Noesis
Level 3 — ProductionGeneral Organic Chemistry (GOC)

General Organic Chemistry (GOC)

⏱ 45 minutes60 marksprintable — key stays hidden on paper

Difficulty: Level 3 — Derive/reason from scratch, explain-out-loud, build arguments from memory Time Limit: 45 minutes Total Marks: 60

Instructions: Answer all questions. Show all reasoning — mechanisms, curved arrows (describe in words where drawing), and rank-orderings must be justified from first principles, not merely stated. Draw structures where asked.

Q1. [12 marks] — Carbocation stability, from scratch

(a) Rank the following carbocations in order of increasing stability, and justify each ranking using the specific electronic effects operating (hyperconjugation counts, resonance structures, inductive): CH3+,CH3CH2+,(CH3)3C+,CH2=CH-CH2+,C6H5-CH2+\text{CH}_3^+,\quad \text{CH}_3\text{CH}_2^+,\quad (\text{CH}_3)_3\text{C}^+,\quad \text{CH}_2=\text{CH-CH}_2^+,\quad \text{C}_6\text{H}_5\text{-CH}_2^+ Count the number of α-C–H hyperconjugative structures for the ethyl and tert-butyl cation, and the number of resonance structures for benzyl. (8)

(b) The neopentyl cation (CH3)3C-CH2+(\text{CH}_3)_3\text{C-CH}_2^+ rearranges. Explain the driving force, name the shift, draw the product cation, and state whether the rearrangement is thermodynamically favourable. (4)

Q2. [10 marks] — IUPAC nomenclature, both directions

(a) Give the correct IUPAC name for each (state locants, cite the priority rule used for numbering): (i) CH3-CH(OH)-CH2-CHO\text{CH}_3\text{-CH(OH)-CH}_2\text{-CHO} (ii) (CH3)2CH-CH2-COOH(\text{CH}_3)_2\text{CH-CH}_2\text{-COOH} (iii) CH2=CH-CH2-C≡CH\text{CH}_2=\text{CH-CH}_2\text{-C}\equiv\text{CH} (6)

(b) Draw the structure of 4-methylpent-3-en-2-one and N,N-dimethylethanamine. (4)

Q3. [12 marks] — Chirality & optical activity, derive the counts

(a) For 2,3-dibromobutane: draw all stereoisomers using wedge/dash, identify each as (R,R)/(S,S)/meso, and explain why the number of distinct stereoisomers is 3 rather than the naive 2n=42^n=4. (6)

(b) A sample of an enantiomerically enriched compound has observed specific rotation [α]obs=+22.0∘[\alpha]_{obs}=+22.0^\circ. The pure (+)(+) enantiomer has [α]=+55.0∘[\alpha]=+55.0^\circ. Compute the enantiomeric excess (ee) and the percentage of each enantiomer present. Show the derivation. (6)

Q4. [10 marks] — Electronic effects, reasoning out loud

(a) Explain, from scratch, why the −NO2-\text{NO}_2 group is deactivating and meta-directing in electrophilic aromatic substitution. Use resonance structures of the intermediate arenium ion to support your argument. (6)

(b) Rank acidity and justify with electronic effects: acetic acid, chloroacetic acid, trichloroacetic acid, formic acid. (4)

Q5. [8 marks] — Mechanism from memory (curved arrows)

Write out the complete mechanism (describe each curved arrow: origin and destination) for the acid-catalysed hydration of propene (CH3CH=CH2+H2O→H+\text{CH}_3\text{CH=CH}_2 + \text{H}_2\text{O} \xrightarrow{\text{H}^+}). Identify heterolysis vs homolysis events, name the intermediate, and use Markovnikov's rule to justify the major product. (8)

Q6. [8 marks] — Isomerism synthesis

For molecular formula C4H10O\text{C}_4\text{H}_{10}\text{O}: (a) Draw and name all structural isomers, classifying each as alcohol or ether. (5) (b) Among your alcohol isomers, identify which one is chiral and assign R/S to its stereocentre. (3)

Answer keyMark scheme & solutions

Q1 (12)

