00:01
This question asks us why the reaction of an alkyl halide with ammonia gives us a low yield of a primary mean for part a.
00:07
So if we have ammonia and i'm going to use methyl bromide as my alkyl halide, if we think about this reaction, this will be an s &2 reaction and our products would look like this.
00:22
This would be positive, but then it would probably be deprotonated, especially if we're in basic conditions.
00:29
And then, but this product is still a nucleophile.
00:35
This end still has a lone pair on it, so this reaction can happen again and again.
00:39
So if we reacted with another equivalent of methyl bromide, we would get now this secondary amine.
00:46
And then if we did it again, we would get a tertiary amine.
00:52
Oops, we don't have h anymore.
00:57
And then if we did it again, we would get a quaternary ammonium salt.
01:04
That would look like this.
01:07
So the reason that we don't get very much of that primary amine is because it will continue to react as long as it is still a nucleophile, as long as it still has that will impair on it until it gets to the point where it can't take on anymore and the nitrogen becomes positive...