00:03
In this problem, the mechanism for a natural product cyclization is to be written.
00:13
The starting machine base is in the structure.
00:34
This will undergo formation of a bromonymione with a bromeliomeo here.
00:57
Then it forms a carbopathyone, a cyclic carbotion and then forms a final product with a cure anipronium and double form.
01:48
Now in order to find out the mechanism of the reaction, we have to first point the bonds which are formed and bonds which are broken.
02:10
So for that purpose we will number these molecules, this leads of the carbon bond, 1, 2, 3, 4, 5, 6, 7, 8, 14, and 15.
02:36
So if i try to identify the same molecules here, you see that the second ring, seems to be intact so this should be 15 it is 12 14 11 9 then this is 7 and 8 here 6 5 4 3 1 and 2 so we'll identify the balls that are formed 3 9 bond is born 3 9 sigma ball and then 6 7 5 5 5 5 5 the bond is broken is 795 bond and this is a double point from the left in this is lost from here...