Hello, good day! We continue studying organic chemistry. Today's topic is alkyl groups, and as an introduction, alkyl groups are formed by replacing one or more hydrogen atoms in a hydrocarbon with one or more halogen atoms. We recall that halogens are elements in group 17, whose main representatives are fluorine, chlorine, bromine, and iodine. Their general formulas are Rx, where R is the alkyl radical, and x is the halogen (fluorine, chlorine, bromine, or iodine). If it's an alkyl radical, then a halogen group is formed. Now, let's analyze the characteristics of the alkyl group. We'll do this using the following table, where the first column shows the compound we 're talking about: alkyl groups. The functional group, in this case, is represented by x, which represents the halogens ( chlorine, fluorine, bromine, or iodine). Iodine, and finally the general formula. In this case, we are given two general formulas: the first when we have, in this case, an aliphatic branch or chain, and the second when we have an aromatic chain. Now, let's analyze the nomenclature for iodine according to IUPAC. In this case, we have two sections. Section one tells us that we must write the name of the halogen. Obviously, its representation is x, as we have said previously, and we will add the ending 'puro' followed by the preposition 'de' and then the name of the radical. Section two tells us that they can also be named considering the halogen as a substituent, indicating its position, and then the name of the corresponding hydrocarbon. Section two will be the section or nomenclature that we will use most often. That is to say, the exercises we do at the end of this session and those exercises that appear on the exam will be more oriented towards that second section. So, let's move on to the examples. As an example, we have our first structure: H3H22. There are two ways to name it. As the first one tells us, it consists of naming the halogen with the ending -pure. That is, we are going to put the two nomenclatures here. So the first is the name of the halogen, which in this case is bromine with the ending -oro, so it will be bromide of the preposition of, and the name of the radical is radical. In this case, it has three carbons, so it would be pro-bromide. The radical will be characterized by having the ending -y, the bromide of propane. This is the first nomenclature. The second nomenclature, according to the nomenclature, would be 1-bromopropane. In the second nomenclature, in this case, since there is no functional group, the numbering will start from the end closest to what is, in this case, the essential tree, which would be the boron and carbons 1, 2, and 3. That is why the second name is 1-bromopropane. Okay, let's continue. Now we are going to the next section. We are going to analyze a rather curious structure, in this case, CH3H3H3, and there is fluorine again. They are going to generate two names. The first name consists, as we have already been told, of naming the halogen with the ending "-oro," in this case, fluorine, so it would be fluoride. The name of the radical, in this case, the radical is this one we have here, which I have enclosed in red, and as we can see, it is a fairly well-known radical for us; it is a useful one. So it would be butyl fluoride. This would be the first name. Let's remember that in the second name, we take what is the halogen as a branch. So the main chain would be this one here, and it is, in this case, an isomer. If we were to regenerate the structure, it would look like this: CH3CH3FLCH3. And by passing the axis of symmetry, we would see that what is on the left and what is on the right is the same. Therefore, we conclude that it is an isomer. Being an isomer, the numbering starts from any side, that is, carbon 1, 2, and 3. And so, observing this, we see that its name in this case would be... and let's check the branches. To determine the name, it's methyl and it's fluorine. Remember that this goes in alphabetical order, so it would be 2 fluorine, all methyl, the chain has three carbons, so it would be, but not... and that's our second structure. Again, if we have doubts about a structure, remember that we can generate a more user- friendly configuration of its structure. In this case, the original structure didn't allow me to clearly see that it was an isomer, so what we did was convert the original structure to a more user- friendly structure by focusing, so to speak, on the two atoms that are on carbon 1 and carbon 3. And now we can clearly see that it's two fluorine 2000 propane. Okay, let's move on to our next example, section c. Again, we generate two names. The first name, as we recall, consists of the name of the halogen with the ending -pure. Its name is chlorine, with that ending it would be chloride. And the name of the radical, the radical in this case is a cyclovinyl group, so it would be cyclovinyl, and that's the first name: cyclopentyl chloride. Remember that... The second name is when we take the halogen in this case as a single element; here it would simply be chlorocyclopentane. We remember that critically, we are told that the carbon to which the branch is attached would be carbon number one. Here, the reason why we don't call it 1-chloropentane is because, in this situation, since there is no other branch, let's say it can be omitted because it is understood that the chlorine is on carbon number 1. That is why we do n't call it 1- chloropentane and we only