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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