Halides and Oxygenated Derivatives

Question 1 (12)

(a) Conditions: dilute base, highly polar protic solvent (aqueous ethanol) → favours ionisation → $\text{S}_\text{N}1$ (secondary substrate can go either way, but low $[\text{OH}^-]$ + high water content favours $\text{S}_\text{N}1$). (2) Mechanism proceeds via a planar carbocation → nucleophile attacks both faces → racemisation (product is $(\pm)$-butan-2-ol, roughly 50:50). (2)

(b) High concentration of strong, small base $\text{NaOCH}_3$ + secondary substrate → E2 dominates over substitution. (2) Zaitsev (more substituted alkene) major product = but-2-ene ($\text{CH}_3\text{-CH=CH-CH}_3$), predominantly trans (E). Minor = but-1-ene. (2)

(c) $\text{S}_\text{N}2$ reactivity order: $I > Br > F$. (1) Leaving-group ability increases down the group ($I^-$ is the most stable/weakest base, best leaving group). C–I bond is weakest, easiest to break; C–F strongest. (2) Since rate-determining $\text{S}_\text{N}2$ step involves C–X breaking, iodide reacts fastest; fluoride is essentially unreactive. (1)

Question 2 (12)

(a) The C–Cl bond has partial double-bond character from lone-pair conjugation of Cl into the ring, shortening/strengthening it. The ring $\pi$ system and $\text{sp}^2$ carbon repel incoming nucleophiles, and there is no low-energy backside pathway. Hence very low reactivity. (3)

(b) Benzyne (elimination–addition) mechanism. (1) $\text{NaNH}_2$ removes the ortho H and $\text{Cl}^-$ leaves, forming a symmetric benzyne triple bond spanning the labelled C and its ortho neighbour. (2) Nucleophile ($\text{NH}_2^-$) adds to either carbon of the (nearly) symmetric triple bond with equal probability → $\text{NH}_2$ ends up 50:50 on labelled and adjacent carbon, explaining the isotopic scrambling. (2)

(c) The two $\text{NO}_2$ groups (ortho + para to Cl) are strong electron-withdrawing groups that stabilise the negative charge of the intermediate. (1) Mechanism = $\text{S}_\text{N}\text{Ar}$ (addition–elimination). (1) Methoxide adds to give the resonance-stabilised Meisenheimer complex (carbanion delocalised onto the nitro oxygens), then $\text{Cl}^-$ is expelled. (2)

Question 3 (12)

(a) Increasing acidity: $\text{ethanol} < \text{p-methoxyphenol} < \text{phenol} < \text{p-nitrophenol}$ (2)

• Ethanol weakest: alkoxide has no resonance stabilisation. (1)
• Phenoxide is resonance-stabilised (charge delocalised into ring) → phenols far more acidic. (1)
• $p\text{-OCH}_3$ donates electron density (resonance) → destabilises phenoxide → less acidic than phenol; $p\text{-NO}_2$ withdraws by resonance → stabilises anion → most acidic. (1)

(b) $\frac{K_a(\text{phenol})}{K_a(\text{ethanol})} = 10^{-(10-16)} = 10^{6}.$ (2) Phenol is about one million times more acidic than ethanol — a large difference driven by resonance stabilisation of phenoxide. (1)

(c) Product = salicylic acid (2-hydroxybenzoic acid): benzene ring bearing $\text{-OH}$ and, ortho to it, $\text{-COOH}$. (2) Phenoxide ion is a far better nucleophile than neutral phenol (higher electron density at ortho/para ring carbons), enabling attack on the electrophilic $\text{CO}_2$; neutral phenol is too unreactive. (2)

Question 4 (12)

(a) Disconnection routes:

• Route A: $(\text{CH}_3)_3\text{C-O}^-$ (tert-butoxide) + $\text{CH}_3\text{CH}_2\text{-Br}$ (ethyl bromide). (2)
• Route B: $\text{CH}_3\text{CH}_2\text{-O}^-$ (ethoxide) + $(\text{CH}_3)_3\text{C-Br}$ (tert-butyl bromide). (2)

(b) Route A works (good yield). The alkyl halide (primary ethyl bromide) undergoes clean $\text{S}_\text{N}2$; tert-butoxide, though bulky/basic, attacks an unhindered primary carbon. (2) Route B fails: tert-butyl bromide is a 3° substrate — $\text{S}_\text{N}2$ impossible (steric), and the strong base ethoxide causes E2 elimination to isobutylene instead of ether. (2)

(c) Heating tert-butyl ethyl ether with excess $\text{HI}$: protonation of the ether O, then cleavage. Products = tert-butyl iodide (or isobutylene) + ethanol → then ethyl iodide (excess HI). (2) The bond that breaks is the C(tert-butyl)–O, because that carbon forms a stable tertiary carbocation ($\text{S}_\text{N}1$-like), whereas the ethyl side would need a much less stable primary cation. So the tert-butyl group leaves as the cation (→ tert-butyl iodide), leaving ethanol. (2)

Question 5 (12)

(a) Reaction = Cannizzaro (benzaldehyde has no α-H). (1) Products: benzyl alcohol ($\text{C}_6\text{H}_5\text{CH}_2\text{OH}$, reduced) + sodium benzoate ($\text{C}_6\text{H}_5\text{COO}^-\text{Na}^+$, oxidised). (2) One aldehyde is oxidised to the carboxylate, the other reduced to the alcohol (disproportionation). (1)

(b) Reaction = Claisen–Schmidt (crossed aldol condensation). (1) Product = cinnamaldehyde, $\text{C}_6\text{H}_5\text{-CH=CH-CHO}$. (2) Benzaldehyde has no α-hydrogen, so it cannot form an enolate — it can only act as the electrophile. Acetaldehyde provides the enolate/nucleophile. Cross-attack of acetaldehyde enolate on benzaldehyde dominates; the resulting product is conjugated (aromatic ring + C=C–C=O) and dehydrates readily, driving the equilibrium. (2)

(c) With $\text{R}_2\text{CuLi}$, 1,4-(conjugate/Michael) addition dominates. (1) Cuprates are soft nucleophiles that prefer the softer $\beta$-carbon of the conjugated system, giving the more stable enolate and ultimately the saturated aldehyde; hard organolithiums/Grignards would give 1,2-addition. (2)

[
  {"claim":"Ka(phenol)/Ka(ethanol) = 10^6 from pKa 10 and 16",
   "code":"pKa_phenol=10; pKa_ethanol=16; ratio=10**(-(pKa_phenol-pKa_ethanol)); result = ratio==10**6"},
  {"claim":"SN2 leaving-group order I>Br>F corresponds to decreasing C-X bond strength; check bond energies kJ/mol CF>CBr>CI",
   "code":"CF=485; CBr=285; CI=213; result = (CF>CBr) and (CBr>CI)"},
  {"claim":"Benzyne symmetric addition gives 50:50 label distribution (probability 0.5 each)",
   "code":"p_labelled=0.5; p_ortho=0.5; result = abs(p_labelled+p_ortho-1)<1e-9 and p_labelled==0.5"},
  {"claim":"pKa difference of 6 units means 1e6 fold Ka ratio (log10)",
   "code":"import sympy as sp; diff=6; result = sp.simplify(sp.log(10**diff,10))==diff"}
]