Organic Chemistry — all inquiry questions. Covers nomenclature, IMF trends, hydrocarbon and alcohol reactions, organic acids and bases, pathways, polymers, and extended synthesis.
1. What is the correct IUPAC name for CH₃CH₂CH(CH₃)COOH?
2. Arrange in order of increasing boiling point: propanamide, propane, propan-1-amine, propanoic acid, propan-1-ol.
3. A haloalkane is treated with aqueous NaOH. What type of reaction occurs and what is the product?
4. The industrial hydration of ethene (CH₂=CH₂ + H₂O → CH₃CH₂OH) runs at ~300 °C and ~65 atm with only ~5% single-pass conversion. How is high overall yield achieved?
5. Butanal and butan-2-one are both C₄H₈O. A student adds Tollens' reagent to each. What is observed, and what structural feature explains the difference?
6. What is the correct product when propan-2-ol is dehydrated with concentrated H₂SO₄ at ~170 °C?
7. Which of the following is NOT a required condition for fermentation to produce ethanol?
8. A condensation polymer has repeat unit [–NH–(CH₂)₅–CO–]ₙ. What monomer produced this polymer and what is the by-product?
9. Which statement correctly identifies the role of conc. H₂SO₄ in esterification?
10. An unknown compound: (i) decolourises bromine water; (ii) gives no colour change with K₂Cr₂O₇/H₂SO₄; (iii) requires TWO additions of Br₂ to fully decolourise. What compound class is most consistent?
Section A — MC Answers: 1-A | 2-A | 3-B | 4-B | 5-B | 6-C | 7-D | 8-B | 9-B | 10-C
MC Explanations:
1-A: CH₃CH₂CH(CH₃)COOH — longest chain through COOH: C1=COOH, C2=CH(CH₃), C3=CH₂, C4=CH₃ = butanoic acid. Methyl branch at C2 → 2-methylbutanoic acid. Option C (2-ethylpropanoic acid) uses a shorter 3C main chain — incorrect IUPAC rule (always use longest chain).
2-A: IMF ranking: propane (dispersion, BP −42 °C) < propan-1-amine (N–H H-bond, 48 °C) < propan-1-ol (O–H H-bond, 97 °C) < propanoic acid (O–H + dimerisation, 141 °C) < propanamide (N–H + C=O network, 213 °C).
3-B: Aqueous NaOH → nucleophilic substitution → alcohol + NaX. Alcoholic NaOH → elimination → alkene.
4-B: Excess steam shifts equilibrium right (Le Chatelier). Unreacted ethene is separated and recycled. Increasing temperature (A) shifts the exothermic reaction LEFT. Catalysts do not change equilibrium position (C).
5-B: Tollens' oxidises aldehydes — the H on the –CHO carbon is removed. Butanal has this H → positive. Butan-2-one has no H on the C=O carbon → cannot be oxidised → negative.
6-C: Dehydration = elimination. Conc. H₂SO₄, 170 °C removes H from C1 and –OH from C2 → propene + H₂O. At lower temperature (~140 °C) and excess alcohol, ether forms instead.
7-D: Conc. H₂SO₄ is used for esterification and alkene hydration — not fermentation. H₂SO₄ would denature yeast enzymes. Fermentation requires yeast, ~35 °C, and anaerobic conditions.
8-B: The repeat unit [–NH–(CH₂)₅–CO–]ₙ shows one amide bond per 6-carbon unit → Nylon 6, from a single bifunctional monomer (6-aminohexanoic acid). Nylon 6,6 uses two monomers. By-product: H₂O.
9-B: Conc. H₂SO₄ serves two roles: (1) acid catalyst — activates carboxylic acid by protonation; (2) dehydrating agent — absorbs H₂O produced, shifting equilibrium right. It is not consumed (it is a catalyst by definition).
10-C: (i) decolourises Br₂ → unsaturation; (ii) no K₂Cr₂O₇ change → no oxidisable group (not alcohol/aldehyde); (iii) requires two Br₂ additions → two π bonds → C≡C. An alkene (one π bond) is fully decolourised with one addition of Br₂.
