00:01
In problem 9 .27, we are asked to create products specified starting with a terminal alkyne.
00:09
Let's look at the first reaction, which will be the top right, and we will work clockwise from there.
00:18
So to get a single chlorine in a markovnikov position, we are first going to need to reduce a triple bond to a double bond.
00:28
We will do that with hydrogen and a lindlark.
00:33
Catalyst.
00:46
This will give us a double bond.
00:50
And now we could either add, we could add hdl, and this will give a copper cadion intermediate, creating a chlorine on the second carbon.
01:11
Now let's look at the second reaction.
01:13
In the second reaction, we have a terminal alcohol.
01:18
There can be multiple ways to do this problem, but i'm going to first reduce this using hydrogen and a lindlar catalyst and then react it with boring and a base.
01:56
So first this adds boring to the terminal end of the now double bond and then we will use peroxide and a base.
02:15
This will give us a terminal alcohol.
02:22
Now for the third reaction.
02:23
In the third reaction, we have to add another methyl group to the end of our triple bond.
02:30
So we're going to do this first by removing the terminal hydrogen.
02:36
We will do this with nanh2 in liquid pneumonia.
02:49
And then we will add the methyl group.
02:52
And i'm going to add with methyl or m -e -i, so iodomethel.
03:05
And that's how we create the third product.
03:11
Now, in the next reaction, we are to create a ketone on the second carbon.
03:19
This is also the markovnikov position for the ketone.
03:24
So in this reaction, we are going to use h2o, h2s .4, and hgso4 to create the markovnikov ketone...