Write an equation for the proton transfer reaction that occurs when each of the following bases reacts with water. In each case, draw curved arrows that show the mechanism of the proton transfer:
Alright guys. In this problem we need to go ahead and figure out what happens in this reaction, where we have a four carbon branched, right? We have a three carbon chain with one carbon coming off in the middle and we have a negative charge there in the middle and we have water in the area in the solution, right? What's going to happen here? Well, we know it's a proton transfer, right? The negative charge is kind of a giveaway but guys, let's go ahead and talk about our pKa's, what's the pKa of a regular alkane which is what we have here? pKa of of the protonated version is 50, right? And that's how we calculate these things 50, what about this? the pKa of this, that's only 15.7, right? So guys, this is a huge difference, right? This is 34.3 difference, so this is 10 to the 34.3 difference, that's insane, so guys, what do you think is going to happen? Well, this is going to come in and grab this hydrogen and kick electrons on to that O, but now guys my question is, is this going to be a reversible reaction? are we going to have a straight arrow like that? where are we going to have equilibrium arrows, we're most slightly just going to have a straight arrow in one direction because it's just the sheer magnitude of difference, so what we're going to end up with it something like this, right? where we just have our, our four carbons here, with no charge, right? And we know that our hydrogen is implied here now and now we also have our OH negative, alright? Let's move on.