Hydrocarbons

Section A (1 mark each)

Q1 — (c) Butane. Wurtz couples two ethyl groups: $2CH_3CH_2Br + 2Na \rightarrow CH_3CH_2CH_2CH_3 + 2NaBr$. (1)

Q2 — (b) Ethane. Anodic decarboxylation gives $CH_3^{\bullet}$ radicals which dimerise: $2CH_3COO^- \rightarrow 2CH_3^{\bullet} + 2CO_2 \rightarrow C_2H_6$. (1)

Q3 — (b). Initiation = homolysis of $Cl_2$ by light forming two chlorine radicals. (a)/(c) are propagation; (d) is termination. (1)

Q4 — (c) Anti. Anti places the two methyls $180°$ apart → minimum steric strain → lowest energy. (1)

Q5 — (c) Cyclopentane. Baeyer assumed planar rings; deviation from $109.5°$ is smallest for the pentagon ($108°$), so he predicted it as least strained. (1)

Q6 — (b) 1-bromopropane. Peroxide effect (Kharasch) gives anti-Markovnikov addition via free-radical mechanism, Br on terminal carbon. (1)

Q7 — (b) propan-1-ol. Hydroboration–oxidation gives anti-Markovnikov, syn addition of $H$–$OH$; OH on less substituted carbon. (1)

Q8 — (c) Cyclopentadienyl anion. 6 $\pi$ electrons, planar, cyclic → $4n+2$ with $n=1$. Others have $4n$ electrons (antiaromatic/non-aromatic). (1)

Q9 — (a) two molecules of ethanal. $CH_3CH=CHCH_3$ cleaves at the double bond; reductive workup gives aldehydes: $2 CH_3CHO$. (1)

Q10 — (b) Deactivating, meta director. $-NO_2$ withdraws electrons (–I, –M), deactivates the ring and directs incoming electrophile to meta. (1)

Section B (1 mark per correct pair) [6]

Q11.

• (i) → (R) — hydrogenation gives alkane
• (ii) → (P) — Markovnikov hydration of alkyne gives methyl ketone
• (iii) → (Q) — cold dilute $KMnO_4$ = syn-dihydroxylation (cis-diol)
• (iv) → (S) — $Br_2/CCl_4$ gives vicinal dibromide (anti addition)
• (v) → (U) — hot conc. $KMnO_4$ cleaves to carbonyls/acids
• (vi) → (T) is a distractor; correct: hydroboration gives anti-Markovnikov alcohol. Accept (vi) → (T) only if labelled "alcohol"; strictly (vi) → anti-Markovnikov alcohol. Award 1 for reasonable mapping to alcohol product.

(Award 1 mark each for i–v; option (vi) mapped to an alcohol product = 1 mark; max 6.)

Section C (1 verdict + 1 justification = 2 each) [14]

Q12 — True. The $sp$ carbon of terminal alkynes has high s-character (50%), holding the lone pair closer to the nucleus → more stable conjugate base ($pK_a \approx 25$ vs alkene ~44, alkane ~50). (2)

Q13 — False. Zaitsev's rule predicts the more substituted alkene, i.e. but-2-ene, as major — not but-1-ene. (2)

Q14 — True. In the equatorial position the substituent avoids 1,3-diaxial steric interactions, so the equatorial chair is lower in energy. (2)

Q15 — False. Pyridine's nitrogen lone pair lies in an $sp^2$ orbital in the ring plane, not in the $\pi$ system; aromaticity comes from the 6 $\pi$ electrons of the ring, not from the lone pair. (2)

Q16 — True. The acylium ion $R\text{-}C^+\!=\!O$ is resonance-stabilised (no rearrangement), and the deactivating ketone product resists further substitution (no polyacylation). (2)

Q17 — True. The proton adds to give the more stable (more substituted) carbocation; nucleophile ($Br^-$) then attaches to that carbon → Markovnikov product. (2)

Q18 — True. $Br^{\bullet}$ is less reactive and more selective; the higher transition-state energy makes it discriminate strongly for the weaker (3° > 2° > 1°) C–H bond. (2)

[
  {"claim":"Wurtz of 2 ethyl halides gives a 4-carbon alkane (butane)","code":"carbons = 2*2; result = (carbons == 4)"},
  {"claim":"Kolbe of 2 acetate (2 carbons each after losing CO2 gives 1 C radical) yields a 2-carbon alkane","code":"c_per_radical = 1; product_c = 2*c_per_radical; result = (product_c == 2)"},
  {"claim":"Cyclopentadienyl anion has 6 pi electrons satisfying 4n+2 with n=1","code":"pi = 6; n = (pi-2)/4; result = (n == 1)"},
  {"claim":"Ozonolysis of symmetric 2-butene (4C) gives two 2-carbon carbonyl fragments","code":"total_c = 4; frag = total_c/2; result = (frag == 2)"}
]