04-BS-12 · May 2014
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
National Exam 04-BS-12, Organic Chemistry — May 2014. 3 hours, closed-book examination; no calculator allowed. Answer ALL FIVE problems; each problem is of equal value (20 points), and the lettered/numbered sub-parts of a given problem may be treated independently.
Reference texts: McMurry, Organic Chemistry, 9th ed. (alkane/carbon classification, alkene stability and E/Z isomerism, electrophilic addition, electrophilic aromatic substitution and benzylic oxidation, Friedel–Crafts acylation, diazonium chemistry, constitutional isomerism).
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) All five constitutional isomers of C6H14. Fixing the longest chain first (6, then 5, then 4 carbons) and placing all remaining branches on each length systematically finds exactly five distinct connectivities (no sixth arrangement exists): the straight chain, two singly-branched pentane-backbone isomers, and two doubly-branched butane-backbone isomers.
| Isomer | Condensed structural formula |
|---|---|
| hexane | CH3CH2CH2CH2CH2CH3 |
| 2-methylpentane | CH3CH(CH3)CH2CH2CH3 |
| 3-methylpentane | CH3CH2CH(CH3)CH2CH3 |
| 2,2-dimethylbutane | CH3C(CH3)2CH2CH3 |
| 2,3-dimethylbutane | CH3CH(CH3)CH(CH3)CH3 |
b) Reaction classification.
(i) CH3CH=CH2 + H2 → CH3CH2CH3: two new atoms (H, H) are added across the double bond and nothing leaves the molecule — this is an addition reaction (catalytic hydrogenation).
(ii) CH3CH2CH2OH $\xrightarrow{H_2SO_4}$ CH3CH=CH2 + H2O: a small molecule (H2O) is removed from the substrate and a new π bond forms in its place — this is an elimination reaction (acid-catalyzed E1 dehydration).
| Part | Classification |
|---|---|
| (i) | addition (hydrogenation) |
| (ii) | elimination (acid-catalyzed dehydration) |