00:01
Hello, in this question we are given a molecule here, that is on the above the plane, br is present here and 1, 2 and methyl group is present here.
00:14
So with the reaction of in the presence of n .a .o .m.
00:17
That is sodium methyl acetate is there.
00:21
So or acetate is there in the presence of how the reaction will occur and what will be our final problem.
00:27
We can see here that oxygen is present here that is the negative charges here.
00:30
And here in this carbon you can see in this carbon one methyl group it is present here one methyl group it is present here right and one substituter group is there and hydrogen is also there so this hydrogen will get removed and then this negative part we are following the we are following is that's up rules here that's up rules we have to follow so now this negative part then when it will, this hydrogen, when it is get deprotonated, then positive charge will be there and then negative charge will attack at this position.
01:14
So now you can see here and again this h and this br will get removed.
01:20
So then hbr get removed and we have the final product as like this.
01:27
This one...