0:00:02.320,0:00:04.450 - [Voiceover] Let's say we[br]were asked to name the molecule 0:00:04.450,0:00:06.120 on the top left. 0:00:06.120,0:00:08.290 We would start by numbering our carbons. 0:00:08.290,0:00:12.490 So this would be carbon[br]one, two, three, and four. 0:00:12.490,0:00:16.149 Notice we have a double[br]bond starting at carbon two. 0:00:16.149,0:00:19.720 So the name of this[br]molecule would be 2-butene. 0:00:19.720,0:00:23.290 Two because we have our double[br]bond starting at carbon two. 0:00:23.290,0:00:26.350 "But" because we have four[br]carbons and "ene" because 0:00:26.350,0:00:29.030 we have a double bond[br]presence in the molecule. 0:00:29.030,0:00:31.980 What about naming the[br]molecule on the right? 0:00:31.980,0:00:36.290 We number our carbons[br]one, two, three, and four. 0:00:36.290,0:00:40.230 And once again we have a double[br]bond starting at carbon two. 0:00:40.230,0:00:44.340 So the name of this[br]molecule would be 2-butene. 0:00:44.340,0:00:47.420 However these are two[br]different molecules and 0:00:47.420,0:00:50.350 the reason why is because[br]there's no free rotation 0:00:50.350,0:00:52.090 around a double bond. 0:00:52.090,0:00:55.750 Single bonds have free rotation[br]but double bonds don't. 0:00:55.750,0:00:58.682 So you couldn't rotate the[br]molecule on the left to 0:00:58.682,0:01:01.150 look like the molecule on the right. 0:01:01.150,0:01:04.379 Therefore they must be[br]isomers of each other 0:01:04.379,0:01:08.290 and we need a way to[br]distinguish between our isomers. 0:01:08.290,0:01:12.550 And so one way to do that is[br]to use cis/trans terminology. 0:01:12.550,0:01:14.620 So if we look at the molecule on the left, 0:01:14.620,0:01:17.520 we can see we have two methyl groups. 0:01:17.520,0:01:20.950 And those two methyl[br]groups are on the same side 0:01:20.950,0:01:22.550 of our double bond. 0:01:22.550,0:01:25.490 So if I draw a line in here,[br]it's easier to see those two 0:01:25.490,0:01:27.950 methyl groups are on the same sides. 0:01:27.950,0:01:30.370 And we call that the cis isomer. 0:01:30.370,0:01:32.720 So we put cis in front of our name here. 0:01:32.720,0:01:34.820 I'm attempting to write it in italics. 0:01:34.820,0:01:37.250 So this would be cis-2-butene. 0:01:37.250,0:01:39.320 On the right when we look[br]at those methyl groups, 0:01:39.320,0:01:42.020 these two methyl groups[br]are on opposite sides 0:01:42.020,0:01:43.050 of the double bond. 0:01:43.050,0:01:45.950 So I draw a line in here[br]to make it easier to see 0:01:45.950,0:01:49.420 those two methyl groups[br]are on opposite sides. 0:01:49.420,0:01:51.290 And we call that trans. 0:01:51.290,0:01:53.120 So this is trans isomer. 0:01:53.120,0:01:56.790 I'm going to write trans here[br]in italics, attempt to anyway. 0:01:56.790,0:02:00.090 So we have cis-2-butene[br]and trans-2-butene. 0:02:00.090,0:02:04.220 These are different molecules[br]with different properties. 0:02:04.220,0:02:06.250 If you want to use cis/trans terminology, 0:02:06.250,0:02:08.820 you're looking for two identical groups 0:02:08.820,0:02:10.490 and you are comparing them. 0:02:10.490,0:02:13.250 So let's look at these next two[br]examples here and figure out 0:02:13.250,0:02:16.350 which one is cis and which one is trans. 0:02:16.350,0:02:18.820 We're looking for identical groups. 0:02:18.820,0:02:21.050 So over here we have[br]an ethyl group attached 0:02:21.050,0:02:22.250 to our double bond 0:02:22.250,0:02:25.220 and on the right we have an[br]ethyl group to our double bond. 0:02:25.220,0:02:28.120 Those two ethyl groups[br]are on the same side of 0:02:28.120,0:02:32.100 our double bond so this[br]must be the cis isomer. 0:02:32.100,0:02:35.620 On the right we have this ethyl[br]group and this ethyl group 0:02:35.620,0:02:38.850 on opposite sides of our double bond. 0:02:38.850,0:02:42.150 So that must be the trans isomer. 0:02:42.150,0:02:43.920 All right, let's do some more examples. 