00:02
Problem 9 .3 asks us to find the products for the following reactions with al -alcines.
00:09
In problem a, we have one mole of pentine reacted with two moles of chloride.
00:17
Now we know that chloride is going to add an antide, but because we have two moles of chloride, this reaction is going to take the triple bond to a single bond.
00:34
This will give us a pentine, a pentane with four chlorides.
00:47
Fluorides on it.
01:05
The name of this molecule is 1 -1 -2 -2 -hetra -chloro -centane.
01:19
Now let's look at reaction b.
01:22
In reaction b, we actually need to take the mechanism into account.
01:26
For these reactions, the nucleophilic triple bonds will attack the electrophilic hydrogen with the bromine...