04-BS-12 · May 2018
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
National Exam 04-BS-12, Organic Chemistry — May 2018. 3 hours, closed-book examination (one Casio/Sharp-approved calculator and one hand-written aid sheet permitted); NOTES on page 1 state that TEN (10) questions constitute a complete exam paper and only the first 10 as they appear in the answer book are marked, but this sitting prints 13 numbered questions — every question and sub-part below is answered in full.
Reference texts: McMurry, Organic Chemistry, 9th ed. (drug acid–base/salt pharmacokinetics, steroid/bile-acid amphiphilicity, arene-oxide metabolism, cyclopropane stereochemistry and CIP assignment, reaction-energy diagrams, ester equilibria and intramolecular effective molarity, SN2 stereochemistry at a common stereocentre, named-drug synthesis design, epoxide ring-opening stereochemistry, mass-spectral formula discrimination, opioid IR/NMR structure elucidation, keto–enol tautomerism and conjugation/acidity, and condensation-polymer monomer identification). Every molecular formula, mass-balance, exact-mass, and stereochemical (R/S) assignment below.
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.
Step 1: Williamson ether synthesis at the epoxide's primary CH2Cl carbon. 1-Naphthol is deprotonated by a mild base to give the naphthoxide, a good nucleophile. This oxygen attacks epichlorohydrin at its primary, unhindered CH2Cl carbon (SN2, chloride is the better leaving group and the least hindered site — the epoxide ring itself is left untouched in this step), displacing chloride and giving the aryl glycidyl ether intermediate.
Step 2: epoxide ring-opening by isopropylamine. Isopropylamine, a good nitrogen nucleophile, opens the strained epoxide ring at its less hindered (terminal) carbon (SN2-like backside attack under these basic/neutral conditions), generating the secondary amine and, simultaneously, the free secondary alcohol from the epoxide oxygen. This delivers the final target directly.
Both steps are nucleophilic substitutions at an sp3 carbon bearing a leaving group (chloride in step 1, the strained epoxide C–O bond in step 2) — exactly the "two successive nucleophilic substitution reactions" the question specifies, and the overall atom-economy is clean: 1-naphthol + epichlorohydrin + isopropylamine → propranolol, with only chloride lost (as HCl, taken up by base) along the way.
