00:01
In this problem, we have a reaction of 2 -pentine, so you have 5 carbons, and at carbon 2, you have a double bond.
00:09
So first, if we react it with hydrogen in the presence of palladium, this is going to be a reduction.
00:19
So instead of our double bond, we'll get a single bond now.
00:23
If we react this with diatomic bromine, we're going to get bromines on opposite sides instead of our double bond.
00:34
So it should look like that.
00:37
If we react with hydrogen chloride, here we want to put the chlorine on the most substituted carbon.
00:47
Here, though, we're equally substituted...