00:01
In this problem, we're asked to synthesize each of these compounds with utilizing just acetylene as a starting product.
00:10
Now, for the first part of this problem, we have to realize that we need to only deprotonate one side of acetylene.
00:18
So we need to use an anh2, one equivalent, followed by whatever else we need with a iodine attached to it, also in one equivalent.
00:33
If used in excess, we would have symmetrical addition for this first problem.
00:37
The second one, we start with the exact same method.
00:41
We deprotonate and add iodide, so we'll abbreviate that.
00:49
And then, as the third step, we need to come from a triple bond down to a double bond, which we can do with the dissolving metal reduction.
00:59
So sodium metal and liquid ammonia.
01:03
The next problem is the exact same idea, except this time.
01:08
We're going to use excess an h2 and excess methyl iodide because we need symmetric addition.
01:19
And then we need to come down to a cis alkene.
01:22
If we need to come to a cis alkyem, we have to use a lindler's reduction using hydrogen gas and lindner's catalyst.
01:31
For the next problem, we need to come to an alvehyde...