00:01
In reaction a, there are two chlorides that can both act as leaving groups, and we have two equivalents of sodium amide, which is a strong base.
00:12
So two elimination reactions take place to make a carbon carbon triple bond.
00:27
Now, it's also possible to get some that would be on the, make it a terminal alkyme, but the major product should have it in.
00:37
The more substituted or in the internal position.
00:45
Reaction b is an example of oxymurcation, demercuration, which converts alkenes to markovnakov alcohols.
01:00
Now, in this case, both of the alkane carbons are equally substituted, so we would expect a mixture of products of the pantane 2 -all and pentane 3 -all.
01:22
The product of c will depend on how much hydrogen is present.
01:28
If it's just one equivalent, then the carbic carbon double bond will be reduced.
01:41
But if there is excess, then the nitriol group can be reduced to an amoeuvre...