1 00:00:02,320 --> 00:00:04,450 - [Voiceover] Let's say we were asked to name the molecule 2 00:00:04,450 --> 00:00:06,120 on the top left. 3 00:00:06,120 --> 00:00:08,290 We would start by numbering our carbons. 4 00:00:08,290 --> 00:00:12,490 So this would be carbon one, two, three, and four. 5 00:00:12,490 --> 00:00:16,149 Notice we have a double bond starting at carbon two. 6 00:00:16,149 --> 00:00:19,720 So the name of this molecule would be 2-butene. 7 00:00:19,720 --> 00:00:23,290 Two because we have our double bond starting at carbon two. 8 00:00:23,290 --> 00:00:26,350 "But" because we have four carbons and "ene" because 9 00:00:26,350 --> 00:00:29,030 we have a double bond presence in the molecule. 10 00:00:29,030 --> 00:00:31,980 What about naming the molecule on the right? 11 00:00:31,980 --> 00:00:36,290 We number our carbons one, two, three, and four. 12 00:00:36,290 --> 00:00:40,230 And once again we have a double bond starting at carbon two. 13 00:00:40,230 --> 00:00:44,340 So the name of this molecule would be 2-butene. 14 00:00:44,340 --> 00:00:47,420 However these are two different molecules and 15 00:00:47,420 --> 00:00:50,350 the reason why is because there's no free rotation 16 00:00:50,350 --> 00:00:52,090 around a double bond. 17 00:00:52,090 --> 00:00:55,750 Single bonds have free rotation but double bonds don't. 18 00:00:55,750 --> 00:00:58,682 So you couldn't rotate the molecule on the left to 19 00:00:58,682 --> 00:01:01,150 look like the molecule on the right. 20 00:01:01,150 --> 00:01:04,379 Therefore they must be isomers of each other 21 00:01:04,379 --> 00:01:08,290 and we need a way to distinguish between our isomers. 22 00:01:08,290 --> 00:01:12,550 And so one way to do that is to use cis/trans terminology. 23 00:01:12,550 --> 00:01:14,620 So if we look at the molecule on the left, 24 00:01:14,620 --> 00:01:17,520 we can see we have two methyl groups. 25 00:01:17,520 --> 00:01:20,950 And those two methyl groups are on the same side 26 00:01:20,950 --> 00:01:22,550 of our double bond. 27 00:01:22,550 --> 00:01:25,490 So if I draw a line in here, it's easier to see those two 28 00:01:25,490 --> 00:01:27,950 methyl groups are on the same sides. 29 00:01:27,950 --> 00:01:30,370 And we call that the cis isomer. 30 00:01:30,370 --> 00:01:32,720 So we put cis in front of our name here. 31 00:01:32,720 --> 00:01:34,820 I'm attempting to write it in italics. 32 00:01:34,820 --> 00:01:37,250 So this would be cis-2-butene. 33 00:01:37,250 --> 00:01:39,320 On the right when we look at those methyl groups, 34 00:01:39,320 --> 00:01:42,020 these two methyl groups are on opposite sides 35 00:01:42,020 --> 00:01:43,050 of the double bond. 36 00:01:43,050 --> 00:01:45,950 So I draw a line in here to make it easier to see 37 00:01:45,950 --> 00:01:49,420 those two methyl groups are on opposite sides. 38 00:01:49,420 --> 00:01:51,290 And we call that trans. 39 00:01:51,290 --> 00:01:53,120 So this is trans isomer. 40 00:01:53,120 --> 00:01:56,790 I'm going to write trans here in italics, attempt to anyway. 41 00:01:56,790 --> 00:02:00,090 So we have cis-2-butene and trans-2-butene. 42 00:02:00,090 --> 00:02:04,220 These are different molecules with different properties. 43 00:02:04,220 --> 00:02:06,250 If you want to use cis/trans terminology, 44 00:02:06,250 --> 00:02:08,820 you're looking for two identical groups 45 00:02:08,820 --> 00:02:10,490 and you are comparing them. 46 00:02:10,490 --> 00:02:13,250 So let's look at these next two examples here and figure out 47 00:02:13,250 --> 00:02:16,350 which one is cis and which one is trans. 48 00:02:16,350 --> 00:02:18,820 We're looking for identical groups. 49 00:02:18,820 --> 00:02:21,050 So over here we have an ethyl group attached 50 00:02:21,050 --> 00:02:22,250 to our double bond 51 00:02:22,250 --> 00:02:25,220 and on the right we have an ethyl group to our double bond. 52 00:02:25,220 --> 00:02:28,120 Those two ethyl groups are on the same side of 53 00:02:28,120 --> 00:02:32,100 our double bond so this must be the cis isomer. 54 00:02:32,100 --> 00:02:35,620 On the right we have this ethyl group and this ethyl group 55 00:02:35,620 --> 00:02:38,850 on opposite sides of our double bond. 56 00:02:38,850 --> 00:02:42,150 So that must be the trans isomer. 57 00:02:42,150 --> 00:02:43,920 All right, let's do some more examples. 