In the reaction, CH3O− acts as a nucleophile, the phosphorus as an electrophilic center, and CH3S as a leaving group. (Don't be concerned that phosphorus has more than an octet of electrons-that's allowed, and it's irrelevant to the problem.)
a. In the equation, draw the curved arrows for the process described.
b. In the box, draw the products of the reaction. Don't forget formal charges.
Correct Answer:
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