00:04
This problem asks us to show the products of both acid catalyzed hydration and hydroboration oxidation for each of these molecules.
00:14
So for a, we have one vutine.
00:17
And if we do acetylized hydration, that means that we're going to put our o on the internal carbon.
00:24
So it'll be here.
00:25
We'll have an enol that will have a double bond and then the oh on this carbon right here.
00:32
And then that will automorize into a ketone on our second carbon.
00:36
So our products will look like this.
00:42
For hydroboration oxidation, this does a very similar, it gives us a similar product to the hydroboration, to the acid catalized hydration, but it does it anti -marcov, which means that the oh goes on the end carbon over here.
00:58
So our enol in this case will look like this.
01:02
We have those other two carbons there, which means that our end product is actually an aldehyde.
01:09
So because it's on the end, it turns into an aldehyde instead of a ketone.
01:13
So it looks like this.
01:17
For b, we have an internal symmetrical alkyne, which means that no matter which side we pick, it's going to look the same in the end for both of them.
01:27
So we'll get another ketone here...