OH and C=O (e.g., a hydroxy ketone): –OH (hydroxyl) and C=O (carbonyl/ketone)
(b) Structure with –NH₂: –NH₂ (amino group)
(c) Structure with –COOH: –COOH (carboxyl group)
Q 8.9: Which is more stable: ⁻OCH=CH– or CH₃CHO⁻?
⁻OCH=CH– (vinyl alkoxide) is more stable. The negative charge can be delocalised over the oxygen and through conjugation with the double bond (resonance stabilisation). CH₃CHO⁻ (acetaldehyde carbanion) has no such delocalisation. Greater resonance stabilisation → greater stability.
Q 8.10: Why alkyl groups act as electron donors when attached to π system
Alkyl groups have C–H σ bonds that can overlap with the π system through hyperconjugation. This electron density donation stabilises carbocations (and partially destabilises carbanions). The more alkyl groups, the more hyperconjugation and the more stable the carbocation.
Q 8.11: Resonance structures
(a) C₆H₅OH (phenol): The lone pair on oxygen delocalises into the ring; resonance structures show negative charges at ortho and para positions of the ring.
(b) C₆H₅NO₂ (nitrobenzene): The nitrogen's electron-withdrawing effect creates resonance structures with positive charges at ortho and para positions in the ring.
(c) CH₃CH=CHCHO: O⁻–CH=CH–CH=CH₂ ↔ O=CH–CH⁻–CH=CH₂ etc. — conjugated system with resonance across the double bonds.
(d) CH₃–CHO: O⁻=CH–CH₃ ↔ O=CH–⁻CH₂ — limited resonance, less stable contributor.
(e) CH₂=C(H)– and (f) involve allylic/vinylic resonance.
Q 8.12: Electrophiles and nucleophiles
Electrophiles are electron-deficient species that accept electron pairs. Examples: H⁺, NO₂⁺, Br⁺, AlCl₃, BF₃, carbocations (R⁺).
Nucleophiles are electron-rich species that donate electron pairs. Examples: OH⁻, CN⁻, NH₃, H₂O, carbanions (R⁻).
Q 8.13: Identify nucleophile or electrophile in each reaction
(a) CH₃COOH + HO⁻ → CH₃COO⁻ + H₂O: HO⁻ is a nucleophile (donates electron pair to H⁺)
(b) CH₃COCH₃ + C⁻N → product: CN⁻ is a nucleophile (attacks carbonyl C)
(c) CH₃CH₂+ ... → product: CH₃CH₂⁺ (ethyl carbocation) is an electrophile
Q 8.14: Classify reaction types
(a) CH₃CH₂Br + HS⁻ → CH₃CH₂SH + Br⁻: Nucleophilic substitution (SN2)
(b) (CH₃)₃C⁺ + CH₃OH → (CH₃)₃COCH₃ + H⁺: Nucleophilic addition/substitution
(c) CH₃CH₂Br + HO⁻ → CH₂=CH₂ + H₂O + Br⁻: Elimination (E2)
(d) (CH₃)₃C–CH₂OH + HBr → product: Electrophilic substitution or SN1
15: Structural/geometrical isomers or resonance contributors?
(a) Two structures that differ in position of a substituent: structural isomers
(b) Two structures differing in configuration around a double bond: geometrical isomers
(c) Two structures connected by arrows showing electron movement with the same connectivity: resonance contributors
Answer
Q 8.16: Homolysis or heterolysis; classify intermediates
(a) Cl–Cl bond broken with one electron to each Cl: Homolysis → free radicals (Cl•)
(b) C–Br bond broken with both electrons to Br: Heterolysis → carbocation (R⁺) and bromide ion (Br⁻)
(c) C–H bond broken with both electrons to C: Heterolysis → carbanion (R⁻) and proton (H⁺)
(d) O–H bond in CH₃OH broken homolytically: Homolysis → free radicals (CH₃O• and H•)
Q 8.17: Inductive and electromeric effects; explain acidic order
Inductive effect (I): The transmission of the polarity of a bond through the σ-bond chain due to electronegativity difference. Electron-withdrawing groups (–I): –F, –Cl, –NO₂; Electron-donating (+I): –CH₃, –C₂H₅.
Electromeric effect (E): Complete transfer of π electrons to one atom under attack of a reagent. Temporary effect.
(a) ClCH₂COOH > ClCH₂CH₂COOH > ClCH₂CH₂CH₂COOH: Explained by inductive effect. Cl is an electron-withdrawing group (–I). The closer Cl is to the COOH, the more it withdraws electron density from O–H, increasing acid strength. Inductive effect decreases with distance.
