Halides and Oxygenated Derivatives

Question 1 (12)

(a) (6)

• SN2 with CN⁻/DMSO (3): backside attack, single concerted step, Walden inversion of configuration. (S)-2-bromobutane → (R)-2-methylbutanenitrile (inversion; note: priority change may alter R/S label but the point is inversion of spatial configuration, giving a single optically active enantiomer). 2 marks mechanism/inversion, 1 mark for stating optically active product.
• SN1 solvolysis (3): rate-determining ionisation to a planar sp² secondary carbocation, then nucleophile attacks from both faces → racemic mixture (racemisation, ~50:50), optically inactive product. 2 marks carbocation + both-face attack, 1 mark racemic/inactive.

(b) (3): SN2 rate $= k[\text{RBr}][\text{Nu}]$ (second-order, bimolecular); SN1 rate $= k[\text{RBr}]$ (first-order, unimolecular). Doubling [Nu] doubles SN2 rate because Nu is in the rate-determining TS; it does not change SN1 rate because Nu enters after the slow ionisation step. 2 marks rate laws, 1 mark explanation.

(c) (3): Protic MeOH H-bonds to and solvates the nucleophile (CN⁻), reducing its reactivity → slows SN2; MeOH also stabilises the developing carbocation/charge-separated TS, favouring SN1. For a secondary substrate the shift is toward more SN1 character and overall slower substitution by the free anion. Award for correct direction + solvation reasoning.

Question 2 (10)

(a) (5):

• KOEt/EtOH → small base → Zaitsev: more substituted alkene 2-methylbut-2-ene major. (2)
• KOtBu → bulky base → Hofmann: less substituted 2-methylbut-1-ene major. (2)
• Explanation: bulky base cannot easily reach internal (more hindered) β-H, so it abstracts the more accessible terminal β-H, giving the less-substituted (Hofmann) product; small base gives the thermodynamically more stable Zaitsev alkene. (1)

(b) (5): E2 is concerted; the C–H and C–LG σ*/σ orbitals must be periplanar to overlap, and anti-periplanar (180°) is preferred (staggered, lower strain, better orbital alignment) over syn. (3) Newman projection: H and Br on opposite sides, dihedral 180°. (1) This stereospecificity means each diastereomer gives a defined alkene geometry (E or Z fixed by which H is anti to the leaving group). (1)

Question 3 (10)

(a) (5):

1. $\text{NH}_2^-$ deprotonates ortho C–H. (1)
2. Loss of Cl⁻ → benzyne (extra π "bond" in plane from sideways sp² overlap). (2)
3. $\text{NH}_2^-$ adds to one benzyne carbon; (1)
4. resulting aryl anion protonated by $\text{NH}_3$ → aniline. (1)

(b) (3): The symmetric benzyne intermediate makes both triple-bond carbons equivalent; nucleophile adds equally to the labelled and adjacent carbon → ~50:50 scrambling of the NH₂ position relative to the ¹⁴C label. This proves a symmetric (benzyne) intermediate, ruling out direct displacement (which would keep NH₂ on the labelled carbon 100%).

(c) (2): (i) Strong electron-withdrawing groups (e.g. –NO₂) ortho/para to the halide — stabilise the anionic Meisenheimer intermediate by resonance. (ii) A good leaving group and the negative charge delocalised onto EWG make addition fast. (1 each).

Question 4 (12)

(a) (6):

1. Base removes α-H of acetone → enolate. (1)
2. Enolate adds to benzaldehyde carbonyl carbon → alkoxide (aldol addition). (1)
3. Protonation → β-hydroxy ketone. (1)
4. E1cb dehydration: α-deprotonation, loss of OH⁻ → benzalacetone (4-phenylbut-3-en-2-one). (2)

• Out loud: the newly formed double bond is conjugated with both the carbonyl and the aromatic ring; this extended conjugation is thermodynamically very stable, so under the basic condensation conditions dehydration is favourable and the enone is isolated. (1)

(b) (4): No α-H, so no enolate. One PhCHO's carbonyl is attacked by OH⁻ → tetrahedral alkoxide; this delivers hydride to a second PhCHO. (2) Hydride donor = the alkoxide adduct (oxidised to benzoic acid/benzoate); hydride acceptor = second benzaldehyde (reduced to benzyl alcohol). Products: benzoate (PhCOO⁻) + benzyl alcohol (PhCH₂OH). (2)

(c) (2): Acetaldehyde has α-hydrogens, so under base it preferentially undergoes aldol self-condensation (faster) rather than Cannizzaro; Cannizzaro requires the absence of α-H.

Question 5 (9)

(a) (5): Increasing acidity (higher $K_a$, lower p$K_a$): ethanol (16) < phenol (10) < p-nitrophenol (7.15) < acetic acid (4.76). (2 for order) Justification (3): ethanol's alkoxide has no delocalisation → weakest acid. Phenoxide is resonance-stabilised into the ring. p-Nitrophenoxide is further stabilised because –NO₂ (para) delocalises the negative charge onto oxygen of nitro group (extra resonance) → more acidic than phenol. Acetate has charge delocalised over two equivalent electronegative oxygens (most effective stabilisation among these) → strongest.

(b) (2): Phenoxide charge is delocalised over ortho/para ring carbons (3 additional resonance structures), lowering the anion energy; ethoxide has localised charge on O with no resonance. Greater conjugate-base stabilisation → larger $K_a$ (~$10^6$ ratio ≈ $16-10 = 6$ p$K_a$ units).

(c) (2): $K_a = 10^{-4.76} = 1.74\times10^{-5}$. (1) $[\text{H}^+]=\sqrt{K_a C}=\sqrt{1.74\times10^{-5}\times0.10}=1.32\times10^{-3}$; pH $=-\log(1.32\times10^{-3})=2.88$. (1)

Question 6 (7)

(a) (5): 1,2-addition = nucleophile adds directly to the carbonyl carbon (C=O) → allylic alcohol/adduct at carbonyl. 1,4-addition (conjugate) = nucleophile adds to the β-carbon; enolate forms and tautomerises → saturated carbonyl product. (2)

• (i) Soft/stabilised nucleophile (cuprate, malonate enolate) → 1,4-addition (conjugate control, thermodynamic). (1.5)
• (ii) Hard organolithium (RLi) → 1,2-addition (kinetic, attacks hard carbonyl carbon). (1.5)

(b) (2): Michael addition. 1,4 favoured because it preserves the strong C=O bond (only the weaker C=C π is sacrificed), giving a more stable saturated carbonyl product (thermodynamic sink). (1 name, 1 reason).

[
  {"claim":"Ka of acetic acid = 10^-4.76 ≈ 1.74e-5","code":"Ka=10**(-4.76); result = abs(Ka-1.74e-5) < 1e-6"},
  {"claim":"pH of 0.10 M acetic acid ≈ 2.88","code":"Ka=10**(-4.76); C=Rational(1,10); H=sqrt(Ka*float(C)); pH=-log(H,10); result = abs(float(pH)-2.88) < 0.02"},
  {"claim":"Phenol/ethanol pKa difference of 6 units = 10^6 in Ka ratio","code":"ratio=10**(16-10); result = ratio == 1000000"},
  {"claim":"SN1 half-face attack gives ~50:50 racemic (0.5 each)","code":"frac=Rational(1,2); result = frac == Rational(1,2)"}
]