00:01
Hello everyone.
00:02
Here in this question, it is given that we have a reaction where an alkyne molecule is reacting with several reactants.
00:12
These reactants are, the first one is sodium amide, second is methyl iodide, third one is dialkyl boron hydride, and then we have hydrogen peroxide and sodium hydroxide.
00:25
We have to predict the product obtained from this reaction.
00:30
Now, the first reaction is between alkyne molecule and sodium amide.
00:37
Let's see what happens in this reaction.
00:41
When alkyne molecule react with sodium amide, it lead to deprotonation of this alkyne molecule and from here the second carbon atom loses its proton atom that is h positive.
00:56
Also, as a byproduct, ammonia is obtained from here.
01:01
Now let's see how this deprotonated alkyne molecule react with the second reactant, that is methyl iodide.
01:13
This we will see in the next reaction.
01:17
Now, as here we can see in this reaction that the deprotonated alkyne molecule react with methyl iodide to form a propine molecule...