04-BS-12 · Undated paper
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
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 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. Every part is a one-step bimolecular nucleophilic substitution: identify the leaving group and the carbon it sits on, confirm that carbon is a good SN2 site (primary or unhindered secondary, not tertiary), then apply the single governing rule — backside attack forces inversion of configuration at the carbon where the bond to the leaving group breaks, while every other stereocentre in the molecule is an untouched spectator and keeps its original configuration exactly.
a) This substrate is drawn as CH3–C*HDCl using a deuterium label (D, mass 2) in place of one of the two otherwise-identical hydrogens on C1 of chloroethane — a classic device: CH3CH2Cl itself has no stereocentre (two identical H's on C1), but swapping one H for its heavier isotope D makes that carbon bear four different groups (Cl, CH3, D, H) and therefore genuinely chiral, letting the exam track the stereochemical outcome of the substitution directly. Backside attack by methoxide on the carbon bearing Cl inverts that stereocentre:
The chlorine (highest CIP priority) is simply replaced by the incoming methoxy oxygen (also highest priority at that carbon), so the descriptor itself flips from the letter it had in the starting material to the opposite letter in the product (S → R here) — a clean, textbook Walden inversion.
b) 1-Iodopentane is a simple primary alkyl halide with no stereocentre anywhere in the molecule, so there is no stereochemistry to indicate — hydroxide simply displaces iodide at the unhindered primary carbon:
c) The wedge/hash stereocentre in this substrate sits at C3 (bearing an ethyl group on one side, a 2-carbon chain to the chloromethyl on the other, plus H and CH3), while the leaving group (Cl) sits on the primary C1, three bonds away. This is the key trap in the question: the reacting carbon and the labelled stereocentre are not the same atom. Ethoxide performs a routine primary SN2 at C1; C3 is never touched by any bond being made or broken, so its configuration is retained exactly as drawn.
d) Here the leaving group (Br) sits directly on the ring stereocentre (a secondary, unhindered ring carbon — still a fine SN2 site), so this time the reacting carbon and a labelled stereocentre are the same atom, and inversion applies there. The methyl-bearing ring carbon (C3) is the spectator and keeps its configuration. The source draws the starting bromide as trans (Br and CH3 on opposite faces of the ring — Br wedge, CH3 hash). Backside attack by cyanide flips only the face at C1: the incoming CN group ends up on the same face the departing Br was not on, i.e. the same face as the (unchanged) methyl group — so the trans bromide becomes the cis nitrile.
