00:01
Ok, we have a couple alkenes and some reagents.
00:05
We need to determine what the products of those reactions will give us.
00:09
First we have br2 in water, and recall that what this will do is install a bromine and an alcohol that are cross, sorry, they are anti to each other with respect to the bond of the carbons that they're bound to respectively, and that the alcohol will end up on the more substituted side of the double bond.
00:38
That is, one substituent will end up here and one will end up here, and since this is the more substituted side, that's the side that will get the alcohol.
00:46
So we can simply redraw the substrate without the double bond, put a bromine on one side and an alcohol on the other, and, oops, i did that backwards, bromine goes on the less substituted side, alcohol on the more substituted side.
01:07
Notice that there in fact are no stereocenters in this molecule.
01:13
This carbon has two ethyl groups off of it, and so it doesn't have four unique substituents, and this carbon has two hydrogens coming off of it, which means there are no stereocenters on that carbon.
01:25
So in fact, this product is simply drawn not with wedges and dashes, but like so, which is why it's necessary for this problem, because they say if you incorrectly show stereochemistry where none exists, you may lose points.
01:41
No stereochemistry exists on this molecule, so we should not show stereochemistry in our products.
01:46
Okay, next, we have bh3 in thf first, and then hydrogen peroxide, water with sodium hydroxide...