00:01
Let's draw the substitution product that results when one bromobutane reacts with each nucleophile.
00:09
For a, we have ch3, ch2, ch2, ch2, ch2, br.
00:22
It's going to react with nosoh, out here.
00:35
So go in here.
00:37
We would produce ch3, th2, ch2, ch2, ch2, ch2, ch2, ch2.
00:45
B -o -h and br -minus.
00:55
For b, we have our nucleophile s -h here.
01:17
This will produce h3, ch2, ch2, h2, h2, h2, s -h, and br -minus.
01:33
For c, our nucleophile is c -n -minus.
01:38
Here, this would come in here.
01:51
The product would be ch3, ch2, ch2, ch2, ch2, cn, n, br minus.
02:03
For a d, our nucleophile is och -ch -ch -3 -2.
02:20
O -c -h -c -3 -2.
02:28
Over here, this would come in here.
02:35
It would be c -h -3 -2.
02:39
Ch2, ch2, ch2, o, ch, ch3, 2, o, ch, ch, ch3, 2, nbr minus.
02:51
For e, there's a reactant...