Question 7 of 13: S N 1 or S N 2? Four Cases

Question 7: S_N1 or S_N2? Four Cases

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. (a) S_N2 — despite a tertiary centre.
   start (tertiary, α to C=O)

   SN2 product (inversion)
   The reacting carbon is tertiary (Br, Me, and two ring carbons), which by substitution count alone would favour S_N1. But it also sits directly α to the lactone carbonyl: a carbocation there would place a full positive charge immediately adjacent to the carbonyl's own partial positive carbon — a strongly destabilising inductive arrangement, with no compensating resonance donation available. S_N1 is therefore disfavoured regardless of the steric substitution pattern. Azide is additionally a small, strongly nucleophilic anion — an excellent S_N2 nucleophile — so the reaction proceeds by backside attack (S_N2) even though the carbon is tertiary; this is the standard "α-halocarbonyl compounds react by S_N2, even when hindered" exception to the usual steric preference.
2. (b) S_N1 — via a resonance-stabilised oxocarbenium ion.
   methyl orthoester (start)

   oxocarbenium intermediate (SN1)
   This is acetal/orthoester exchange chemistry: protonation of the leaving methoxy oxygen, followed by loss of methanol, generates a cyclic oxocarbenium ion flanked by two ring oxygens, each donating a lone pair into the empty p-orbital — an exceptionally stabilised cationic intermediate (far more stable than any ordinary secondary/tertiary carbocation). The pendant hydroxyl then closes intramolecularly onto this flat, resonance-stabilised cation. Acetal/orthoester hydrolysis and exchange always proceeds through this stabilised cation (S_N1-type), never by a backside S_N2 displacement — there is no accessible backside trajectory in a cyclic system, and none is needed given how stable the cation is.
3. (c) S_N1-like — acid-catalysed epoxide opening at the more-substituted (benzylic/tertiary) carbon.
   model epoxide

   (c) acid: OPr at 3° C

   (d) base: OPr at 1° C
   Protonating the epoxide oxygen first makes both ring carbons electrophilic, but the C–O bond to the more substituted carbon (here tertiary and benzylic) breaks preferentially, because that bond-breaking develops a partial positive charge best stabilised at that position (benzylic conjugation with the aromatic ring, plus greater alkyl substitution). The nucleophile (n-PrOH) then attacks that same, more electrophilic (more cation-like) carbon — giving the ether at the tertiary/benzylic position and leaving the new –OH at the less-hindered carbon. This Markovnikov-type regiochemistry is diagnostic of substantial cationic character in the transition state, i.e. an S_N1-like mechanism (a true, fully discrete cation is not required — only enough charge development to control regiochemistry).
4. (d) Clean S_N2 — base-catalysed epoxide opening at the less-hindered carbon. Without acid to activate/protonate the epoxide oxygen, the strained three-membered ring is opened directly by the incoming alkoxide nucleophile via straightforward backside attack, which is only sterically accessible at the less hindered (primary/secondary) carbon. This gives the opposite regiochemistry to part (c): the ether now forms at the less-substituted carbon, inversion of configuration occurs at the carbon attacked, and the tertiary/benzylic carbon retains its original oxygen as free –OH.