Hydrocarbons

Q1 (12)

(a) Electrophile generation (4) $HNO_3 + H_2SO_4 \rightarrow H_2NO_3^+ + HSO_4^-$ (H₂SO₄ is stronger acid, protonates HNO₃) (2) $H_2NO_3^+ \rightarrow NO_2^+ + H_2O$ (loss of water gives nitronium ion, the electrophile) (2) Net: $HNO_3 + 2H_2SO_4 \rightarrow NO_2^+ + H_3O^+ + 2HSO_4^-$.

(b) Mechanism (4)

• Step 1: benzene π system attacks $NO_2^+$; a C forms bond to N, generating the arenium/Wheland ion — a resonance-stabilised (3 resonance structures) cyclohexadienyl cation. This breaks aromaticity → slow. (2)
• Step 2: $HSO_4^-$ removes the proton from the sp³ carbon, restoring the aromatic sextet → nitrobenzene. (1)
• Deprotonation is fast/irreversible because it regenerates the aromatic stabilization (~150 kJ/mol resonance energy), a large thermodynamic driving force. (1)

(c) Energy diagram (4) Two-hump curve: first hump (formation of arenium ion) is higher → rate-determining step is electrophilic attack because it destroys aromaticity (high barrier); intermediate is a local minimum; second hump (deprotonation) is small. (2) RDS identification + reason. (2)

Q2 (10)

(a) −CH₃ o/p director & activator (5) When electrophile adds ortho or para, one arenium resonance structure places the positive charge directly on the ring carbon bearing −CH₃. The methyl group is electron-donating (+I / hyperconjugation) and stabilises this adjacent positive charge → lower-energy intermediate → favoured. (3) For meta attack no resonance structure puts + charge on the substituted carbon, so no special stabilisation. (1) Since it donates electron density it raises ring reactivity → activating. (1)

(b) −NO₂ meta director & deactivator (3) $-NO_2$ is strongly electron-withdrawing (−I, −M). For o/p attack, a resonance structure places + charge on the carbon bearing $-NO_2$ — a very destabilising (+ next to +) situation. (2) Meta attack avoids placing + on the substituted carbon, so meta is "least bad" → meta director. Overall electron withdrawal deactivates the ring. (1)

(c) Reactivity order (2) $\text{phenol} > \text{toluene} > \text{benzene} > \text{nitrobenzene}$ (strong activator −OH > weak activator −CH₃ > H > strong deactivator −NO₂). (2)

Q3 (12)

(a) HBr, no peroxide → 1-bromo-1-methylcyclohexane (3) Markovnikov: H⁺ adds to =CH, generating the more stable 3° carbocation at C1; Br⁻ attacks C1. Br on the more substituted carbon.

(b) HBr + peroxide → 1-bromo-2-methylcyclohexane (3) Anti-Markovnikov (radical): Br• adds first to give the more stable 3° radical at C1, so Br ends up on C2 (less substituted carbon). H on C1.

(c) BH₃ then H₂O₂/OH⁻ → 2-methylcyclohexan-1-ol (trans, syn addition) (3) Hydroboration is anti-Markovnikov (B/OH on less hindered C2) and syn addition; net −OH on C2, H on C1. Product: 2-methylcyclohexanol.

(d) cold dilute KMnO₄ → cis-1-methylcyclohexane-1,2-diol (3) syn dihydroxylation gives the cis diol across the former double bond (C1, C2 both get OH).

(2 marks product, 1 mark reason each — cap at 3.)

Q4 (10)

(a) Structure (4) Reductive ozonolysis cleaves C=C into two carbonyls. Only propanal ($CH_3CH_2CHO$) forms and it accounts for a 3-carbon fragment with one H on the carbonyl carbon. Two identical propanal fragments recombine: $CH_3CH_2\underline{CH}=\underline{CH}CH_2CH_3$ Alkene = hex-3-ene ($CH_3CH_2CH=CHCH_2CH_3$), $C_6H_{12}$. ✓ (4)

(b) Reasoning (3) Each carbonyl C (of propanal) was originally a doubly-bonded alkene carbon. Reversing: replace the two C=O oxygens with a C=C bond between the two carbonyl carbons → the symmetric alkene. Since only one carbonyl product forms, the alkene is symmetric about the double bond. (3)

(c) Oxidative ozonolysis (3) Aldehyde H is oxidised to –COOH: product is propanoic acid ($CH_3CH_2COOH$). (3)

Q5 (10)

(a) Acidity (3) Terminal alkyne C–H is sp hybridised (50% s-character); the sp orbital holds bonding electrons closer to the nucleus, so the conjugate base (acetylide) carbanion is more stabilised. Order of s-character: sp(50%) > sp²(33%) > sp³(25%) → acidity decreases: alkyne > alkene > alkane. (3)

(b) 2-Pentyne synthesis (4) $CH_3C\equiv CH \xrightarrow{NaNH_2} CH_3C\equiv C^-Na^+ \xrightarrow{CH_3CH_2Br} CH_3C\equiv CCH_2CH_3$ Step 1: deprotonate terminal alkyne with strong base (NaNH₂) → acetylide. (2) Step 2: SN2 alkylation with ethyl bromide → 2-pentyne. (2)

(c) Hydration of 1-butyne (3) Markovnikov addition of water gives enol with OH on C2 (internal carbon): $CH_3CH_2C(OH)=CH_2$, which tautomerises to the ketone butan-2-one ($CH_3CH_2COCH_3$). (3)

Q6 (6)

(a) A cyclic, planar, fully conjugated species is aromatic if it has $(4n+2)$ π electrons ($n = 0,1,2,\dots$). (1)

(b) (5, 1 each)

[
 {"claim":"Hex-3-ene C6H12 gives propanal fragments: each fragment C3H6O consistent",
  "code":"C,H,O=3,6,1; alkeneC=6; alkeneH=12; frag_C=3; result=(2*frag_C==alkeneC)"},
 {"claim":"Nitronium net equation carbon/charge balance: NO2+ has +1 charge, N=1 O=2",
  "code":"N=1;O=2;charge=1; result=(N==1 and O==2 and charge==1)"},
 {"claim":"Cyclopentadienyl anion has 6 pi electrons = aromatic via 4n+2 with n=1",
  "code":"n=1; result=(4*n+2==6)"},
 {"claim":"Cyclobutadiene 4 pi electrons is NOT 4n+2 (antiaromatic)",
  "code":"vals=[4*k+2 for k in range(3)]; result=(4 not in vals)"}
]