00:01
In problem 9 .23, we are told to start with 5 dekine and react with the reagents given for each part and give the products.
00:12
In part a, we are reducing the triple bond to a double bond with hydrogen and a limelac catalyst.
00:21
And this will give us double bond in cis position.
00:26
So now we have cis 5 dequine.
00:52
Now let's look at part b.
00:54
In part b, the lithium is going to donate electrons, and the ammonia is going to donate a proton, and this is how we will reduce this triple bond to a double bond.
01:09
The double bond will be in transposition.
01:32
And our double bond is on carbon 5.
01:35
So now we have trans to we have trans five decane.
01:44
And just remember that the linlar catalyst always gives you a cyst double bond, and the lithium ammonia will always give you trans.
01:57
And now for part c, we have one equivalent of bromine, and we are reacting this with our five dechine.
02:09
So this will give us a trans molecule with bromine on each side of the double bond.
02:47
So here we have trans, 5, 6, dibromo, 6, dechene...