Chapter Practice

Correct: B. Phenoxide ion (C₆H₅O⁻) is stabilised by delocalisation of the negative charge over the benzene ring (5 resonance structures). Ethoxide ion (C₂H₅O⁻) has no such resonance stabilisation. Greater stabilisation of conjugate base → stronger acid. pKa(phenol) ≈ 10, pKa(ethanol) ≈ 16.

Correct: A. Lucas test: 3° alcohol → immediate turbidity (SN1, stable carbocation); 2° alcohol → turbidity after ~5 minutes; 1° alcohol → no turbidity at room temperature (SN1 too slow, SN2 hindered by ZnCl₂). The turbidity is due to insoluble alkyl chloride formed.

Correct: B — 2,4,6-tribromophenol. The –OH group is strongly o,p-directing (activating). All three ortho and para positions are substituted in presence of excess bromine water. A white precipitate forms immediately — this is used as a test for phenol. With CS₂ (non-polar solvent), only 4-bromophenol forms.

Correct: B — 140°C. At 140°C with conc. H₂SO₄: 2CH₃CH₂OH → CH₃CH₂–O–CH₂CH₃ + H₂O (ether). At 170°C: CH₃CH₂OH → CH₂=CH₂ + H₂O (alkene, elimination dominates). Temperature controls which product forms.

Correct: A — Salicylic acid. Kolbe's reaction: C₆H₅ONa + CO₂ → C₆H₄(OH)(COONa) → (after acidification) C₆H₄(OH)(COOH) = salicylic acid. CO₂ undergoes electrophilic substitution at the ortho position. Salicylic acid is used to make aspirin (acetylsalicylic acid).

Correct: C — Ethanol. Iodoform test is positive for: (i) Alcohols with CH₃CH(OH)– group (secondary alcohols with CH₃ group), (ii) CH₃CHO (ethanal), (iii) Methyl ketones (CH₃COR). Ethanol (CH₃CH₂OH) is oxidised to CH₃CHO which undergoes iodoform reaction. CH₃OH, 1-propanol, 1-butanol → negative.

Correct: D — Silver mirror. Tollens' reagent = [Ag(NH₃)₂]⁺ (ammoniacal silver nitrate). Aldehyde reduces Ag⁺ to Ag (silver mirror). Ketones do NOT give this test. Fehling's test gives a brick-red precipitate of Cu₂O with aldehydes (not with aromatic aldehydes like benzaldehyde).

Correct: C — CH₃CHO. Aldol condensation requires an α-hydrogen (H on the carbon adjacent to the C=O). HCHO has no α-carbon. C₆H₅CHO has no α-H (the adjacent carbon is part of the benzene ring, no H available). CCl₃CHO has no α-H (all replaced by Cl). CH₃CHO has 3 α-H atoms.

Correct: A. Cannizzaro reaction requires aldehydes with NO α-H (cannot undergo aldol). Examples: HCHO, C₆H₅CHO, (CH₃)₃CCHO. In conc. NaOH, disproportionation occurs: one molecule is oxidised (to carboxylate) and another is reduced (to alcohol). HCHO + HCHO → HCOO⁻ + CH₃OH.

Correct: B. Nucleophilic addition reactivity depends on: (i) electrophilicity of carbonyl C — alkyl groups donate electrons, reducing electrophilicity; (ii) steric hindrance. HCHO (no alkyl groups) > CH₃CHO (1 alkyl group) > (CH₃)₂CO (2 alkyl groups). Aldehydes are more reactive than ketones.

Correct: A — Lactic acid. CH₃CHO + HCN → CH₃CH(OH)CN (cyanohydrin) → acid hydrolysis (H₃O⁺, heat) → CH₃CH(OH)COOH = lactic acid (2-hydroxypropanoic acid). This is an important method to increase the carbon chain length by one.

Correct: B — –CH₂–. Clemmensen reduction: Zn(Hg)/conc. HCl + heat. C=O → –CH₂–. Used for acid-sensitive compounds. Wolff-Kishner reduction (NH₂NH₂, KOH, ethylene glycol): also C=O → –CH₂– but used in basic conditions. Both reduce the carbonyl completely to a methylene group.

Correct: D — CCl₃COOH. Electron-withdrawing groups (–I effect) on the α-carbon stabilise the carboxylate anion → increase acidity. More Cl atoms → stronger acid. pKa: CH₃COOH (4.76) > ClCH₂COOH (2.86) > Cl₂CHCOOH (1.48) > Cl₃CCOOH (0.65). Each Cl added increases acid strength.

