00:01
Now, in this problem, we're asked to predict the products of two butyne with one equivalent of molecular bromine and dichloromethane and one equivalent of hydrobromic acid in peroxides.
00:17
Now, normally, with molecular halogens and dichloromethane, we might expect some kind of alkane reducing all the way down to a single bond with halogens on either side.
00:32
Of the original location of the triple bond.
00:34
But because we only have one equivalent, we can't go all the way here.
00:39
There isn't enough bromine present.
00:41
What we're actually going to result in is the creation of a double bond.
00:47
Now, this will always result in anti -edition across that triple bond because of the mechanism.
00:57
So we might name this product.
01:00
Clearly, it is a e -configuration.
01:04
Because the bromines, the high priority groups, are far apart.
01:08
It is 2 .3 dibromo, and it is a but we must name where the butene is, of course, at the 2 position...