Everyone, in this example question, it says, draw the mechanism for the acid catalyzed hydration under specific and general acid catalysis. Alright, let's work on specific catalysis first. Remember, this happens in a non-concerted step, meaning it happens in more than one step.

So let's start with our specific catalysis. We have our structure here, and here's our OH group. We're going to react it such that this π bond will break and grab an H, then the A flies away for now. We're going to do this by Markovnikov's rule, which means that the H will go to the less substituted alkene carbon, the less substituted alkene carbon. So it will go to this one here that's a CH₂, making it a CH₃, which would mean this carbon here becomes positive. Because that carbon is positive, that is the carbon we are going to attack with this hydroxyl group. We'll use one of its lone pairs and attach right there.

So let's draw this cyclic portion first. The oxygen is connecting to this carbon here. So, let’s count: 1, 2, 3, 4, 5. We're going to make a 5-membered ring and connect it here. We still have this methyl group over here. The oxygen is making 3 bonds, so it's positively charged here. The A- that left comes back to deprotonate that oxygen to make it neutral. So this comes in here, grabs this H here. So what we get at the end is this: Here goes our neutral oxygen. We have our 5-membered ring that we created as a result, and the methyl group that's on it.

This is what we get in the end through our specific catalysis. Now, if we're going to do general catalysis, remember this happens by concerted steps, so it's all at one time. So let's look at general now. Here we're going to draw our structure again. And so then we're going to have this. And then we’re going to have our reaction all in one step. We can have this hydroxyl group hitting here at the same time this pi bond is breaking to grab this, and this bond is breaking here. So we have quite a few arrows moving here.

So then we're going to get this structure; again, here goes our oxygen making three bonds and being positively charged. And then A- comes back deprotonating that oxygen. So now we're going to have our neutral structure at the end, and we have lone pairs here. I didn't show them in the top one, but they're there. Right. So both processes happen by nonconcerted and concerted effort. But, at the end, we're still going to get our same product.

This is what happens if we're following our specific catalysis versus our general catalysis.