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04-BS-12 · Undated paper

Question 5 of 13

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Notes on this paper

National Exam 04-BS-12, Organic Chemistry — May 2019 sitting (the page-1 header and the running footer, "04-BS-12/May 2019", both give the date). 3 hours, closed-book examination; one Casio/Sharp-approved calculator 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. (acid/base theory, functional-group identification, SN1/SN2 mechanisms and stereochemistry, alkyne/acetylide synthesis, electrophilic aromatic substitution, IR/NMR/mass-spectral structure elucidation, named-drug synthesis design, and step-growth polymer chemistry). Every molecular formula, exact mass, and stereochemical (R/S, cis/trans) assignment below.

Question 5 (5/13)

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.

Approach. Two SN2 steps are specified, so the retrosynthesis is essentially forced: 1-naphthol must first be deprotonated (a strong base such as NaH, or simply NaOH) so the resulting naphthoxide can act as a nucleophile, and epichlorohydrin has exactly one electrophilic site accessible to a bulky phenoxide — the less hindered epoxide carbon (attack at the more substituted carbon is sterically and electronically disfavoured). That opens the epoxide (intramolecular displacement of the ring oxygen by chloride's leaving-group character is not what happens here; rather the epoxide oxygen becomes the new secondary alcohol) to leave a fresh primary alkyl chloride dangling from the other end of the three-carbon bridge — exactly the electrophile isopropylamine needs for the second, independent SN2.

+ NaH, then
Step 1: sodium 1-naphthoxide opens epichlorohydrin's epoxide at the less-hindered CH2 (SN2).
+ SN2
Step 2: isopropylamine displaces the remaining primary chloride, giving propranolol.
Propranolol (target).

Step 1 — 1-naphthol (pKa ≈ 9.4) is deprotonated to the naphthoxide, which attacks epichlorohydrin's less substituted epoxide carbon (a terminal CH2) in an SN2 fashion, opening the three-membered ring to give a secondary alcohol and leaving the original C–Cl bond of epichlorohydrin untouched: 1-chloro-3-(1- naphthyloxy)propan-2-ol.

Step 2 — isopropylamine, a good nucleophile, displaces the remaining primary chloride in a second, independent SN2 to install the final amine and complete propranolol.