call it chloropentane. Let's continue, let's now analyze the next figure, section d. Now we notice the presence of more than one halogen. Well, in the first nomenclature, remember that it is the name of the halogen followed by the ending -i. That is usually when there is only one halogen. When there is more than one halogen, we will resort to the prefixes -i, -x, -i, etc., depending on the halogens that there are. In this case, there are four halogens, so it would be chlorochloride. Since there is only one carbon, remember that That single carbon in this case comes from methane, which, when modified, becomes methyl, and here we're going to call it methyl tetramethyl chloride. Other names it can receive in this case are tetrachloromethane. A third name is carbon tetrachloride, and then carbon chloride. Now we move on to our last exercise, one last structure, and we have 3 shh 3 73 fluor si h 2 CH CH 3 L CH 3 iniestra. This is our last structure to analyze. And here, in these structures, which in this case are quite long, is where our halogen behaves like a tree of essence. So, since our halogen behaves like a tree of essence, it becomes another branch that needs to be named. The first thing we do is identify the main chain. Since it's a hydrocarbon chain with single bonds, it's a branch, so we simply enclose the longest chain, and that's it. Remember that these branches can appear both in the carbon we're discussing and in any other compound. Hydrocarbons that have a functional group, that is, alcohol or aldehyde, become carbonic acid, to name a few. Okay, we now mark the extremes: left extreme, right extreme, and we're going to see which extreme we start from. In this case, if the number from left to right is 12, on carbon number 2 there opened the first essential tree. If I number from right to left 12, on carbon number 2 there is also an essential tree, then, since the first tree is found at the same distance from both the left and right extremes, it is said that there is a tie. So we have to resort to a tiebreaker. The tiebreaker in this case, in the case of the arcana, consists of looking for a third essential tree. We can see that in the case of the numbering that runs from left to right, we have to go to carbon number 3 to find a third tree. In the case of the route that goes from right to left, we don't have to go beyond carbon number 2 where we find a second essential tree. So the correct numbering is from right to left, so it is carbon 1, 2, 3, 4, 5, and 6 once. We have already identified the main chain and numbered it. Now we identify the branches it has. At carbon number 2, at the top, it has a chlorine atom; at the bottom, a metal. At carbon number 4, at the top, it has a fluorine atom; at the bottom, it also has a metal. Finally, at carbon 5, it also has a metal. The letters in this case are M and F, and the first one is chlorine. So, the name of this structure is 2-chloro- 4-fluorine. If you observe, there are three metals present at carbon 2, carbon 4, and carbon 5, so it will be 3-methyl. The name of the main alkane chain with 6 carbons, in this case, will be hexane. And we have the name of our structure: 2-chloro-4-fluorine-2- methyl. Exactly. And we have our exercise, which is the last exercise on carbon- hydrogen structures. Now we will move on to what is our last exercise of the session. We're going to do this with a linear structure. Again, we're going to involve the hydroxyl functional group here. That is, we're dealing with an alcohol, which has three hydroxyl functional groups. We're not dealing with just any alcohol; we have an alcohol with the "-tri" ending, which refers to those three functional groups. That's our base structure. Now we're going to start adding branches, and we already have our structure here. Again, it's an alcohol from, in this case, wheat. We're going to mark the points of the main chain in blue because it's very important that we don't get confused. Remember that the main chain must be made up of carbons, so please don't count the oxygens as part of the main chain. The main chain is this one that I'm marking with the blue dots, which correspond to the carbons. Now, if you look, if I were to name carbon 1 and number it from left to right, I could name it carbon 1 and number it from right to left. Now, let's analyze which is the correct numbering on carbon 1 from left to right. There is a functional group on the carbon From right to left, there is a functional group, so it tells you that in that sense there is a tie. The tiebreaker criterion regarding functional groups is, as the main number one, showing the presence of a third functional group that causes a tiebreaker. In case there is no third functional group, then I would have to resort to some license, the resistance closest to this extreme. Here we are left with the tiebreaker criterion of the third functional group. As we can see in carbon number one, which goes from left to right, there are some hydroxyl groups. So with this, I already guarantee that the correct numbering is the one that goes from left to right. Therefore, this numbering is erased. We erase the little arrow and we are going to continue with the numbering, which is carbon 2, 4, 5, 6, 7 and 8, 9, 10, 11 and 12. A hydrocarbon chain of 12 carbons. That normally, in the case of an arcane, would be a volcano, or in the case of a tree, it would be a cannon. And in the case of this situation, which has three hydroxyl functional groups, it