04-BS-12 · December 2015
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
National Exam 04-BS-12, Organic Chemistry — December 2015. 3 hours, closed-book examination; any 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 identification, degree of unsaturation, acid-catalysed alcohol dehydration and alkene/arene hydration, alkane nomenclature, catalytic hydrogenation, ester/disulfide isomers, acid-catalysed transesterification mechanism, Markovnikov addition, nitration, carbon classification, benzylic oxidation, diazonium/Sandmeyer 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(i) Methylenecyclohexane + H2. Pd hydrogenates the exocyclic C=C only (the ring itself is already fully saturated — there is no aromatic system here to reduce):
$$\mathrm{C_7H_{12} + H_2 \xrightarrow{Pd} \boxed{C_7H_{14}}}$$
Product: methylcyclohexane.
a(ii) 3-Methylpent-2-ene + HCl. The alkene carbons are unevenly substituted: the left carbon (bearing an ethyl chain and a methyl branch) carries two alkyl substituents, the right carbon (bearing only a terminal methyl) carries one plus an H. Protonation adds H+ to the less substituted right-hand carbon, generating the more stable tertiary carbocation on the left; chloride then captures that cation:
$$\mathrm{CH_3CH_2C(CH_3){=}CHCH_3 + HCl \longrightarrow \boxed{CH_3CH_2C(CH_3)(Cl)CH_2CH_3}}$$
Product: 3-chloro-3-methylpentane.
a(iii) Benzene + HNO3/H2SO4. H2SO4 protonates nitric acid to generate the nitronium electrophile NO2+, which substitutes onto the ring via the standard attack→arenium→deprotonation electrophilic aromatic substitution sequence:
$$\mathrm{C_6H_6 + HNO_3 \xrightarrow{H_2SO_4} \boxed{C_6H_5NO_2} + H_2O}$$
Product: nitrobenzene.
| Part | Product |
|---|---|
| a(i) | methylcyclohexane |
| a(ii) | 3-chloro-3-methylpentane |
| a(iii) | nitrobenzene |
b) 1-Ethyl-3-methylbenzene + H2CrO4, heat. Both side chains — the ethyl group and the methyl group — carry a benzylic C–H, and hot chromic acid oxidises any benzylic side chain all the way to a ring-attached carboxylic acid, regardless of how many carbons the chain started with (every carbon beyond the first is progressively oxidised and cleaved off, ultimately as CO2/H2O, until only the ring-attached carbon survives as –COOH):
$$\mathrm{\text{1-ethyl-3-methylbenzene} \xrightarrow{H_2CrO_4,\ \Delta} \boxed{\text{isophthalic acid}}}$$
Product: isophthalic acid (benzene-1,3-dicarboxylic acid). Note that the ethyl group's two carbons do not survive as –CH2COOH — forcing conditions oxidise straight through to the single-carbon acid, exactly the same end product a starting methyl group gives.
| Part | Result |
|---|---|
| a(i)–(iii) | see table above |
| b | isophthalic acid |
