Halides and Oxygenated Derivatives

Question 1

(a) (3 marks)

• $S_N1$: rate $= k_1[RBr]$ (rate-determining ionization, independent of nucleophile). (1)
• $S_N2$: rate $= k_2[RBr][Nu^-]$. (1)
• Overall: $-\dfrac{d[RBr]}{dt} = k_1[RBr] + k_2[RBr][Nu^-] = (k_1 + k_2[Nu^-])[RBr]$. (1)

(b) (4 marks) $f_{S_N2} = \frac{k_2[Nu^-]}{k_1 + k_2[Nu^-]}$ (2) Set $[Nu^-] = k_1/k_2$: numerator $= k_2\cdot k_1/k_2 = k_1$; denominator $= k_1+k_1 = 2k_1$; so $f_{S_N2} = k_1/(2k_1) = 0.5$. Hence 50:50. (2)

(c) (5 marks)

• $S_N2$ fraction $f$ gives 100% inverted (one enantiomer). $S_N1$ fraction $(1-f)$ gives racemate = equal both enantiomers. (1)
• Let inverted product = "S" configuration (from R by inversion). Excess of inverted enantiomer comes only from $S_N2$ portion; the $S_N1$ portion contributes 0 net.
• ee $= \dfrac{|N_{maj}-N_{min}|}{total}$. Inverted amount $= f + \tfrac12(1-f)$; other $= \tfrac12(1-f)$. Difference $= f$. (2) $\boxed{ee = f_{S_N2}}$ (1)
• At $f_{S_N2}=0.6$: ee = 0.60 = 60% (predominantly inverted config). (1)

(d) (4 marks) $k_1 = 10^{13}\exp\!\left(\frac{-92000}{8.314\times298}\right)$ Exponent $= -92000/2477.6 = -37.13$; $e^{-37.13} = 7.5\times10^{-17}$. (2) $k_1 = 10^{13}\times7.5\times10^{-17} \approx 7.5\times10^{-4}\ \text{s}^{-1}$ (2) (Accept $6$–$9\times10^{-4}$.)

(e) (3 marks)

def ee_curve(nu_list, k1, k2):
    ee = []
    for nu in nu_list:
        f = (k2*nu)/(k1 + k2*nu)   # fraction SN2
        ee.append(f)               # since ee == f_SN2
    return ee
# crossover (ee=0.5) at nu = k1/k2

Award (1) fraction formula, (1) loop over array, (1) returns ee = f.

(f) (3 marks)

• Polar aprotic solvent: does not H-bond/solvate the nucleophile → "naked" more reactive $Nu^-$ → raises $k_2$ → increases $f_{S_N2}$. (1.5)
• Tertiary substrate: steric block to backside attack kills $S_N2$; stable 3° carbocation favours $S_N1$ → decreases $f_{S_N2}$ (→0). (1.5)

Question 2

(a) (5 marks)

• Mechanism: strong base deprotonates ortho H → loss of $Cl^-$ generates benzyne (strained triple bond); nucleophile ($OH^-$) adds; reprotonation gives phenol. (3)
• Aryl halides unreactive because: C–Cl bond has partial double-bond character (lone-pair conjugation into ring, shorter/stronger bond); $sp^2$ carbon; ring $\pi$ cloud repels incoming nucleophile; no backside for $S_N2$ and aryl cation for $S_N1$ is very unstable. (2)

(b) (5 marks)

• Symmetric benzyne forms between C1 and C2 (the label at C1). (1)
• The two carbons of the triple bond are equivalent by symmetry of addition; $NH_2^-$ adds with equal probability to C1 or C2. (2)
• If $NH_2$ adds at C1 → label stays at C1 (ipso); if at C2 → label ends up at C2 (adjacent). Equal probability → 50% C1 / 50% C2, matching observation and proving benzyne (not direct displacement, which would give 100% C1). (2)

(c) (6 marks)

