00:01
So how might you prepare each of the following molecules using a nucleophilic substitution reaction at some step in the process? so to start off with a, we have ch3br plus n -a plus minus c is triple bonded to c -c -h -ch -c -3 -2.
00:26
And this reacts and goes to ch3c triple bonded to cchh ch32 plus n -a -br so the strong base ch3c 2211 to the c brings out the elimination and this produces two separate products for part b we have h3c attached to the bromine plus o negative attached to c, c, c -h -3 -3, goes to the h -3c, bonded to the oxygen, bonded to the c, c -h -33, plus the n -a -b -r.
01:23
And so then you notice here that the n -a -b -r is the common nucleophilic reaction here.
01:28
So for c, we have c .h3, c .h2, c .h2, c .h2, c .r plus naccn.
01:41
And that reacts, and you get c .h3, c .h2, c .n.
01:49
Plus, nabr...