NivaarExam PrepOfficial exam papers ↗

04-BS-12 · December 2019

Question 8 of 13: Propranolol Total Synthesis

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Exam 04-BS-12, Organic Chemistry — December 2019. 3 hours, closed-book examination (no calculator; 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. (functional groups, stereochemistry, SN1/SN2 mechanisms and stereochemistry, steroid/bile-acid amphiphilicity, named-drug synthesis design, reaction-energy diagrams, polymer chemistry, arene-oxide metabolism, and mass-spectral/IR/NMR structure elucidation). Every molecular formula, exact mass, and stereochemical (R/S, cis/trans, meso/chiral) assignment below.

Question 8: Propranolol Total Synthesis (equal value)

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.

1-naphthol
epichlorohydrin
isopropylamine
propranolol

Step 1 (first SN2): Deprotonate 1-naphthol (base, e.g. NaOH) to give the naphthoxide, a good nucleophile. The phenoxide oxygen performs a back-side SN2 attack on the primary C–Cl carbon of epichlorohydrin (the epoxide ring itself is untouched, since it is not the electrophilic site under basic conditions), displacing chloride and giving the aryl glycidyl ether intermediate.

Step 2 (second SN2): Isopropylamine, a good nucleophile, opens the strained epoxide ring by attacking its less hindered, terminal carbon (SN2, backside attack). This installs the –NH–CH(CH3)2 group on the terminal carbon and, because the ring oxygen becomes a free alkoxide/alcohol on the adjacent (internal) carbon, generates the secondary –OH exactly at the middle carbon — giving propranolol directly.

intermediate: 1-(naphthalen-1-yloxy)-2,3-epoxypropane