0:02:43.920,0:02:45.890 I'll go down to here. 0:02:45.890,0:02:49.320 On the left we have[br]cinnamaldehyde molecule. 0:02:49.320,0:02:52.580 We're looking for two[br]identical groups so we can use 0:02:52.580,0:02:54.410 cis or trans. 0:02:54.410,0:02:56.590 You can also use hydrogens, right. 0:02:56.590,0:02:59.050 You don't have to use a[br]methyl group or an ethyl group 0:02:59.050,0:03:01.750 so if we look at our double[br]bond we know there's a 0:03:01.750,0:03:03.850 hydrogen attached to this carbon 0:03:03.850,0:03:06.790 and we know there's a hydrogen[br]attached to this carbon. 0:03:06.790,0:03:10.550 And those two hydrogens are[br]on opposite sides of our 0:03:10.550,0:03:11.650 double bonds. 0:03:11.650,0:03:14.520 And I'm drawing a line here[br]to make it easier to see. 0:03:14.520,0:03:17.070 Right, these two hydrogens[br]are on opposite side 0:03:17.070,0:03:19.690 so we're talking about trans here. 0:03:19.690,0:03:23.070 Those hydrogens are[br]across from each other. 0:03:23.070,0:03:26.870 What about the tetra-substituted alkene 0:03:26.870,0:03:29.000 on the right? 0:03:29.000,0:03:32.720 We need two identical[br]groups to use our cis/trans 0:03:32.720,0:03:35.550 and here we have an ethyl[br]group, and here we have 0:03:35.550,0:03:36.820 an ethyl group. 0:03:36.820,0:03:38.790 All right over here we[br]have a methyl group and an 0:03:38.790,0:03:40.150 isopropyl group. 0:03:40.150,0:03:43.320 But the two methyl groups[br]are on the same side of 0:03:43.320,0:03:44.450 our double bond. 0:03:44.450,0:03:49.450 So I draw a line in here and[br]we see that these two groups 0:03:49.590,0:03:51.190 are on the same side, 0:03:51.190,0:03:54.690 therefore we're talking about cis here. 0:03:54.690,0:03:58.250 So this double bond has[br]a cis configuration. 0:03:58.250,0:04:00.320 Let's compare the drawing[br]on the left to the 0:04:00.320,0:04:02.010 drawing on the right. 0:04:02.010,0:04:03.820 The first time you look[br]at these two drawings 0:04:03.820,0:04:06.090 you might think these are two isomers, 0:04:06.090,0:04:09.190 and I could use cis/trans[br]terminology to distinguish 0:04:09.190,0:04:10.380 between them. 0:04:10.380,0:04:12.850 However, you can't because these are 0:04:12.850,0:04:16.290 just two ways to represent[br]the same molecule. 0:04:16.290,0:04:18.420 If you picked up this[br]molecule on the left and 0:04:18.420,0:04:19.930 you flipped it up, 0:04:19.930,0:04:22.820 you would get the drawing on the right. 0:04:22.820,0:04:24.580 So they're not isomers of each other. 0:04:24.580,0:04:27.080 This is the same molecule. 0:04:27.080,0:04:31.190 And a fast way to figure that[br]out is to look at this carbon. 0:04:31.190,0:04:34.850 And you can see you have[br]two identical groups bonded 0:04:34.850,0:04:36.750 to that carbon. 0:04:36.750,0:04:39.640 So you can't use cis/trans terminology. 0:04:39.640,0:04:42.690 That's different from the[br]example we did a minute ago. 0:04:42.690,0:04:44.610 We had two identical groups, 0:04:44.610,0:04:46.590 right these two ethyl groups here. 0:04:46.590,0:04:49.120 However those two ethyl[br]groups weren't bonded 0:04:49.120,0:04:50.550 to the same carbon. 0:04:50.550,0:04:53.620 Those two ethyl groups are[br]bonded to different carbons. 0:04:53.620,0:04:56.490 So this ethyl group is bonded[br]to this carbon, and this 0:04:56.490,0:04:59.080 ethyl group is bonded to this carbon. 0:04:59.080,0:05:02.610 So we were able to use[br]cis/trans terminologies. 0:05:02.610,0:05:04.820 So we looked at our[br]double bond and we said 0:05:04.820,0:05:06.990 those two ethyl groups[br]are on the same side of 0:05:06.990,0:05:08.120 our double bond, 0:05:08.120,0:05:12.170 so this represents a cis[br]configuration of the double bond. 0:05:12.170,0:05:14.680 So we can't do that up here because 0:05:14.680,0:05:17.390 while we do have two identical groups, 0:05:17.390,0:05:21.620 those identical groups are[br]bonded to the same carbon.