Step 1: Ethanol → ethanoic acid
CH₃CH₂OH + 2[O] → CH₃COOH + H₂O. Reagent: K₂Cr₂O₇/H₂SO₄ (excess); conditions: REFLUX. Orange → green. Why reflux: keeps the aldehyde intermediate (ethanal) in contact with excess oxidant → over-oxidation to carboxylic acid (ethanoic acid). If distillation were used, ethanal would be removed before further oxidation. [2 marks]
Step 2: Ethanoic acid + ethanol → ethyl ethanoate
CH₃COOH + CH₃CH₂OH ⇌ CH₃COOC₂H₅ + H₂O. Catalyst: conc. H₂SO₄ (a few drops); conditions: reflux; ⇌ arrow required. Why H₂SO₄: (1) protonates C=O to activate it for nucleophilic attack by ethanol; (2) dehydrating agent — absorbs H₂O produced, shifting equilibrium right. Why reflux: prevents loss of volatile ethanol (BP 78 °C) and ethyl ethanoate (BP 77 °C). [2 marks]
Ethanol used as BOTH the starting material for Step 1 AND the alcohol for Step 2 — the batch from Step 1 must be split. [1 mark] Intermediate named: ethanoic acid. [1 mark]
(a) Compound A (C₄H₈O): positive Tollens' → aldehyde. Positive Br₂ water → consistent with aldehyde (can reduce Br₂ by oxidation). A = butanal (CH₃CH₂CH₂CHO). [1 mark]
Compound B (C₄H₈O): negative Tollens', negative K₂Cr₂O₇ → cannot be oxidised → ketone. B = butan-2-one (CH₃COCH₂CH₃). [1 mark]
Compound C (C₄H₁₀O): negative Tollens' → not aldehyde. Positive K₂Cr₂O₇ (orange → green) under reflux → oxidised → must be a primary or secondary alcohol; Tollens' negative → not oxidised to aldehyde first → secondary alcohol. C = butan-2-ol (CH₃CH(OH)CH₂CH₃). [1 mark]
(b) CH₃CH₂CH₂CHO + [O] → CH₃CH₂CH₂COOH. Product: butanoic acid. Colour: orange → green. [2 marks]
(c) Butan-2-one is a ketone. The carbonyl carbon (C2) has no H atom attached — it is bonded to CH₃ (C1) and CH₂CH₃ (C3). Oxidation requires removal of H from the carbonyl carbon; without this H, the oxidant cannot react → K₂Cr₂O₇ remains orange (Cr₂O₇²⁻ not reduced to Cr³⁺). [2 marks]
(a) Soap = sodium/potassium salt of a long-chain fatty acid (e.g. sodium stearate, C₁₇H₃₅COO⁻Na⁺). Two regions: (1) long non-polar hydrocarbon chain (C₁₁–C₁₇) = hydrophobic tail — compatible with oils/grease via dispersion forces; (2) ionic carboxylate head group (–COO⁻Na⁺) = hydrophilic head — attracted to water via ion–dipole interactions. This amphiphilic structure enables cleaning. [1 mark]
(b) Above the critical micelle concentration, soap molecules aggregate into spheres: hydrophobic tails point inward (away from water), hydrophilic heads point outward (solvated by water). Negatively charged heads repel adjacent micelles, preventing coalescence. Interior is effectively non-polar. [1 mark]
(c) Hydrophobic tails insert into grease droplet (dispersion forces with non-polar grease); ionic heads remain in water. Mechanical agitation lifts the grease off the surface, surrounded by a soap shell (tails in grease, heads in water) → emulsified micelle. Negatively charged surface repels the (also negative) dish surface → grease stays dispersed → rinsed away. [2 marks]
(d) Hard water contains Ca²⁺ and Mg²⁺ → react with carboxylate head: 2RCOO⁻(aq) + Ca²⁺(aq) → (RCOO)₂Ca(s)↓. Soap is consumed as insoluble scum; unavailable for cleaning; deposits on surfaces. Synthetic detergents (e.g. sodium lauryl sulfate) use a sulfate head group (–OSO₃⁻): calcium/magnesium salts of sulfates ARE soluble → no precipitation in hard water → detergent remains effective. [3 marks: ionic equation + explanation of scum + detergent solution]
(a) Addition polymerisation: monomers must have one C=C per monomer — pi bond opens to form new C–C. No by-product — all atoms appear in polymer. Linkage: C–C only (non-polar, not hydrolysable). Example: ethene → polyethylene [–CH₂–CH₂–]ₙ. [1 mark]
Condensation polymerisation: monomers must have two reactive functional groups each (or one bifunctional monomer). Small molecule by-product released at each bond (usually H₂O). Linkage: ester (–COO–) or amide (–CO–NH–) — polar, hydrolysable. Example: hexane-1,6-diamine + hexanedioic acid → Nylon 6,6. [1 mark] By-product contrast: addition = none; condensation = H₂O per bond. [1 mark]
(b) HDPE: linear (unbranched) chains → close parallel packing → semi-crystalline domains → maximises surface area of contact → strong cumulative dispersion forces → high melting point (~135 °C), rigid at room temperature. [1 mark]
Amorphous PET: benzene ring from terephthalic acid creates steric bulk → prevents regular chain packing when cooled rapidly → amorphous (randomly arranged chains). Amorphous structure has no crystal grain boundaries to scatter light → transparent. (Slowly cooled PET forms crystalline regions → becomes opaque.) [1 mark]
(c) Cellulose: β-1,4 glycosidic bonds — the –OH at C1 of each glucose is in the axial (β) orientation → adjacent rings rotate 180° relative to each other → straight extended chains → H-bonded microfibrils. Human digestive enzymes (amylase) are stereospecific — the active site is shaped for α-geometry only → cannot cleave β-1,4 bonds → cellulose passes through the gut intact as dietary fibre. [3 marks: β/α bond geometry + chain shape consequence + enzyme specificity]
Starch: α-1,4 glycosidic bonds — –OH at C1 in equatorial (α) orientation → chains curve into helical coil → less tightly packed than cellulose microfibrils. Salivary and pancreatic amylase active sites fit the α-1,4 geometry → efficient hydrolysis → glucose released → absorbed and used for energy.
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