(a) Order of increasing stability: CH3+<CH3CH2+<(CH3)3C+<CH2=CH-CH2+≈C6H5CH2+\text{CH}_3^+ < \text{CH}_3\text{CH}_2^+ < (\text{CH}_3)_3\text{C}^+ < \text{CH}_2=\text{CH-CH}_2^+ \approx \text{C}_6\text{H}_5\text{CH}_2^+

  • CH₃⁺ — no α-C–H, no +I, no resonance ⇒ least stable. (1)
  • CH₃CH₂⁺ — 3 α-C–H hyperconjugative structures; small +I from one methyl. (2 for count + reasoning)
  • (CH₃)₃C⁺ — 9 α-C–H hyperconjugative structures; +I from three methyls ⇒ most stable purely-alkyl cation. (2)
  • Allyl (CH₂=CH–CH₂⁺) — stabilised by resonance: 2 equivalent resonance structures delocalising + charge over two carbons. Resonance > hyperconjugation. (1.5)
  • Benzyl (C₆H₅CH₂⁺) — 5 resonance structures (positive charge delocalised onto ring ortho/para positions + benzylic C); comparable/greater than allyl. (1.5)

Counts required: ethyl = 3, tert-butyl = 9, benzyl resonance structures = 5 (or 4 excluding the parent), allyl = 2. (marks embedded above)

(b) Neopentyl cation is a primary cation. A 1,2-methyl shift (methanide shift) moves a CH₃ (with bonding pair) from the adjacent quaternary carbon to the cationic carbon: (CH3)3C-CH2+⟶(CH3)2C+-CH2CH3(\text{CH}_3)_3\text{C-CH}_2^+ \longrightarrow (\text{CH}_3)_2\text{C}^+\text{-CH}_2\text{CH}_3 Product = 2-methylbutan-2-yl cation (tertiary). (2) Driving force: conversion of unstable 1° cation → stable 3° cation. Thermodynamically favourable (ΔG < 0). (2)

Q2 (10)

(a) (i) CH3CH(OH)CH2CHO\text{CH}_3\text{CH(OH)CH}_2\text{CHO} → 3-hydroxybutanal. Principal characteristic group = –CHO (aldehyde, C1); –OH as hydroxy prefix. (2) (ii) (CH3)2CHCH2COOH(\text{CH}_3)_2\text{CHCH}_2\text{COOH} → 3-methylbutanoic acid. –COOH gets C1; longest chain = 4 C. (2) (iii) CH2=CHCH2C≡CH\text{CH}_2=\text{CHCH}_2\text{C}\equiv\text{CH} → pent-1-en-4-yne. Lowest locants to the set {ene, yne}; when tie, double bond gets lower number. (2)

(b)

  • 4-methylpent-3-en-2-one: CH3-CO-CH=C(CH3)2\text{CH}_3\text{-CO-CH=C(CH}_3)_2 (mesityl oxide). (2)
  • N,N-dimethylethanamine: CH3CH2-N(CH3)2\text{CH}_3\text{CH}_2\text{-N(CH}_3)_2. (2)

Q3 (12)

(a) 2,3-dibromobutane CH3CHBr-CHBr-CH3\text{CH}_3\text{CHBr-CHBr-CH}_3, two identical stereocentres.

  • (2R,3R) and (2S,3S) — a pair of enantiomers (chiral). (2)
  • (2R,3S) = meso — has internal mirror plane / Ci; superimposable on its mirror image (2R,3S ≡ 2S,3R). (2) Total distinct = 3. Naive 22=42^2 = 4 overcounts because the meso form's two "mirror images" are the same molecule (internal symmetry), collapsing two into one. (2)

(b) ee = ratio of observed to maximum rotation: ee=[α]obs[α]pure×100=22.055.0×100=40%\text{ee} = \frac{[\alpha]_{obs}}{[\alpha]_{pure}}\times100 = \frac{22.0}{55.0}\times100 = 40\% (2) ee = (%major − %minor); with %major + %minor = 100: %(+)=100+402=70%,%(−)=100−402=30%.\%(+) = \frac{100+40}{2}=70\%,\qquad \%(-) = \frac{100-40}{2}=30\%. (4)

Q4 (10)

