- #1
jfeyen
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Homework Statement
I'm having a hard time deciding exactly what the mechanism would be for
CH3CH=CHCH3→CH3C[itex]\equiv[/itex]CCH3
Homework Equations
The Attempt at a Solution
My professor only explained half of the mechanism quickly assuming the rest would be obvious, so I know that it will start with a halogenation breaking the double bond before reforming the triple bond. My problem is whether only one equivalent of Br2 is needed or two before you are able to dehalogenate and form the triple bond. I'm also having a hard time figuring out why the bromines have be be added in the first place only to be taken off again. Also it seems that after the bromination there is a second step adding a catalyst (?) of either Zn or something that looks like +o- (t-butoxide?). Does this have any effect on the whether we need 2 or 4 bromines on the molecule before reducing? And what exactly makes the bromines come off to form the triple bond?