Correct: C — Aldehydes and ketones. 2,4-DNP reacts with both aldehydes and ketones to form orange or yellow 2,4-dinitrophenylhydrazone (a precipitate). This confirms the presence of a C=O group but does NOT distinguish aldehydes from ketones. Tollens' test distinguishes them further.

Correct: D. In gas phase: (CH₃)₃N > (CH₃)₂NH > CH₃NH₂ > NH₃ (purely inductive effect). In water, solvation of the conjugate acid matters. (CH₃)₃NH⁺ has bulky groups → poor solvation → less stable → 3° amine appears less basic than 2°. So aqueous order: (CH₃)₂NH > CH₃NH₂ > (CH₃)₃N > NH₃.

Correct: A. In aniline, the lone pair on N is in conjugation (resonance) with the π system of benzene ring → delocalised over the ring. This makes the lone pair less available to donate to H⁺. pKb(aniline) ≈ 9.4 vs pKb(methylamine) ≈ 3.4. Aniline is sp² hybridised (different from methylamine which is sp³).

Correct: C — Methylamine (CH₃NH₂). Hoffmann degradation: RCONH₂ → RNH₂ (one less carbon). CH₃CONH₂ (2 carbons, ethanamide) → CH₃NH₂ (1 carbon, methylamine). Reagents: Br₂ + NaOH. Key intermediate is isocyanate (CH₃–N=C=O) which hydrolyses to give the amine with loss of CO₂.

Correct: B — NaNO₂ + HCl at 0–5°C. Diazotisation: ArNH₂ + NaNO₂ + 2HCl → ArN₂⁺Cl⁻ + NaCl + 2H₂O. Temperature must be kept low (0–5°C) to prevent decomposition of the diazonium salt. Diazonium salts are versatile intermediates for making ArX (Sandmeyer), ArOH, ArCN, azo dyes, etc.

Correct: B — Secondary amine. R₂NH + HNO₂ → R₂N–NO (N-nitrosamine, yellow oily liquid). Primary amines: react with HNO₂ to form unstable diazonium compounds (aliphatic) or stable diazonium salts (aromatic). Tertiary aliphatic amines: form ammonium salts. This is used to distinguish 1°, 2°, 3° amines.

Correct: C — Aliphatic primary amines. Gabriel synthesis: phthalimide + KOH → potassium phthalimide; then + RX → N-alkyl phthalimide; then hydrolysis → RNH₂ (1° amine) + phthalic acid. Cannot be used for aromatic primary amines (ArX does not react with phthalimide anion — poor SN2 reactant).

Correct: C. In sucrose, the glycosidic linkage uses C-1 of α-glucose and C-2 of β-fructose (both anomeric positions). This means there is no free –CHO or potential –CHO in open chain form → cannot reduce Fehling's or Tollens' reagent → non-reducing sugar. Maltose and lactose are reducing (one free anomeric –OH).

Correct: C — Hydrogen bonds. In α-helix: H-bonds between C=O of one peptide bond and N–H of the peptide bond 4 residues away (within the same chain). In β-pleated sheet: H-bonds between parallel or antiparallel chains. Disulphide bonds stabilise tertiary structure. Hydrophobic interactions also stabilise 3° structure.

Correct: D — All of the above. DNA vs RNA: (i) Sugar — deoxyribose vs ribose; (ii) Bases — Adenine, Thymine, Guanine, Cytosine vs A, Uracil, G, C; (iii) Strands — double vs single; (iv) Base pairing in DNA: A=T (2 H-bonds) and G≡C (3 H-bonds). Purines (A, G) pair with Pyrimidines (T/U, C).

Correct: B. Both starch and cellulose are made of glucose units (C₆H₁₂O₆). Starch has α-1,4 glycosidic linkages (amylose) and α-1,6 branches (amylopectin) → coiled structure → energy storage → digestible. Cellulose has β-1,4 linkages → straight chains → structural rigidity → not digestible by humans.

Correct: A — Glucose. Mutarotation: the change in optical rotation when α-D-glucose (+112°) or β-D-glucose (+18.7°) dissolves in water, until an equilibrium mixture (+52.7°) is reached. This occurs through the open-chain aldehyde form as an intermediate. Sucrose shows no mutarotation (no free anomeric OH).

Correct: B. Denaturation disrupts secondary, tertiary and quaternary structure — but NOT the primary structure (peptide bonds remain intact). H-bonds, disulphide bonds (–S–S–), hydrophobic interactions, and ionic bonds are broken by heat, acids, bases, organic solvents. The protein loses its biological activity. Example: boiling egg white (albumin denaturation).