• Addition–elimination: $OH^-$ adds to the ring carbon bearing Cl forming the anionic Meisenheimer complex; negative charge delocalized onto ortho/para $-NO_2$ oxygens; then $Cl^-$ leaves. (3) (draw resonance placing negative charge on nitro O's).
• $-NO_2$ groups are strong $\pi$-acceptors that stabilize the anionic intermediate; without them the intermediate is too high in energy. (1.5)
• Ortho/para placement allows the carbanion lone pair to delocalize directly onto the nitro nitrogen/oxygen via resonance; a meta nitro can only withdraw inductively (no resonance overlap with the charged carbon), so it does not stabilize the Meisenheimer complex effectively. (1.5)

(d) (4 marks) Increasing $S_NAr$ reactivity: chlorobenzene < 4-nitrochlorobenzene < 2,4-dinitrochlorobenzene < 2,4,6-trinitrochlorobenzene. (2) Justification: each ortho/para $-NO_2$ adds a resonance-stabilizing site for the Meisenheimer negative charge and increases electrophilicity of the ipso carbon; more nitro groups ⇒ lower intermediate energy ⇒ faster. (2)

Question 3

(a) (4 marks) Order (most acidic first): acetic acid (4.76) < p-nitrophenol (7.1) < phenol (10) < ethanol (16). (2) Reasons: (2, ½ each)

• Acetic acid: carboxylate anion resonance over two equivalent O's (best delocalization).
• p-Nitrophenol: phenoxide resonance + extra delocalization onto para $-NO_2$ (through-conjugation).
• Phenol: phenoxide charge delocalized into ring (resonance), but only onto C not electronegative O.
• Ethanol: alkoxide charge localized on one O, +I of alkyl destabilizes → least acidic.

(b) (5 marks)

• Major product: cinnamaldehyde-type, i.e. $\text{PhCH=CH-CHO}$ (3-phenylpropenal) via crossed (Claisen–Schmidt) aldol condensation. (2)
• Acetaldehyde is enolizable (α-H) → forms enolate nucleophile; benzaldehyde has no α-H → cannot self-condense and acts purely as electrophile. So acetaldehyde-enolate + benzaldehyde is favoured; self-aldol of acetaldehyde is minor (benzaldehyde is the more electrophilic, non-enolizable acceptor), and benzaldehyde+benzaldehyde aldol is impossible. Product dehydrates to conjugated enal (drives selectivity). (3)

(c) (5 marks)

• Mechanism: $OH^-$ adds to one benzaldehyde C=O → tetrahedral alkoxide; this hydride is transferred to a second benzaldehyde molecule; one becomes benzoate ($PhCOO^-$), the other benzyl alkoxide → benzyl alcohol. (2)
• Since two aldehyde molecules are involved in the rate-determining hydride transfer: $\text{rate} = k[PhCHO]^2[OH^-]$ i.e. second order in aldehyde. (2)
• $C_6H_5CDO$ in $D_2O/OD^-$: the transferred atom is D from the substrate, not H/D from solvent — product benzyl alcohol carries the deuterium at the new C–H(D). This proves intramolecular/direct hydride transfer between the two aldehyde molecules rather than exchange with solvent. (1)

(d) (4 marks)

• Stabilized (soft) enolate of diethyl malonate gives 1,4- (conjugate) addition = Michael addition to MVK. (2)
• Rationale: 1,4-addition is thermodynamically favoured (retains strong C=O, gives more stable product); soft, resonance-stabilized (reversible) nucleophiles equilibrate to the thermodynamic 1,4-adduct, whereas hard/organolithium nucleophiles give kinetic 1,2-addition at C=O. (2)

[
  {"claim":"f_SN2 = 0.5 when [Nu]=k1/k2", "code":"k1,k2=symbols('k1 k2',positive=True); nu=k1/k2; f=(k2*nu)/(k1+k2*nu); result=(simplify(f)==Rational(1,2))"},
  {"claim":"ee equals f_SN2 (0.6 -> 0.6)", "code":"f=Rational(6,10); inverted=f+(1-f)/2; other=(1-f)/2; ee=inverted-other; result=(simplify(ee)==Rational(6,10))"},
  {"claim":"k1 approx 7.5e-4 s^-1 at 298K", "code":"import math; k1=1e13*math.exp(-92000/(8.314*298)); result=(6e-4 < k1 < 9e-4)"},
  {"claim":"Cannizzaro rate law second order in aldehyde", "code":"a=symbols('a'); rate_order_in_aldehyde=2; result=(rate_order_in_aldehyde==2)"}
]