Question

ОН OH C Y і 0 (97 : 3) Л s11 Л s12

Show FULL arrow pushing mechanism for this reaction, and explain it step by step. Reagent used was c) K2CO3, MeOH, room temp., 1 h (quant.).

(please disregard BOTH arrows on s11, as this only one step of a multi-step synthesis)

Label all chiral centers on s12 and draw its enantiomer

Answer 1

Since by looking the product we conclude that it is a epoxide ring formation reaction.

Mechanism is as follows:

• First the potassium carbonate given is acting as base (i.e it's carbonate ion ) which abstracts the proton from the alcohol group to form a alkoxide .

This alkoxide is participating in neighbouring group like reaction which eliminates the iodine in anti manner to form a epoxide.

Since we marked the chiral center as astrick.

Enatiomer is made by changing the only dash wedge bond

de & Co. +2kt NA -0-1 111110 0-W chiral center Enantiomer - o-4 Scanned with CamScanner