 OK, so good job, Renee, for figuring out this mechanism. But let's go through it. So this, actually, should be access. OK, so what's going to happen first is the Grignard addition to the acid chloride making the ketone intermediate, then addition again to that ketone making the alkoxide. In this case, it'll be a tertiary alkoxide. Then when you add acid to that, that's going to protonate that alkoxide. And then that resulting alcohol will get protonated. And then, like Renee said, you could do an E1 or an E2 reaction in this case. Probably you're going to do the E1 because you're going to get that tertiary carbocation. It's very stable. But you can imagine either one of them going. So if you don't remember from organic one, you've got to remember all this stuff, OK? So let's just go ahead and do this mechanism. Does that hold up where I want to do it? So like we said, ethyl magnesium bromide, I like to put the arrow through the bit. And in this case, since we're going to further react it, I'm going to actually put instead of just the abbreviation for ethyl, the actual ethyl group, like that. Of course, that's going to collapse back down, knocking the CL out because your ketamine intermediate that we were talking about. Next thing that's going to happen, well, we've got excess of that ethyl magnesium bromide. So remember, when we react excess Griniard with an acid chloride, it goes to the tertiary alcohol. So we're going to get, of course, not the alcohol yet, but the tertiary alcohol side. Because Griniard reagents make a basic solution. One of these things that we're always constantly trying to harp on is that in a basic solution, you've got to have negative charges only or neutral stuff. In an acidic solution, you've got to have positive stuff and neutral stuff. If you've got negative charges in an acidic solution, you know you've done something wrong or the opposite. So there's Sharon's little man, right? And so the H2SO4 reaction, it's got some water in there. So we'll just say we're using the hydronium ion. And even if you were using sulfuric acid, of course, that's going to prognate the alkoxide. We're going to have another prognation of that, but this time the alcohol. So here's the tertiary alcohol. So we're going to prognate again. And like we were saying, the next step could be in dispute. For me, I would say it's probably reasonable to say that it would come off before the deprognation for a couple of reasons, because this hydrogen here is hard to get to. And this will make a stable carbocation. So the elimination here is going to take two steps more than likely. So that's an E1 mechanism. So leaving up a leaving group, can I erase the top part? That tertiary carbocation. And of course, we've got water, also the product of that reaction. And maybe that water molecule, probably not. But since that's the water molecule I just wrote, that's the one I'm going to use for the elimination or the deprognation step, I guess, organic products. So there's another organic synthesis thing. So and I guess we should probably write that we get the hydronium ion back, so there's no confusion about where those are. Are there any questions on this one? So good job bringing it.