00:01
This is the answer to chapter 1, problem number 77, from the smith organic chemistry textbook.
00:10
And this problem gives us six molecules and asks us to identify the dipoles in each molecule.
00:19
Or sorry, well, yeah, so it asks us to label the polar bonds in each molecule.
00:25
So that's essentially the same thing as i said at first.
00:28
And then also we're asked to indicate the direction of the net dipole if there is a net dipole in the molecule.
00:35
And so i remember the way that we draw the arrow to show polarity in a bond is the arrowhead going towards the electronegative atom and then a little plus sign at the tail of the arrow to show that that's the possible.
00:58
Part of the polar bond.
01:02
And so for a, bromine is much more electronegative than carbon.
01:08
And so bromine is going to be the negative portion of these polar bonds.
01:14
And so the plus sign toward the tail will be at the carbon end.
01:20
And then so we're also asked to identify the net dipole.
01:24
And so in this instance, it's down because each of these these bromines is pulling a little bit down.
01:31
Obviously, the bottom one is pulling totally down.
01:34
And so that's the net dipole there to the right.
01:39
Okay, and so we'll follow this same pattern for the remaining five parts of this problem.
01:45
So in b, this is diethyl ether.
01:49
Oxygen, of course, is much more electronegative than carbon.
01:54
So each carbon bound to the oxygen is going to have a dipole that looks.
01:59
Looks like that, polar bond that looks like that.
02:03
And the net dipole moment here is also going to be down because the oxygen is pulling each of these a little towards the center and down.
02:16
And so the net, the overall, is a downward effect.
02:20
So then looking at c, so in c for the first time, we have a symmetrical molecule.
02:28
So in c, um, so in c, um, each of these carbon bromine bonds is, of course, polarized, as we would expect, with the bromine being negative and the carbon being positive...