04-BS-12 · May 2015
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
National Exam 04-BS-12, Organic Chemistry — May 2015. 3 hours, closed-book examination; any non-communicating (non-programmable) calculator permitted. Answer ALL FIVE problems; each problem is of equal value (20 points), and the lettered sub-parts of a given problem may be treated independently.
Reference texts: McMurry, Organic Chemistry, 9th ed. (functional-group nomenclature, electrophilic addition and Markovnikov's rule, alkene stability/substitution, catalytic hydrogenation, electrophilic aromatic substitution and the Friedel–Crafts acylation mechanism, diazonium chemistry, combustion).
Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.
All four reactions are acid- or metal-catalysed additions across a C=C. Parts (i), (ii) and (iv) are Markovnikov additions of an unsymmetrical reagent (H–I or H–OH) across an unsymmetrically substituted alkene: the proton always adds to the alkene carbon that already carries more hydrogens, because that leaves the positive charge on the more highly substituted carbon, where it is stabilised as the more substituted (here, tertiary) carbocation before the nucleophile (I− or H2O) closes in. Part (iii) is a straightforward catalytic hydrogenation of a plain, unsubstituted ring alkene.
(i) Methylenecyclohexane + HI. The exocyclic =CH2 carbon has two H's already; protonating it places the new positive charge on the ring carbon, which is tertiary (bonded to two ring CH2's and the former =CH2, now –CH3) — far more stable than the alternative primary cation on the exocyclic carbon. Iodide then closes onto that tertiary ring carbon:
$$\mathrm{C_7H_{12} + HI \longrightarrow C_7H_{13}I}$$
Product: 1-iodo-1-methylcyclohexane.
(ii) 1-Methylcyclohex-1-ene + H2O. The ring's C1 (bearing the methyl group) is already the more substituted alkene carbon; protonating C2 places the cation on C1, which is tertiary (two ring bonds + the methyl group), so water adds its oxygen there:
$$\mathrm{C_7H_{12} + H_2O \xrightarrow{H_2SO_4} C_7H_{14}O}$$
Product: 1-methylcyclohexan-1-ol (a tertiary alcohol).
(iii) Cyclohexene + H2. With no substituents to create a regiochemical choice, Pd simply delivers both new C–H bonds syn across the one ring double bond; the mild stated conditions (25°C, 3 atm) are entirely sufficient for an isolated, unhindered cycloalkene (contrast with the much harsher forcing needed to hydrogenate a fully aromatic ring, which must overcome its resonance stabilisation):
$$\mathrm{C_6H_{10} + H_2 \xrightarrow{Pd,\ 25^\circ C,\ 3\ atm} C_6H_{12}}$$
Product: cyclohexane.
(iv) 2-Methylbut-2-ene + H2O. The two alkene carbons are (CH3)2C= (bearing two methyls, no H) and =CHCH3 (bearing one H). Protonation occurs at the CH carbon (the one with more H's already), leaving the cation on the (CH3)2C carbon — tertiary, the most stable option — so –OH ends up there:
$$\mathrm{C_5H_{10} + H_2O \xrightarrow{H_2SO_4} C_5H_{12}O}$$
Product: 2-methylbutan-2-ol (a tertiary alcohol).
| Part | Product | Mechanism |
|---|---|---|
| (i) | 1-iodo-1-methylcyclohexane | Markovnikov HI addition, tertiary cation |
| (ii) | 1-methylcyclohexan-1-ol | Markovnikov hydration, tertiary cation |
| (iii) | cyclohexane | syn catalytic hydrogenation, Pd |
| (iv) | 2-methylbutan-2-ol | Markovnikov hydration, tertiary cation |
