alright, guys. So let's try to attempt to answer this question. I was left on the bottom dealing with hydrogenation. So remember, we need one mole off H two to remove one pie bond. And remember, a double bond has one pie bond and one Sigma bond. We don't care about the Sigma because we can't get rid of Sigma's. A triple bond has two PI bonds and one Sigma bond. So if we look, what do we have? We have one double bond, so that's one pie. We have another double bond, so that's a second. And then we have a triple bond, which adds another two pi bonds. So we have a total off four pi bonds. And remember, for every pi bon, we need one mole of H two. Since we have four pi bonds, that means we need four moles of H two to completely reduced this compound. So our answer here would be four moles of H two and using formals of H two on this. It would give us this as our answer. We get rid of that double bond in the ring, these two carbons, each of them would get another hydrogen the double bomb would disappear. You can draw the hydrogen anyway you want, and then here there's a carbon there. But there's also a carbon here that's not shown. So they gained two. Hydrogen is there, and that would be your answer at the end, where we've completely reduced it by adding four moles of H two.