00:01
So we have the following organic compound.
00:03
We have two iota, two methylbutane, and we want to know, we're going to react it with ethanol, and we want to know what would happen to the rate of this reaction, given the following reaction conditions.
00:13
Well, first, we have to know what kind of reaction it is in order to determine how the rate will be affected.
00:18
And we see that we have a iota, tertiary iota group, and therefore it's likely, since our ethanol is a weak movement file, that this is going to occur via an sn1 type reaction.
00:34
However, ethanol may also act as a weak base, depending on how much heat is added into the reaction, we may get an e1 elimination reaction as well.
00:44
So we can predict the products to look something like this, where we will go ahead and replace our iota group with the hydroxide group for our sn1, and for our elimination, it's going to look something like this.
01:10
Where we get the major product is this because of zyte -sebs rule.
01:19
So what's important to realize here is that we have, whether it's an s -n -1 or e -1 reaction, is that they're both the unimolecular.
01:30
So that the rate of this reaction only depends on our alkali, because we have to remember that these reactions proceed via a carbocadion intermediate, and the rate limiting step is our leaving group, which is how fast our leaving group leaves.
01:47
And therefore our rate is going to depend on the concentration of our iota group or our alkylo halide.
02:00
So we would say that at the rate depends only on the concentration of alkyl halide, which we go ahead and write rx here...