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.
i) Benzylic oxidation of sec-butylbenzene. Hot H2CrO4 is a powerful oxidant that attacks any benzylic C–H (a C–H on the carbon directly attached to the ring); regardless of how long or branched the rest of the chain is, the entire side chain is oxidatively degraded down to a single carbon, which ends up as –COOH directly on the ring:
$$\mathrm{C_6H_5CH(CH_3)CH_2CH_3 \xrightarrow{H_2CrO_4,\ \Delta} C_6H_5COOH}$$
Product: benzoic acid. (The three carbons beyond the ipso/benzylic position are lost as smaller oxidation fragments; this is not a mass-balanced single-step equation, the same convention used for every benzylic-oxidation question in this subject.)
ii) Benzylic oxidation of 1-methyl-3-(butan-2-yl)benzene (meta-sec-butyltoluene). Both substituents — the methyl and the sec-butyl group — have at least one benzylic C–H, so both are fully oxidized to –COOH, meta to each other:
$$\mathrm{CH_3\text{-}C_6H_4\text{-}CH(CH_3)CH_2CH_3 \xrightarrow{H_2CrO_4,\ \Delta} HOOC\text{-}C_6H_4\text{-}COOH}$$
Product: isophthalic acid (benzene-1,3-dicarboxylic acid).
iii) 2-Methylpropene + Br2. Br2 adds across the double bond (anti addition via a bromonium-ion intermediate, no regiochemical ambiguity since both new C–Br bonds land on the two alkene carbons):
$$\mathrm{(CH_3)_2C{=}CH_2 + Br_2 \longrightarrow (CH_3)_2CBr\text{-}CH_2Br}$$
Product: 1,2-dibromo-2-methylpropane (verified atom-balanced, C4H8Br2).
iv) Cyclohexene + HBr. Markovnikov addition, but cyclohexene's two alkene carbons are equivalent by the ring's symmetry, so there is no regiochemical choice to make:
$$\mathrm{C_6H_{10} + HBr \longrightarrow C_6H_{11}Br}$$
Product: bromocyclohexane.
| Part | Product |
|---|---|
| (i) | benzoic acid |
| (ii) | isophthalic acid (benzene-1,3-dicarboxylic acid) |
| (iii) | 1,2-dibromo-2-methylpropane |
| (iv) | bromocyclohexane |