(b) CH₃CH₂COOH > (CH₃)₂CHCOOH > (CH₃)₃CCOOH: Explained by +I effect. Methyl groups donate electrons (+I effect), increasing electron density on COOH and destabilising the conjugate base (carboxylate anion). More methyl groups = stronger +I effect = weaker acid.
Q 8.18: Principles of separation techniques
(a) Crystallisation: A substance is dissolved in minimum hot solvent, then cooled. The substance precipitates as crystals while impurities remain in solution. The process exploits difference in solubility at different temperatures. Example: purification of impure sugar.
(b) Distillation: A liquid mixture is heated; the more volatile component vaporises first. Vapours are condensed and collected. Example: separation of ethanol and water (ethanol b.p. 78°C, water 100°C).
(c) Chromatography: Mixture is carried by a mobile phase over a stationary phase. Components travel at different rates depending on their affinity for stationary vs. mobile phase. Example: TLC separates amino acids — polar ones travel less far.
Q 8.19: Method to separate compounds with different solubility
Fractional crystallisation: Dissolve the mixture in a minimum of hot solvent. On cooling, the less soluble compound crystallises first. Filter to separate it. Cool the filtrate further to crystallise the more soluble compound. Repeat if necessary to get pure fractions.
Q 8.20: Difference between distillation, distillation under reduced pressure, steam distillation
Simple distillation: Liquid is boiled at normal atmospheric pressure; used for liquids with significantly different boiling points.
Distillation under reduced pressure (vacuum distillation): Pressure is reduced so the liquid boils at a lower temperature. Used for compounds that decompose at their normal boiling point.
Steam distillation: Steam is passed into the liquid. The compound codistils with water at a temperature below 100°C (sum of partial pressures = 1 atm). Used for temperature-sensitive compounds immiscible with water.
Q 8.21: Chemistry of Lassaigne's test
Organic compound is fused with sodium metal, converting N, S, X (halogens) to ionic forms: NaCN, Na₂S, NaX. These are then detected:
Nitrogen: NaCN + FeSO₄ → Na₄[Fe(CN)₆]; heated with Fe³⁺ → Fe₄[Fe(CN)₆]₃ (Prussian blue)
Sulphur: Na₂S + Na₂[Fe(CN)₅NO] → violet coloured complex; or Na₂S + Pb(CH₃COO)₂ → PbS (black precipitate)
Halogen: NaX + AgNO₃ → AgX precipitate (white for Cl, yellow for Br, pale yellow for I)
Q 8.22: Difference between Dumas and Kjeldahl's methods for nitrogen
Dumas method: Compound heated with CuO in CO₂ atmosphere → gaseous products including N₂. N₂ collected over KOH solution. Volume of N₂ measured.
Kjeldahl's method: Compound digested with conc. H₂SO₄ → (NH₄)₂SO₄. Then made alkaline with NaOH → NH₃ liberated. NH₃ absorbed in standard acid, then back-titrated.
Kjeldahl's not applicable to: nitro/azo/ring nitrogen compounds (pyridine). Dumas is more general.
Q 8.23: Principle of estimation of halogens, sulphur, phosphorus
Halogens (Carius method): Compound heated with fuming HNO₃ + AgNO₃ in Carius tube → AgX precipitate. Weighed to calculate % halogen.
Sulphur (Carius): Compound heated with fuming HNO₃ or Na₂O₂ → H₂SO₄ → BaSO₄ precipitate with BaCl₂. Weigh BaSO₄ to find % S.
Phosphorus: Compound heated with fuming HNO₃ → H₃PO₄ → (NH₄)₃PO₄·12MoO₃ (yellow ppt) with ammonium molybdate. Alternatively precipitated as MgNH₄PO₄ → Mg₂P₂O₇ on ignition.
Q 8.24: Principle of paper chromatography
Paper chromatography uses cellulose paper (stationary phase, which holds water) and an organic solvent as mobile phase. The mixture is spotted near the bottom of the paper and placed in solvent. As solvent rises by capillary action, components separate based on their different solubilities in the stationary water film vs. the mobile organic phase. Rf = distance moved by substance / distance moved by solvent front.
Q 8.25: Why HNO₃ added before AgNO₃ in halogen test?
Nitric acid is added to decompose Na₂S and NaCN (cyanide and sulphide) that were formed during Lassaigne's test. These would react with AgNO₃ to give Ag₂S (black) and AgCN (white), giving false positives for halogen test. HNO₃ decomposes them before AgNO₃ is added so only AgX forms for genuine halogen detection.
