Hydrocarbons

Q1 (10)

(a) Degree of unsaturation of $C_5H_8 = (2\times5+2-8)/2 = 2$. [1]

• Full hydrogenation → $n$-pentane ⇒ unbranched 5-carbon skeleton. [1]
• Ag precipitate with ammoniacal $AgNO_3$ ⇒ terminal alkyne (acidic ≡C–H)... but 2 degrees of unsaturation with a straight chain and two ozonolysis fragments containing C=C bonds suggests a diene, not an alkyne.

Reconciling: The Ag test indicates terminal alkyne would give 1 degree from triple bond = 2 degrees total (a triple bond counts as 2). But ozonolysis giving HCHO + malondialdehyde ($OHC-CH_2-CHO$) accounts for a chain broken at two C=C double bonds: fragments total carbons = 1 (HCHO) + 3 (malondialdehyde) = 4... plus the required 5th carbon.

Correct structure: penta-1,4-diene, $CH_2{=}CH{-}CH_2{-}CH{=}CH_2$. [2] Ozonolysis cleaves both terminal double bonds:

• Each terminal $=CH_2$ → $HCHO$ (two molecules, "only HCHO" as one product type). [1]
• The central $CH-CH_2-CH$ unit → $OHC-CH_2-CHO$ (malondialdehyde). ✓ [1]

Note: Penta-1,4-diene has no terminal ≡CH, so it would NOT give a silver precipitate. The consistent structure satisfying all four clues (Ag test + malondialdehyde) is pent-1-en-4-yne, $CH{\equiv}C{-}CH_2{-}CH{=}CH_2$.

• Terminal alkyne → Ag precipitate ✓ [count within reasoning]
• Hydrogenation → $n$-pentane ✓
• Ozonolysis of the terminal C=C → $HCHO$; the alkyne on oxidative-type cleavage is not the reductive route, so the clean double-fragment pattern best matches penta-1,4-diene.

Accepted answer: X = pent-1-en-4-yne $CH{\equiv}C{-}CH_2{-}CH{=}CH_2$ (satisfies Ag test, hydrogenation, and gives HCHO from the alkene end). Full marks for a structure consistent with the terminal-alkyne (Ag) clue + $n$-pentane skeleton + HCHO fragment. [total 6]

(b) HBr (no peroxide) adds Markovnikov to the alkene (the more nucleophilic π-bond) in preference to the alkyne. Proton adds to terminal $CH_2$; $Br$ goes to the more substituted internal carbon (more stable 2° carbocation). [2] Product: $CH{\equiv}C{-}CH_2{-}CHBr{-}CH_3$ (2-bromo substitution at the former internal alkene carbon). [2]

Q2 (10)

(a) 2-methylbutane: $(CH_3)_2CH{-}CH_2{-}CH_3$. H-types: [4]

• Two equivalent methyls on C2 → 6 primary H (type A)
• C1... same as above; the terminal $CH_3$ of ethyl → 3 primary H (type B)
• C2 tertiary CH → 1 tertiary H (type C)
• C3 $CH_2$ → 2 secondary H (type D)

Products:

1. 1-chloro-2-methylbutane (from 6 gem-dimethyl H)
2. 2-chloro-2-methylbutane (from 1 tertiary H)
3. 3-chloro-2-methylbutane (from 2 secondary H)
4. 1-chloro-3-methylbutane / (2-chloromethyl...) — from the ethyl terminal 3 primary H → gives 1-chloro-2-methylbutane's isomer = 1-chloro-2-methylbutane actually the CH₃-CH₂ end gives a different primary product.

(b) Weighting = (no. of H) × (relative reactivity): [6]

• 6 primary (dimethyl) × 1 = 6.0
• 3 primary (ethyl CH₃) × 1 = 3.0 → total primary = 9.0
• 1 tertiary × 5 = 5.0
• 2 secondary × 3.8 = 7.6

Sum = 6.0 + 3.0 + 5.0 + 7.6 = 21.6

Percentages:

• 1-chloro-2-methylbutane (dimethyl primary): $6.0/21.6 = 27.8\%$
• 1-chloro-3-methylbutane (ethyl primary): $3.0/21.6 = 13.9\%$
• 2-chloro-2-methylbutane (tertiary): $5.0/21.6 = 23.1\%$
• 3-chloro-2-methylbutane (secondary): $7.6/21.6 = 35.2\%$

(±0.5% acceptable)

Q3 (10)

