 Okay, so here we've got this ketone, methylphenyl ketone, and we're reacting it with bromine in the solution of acetic acid, okay, and we get this product here, alpha bromide, okay. So what we're going to do is draw the mechanism of this reaction, okay. Remember it's got to go through the enol, okay. So you can try it on your own while I'm doing it, or you can wait for me to do draw the arrows and pretend that you did them and see if you can get the right answer, okay. So I'm going to erase this stuff. Everybody got that copy done? The first thing we want to remember, the acetic acid is a solvent, it's an acid. If we don't know the structure of the acetic acid, it could just be, I guess, but I'll draw it out for us. So that's acetic acid. For these long pairs here, they're slightly basic, and that's what's going to induce the formation of the enolene. We're going to get something like that. Is that what you wrote down? So I think this pen might die too, unfortunately. Remember there's three protons there? Deep protonate. Deep protonate, yes. So we have to make first the enolene. Okay, so remember that. So that's the intermediate in this reaction. And then, of course, we get this, I guess. So that's like the solvent slash catalyst for this reaction. So now we've got that other thing that we had in there. Do you guys remember what that was? Yeah, the R2, to draw it out like that, and kick back down like that, attack that bromine like that, so that's going to be a forward arrow. So that's the driving force of the reaction. Intermediate. You also have the Br minus out there. And we can just say that's going to be the phase plus you. Deep protonate attack. Giving us our final product. Alpha brominated. Does it make sense? Oh, sorry, yeah. Is the enol or one right before it? Are those both considered intermediates? So this is an intermediate. This is an intermediate. This is an intermediate. The one that's the enol is this one. This is just a protonated key. This is also a protonated key. Any other questions? Okay, cool.