WEBVTT 00:00:02.320 --> 00:00:04.450 - [Voiceover] Let's say we were asked to name the molecule 00:00:04.450 --> 00:00:06.120 on the top left. 00:00:06.120 --> 00:00:08.290 We would start by numbering our carbons. 00:00:08.290 --> 00:00:12.490 So this would be carbon one, two, three, and four. 00:00:12.490 --> 00:00:16.149 Notice we have a double bond starting at carbon two. 00:00:16.149 --> 00:00:19.720 So the name of this molecule would be 2-butene. 00:00:19.720 --> 00:00:23.290 Two because we have our double bond starting at carbon two. 00:00:23.290 --> 00:00:26.350 "But" because we have four carbons and "ene" because 00:00:26.350 --> 00:00:29.030 we have a double bond presence in the molecule. 00:00:29.030 --> 00:00:31.980 What about naming the molecule on the right? 00:00:31.980 --> 00:00:36.290 We number our carbons one, two, three, and four. 00:00:36.290 --> 00:00:40.230 And once again we have a double bond starting at carbon two. 00:00:40.230 --> 00:00:44.340 So the name of this molecule would be 2-butene. 00:00:44.340 --> 00:00:47.420 However these are two different molecules and 00:00:47.420 --> 00:00:50.350 the reason why is because there's no free rotation 00:00:50.350 --> 00:00:52.090 around a double bond. 00:00:52.090 --> 00:00:55.750 Single bonds have free rotation but double bonds don't. 00:00:55.750 --> 00:00:58.682 So you couldn't rotate the molecule on the left to 00:00:58.682 --> 00:01:01.150 look like the molecule on the right. 00:01:01.150 --> 00:01:04.379 Therefore they must be isomers of each other 00:01:04.379 --> 00:01:08.290 and we need a way to distinguish between our isomers. 00:01:08.290 --> 00:01:12.550 And so one way to do that is to use cis/trans terminology. 00:01:12.550 --> 00:01:14.620 So if we look at the molecule on the left, 00:01:14.620 --> 00:01:17.520 we can see we have two methyl groups. 00:01:17.520 --> 00:01:20.950 And those two methyl groups are on the same side 00:01:20.950 --> 00:01:22.550 of our double bond. 00:01:22.550 --> 00:01:25.490 So if I draw a line in here, it's easier to see those two 00:01:25.490 --> 00:01:27.950 methyl groups are on the same sides. 00:01:27.950 --> 00:01:30.370 And we call that the cis isomer. 00:01:30.370 --> 00:01:32.720 So we put cis in front of our name here. 00:01:32.720 --> 00:01:34.820 I'm attempting to write it in italics. 00:01:34.820 --> 00:01:37.250 So this would be cis-2-butene. 00:01:37.250 --> 00:01:39.320 On the right when we look at those methyl groups, 00:01:39.320 --> 00:01:42.020 these two methyl groups are on opposite sides 00:01:42.020 --> 00:01:43.050 of the double bond. 00:01:43.050 --> 00:01:45.950 So I draw a line in here to make it easier to see 00:01:45.950 --> 00:01:49.420 those two methyl groups are on opposite sides. 00:01:49.420 --> 00:01:51.290 And we call that trans. 00:01:51.290 --> 00:01:53.120 So this is trans isomer. 00:01:53.120 --> 00:01:56.790 I'm going to write trans here in italics, attempt to anyway. 00:01:56.790 --> 00:02:00.090 So we have cis-2-butene and trans-2-butene. 00:02:00.090 --> 00:02:04.220 These are different molecules with different properties. 00:02:04.220 --> 00:02:06.250 If you want to use cis/trans terminology, 00:02:06.250 --> 00:02:08.820 you're looking for two identical groups 00:02:08.820 --> 00:02:10.490 and you are comparing them. 00:02:10.490 --> 00:02:13.250 So let's look at these next two examples here and figure out 00:02:13.250 --> 00:02:16.350 which one is cis and which one is trans. 00:02:16.350 --> 00:02:18.820 We're looking for identical groups. 00:02:18.820 --> 00:02:21.050 So over here we have an ethyl group attached 00:02:21.050 --> 00:02:22.250 to our double bond 00:02:22.250 --> 00:02:25.220 and on the right we have an ethyl group to our double bond. 00:02:25.220 --> 00:02:28.120 Those two ethyl groups are on the same side of 00:02:28.120 --> 00:02:32.100 our double bond so this must be the cis isomer. 00:02:32.100 --> 00:02:35.620 On the right we have this ethyl group and this ethyl group 00:02:35.620 --> 00:02:38.850 on opposite sides of our double bond. 00:02:38.850 --> 00:02:42.150 So that must be the trans isomer. 00:02:42.150 --> 00:02:43.920 All right, let's do some more examples. 