04-BS-12 · December 2013
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
National Exam 04-BS-12, Organic Chemistry — December 2013. 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. (nomenclature, isomerism, alkene/alkyne addition reactions, electrophilic aromatic substitution, alcohols and ethers, esters, oxidation-reduction of alcohols, Grignard synthesis).
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) Iso-octane. 2,2,4-Trimethylpentane is a five-carbon (pentane) backbone carrying two methyl branches on C2 and one methyl branch on C4:
Condensed structural formula: CH3–C(CH3)2–CH2–CH(CH3)–CH3 (molecular formula C8H18). This branched isomer of octane resists auto-ignition far better than the straight-chain n-octane, which is why it defines the top of the 0–100 octane-rating scale.
(b) Primary / secondary / tertiary classification. A carbon's class is set purely by how many other carbons it is bonded to, not by how many hydrogens it carries: primary = 1 C neighbour, secondary = 2, tertiary = 3, and a carbon bonded to 4 other carbons (no H at all) is quaternary — a category the question's three choices don't literally include, so it is flagged explicitly below where it occurs.
i) 2,2-dimethylbutane, CH3–C(CH3)2–CH2–CH3: C1 (a methyl on the backbone) and both branch methyls on C2 are each bonded to only C2 → primary (3 primary carbons total). C2 is bonded to C1, C3, and both branch methyls — four carbons, no hydrogen → quaternary (not primary/ secondary/tertiary). C3 is bonded to C2 and C4 → secondary. C4 is bonded only to C3 → primary.
ii) CH3CH2CH2CH2CH(CH3)CH(CH3)CH3 (a 2,3-dimethylheptane skeleton numbered from the other end). Labelling the seven backbone carbons C1–C7 left to right, with branch methyls Ca on C5 and Cb on C6: C1 (1 neighbour) → primary; C2, C3, C4 (each 2 neighbours) → secondary; C5 (bonded to C4, C6, Ca — 3 neighbours) → tertiary; C6 (bonded to C5, C7, Cb — 3 neighbours) → tertiary; C7 (1 neighbour) → primary; both branch methyls Ca, Cb (1 neighbour each) → primary.
iii) Hexane, CH3CH2CH2CH2CH2CH3: a straight chain has no branch point, so there is no tertiary or quaternary carbon at all. C1 and C6 (1 neighbour each) → primary; C2, C3, C4, C5 (2 neighbours each) → secondary.
c) Family identification. i) The structure is a four-carbon chain with an –OH bonded to the second (sp3, two-carbon-neighbour) carbon: butan-2-ol, an alcohol (specifically secondary, C4H10O). ii) The structure is a benzene ring with one –CH2CH3 substituent: ethylbenzene, an aromatic hydrocarbon (arene) (C8H10) — the ring's six delocalised π electrons (Hückel 4n+2, n=1) make it aromatic rather than a cyclic triene.
| Part | Answer |
|---|---|
| (a) | CH3-C(CH3)2-CH2-CH(CH3)-CH3, C8H18 |
| (b)(i) | 3 primary, 1 quaternary, 1 secondary, 1 primary |
| (b)(ii) | 3 primary (incl. 2 branch methyls), 3 secondary, 2 tertiary |
| (b)(iii) | 2 primary, 4 secondary, 0 tertiary |
| (c)(i) | butan-2-ol — alcohol |
| (c)(ii) | ethylbenzene — aromatic hydrocarbon |