58 00:02:43,920 --> 00:02:45,890 I'll go down to here. 59 00:02:45,890 --> 00:02:49,320 On the left we have cinnamaldehyde molecule. 60 00:02:49,320 --> 00:02:52,580 We're looking for two identical groups so we can use 61 00:02:52,580 --> 00:02:54,410 cis or trans. 62 00:02:54,410 --> 00:02:56,590 You can also use hydrogens, right. 63 00:02:56,590 --> 00:02:59,050 You don't have to use a methyl group or an ethyl group 64 00:02:59,050 --> 00:03:01,750 so if we look at our double bond we know there's a 65 00:03:01,750 --> 00:03:03,850 hydrogen attached to this carbon 66 00:03:03,850 --> 00:03:06,790 and we know there's a hydrogen attached to this carbon. 67 00:03:06,790 --> 00:03:10,550 And those two hydrogens are on opposite sides of our 68 00:03:10,550 --> 00:03:11,650 double bonds. 69 00:03:11,650 --> 00:03:14,520 And I'm drawing a line here to make it easier to see. 70 00:03:14,520 --> 00:03:17,070 Right, these two hydrogens are on opposite side 71 00:03:17,070 --> 00:03:19,690 so we're talking about trans here. 72 00:03:19,690 --> 00:03:23,070 Those hydrogens are across from each other. 73 00:03:23,070 --> 00:03:26,870 What about the tetra-substituted alkene 74 00:03:26,870 --> 00:03:29,000 on the right? 75 00:03:29,000 --> 00:03:32,720 We need two identical groups to use our cis/trans 76 00:03:32,720 --> 00:03:35,550 and here we have an ethyl group, and here we have 77 00:03:35,550 --> 00:03:36,820 an ethyl group. 78 00:03:36,820 --> 00:03:38,790 All right over here we have a methyl group and an 79 00:03:38,790 --> 00:03:40,150 isopropyl group. 80 00:03:40,150 --> 00:03:43,320 But the two methyl groups are on the same side of 81 00:03:43,320 --> 00:03:44,450 our double bond. 82 00:03:44,450 --> 00:03:49,450 So I draw a line in here and we see that these two groups 83 00:03:49,590 --> 00:03:51,190 are on the same side, 84 00:03:51,190 --> 00:03:54,690 therefore we're talking about cis here. 85 00:03:54,690 --> 00:03:58,250 So this double bond has a cis configuration. 86 00:03:58,250 --> 00:04:00,320 Let's compare the drawing on the left to the 87 00:04:00,320 --> 00:04:02,010 drawing on the right. 88 00:04:02,010 --> 00:04:03,820 The first time you look at these two drawings 89 00:04:03,820 --> 00:04:06,090 you might think these are two isomers, 90 00:04:06,090 --> 00:04:09,190 and I could use cis/trans terminology to distinguish 91 00:04:09,190 --> 00:04:10,380 between them. 92 00:04:10,380 --> 00:04:12,850 However, you can't because these are 93 00:04:12,850 --> 00:04:16,290 just two ways to represent the same molecule. 94 00:04:16,290 --> 00:04:18,420 If you picked up this molecule on the left and 95 00:04:18,420 --> 00:04:19,930 you flipped it up, 96 00:04:19,930 --> 00:04:22,820 you would get the drawing on the right. 97 00:04:22,820 --> 00:04:24,580 So they're not isomers of each other. 98 00:04:24,580 --> 00:04:27,080 This is the same molecule. 99 00:04:27,080 --> 00:04:31,190 And a fast way to figure that out is to look at this carbon. 100 00:04:31,190 --> 00:04:34,850 And you can see you have two identical groups bonded 101 00:04:34,850 --> 00:04:36,750 to that carbon. 102 00:04:36,750 --> 00:04:39,640 So you can't use cis/trans terminology. 103 00:04:39,640 --> 00:04:42,690 That's different from the example we did a minute ago. 104 00:04:42,690 --> 00:04:44,610 We had two identical groups, 105 00:04:44,610 --> 00:04:46,590 right these two ethyl groups here. 106 00:04:46,590 --> 00:04:49,120 However those two ethyl groups weren't bonded 107 00:04:49,120 --> 00:04:50,550 to the same carbon. 108 00:04:50,550 --> 00:04:53,620 Those two ethyl groups are bonded to different carbons. 109 00:04:53,620 --> 00:04:56,490 So this ethyl group is bonded to this carbon, and this 110 00:04:56,490 --> 00:04:59,080 ethyl group is bonded to this carbon. 111 00:04:59,080 --> 00:05:02,610 So we were able to use cis/trans terminologies. 112 00:05:02,610 --> 00:05:04,820 So we looked at our double bond and we said 113 00:05:04,820 --> 00:05:06,990 those two ethyl groups are on the same side of 114 00:05:06,990 --> 00:05:08,120 our double bond, 115 00:05:08,120 --> 00:05:12,170 so this represents a cis configuration of the double bond. 116 00:05:12,170 --> 00:05:14,680 So we can't do that up here because 117 00:05:14,680 --> 00:05:17,390 while we do have two identical groups, 118 00:05:17,390 --> 00:05:21,620 those identical groups are bonded to the same carbon.