00:02:43.920 --> 00:02:45.890 I'll go down to here. 00:02:45.890 --> 00:02:49.320 On the left we have cinnamaldehyde molecule. 00:02:49.320 --> 00:02:52.580 We're looking for two identical groups so we can use 00:02:52.580 --> 00:02:54.410 cis or trans. 00:02:54.410 --> 00:02:56.590 You can also use hydrogens, right. 00:02:56.590 --> 00:02:59.050 You don't have to use a methyl group or an ethyl group 00:02:59.050 --> 00:03:01.750 so if we look at our double bond we know there's a 00:03:01.750 --> 00:03:03.850 hydrogen attached to this carbon 00:03:03.850 --> 00:03:06.790 and we know there's a hydrogen attached to this carbon. 00:03:06.790 --> 00:03:10.550 And those two hydrogens are on opposite sides of our 00:03:10.550 --> 00:03:11.650 double bonds. 00:03:11.650 --> 00:03:14.520 And I'm drawing a line here to make it easier to see. 00:03:14.520 --> 00:03:17.070 Right, these two hydrogens are on opposite side 00:03:17.070 --> 00:03:19.690 so we're talking about trans here. 00:03:19.690 --> 00:03:23.070 Those hydrogens are across from each other. 00:03:23.070 --> 00:03:26.870 What about the tetra-substituted alkene 00:03:26.870 --> 00:03:29.000 on the right? 00:03:29.000 --> 00:03:32.720 We need two identical groups to use our cis/trans 00:03:32.720 --> 00:03:35.550 and here we have an ethyl group, and here we have 00:03:35.550 --> 00:03:36.820 an ethyl group. 00:03:36.820 --> 00:03:38.790 All right over here we have a methyl group and an 00:03:38.790 --> 00:03:40.150 isopropyl group. 00:03:40.150 --> 00:03:43.320 But the two methyl groups are on the same side of 00:03:43.320 --> 00:03:44.450 our double bond. 00:03:44.450 --> 00:03:49.450 So I draw a line in here and we see that these two groups 00:03:49.590 --> 00:03:51.190 are on the same side, 00:03:51.190 --> 00:03:54.690 therefore we're talking about cis here. 00:03:54.690 --> 00:03:58.250 So this double bond has a cis configuration. 00:03:58.250 --> 00:04:00.320 Let's compare the drawing on the left to the 00:04:00.320 --> 00:04:02.010 drawing on the right. 00:04:02.010 --> 00:04:03.820 The first time you look at these two drawings 00:04:03.820 --> 00:04:06.090 you might think these are two isomers, 00:04:06.090 --> 00:04:09.190 and I could use cis/trans terminology to distinguish 00:04:09.190 --> 00:04:10.380 between them. 00:04:10.380 --> 00:04:12.850 However, you can't because these are 00:04:12.850 --> 00:04:16.290 just two ways to represent the same molecule. 00:04:16.290 --> 00:04:18.420 If you picked up this molecule on the left and 00:04:18.420 --> 00:04:19.930 you flipped it up, 00:04:19.930 --> 00:04:22.820 you would get the drawing on the right. 00:04:22.820 --> 00:04:24.580 So they're not isomers of each other. 00:04:24.580 --> 00:04:27.080 This is the same molecule. 00:04:27.080 --> 00:04:31.190 And a fast way to figure that out is to look at this carbon. 00:04:31.190 --> 00:04:34.850 And you can see you have two identical groups bonded 00:04:34.850 --> 00:04:36.750 to that carbon. 00:04:36.750 --> 00:04:39.640 So you can't use cis/trans terminology. 00:04:39.640 --> 00:04:42.690 That's different from the example we did a minute ago. 00:04:42.690 --> 00:04:44.610 We had two identical groups, 00:04:44.610 --> 00:04:46.590 right these two ethyl groups here. 00:04:46.590 --> 00:04:49.120 However those two ethyl groups weren't bonded 00:04:49.120 --> 00:04:50.550 to the same carbon. 00:04:50.550 --> 00:04:53.620 Those two ethyl groups are bonded to different carbons. 00:04:53.620 --> 00:04:56.490 So this ethyl group is bonded to this carbon, and this 00:04:56.490 --> 00:04:59.080 ethyl group is bonded to this carbon. 00:04:59.080 --> 00:05:02.610 So we were able to use cis/trans terminologies. 00:05:02.610 --> 00:05:04.820 So we looked at our double bond and we said 00:05:04.820 --> 00:05:06.990 those two ethyl groups are on the same side of 00:05:06.990 --> 00:05:08.120 our double bond, 00:05:08.120 --> 00:05:12.170 so this represents a cis configuration of the double bond. 00:05:12.170 --> 00:05:14.680 So we can't do that up here because 00:05:14.680 --> 00:05:17.390 while we do have two identical groups, 00:05:17.390 --> 00:05:21.620 those identical groups are bonded to the same carbon.