(a) –NO₂ is strongly electron-withdrawing by –I and –M (empty π* / positive N accepts electron density). In EAS the arenium (σ-complex) intermediate carries a + charge. For ortho/para attack, a resonance structure places the + charge on the carbon bearing –NO₂ — placing + adjacent to the already electron-poor N⁺ is highly destabilising. (3) For meta attack, no resonance structure puts + on the substituted carbon, so meta intermediate is less destabilised ⇒ meta-directing. Overall electron withdrawal raises the activation energy relative to benzene ⇒ deactivating. (3)

(b) Acidity increasing: CH3COOH<HCOOH<ClCH2COOH<Cl3CCOOH\text{CH}_3\text{COOH} < \text{HCOOH} < \text{ClCH}_2\text{COOH} < \text{Cl}_3\text{CCOOH}

  • Acetic weakest: CH₃ is +I, destabilises carboxylate. (1)
  • Formic: H has no +I (less electron-donating than CH₃) ⇒ stronger than acetic. (1)
  • Chloroacetic: Cl is –I, stabilises carboxylate. (1)
  • Trichloroacetic: three –I Cl atoms, strongest –I ⇒ most acidic (pKa ≈ 0.7). (1)

Q5 (8)

Acid-catalysed hydration of propene, Markovnikov:

  1. Protonation of alkene: curved arrow from the C=C π bond → to H of H₃O⁺ (or H⁺). H–O bond breaks heterolytically, electrons to O. Forms carbocation. (2)
  2. Markovnikov regiochemistry: H adds to terminal CH₂ so + charge forms on the secondary (central) carbon: CH3-CH+-CH3\text{CH}_3\text{-CH}^+\text{-CH}_3 (2° more stable than 1°). Intermediate = secondary carbocation (isopropyl cation). (2)
  3. Nucleophilic attack: lone pair on O of H₂O → cationic carbon (arrow O→C). Forms oxonium CH3CH(OH2+)CH3\text{CH}_3\text{CH(OH}_2^+)\text{CH}_3. (1)
  4. Deprotonation: water base removes proton from oxonium O (arrow O–H → base) regenerating H⁺ catalyst. Product = propan-2-ol CH3CH(OH)CH3\text{CH}_3\text{CH(OH)CH}_3. (2)

All bond-breaking/forming steps are heterolytic; no homolysis (this is an ionic, not radical, mechanism). (1)

Q6 (8)

(a) C₄H₁₀O isomers (7 total): Alcohols:

  1. Butan-1-ol CH3CH2CH2CH2OH\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH}
  2. Butan-2-ol CH3CH2CH(OH)CH3\text{CH}_3\text{CH}_2\text{CH(OH)CH}_3
  3. 2-methylpropan-1-ol (CH3)2CHCH2OH(\text{CH}_3)_2\text{CHCH}_2\text{OH}
  4. 2-methylpropan-2-ol (CH3)3COH(\text{CH}_3)_3\text{COH} Ethers:
  5. Ethoxyethane (diethyl ether) CH3CH2OCH2CH3\text{CH}_3\text{CH}_2\text{OCH}_2\text{CH}_3
  6. 1-methoxypropane CH3OCH2CH2CH3\text{CH}_3\text{OCH}_2\text{CH}_2\text{CH}_3
  7. 2-methoxypropane CH3OCH(CH3)2\text{CH}_3\text{OCH(CH}_3)_2 (5 marks: ~4 alcohols + 3 ethers, correct naming/classification)

(b) Only butan-2-ol has a stereocentre (C2 bonded to OH, H, CH₃, C₂H₅ — four different groups). Priorities: OH > CH₂CH₃ > CH₃ > H. Both (R) and (S) exist; the enantiomers are (R)-butan-2-ol and (S)-butan-2-ol. (3)

[
  {"claim":"tert-butyl cation has 9 alpha C-H hyperconjugative structures", "code":"result = (3*3 == 9)"},
  {"claim":"ee = 22/55*100 = 40 percent", "code":"result = (Rational(22,55)*100 == 40)"},
  {"claim":"percent plus enantiomer = 70, minus = 30 for ee 40", "code":"ee=40; plus=(100+ee)/2; minus=(100-ee)/2; result = (plus==70 and minus==30)"},
  {"claim":"2,3-dibromobutane gives 3 distinct stereoisomers not 4", "code":"naive=2**2; distinct=naive-1; result = (distinct==3)"}
]