00:01
So this one is going to include four intermediates in a final major product.
00:07
Some of these that are given try to trick you with some intermediate products and some incorrect elimination substitutions and reactions.
00:14
But i'm going to take you to step by step.
00:18
So we're first started with toluene, which is a methyl attached to a benzene ring.
00:23
And it is being attacked by hydrofluic acid and propene.
00:31
And what this propene does in the hydrologic acid is it creates a carbocation like this.
00:40
That's at the second position.
00:42
I don't know why it's not going to write a plus sign.
00:44
There's a plus sign at that second position.
00:48
And because the methyl group is an ortho -paro -director, opie, is going to attack the ortho or paraposition.
00:58
As i can show you, the two closest ones are the ortho positions.
01:02
And for this one is the parapetition.
01:04
We're going to stick with the para product here because it's usually the most preferred when there's open spots.
01:10
This is the para product, but it's obviously going to be 50 -50.
01:13
So when we do that, we form our second product right here.
01:20
And when we react any type of strong acid with na2, cr2 -07, is going to turn any ch bond that's attached off of the benzene ring into a carboxic acid.
01:32
So c -o -o -h, basically.
01:34
A carbonyl and an alcohol, it creates a carboxic acid with any h bond.
01:39
So we have an h here and we have an h here.
01:43
So that will turn into two coohs, as we can see right here.
01:49
Then we're reacting hno3 with h2so4.
01:52
Because both of the carboxic acids are both metered directors and it's symmetric at this point, it will attack any of the four circles you can see in this position...