00:01
Okay, so for this problem, you have pairs of reactions that differ by one key component.
00:10
And based on that one single difference, you want to figure out which of the reactions in each of these pairs is going to occur at a faster rate.
00:23
And, okay, so let's look at this first one.
00:28
Right? so this first one, you have a methyl bromide.
00:36
And so because you have a methyl attached to a leaven group, this is going to tell you that you have an sn2 reaction.
00:43
Okay, so everything that we want to consider, and actually each of these problems is sn2, we want to think about what's going to give us a faster reaction for an sn2 over an sn1 reaction.
00:58
So for this one, if we look, the only difference is between our nucleophile.
01:05
So we have a hydropside versus water.
01:09
And so in this case, hydroxide is actually a better nucleophile because it has a negative charge concentrated on that oxygen.
01:22
So because this is a better nucleophile in our sn2 reaction, this reaction rate is going to be faster because oh is a better nucleophile when you're comparing that versus water.
01:48
So now let's look in our second reaction.
01:50
So in this case, our nucleophiles are the same.
01:53
We have hydroxide.
01:55
Now the difference is in leaving group ability.
01:59
Right.
01:59
So if you remember back to acidity, when you're talking about, or when you're talking about leaving groups, your better leaving group is going to have a weaker, is going to be a weaker conjugate base.
02:23
And so its conjugate acid is going to be stronger.
02:26
So if we compare h .i.
02:31
Versus h .c .l.
02:34
H .i.
02:35
Is a stronger acid.
02:39
So it's going to have a weaker conjugate base.
02:43
I'm just abbreviate at cb...