Hello all,
I had a quick question.
As mentioned, 3-cyclohexen-1-carboxylic acid goes through reactions in 2 steps.
First reaction is with
I2, Sodium bicarbonate in CH3CN-water
then this product reacts with
Sodium Hydroxid in MeOH-water
What would be the product?
I assume in the first step, you would halogenate the double bond in Cyclohexene and bicarbonate binds to the other of the doulbe bond?
-This i'm not quite sure, because I learned that Iodine doesn't really react with double bonds... and since bicarbonate is a weak base, wouldn't it compete with Iodine to act as a nucleophile to attach to the other side of the double bond?
And for the second step, I guess the carboxylic acid group would be attacked through
acid catalyzed mechanism to form an ester with methoxy group?...
Hmm.. I'm confused.
please help me out
Thank you!