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.
a) Ranking three diethylcyclohexene positional isomers. All three compounds share the same molecular formula (C10H18, one ring + one C=C = DoU 2) and the same two ethyl substituents on the same pair of adjacent ring carbons — the only difference is which ring edge carries the C=C double bond. Alkene stability increases with the number of alkyl groups attached directly to the two sp2 (double-bond) carbons — hyperconjugative donation from each attached C–H/C–C bond stabilises the π-system, so more substituents on the alkene carbons themselves means a lower-energy, more stable alkene. An ethyl group sitting on a carbon that is not part of the C=C (merely allylic or further away) contributes nothing to this count.
(A) Both ethyls on the alkene carbons — tetrasubstituted. The ring double bond falls exactly on the two ethyl-bearing carbons, so each sp2 carbon carries one ring C–C bond plus the ethyl group (zero H's on either alkene carbon): four alkyl substituents on the C=C in total.
(B) One ethyl on an alkene carbon, one allylic — trisubstituted. The double bond has shifted one ring position over, so only one of the two ethyl-bearing carbons is still sp2; the other ethyl now sits on an sp3 (allylic) carbon that does not touch the π-system at all: three alkyl substituents on the C=C.
(C) Neither ethyl on an alkene carbon — disubstituted. The double bond sits on the far side of the ring, on two carbons that carry only ring C–C bonds (each still has one H); both ethyl groups are now allylic/homoallylic and contribute nothing to the alkene's own substitution: the bare two ring C–C bonds are the only alkyl substituents on the C=C — the ordinary disubstituted pattern of any plain cycloalkene.
(i) Most stable ⇒ (A), the tetrasubstituted isomer (both ethyls directly
on the double bond).
(ii) Least stable ⇒ (C), the disubstituted isomer
(both ethyls off the double bond entirely). (B) sits in between: tetra > tri > di.
| Isomer | Alkyl groups on C=C | Stability rank |
|---|---|---|
| (A) 1,2-diethylcyclohex-1-ene | 4 (tetrasubstituted) | Most stable |
| (B) 1,2-diethylcyclohex-2-ene | 3 (trisubstituted) | Middle |
| (C) 4,5-diethylcyclohex-1-ene | 2 (disubstituted) | Least stable |
b) Mono-hydrochlorination of 2-methylpropene. "Mono-chlorination" here refers to installing a single chlorine by the reagent actually shown, HCl — an electrophilic addition (hydrohalogenation) across the alkene, not a radical Cl2 substitution. The mechanism runs in two steps through the more stable carbocation:
Product: 2-chloro-2-methylpropane (tert-butyl chloride). Overall this is a two-step ionic addition (protonation then anion capture), never a single concerted step, and it is the more-stable tertiary cation — not the alkene's geometry alone — that dictates which carbon ends up bearing the chlorine.
