Okay. This problem is describing a situation in which we have isolated the are I summer of two core Penton and we're going to react it with seal to on HV, which are the regents needed for radical coronation to make 24 die for plantain. So before we start, let's just ensure that this is the r I summer. So we have a hydrogen in the back that's in the bank. My Korean is Priority one. My, this group is number two, and this group is priority number three that is going in a clockwise direction, which corresponds to the R I summer. So that is correct. As for the reaction, if we erect Seal two and HB with this compound and we should get a situation in which my chlorine in the front is unaffected, my ah, right summer is unaffected and we're just adding a Korean to the fourth position. The reason why it's unaffected is because the this chlorine, the one that position for, is not affecting carbon number two. So there's no changing sterile chemistry. As for the stair chemistry of the Korean at position over four, that is up for grabs because it could be in the front like this. Pour it could be in the back. But no matter what, my chlorine at position number two remains unchanged. It does not. It is not affected by radical coronation by carbon number four because it's uncover number two there unassociated with each other. So you should be. My two answers my two different products. So the question is also asking. And it what? Whether they're president, different ratios. And the answer to that is probably no, because we have a radical coronation in car number four, there is no difference in whether it was on a secondary treasure primary carbon cut on. It was just what whether it was on in the front or in the back. So as far as which one to be president more. I would say that this one would probably be present in a little bit more than this one, but not by much. So it might be approximately 51 49 but more so 50 50. So the reason why I would say this one might be president little bit more is because if this is in the front, this might be taking up more space in the front, making it harder for a radical coronation to be happening in the front as well. So it would occur in the back just because there's more space. Okay, so as for whether these are optically active, that is just a fancy name to say. Are they Cairo? So let's analyze thes. Okay, I'm gonna race to use. Okay, So are these Cairo or not? Let's analyze this one first. So Kyra ality is basically the assumption that a carbon has four different multi four different atoms or molecules attached to that central carbon. So let's analyze this carbon right here. We have 1234 different groups attached to that central carbon. Okay? And we also have four different groups attached to this central carbon. OK, But as as for whether or not their cargo, let's say that we were to marry this image so that we would end up with these Koreans in the back, whether I have the corns in the back or have them in the front as long as they're both in the front or both in the back there considered identical. So this would be even though they have caramel parts, they are identical whether we mirror image them or not. So these would be me. So compounds so not optically active. As for this one, it has Carol parts because we have the four different atoms slash molecules attached to my central carbon and the same for this site. But if we were to mirror image this, I would have a corn in the back and corn in the front. These are not the same. So these this molecule will be considered Cairo and this one would be considered me so or not Carol. So this one is optically active.