00:01
Okay, we have methanol and a secondary bromoalkane.
00:12
And we are reacting this in the presence of methanol.
00:19
So recall that secondary bromoalkanes can undergo all four reaction types, but that methanol, our reagent here, is a weak nucleophile and a weak base.
00:39
Weak nucleophiles and weak bases will rule out sn2 and e2, leaving sn1 and e1 as our remaining options.
00:52
Now note that this is a solvolysis, meaning this reaction is not just methanol as a reactant, but in fact this reaction is taking place in methanol.
01:05
This tells us that we are in a polar protic solvent.
01:15
And due to the hydrogen bonding between methanol and other methanol molecules that are the solvent, the nucleophilicity of methanol is decreased, whereas the basicity is less hindered or less affected by the hindrance of the hydrogen bonding networks, which means that we will favor e1, sn1.
01:39
Therefore we can determine that the major product of this reaction would be an e1 reaction.
01:46
What does the e1 reaction of this look like? well, we've been asked to show what the appropriate flow of electrons is in the starting materials.
01:56
The first step in e1 is that bromine leaves, generating the following carbocation, at which point the methanol will attack, leaving the final step to be deprotonation, either by a second methanol molecule or by the bromine...