Although useful in agriculture as a soil fumigant, methyl bromide is an ozone-depleting chemical, and its production is being phased out. The industrial preparation of methyl bromide is from methanol, by reaction with hydrogen bromide. Write a mechanism for this reaction and classify it as SN1 or SN2.
Hey everyone. So, here we're going to be drawing out a reaction and it's going to be between our methanol. So, go ahead and draw them out, we have methanol, we know it's going to be simply a CH3 attached to an OH that's going to be reacting with hydrogen bromide. So, hydrogen bromide we know that simply just going to be HBr and what do you think is going to happen in our first step? Well, we're reacting in alcohol with the strong acid. So, in our first step we should expect a protonation. So, just remember that this oxygen is going to have two lone pairs, it's going to act as a nucleophile, come in and grab a proton, since we made a bond we need to break the bond between our hydrogen and our bromine, so this bond is going to break. Now, originally we had three lone pairs on our bromine, okay? After this happens what we're going to get is we're going to get a methyl group, we're going to have our O to our H, this is going to be our new proton, this bridge is going to be positively charged and then what we're also going to get is a bromide ion. Now, that's going to have a negative charge because originally it had those three sets of lone pairs but now it's going to have an extra one, okay? So it's going to be negatively charged, so what do you guys think is going to happen now? well, here notice that, maybe you would think that this could break off on its own and we get a carbocation that looks like this but this is very unstable. So, just remember that SN1 will take place more on a secondary or a tertiary leaving group, here we have a methyl leaving group. So, definitely SN2 is going to take place, so what happens in SN2 is we can get rid of this, this negative charge is going to come in and attack this carbon and kick off this leaving group what we get after that, let's go down here, is we're not going to have simply a CH3 attached to our bromine, okay? And we're also going to have is we're going to have plus water because it left, okay? So this will be our fast step and our slow rate determining step will attack our nucleophile, okay? And this is our major expected product. Alright guys, so I hope this made sense and let me know if you have any questions on our SN2 reaction.