1 00:00:02,140 --> 00:00:09,500 Hi guys and welcome to this new video about how to name organic compounds following aupac rules. 2 00:00:13,660 --> 00:00:20,539 In this video we're going to focus on branched chain alkanes. If we want to name a branched 3 00:00:20,539 --> 00:00:28,219 alkane we need to know how to name the branches. The branches are radical which are nothing but 4 00:00:28,859 --> 00:00:35,100 the molecular compound which comes from a regular hydrocarbon without one hydrogen. 5 00:00:35,259 --> 00:00:52,420 How can we name it? Firstly, we need to take the name of the regular hydrocarbon. For example, for methane, the radical is going to be CH3. It is what? Methyl. The wood, meth, and the engine, yyl. 6 00:00:52,420 --> 00:01:03,420 From ethane we have acyl, from propane propyl, from butane butyl or butyl. 7 00:01:03,420 --> 00:01:14,180 Some radicals have special names which are widely used, partly with the prefix iso. 8 00:01:14,180 --> 00:01:20,180 It is a muscle radical from a radical chain. 9 00:01:20,180 --> 00:01:42,859 For example, isopropyl if it has three carbons, isobutyl if it has four carbons, and isopentyl with five carbons. Remember, carbon-1 for any radical is the nearest carbon to the end of the chain. 10 00:01:42,859 --> 00:01:57,859 Another common prefix is sec, secbutyl, tacbutyl, tacpantyl if they have a tinary carbon atom 11 00:01:57,859 --> 00:02:04,859 and neopantyl if the tinary carbon atom is in between. 12 00:02:04,859 --> 00:02:11,860 Finally, with double bones we have benyl and allyl radicals and isopropanol. 13 00:02:11,860 --> 00:02:19,240 just learn them by heart they are very useful let's try to solve some examples 14 00:02:19,240 --> 00:02:28,169 we need to identify the longest chain in this case it has eight carbon atoms and 15 00:02:28,169 --> 00:02:36,389 then we need to identify all the substituents in an ethyl and methyl 16 00:02:36,389 --> 00:02:43,770 We're going to choose numbering the chain from the 17 00:02:43,770 --> 00:02:50,689 end that provides the lowest possible log count for the substituent in this 18 00:02:50,689 --> 00:02:57,189 case 3-meryl-5- ethyl then we're gonna put the name in order using the 19 00:02:57,189 --> 00:03:05,370 alphabet so 5- ethyl-3-methyl-octane. What happens if there are many 20 00:03:06,389 --> 00:03:14,629 if the same radicals appear many times, we can use Greek prefixes to indicate the number of 21 00:03:14,629 --> 00:03:25,909 equal substituents. We have here three, three, five, seven, seven, meta, penta, mthil, decade. 22 00:03:28,550 --> 00:03:37,590 Here we have some examples. Let's identify the longest chain and let's try to identify 23 00:03:38,150 --> 00:03:43,830 the substituents methyl and ethyl. 24 00:03:43,830 --> 00:03:51,689 Now we're going to compare the lockends that we have in both directions. 25 00:03:51,689 --> 00:03:56,530 When we start from left to right, methyl is in 4, ethyl is in 6. 26 00:03:56,530 --> 00:04:05,830 However, when we start from right to left, ethyl is in 4 and methyl is in 6. 27 00:04:05,830 --> 00:04:13,710 So, we're going to choose from right to left because of the alphabetical order, simply. 28 00:04:14,330 --> 00:04:17,149 So, for example, six must feel no name. 29 00:04:19,490 --> 00:04:23,670 In this case, we have a challenging decision to make. 30 00:04:23,990 --> 00:04:26,910 What is the longest possible chain? 31 00:04:27,449 --> 00:04:35,149 If we start from left to right and horizontally, one, two, three, four, five, six, seven. 32 00:04:35,149 --> 00:04:57,689 There are seven carbon atoms. It will start from the branch on the top. One, two, three, four, five, six, seven. Exactly the same number of carbon atoms. Which chain do I choose? The one that has more branches. If they all have exactly the same number of carbon atoms. 33 00:04:57,689 --> 00:05:23,589 So, the horizontal one. Now let's identify all the branches. Methyl in 2, methyl in 3 and in 5. And then on apropyl in 4. Let's put them all in alphabetical order using prefixes to indicate the 3 methyl groups. 34 00:05:23,589 --> 00:05:35,910 2, 3, 5 trimethyls for propyl obtained. 35 00:05:35,910 --> 00:05:41,410 In this case we need to identify the longest chain as well. 36 00:05:41,410 --> 00:05:47,029 It is the horizontal one with 7 carbon atoms, so hept. 37 00:05:47,029 --> 00:05:58,670 are two methyl groups in two and five in an isobutyl one in four let's put them 38 00:05:58,670 --> 00:06:10,279 in alphabetical order four isobutyl two five dimethyl obtain we're gonna try to 39 00:06:10,279 --> 00:06:19,120 solve a bit more challenging case the first step is to identify the longest 40 00:06:19,120 --> 00:06:27,480 chain which coincides with the horizontal case it is there are it has 41 00:06:27,480 --> 00:06:33,220 10 carbon atoms so it is the chain and we're gonna start numbering by numbering 42 00:06:33,220 --> 00:06:37,740 the main chain by the nearest substituent so from left to right 43 00:06:37,740 --> 00:06:46,709 because we have the method group identify all the substituents and name 44 00:06:46,709 --> 00:06:57,110 them individually. Methyl in 2, Propyl in 3, Isobutyl in 4, Ethyl in 7 and Isopropyl in 7. 45 00:06:57,750 --> 00:07:13,209 Finally, let's try to examine the alphabetical order. For Isobutyl, why? Because iso, 46 00:07:13,209 --> 00:07:19,209 the prefix iso is not considered, so we will be alphabetizing butyl. 47 00:07:20,730 --> 00:07:29,129 Then, 4-isobutyl, 7-acyl, 7-isopropyl, 2-macyl, 3-propyl decay. 48 00:07:31,850 --> 00:07:38,490 Here you can see the three-dimensional model of the molecule we have just named. 49 00:07:38,490 --> 00:07:45,259 I hope you have enjoyed this video, see you next time, bye.