For each of the following molecules, draw a 3-D representation...

Question

For each of the following molecules, draw a 3-D representation.

(d) CHClBrI

Accepted Answer

Hi, everyone. And welcome back. Our next problem says, draw the 3D representation of CHDFC. And then we have four possible structures drawn here la A through D. We'll look at them a little more closely as we think about what our structure should be. So this definitely looks a little weird as molecular formulas go. But let's recognize this as a five atom molecule that has a carbon as its central atom. So that first atom of C give us our central atom of carbon carbon likes to form four bonds. And then we have four other atoms that could bond to. So it has four bonds because we have four additional atoms. Now, we don't always have the same number of atoms and bonds. We could have things like lone pairs. But we know here carbon has four valence electrons. It likes to form four bonds and with those four bonds and no lone pairs, since there's no other valence electrons, we have tetrahedral geometry. So that tetrahedral arrangement means our fore bonds are all 109.5 degrees apart as a bond angle. Now, the trick is that's in three dimensional space and we need to draw two dimensional structure. So we do that, as you can see in our questions with solid lines and then wedges and dashed wedges indicating the three axes of our 3d arrangement. So that would be one is within the plane of our screen and one is coming up out of the plane so that our plane has the X and y axis and then the z axis is coming up out of the plane and then back behind it, and a solid wedge indicates the bond coming up towards us and the dash wedge going back away from us. So a solid wedge going up towards the viewer, a dash wedge back away from the viewer. So both of those out of the plane of the screen or the paper if you're drawing on paper. So first we draw two bonds within the plane and we usually draw them about 100 20 degrees apart, not at the actual bond lang angle, just as a factor of the fact that we're drawing in a two D space. So I personally like to draw a sort of tripod arrangement that helps me visualize my tetrahedral geometry. And in that arrangement, I draw my two bonds within the plane, one uh as a vertical line and then one kind of going down into the right again at about 100 20 degree angle. And then I draw my solid wedge sort of directly underneath and I'm sort of off to the side at a 180 degree angle for my vertical line and then my dash wedge off to the left side. So that also we kind of envision a tripod type shape. But we can see when we look, we're comparing it to our multiple choice answers. And they're drawn a bit differently with the bonds within the plane being drawn as sort of a stretched out V again about 100 20 degrees apart. And then we would need a solid wedge and a dashed wedge. So we often put the solid wedge off to the front left and the dash wedge off to the front. Right. Now, you see again, those are not representing actual bond angles, but in the actual molecule, if it were arranged like that, those two bonds would overlap each other. It's all right, in actual mo molecule. So we draw them sort of staggered just so that we can see what's on those bonds. It wouldn't be useful to us if they were actually overlapping. So we have our four bond angles drawn. Now, we just need to fill in the atoms. Well, again, the central atom is just the center of these lines being a carbon. When we draw a skeletal line structure, like this carbons are just the ends of the line segments, they're not actually indicated directly. So it, it is not labeled on here. And then we have four other atoms. So h we have hydrogen, the D here, we as it looks a little weird, that's actually deuterium, which is an isotope of hydrogen. So, isotope of hydrogen with two neutrons. So it has the same number of protons and electrons, but it has two neutrons. So it's heavier than hydrogen. It's used when you want to label or track the fate of a specific hydrogen as you go through reaction because it will act the same way as hydrogen. It's just got one electron, it will form one bond. But being heavier, you can look at it by a spectroscopy or other methods and always identify it. So you can trace the fate of one particular hydrogen atom through a reaction. So we've got hydrogen deuterium, our at is fluorine and our c chlorine, we're only given a molecular formula, no indication of the bonds. So we can put these in any order. So again, I'm looking to make them match with one of my multiple choice. And all my multiple choices have H and D on the top F and cl on the bottom. But again, it, it doesn't matter. We have this tetrahedral arrangement and since we were not given any other indication, they can go anywhere around on those bonds. So let's compare them to our multi choice option in choice. A, we have the two planar bonds as that sort of stretched out. B we have a solid wedge, we have a dashed wedge and our atoms on their H DC LF as we go around clockwise. So choice A is correct. It corresponds to what we've drawn with that tetrahedral arrangement with the four atoms around a central carbon. When we look at Choice B, well, we've got our planer bonds, but then we have two solid wedges, there's no dashed wedge to indicate a bond going back behind the plane of the screen. So choice B is incorrect. Choice C, we have two solid wedges and two dashed wedges. So we've got no planar bonds. So that's incorrect. And then in choice D we have our planar bonds, but then we have two dashed wedges. So we have no solid wedge. So that's what's wrong with structure G. So once again, our proper 3d 3d representation of this CHDFCL molecular formula is choice A. See you in the next video.

For each of the following molecules, draw a 3-D representation.
(d) CHClBrI

Key Concepts

Molecular Geometry

VSEPR Theory

Stereochemistry

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