Discussion
In the cyclization reaction given below, the most probable product formed is
- A.
- B.
- C.
- D.
Answer: C
The product is the epoxide formed with inversion at the C-Br carbon, 2-(oxiran-2-yl)ethanol with the ring C-O bond wedged (C).
NaH deprotonates a hydroxyl group, and the alkoxide displaces bromide intramolecularly by an SN2 reaction. The CH2O- on the carbon next to C-Br closes a three-membered ring, and 3-exo ring closure is much faster than the 4-exo closure the other OH would need to make an oxetane. So the epoxide forms, not the oxetane of B and D.
Because SN2 attack comes from the side opposite the leaving group, the hashed C-Br bond is replaced by a wedged C-O bond: the configuration at that carbon is inverted. A shows the same epoxide with retention, which an SN2 closure cannot give.
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