00:01
Okay, so we have to write out the mechanism to show how the reaction occurs.
00:07
So we have tri -ethyl bromide.
00:10
Again, tri -meat, and we have three ethels, so ch2, c -h -3, attached to the carbon and the bromine, reacting with methanol, which is a polar prudic solvent.
00:36
So because we have a tertiary aquilite, and it's a polar prudic solvent, this is going to proceed with the sn1 -e -1 mechanism, and they occur together.
00:57
So one's the major product, and one is the minor product, and s -n -1 occurs in two steps.
01:05
The first step is for the aquil halide to leave, because the carbon is very hysterically hindered by the ethels and the bromide, and it would make it very difficult for the oxygen to attack.
01:21
So we have a tertiary carbon -cadion, which is very stable, because these three r groups are electron donating groups that help to stabilize the carbocation.
01:36
The next step is for the oxygen, all methanol, to donate a pair of electrons.
01:43
And the attack only occurs in the front because the back is hysterically hindered.
01:49
It leads to this.
01:56
And oxygen now has a positive charge because it donated a pair of electrons that were negative...