Halides and Oxygenated Derivatives

Q1. (4 marks) (a) $CH_3Br$ → $S_N2$; methyl carbon is unhindered, no stable carbocation possible. (2) (b) $(CH_3)_3C{-}Br$ → $S_N1$; forms a stable 3° carbocation, and protic solvent stabilises ions. (2) Why: $S_N2$ needs steric accessibility; $S_N1$ needs a stable carbocation.

Q2. (4 marks) (a) Rate $= k[\text{R-Br}][OH^-]$ — second order overall. (1) (b) Inversion of configuration at the stereocentre (Walden inversion); backside attack. Product is $(R)$-butan-2-ol. (2) (c) Doubling $[OH^-]$ doubles the rate (first order in nucleophile). (1)

Q3. (4 marks) Increasing leaving-group ability: $F^- < Cl^- < Br^- < I^-$ (2) Why: Larger, more polarizable ions are weaker conjugate bases of stronger acids (HI strongest); the best leaving group is the most stable anion → $I^-$. (2)

Q4. (5 marks) Substrate $CH_3CH_2C(CH_3)_2Br$. (a) Products: 2-methylbut-2-ene (more substituted) and 2-methylbut-1-ene (less substituted). (2) (b) Major = 2-methyl-2-butene, by Zaitsev's rule (more substituted, more stable alkene). (2) (c) A strong bulky base favours E2 giving the Hofmann (less substituted) product. (1)

Q5. (5 marks) (a) The C–Cl bond has partial double-bond character due to resonance/conjugation of the halogen lone pair with the ring; the carbon is $sp^2$ and the ring's $\pi$-electrons repel incoming nucleophile → low reactivity. (2) (b) Benzyne (elimination–addition) mechanism; the reactive intermediate is benzyne (dehydrobenzene) containing a strained additional π "triple-bond" between two ring carbons. (3)

Q6. (4 marks) (a) (i) PCC → aldehyde (stops at partial oxidation). (ii) Jones ($CrO_3/H_2SO_4$) → carboxylic acid (full oxidation). (2) (b) Lucas test distinguishes 1°, 2°, 3° alcohols by rate of turbidity (formation of alkyl chloride) with $HCl/ZnCl_2$; tertiary alcohol reacts fastest (immediate turbidity, $S_N1$). (2)

Q7. (4 marks) (a) The phenoxide ion is resonance-stabilised — negative charge delocalised onto the ring (ortho/para carbons), so phenol loses $H^+$ more readily. Ethoxide has no such delocalisation. (2) (b) Kolbe–Schmidt reaction. (2)

Q8. (5 marks) (a) Williamson: sodium alkoxide + alkyl halide ($S_N2$). E.g. $CH_3ONa + CH_3CH_2Br \rightarrow CH_3{-}O{-}CH_2CH_3 + NaBr$. Best combination uses less hindered halide (ethyl bromide + methoxide, or methyl iodide + ethoxide). (3) (b) Excess hot HI cleaves both C–O bonds: products are $CH_3I$ and $CH_3CH_2I$ (with excess HI both give iodides). (2) Why: $I^-$ attacks the less hindered carbon; with excess HI the alcohol formed is also converted to iodide.

Q9. (5 marks) (a) Cannizzaro reaction; products are benzyl alcohol ($C_6H_5CH_2OH$) and sodium benzoate ($C_6H_5COONa$) — disproportionation. (3) (b) Product is 3-hydroxybutanal (aldol); a new C–C bond forms between the α-carbon of one acetaldehyde and the carbonyl carbon of the other. (2)

[
  {"claim":"SN2 rate doubles when nucleophile concentration doubles (first order)",
   "code":"k, N = symbols('k N', positive=True); rate1 = k*N; rate2 = k*(2*N); result = bool(simplify(rate2/rate1 - 2) == 0)"},
  {"claim":"Phenol pKa 10 corresponds to smaller Ka than ethanol pKa 16, i.e. phenol more acidic",
   "code":"Ka_phenol = Rational(10)**(-10); Ka_ethanol = Rational(10)**(-16); result = bool(Ka_phenol > Ka_ethanol)"},
  {"claim":"2-methyl-2-butene (C5H10) is the Zaitsev product molecular formula from C5H11Br elimination",
   "code":"C_substrate, H_substrate, Br = 5,11,1; C_prod, H_prod = 5,10; result = bool(C_prod==C_substrate and H_prod==H_substrate-1)"},
  {"claim":"Leaving group order by conjugate-acid strength: HI acid dissociation largest so I- best leaving group (pKa ordering)",
   "code":"pKa = {'HF':3.2,'HCl':-7,'HBr':-9,'HI':-10}; order = sorted(pKa, key=lambda x: pKa[x]); result = bool(order[0]=='HI')"}
]