(a) Newman along C2–C3: [4]

• Anti: two $CH_3$ groups at 180° (dihedral) — front CH₃ up, back CH₃ down.
• Gauche: two $CH_3$ at 60°. Anti is more stable: methyl groups maximally separated ⇒ minimum steric (van der Waals) strain. Gauche has a destabilising gauche interaction (~3.8 kJ/mol). [2 for drawings, 2 for reasoning]

(b) In the chair, equatorial substituents point outward, away from the ring; axial substituents point parallel to the ring axis and suffer 1,3-diaxial interactions — steric repulsion with the two other axial H's on the same face (C3 and C5). A bulky t-Bu group axial has severe 1,3-diaxial clashes, so it locks the ring in the conformation placing t-Bu equatorial, minimising strain. [3]

(c) Baeyer assumed all rings are planar, so angle strain = deviation of internal angle from 109.5°. In reality rings pucker (cyclohexane adopts a non-planar chair) where each carbon retains ~109.5° tetrahedral angles ⇒ essentially zero angle strain. Baeyer's flaw was the planarity assumption. [3]

Q4 (10)

(a) Acetone: Markovnikov hydration. [3] Reagents: $H_2O$, $H_2SO_4$, $HgSO_4$. Water adds Markovnikov → enol at C2 → tautomerises to $CH_3{-}CO{-}CH_3$ (acetone). ✓

(b) Propanal: anti-Markovnikov hydration via hydroboration–oxidation. [3] Reagents: (i) $BH_3\cdot THF$ (or disiamylborane for terminal alkynes), (ii) $H_2O_2$, $NaOH$. Boron adds to terminal C → enol at C1 → tautomerises to $CH_3CH_2CHO$ (propanal). ✓

(c) Pent-2-yne: [4]

1. Propyne + $NaNH_2$ → sodium propynide $CH_3{-}C{\equiv}C^-Na^+$ (deprotonation of terminal acidic ≡C–H, $pK_a\approx25$). [2]
2. Alkylate with $CH_3CH_2Br$ (ethyl bromide, the added electrophile) via $S_N2$: [2] $CH_3{-}C{\equiv}C^- + CH_3CH_2Br \rightarrow CH_3{-}C{\equiv}C{-}CH_2CH_3$ = pent-2-yne. ✓

Q5 (10)

(a) [4]

• Nitrobenzene: $-NO_2$ is strongly deactivating and meta-directing (electron-withdrawing, destabilises o/p σ-complex). Major product: 1,3-dinitrobenzene (meta). [2]
• Anisole: $-OCH_3$ is activating (lone-pair donation) and ortho/para-directing. Major product: para-nitroanisole (with some ortho). [2]

(b) Decreasing EAS reactivity: [3] $\text{anisole} > \text{toluene} > \text{benzene} > \text{chlorobenzene} > \text{nitrobenzene}$ Reasoning: $-OCH_3$ strong activator (resonance donation) > $-CH_3$ weak activator (hyperconjugation/induction) > H (reference) > $-Cl$ weakly deactivating (–I dominates) > $-NO_2$ strongly deactivating (–I and –R). [full marks for correct order + justification]

(c) Chlorobenzene: $Cl$ withdraws electron density inductively (–I) deactivating the ring, so overall slower than benzene. But in the o/p σ-complex the chlorine lone pair can donate by resonance to stabilise the positive charge (only possible for o/p attack), so it directs o/p. Deactivation (inductive) and o/p direction (resonance) arise from two different effects — no contradiction. [3]

[
  {"claim":"Q2 total weighting = 21.6","code":"prim=6*1+3*1; tert=1*5; sec=2*3.8; total=prim+tert+sec; result = abs(total-21.6)<1e-9"},
  {"claim":"Q2 tertiary product percentage = 23.15%","code":"total=6*1+3*1+1*5+2*3.8; pct=100*5/total; result = abs(pct-23.148148)<0.01"},
  {"claim":"Q2 secondary product percentage = 35.19%","code":"total=6*1+3*1+1*5+2*3.8; pct=100*7.6/total; result = abs(pct-35.185185)<0.01"},
  {"claim":"C5H8 degree of unsaturation = 2","code":"C=5; H=8; dou=(2*C+2-H)/2; result = dou==2"},
  {"claim":"Q2 all percentages sum to 100","code":"total=6*1+3*1+1*5+2*3.8; s=(6+3+5+7.6)/total*100; result = abs(s-100)<1e-9"}
]