Q 8.26: Reason for fusing with metallic sodium
Organic compounds have C–N, C–S, C–X covalent bonds that are not directly detectable by simple ionic tests. Fusion with sodium converts these covalent bonds into ionic compounds (NaCN, Na₂S, NaX) that dissolve in water and can be detected by standard ionic reactions.
Q 8.27: Technique to separate calcium sulphate and camphor
Sublimation. Camphor sublimes (directly converts to vapour) on gentle heating, while CaSO₄ (ionic salt) does not. Camphor vapours are collected on a cold surface while CaSO₄ remains behind.
28: Why an organic liquid vaporises below its boiling point in steam distillation?
Answer
In steam distillation, two immiscible liquids (water and organic compound) are present. The total pressure = sum of partial pressures of each component. When the sum of their individual vapour pressures equals atmospheric pressure, the mixture boils. Since each contributes its own partial pressure, the mixture boils at a temperature lower than the boiling point of either pure liquid.
Q 8.29: Will CCl₄ give AgCl precipitate with AgNO₃?
No. CCl₄ has covalent C–Cl bonds and no ionisable chloride ions. AgNO₃ detects only free Cl⁻ (ionic chlorides). CCl₄ does not ionise in water, so no AgCl forms. (The Lassaigne test is needed to first convert C–Cl to NaCl.)
Q 8.30: Why KOH absorbs CO₂ in carbon estimation?
In carbon estimation by combustion, the CO₂ produced must be quantitatively absorbed and weighed. KOH solution absorbs CO₂: CO₂ + 2KOH → K₂CO₃ + H₂O. This allows exact determination of carbon content from mass increase of the KOH tube. CaCl₂ tube is used before to absorb water, so the KOH tube absorbs only CO₂.
31: Why acetic acid (not H₂SO₄) for lead acetate sulphur test?
Answer
If H₂SO₄ were used, it would react with Na₂S to produce SO₄²⁻ ions or provide its own S, confusing the test for sulphur. Also, H₂SO₄ might form PbSO₄ (not PbS), giving a false result. Acetic acid is used because it just acidifies the solution without interfering with the sulphide test (CH₃COOH + Na₂S does not form any precipitate with Pb²⁺).
32: Masses of CO₂ and H₂O from combustion of compound (40% C, 6.72% H, rest O)
Answer
Percentage O = 100 − 40 − 6.72 = 53.28%. Take 1 g sample: C = 0.40 g, H = 0.0672 g, O = 0.5328 g
C + O₂
⇒ CO₂: 0.40 g C produces (44/12) × 0.40 = 1.467 g CO₂
2H₂ + O₂
⇒ 2H₂O: 0.0672 g H produces (18/2) × 0.0672 = 0.604 g H₂O
33: % nitrogen by Kjeldahl's; 0.5 g compound, excess H₂SO₄ absorbed, residual H₂SO₄ titrated with 0.5 M NaOH, 30 mL used; 80 mL of 0.5 M H₂SO₄ taken
Answer
H₂SO₄ consumed by NH₃: V(H₂SO₄) − V(NaOH)/2 = 80 × 0.5 mmol − 30 × 0.5/2 mmol
Actually: mmol H₂SO₄ taken = 80 × 0.5 = 40 mmol; mmol NaOH for excess H₂SO₄ = 30 × 0.5 = 15 mmol; mmol H₂SO₄ unreacted = 15/2 = 7.5 mmol (since H₂SO₄ + 2NaOH)
mmol H₂SO₄ reacted with NH₃ = 40 − 7.5 = 32.5 mmol; mmol NH₃ = 2 × 32.5 = 65 mmol
g N = 65 × 10⁻³ × 14 = 0.91 g; % N = 0.91/0.5 × 100 = 182% (re-check given values)
Using the standard formula: % N = 1.4 × M × (2V − V₁) / m where M = molarity, V = mL H₂SO₄, V₁ = mL NaOH, m = mass
Q 8.34: % Cl in compound; 0.15 g compound gives 0.12 g AgCl; M(AgCl) = 143.5
g Cl in 0.12 g AgCl = (35.5/143.5) × 0.12 = 0.02967 g
% Cl = (0.02967/0.15) × 100 = 19.78%
Q 8.35: % S in compound; 0.468 g compound gives 0.668 g BaSO₄; M(BaSO₄) = 233
g S = (32/233) × 0.668 = 0.0917 g
% S = (0.0917/0.468) × 100 = 19.6%
Q 8.36: Hybridised orbitals in CH₂=CH–CH=C(CH₃)
C–C bond between sp² (alkene) and sp³ (CH₃): (c) sp² – sp³
Q 8.37: Prussian blue in Lassaigne's test
(b) Fe₃[Fe(CN)₆]₂ (ferrous ferricyanide) — Wait, Prussian blue is Fe₄[Fe(CN)₆]₃ or Fe3+[Fe2+(CN)₆]³⁻ = iron(III) hexacyanoferrate(II). Answer: (b) Fe₄[Fe(CN)₆]₃
Q 8.38: Most stable carbocation
Stability: tertiary > secondary > primary. (CH₃)₃C⁺ is tertiary — (a) (CH₃)₃C⁺ most stable
Q 8.39: Best technique for isolation/purification
(d) Chromatography — most versatile modern technique, can separate micro-quantities with high precision.
Q 8.40: Classification of CH₃CH₂I + KOH(aq) → CH₃CH₂OH + KI
(b) Nucleophilic substitution — OH⁻ is the nucleophile, attacks the electrophilic carbon bonded to iodine, displacing I⁻.
Important Extra Questions
Q A: Explain sp, sp², and sp³ hybridisation with examples from organic chemistry.
sp³ hybridisation: One s + three p orbitals combine → 4 sp³ hybrid orbitals at 109.5°. Geometry: tetrahedral. Example: methane (CH₄), all C–H bonds are sp³–s sigma bonds.
sp² hybridisation: One s + two p orbitals → 3 sp² orbitals at 120°, leaving one unhybridised p. Geometry: trigonal planar. Example: ethene (CH₂=CH₂), C=C has one σ (sp²–sp²) and one π bond (p–p sideways).
sp hybridisation: One s + one p → 2 sp orbitals at 180°, leaving two unhybridised p orbitals. Geometry: linear. Example: ethyne (HC≡CH), C≡C has one σ and two π bonds.
Q B: What is resonance? Explain with reference to benzene. State the conditions for resonance structures.
Resonance: When a molecule cannot be represented by a single Lewis structure, the true structure is a resonance hybrid — an average of all contributing structures.
Benzene (C₆H₆): Two Kekulé structures (alternating single/double bonds) are resonance structures. The real benzene has all bonds equal (bond length 1.40 Å, between C–C 1.54 Å and C=C 1.34 Å), with electrons delocalised around the ring.
Conditions: (i) Same atomic arrangement; (ii) Same number of electrons; (iii) Resonance structures differ only in position of electrons (not atoms); (iv) Resonance structures should have minimum charge separation; (v) Negative charge on electronegative atom increases stability.
Q C: Distinguish between inductive effect and mesomeric effect (resonance effect).
Inductive effect: Permanent polarisation of σ bonds due to electronegativity difference. Transmitted through chain but decreases rapidly with distance. Examples: –F, –Cl (−I); –CH₃ (+I).
Mesomeric/Resonance effect: Delocalisation of π electrons or lone pairs through a conjugated system. Operates through π bonds. Examples: –OH in phenol (+M, electron donor to ring); –NO₂ (−M, withdraws from ring).
Key difference: Inductive acts through σ bonds over short distances; mesomeric acts through π system over the whole conjugated system.
Q D: What is carbocation? Explain its stability order with hyperconjugation.
A carbocation is a species with a positively charged carbon (trivalent, sp² hybridised, with empty p orbital).
Stability order: 3° > 2° > 1° > CH₃⁺
Hyperconjugation: The C–H sigma bonds of adjacent alkyl groups can overlap with the empty p orbital of the carbocation, delocalising electron density. More α-hydrogen atoms → more hyperconjugation structures → more stable carbocation.
CH₃⁺: 0 hyperconjugation structures; CH₃CH₂⁺: 3 (from CH₃); (CH₃)₂CH⁺: 6; (CH₃)₃C⁺: 9 structures → most stable.
Q E: Explain the principle and use of column chromatography.
Principle: Different components of a mixture have different affinities for stationary phase (silica gel or alumina) vs. mobile phase (solvent). Components with higher affinity for stationary phase move slower; less affinity = move faster.
Setup: A glass column packed with stationary phase (silica gel). The mixture is loaded at top. Solvent (eluent) is passed through. Components elute at different rates and are collected in separate fractions.
Use: Separation and purification of compounds (including pharmaceuticals, natural products), preparative chromatography for larger quantities.