would be a dean of rivers. Ultimately, that's the name we're going to assign to the end where Cano Trebol. Remember that when we analyzed its nomenclature, it told us that if there were two or more functional groups, we had to name the name of the compound followed by the ending -al with the prefix, in this case multiplicative, which will indicate the presence of the hydroxyl groups or the number of hydroxyl groups. In this case, there are three hydroxyl groups, that's why the tri and the zero of hydroxyl have the ending -al. So, this Kano Trebol is the final name we're going to assign to our trunk or main chain. The chain is already numbered, so now what remains is to identify the branches. At carbon 2 there's a metal. At carbon 2 there's also a specific use. At carbon 3, it's very important to see how many carbons constitute this branch: 1, 2, 3, 4, 5, 6. So it's an exit. Carbon 4, the top part, the bottom part is also a metal. Carbon 5, the top part is a steel, 2 carbons, the bottom part is a new pen till. Here I won't count the carbons because I already know that the curved little cross is a vent if it were a cross it's normal I interrupted it at carbon 6 there is chlorine at carbon 7 and bromine carbon 8 and iodine carbon 9 and fluorine in this case I'm going to omit the names thinking it's easy to identify each of us as number 10 useful cycle carbon number 11 carbon cycle number 12 fertile cycle and letters that we are going to contrast or compare and m h n at the chlorine site we are going to put in this case here the names of the and halogens that we are in this case comparing because we were going to need to have in this case the evidence would be of chlorine, bromine and iodine and fluorine from there we have that of useful cycle this of cycles and I did it of fertile cycle and then things are already put because there are too many different branches that I in this case have to take into account we analyze quickly and we see that the first branch in this case corresponds to bromine bromine is Number one, so it would be 77, and bromine. Remember that this is given in alphabetical order. So bromine is already there. Let's go now and find that to give the next name, we have a rather interesting situation because chlorine is competing against cyclohexile, cyclohexile, and cyclopentile. All of you, in this case, the four names that are competing, which I repeat are chlorine, cyclohexile, and cyclopentile, begin with the letter c. Yes, in this case, those names begin with the same letter, and we put them again in order: chlorine, cyclohexile, cyclopentile. So we resort to the second letter, which would be l, against the ones in cyclo. So let's see, for the first one again, done. We're going to run the alphabet: a b c d e f g h i. So, if it goes down and comes before it because it's y j k l, then chlorine is going to be the last one I'm going to name of these four. It's going to be number 4 that I name, the last one. Now things get more interesting because we have to define which of these three names goes first. Obviously, we go through the whole word 'cyclo' because all three have the word 'cyclo' in them. Let's go directly to the seven letters, which would be the useful cycle. Apologies for having deleted it, but the key to "useful" would be the h and the p. Remember that in this case, in the word "cycles," it counts in alphabetical order, unlike in the case of " Thermo Team" and "useful." Here, it does count, but when comparing these names, they have the similarity of containing the word "cycle." So, obviously, I'll go to the next letter where they differ in order to identify which one comes first. So we have "useful cycle, 21st century" and " Cycle Pentile." And here we quickly identify that in "useful cycle," the letter "beba" comes before the "ch" and the "p," so it would be 10, " useful cycle." We continue: the h comes before the p, so now it would be 11, 21st century. Finally, as the last one, it would be 12, " Cycle Pentile." Let's write it correctly in one name: " Cycle Pentile" and "Cycle Pentile." All of these are missing; only the "chlorine" is missing, which is 66, and "chlorine." So, I've already worked with the useful cycle, the profile, and the useful cycle. I'm going to delete the name of the useful cycle because, in this case, it's interfering with my main name. And once I finish naming, I'll repeat later, so I do n't offer you my structure. Okay, we delete this too so it doesn't interfere with our structure, with the name of our structure, and we continue. Okay, so what's left? I already used the letter c. Now let's see what work abc of. Is there anyone with d? No, and is there anyone with e? Yes, who is the one with 5? Useful, 5 profile. Ready, goodbye. To beat a b c d e f. Is there anyone with f? Yes, who is the fluorine 99? And fluorine. Here's the concept because in this kind of exercise you should do it with all the calm in the world, even though it sounds quite tedious and strange. Repeat the alphabet as many times as necessary so you don't get confused, because remember that this part is very delicate. In this case, putting one name before another would be wrong. So we repeat in our alphabet abc of f. Of that lineage h. Is there anyone with h? Yes, there is. Who is the exile that is in carbon 3? So which is 3 Pepsi and we eliminate it too and continue with our neighborhood abc of f g h and there is someone with y if who he made his own so it would be 2 made his own and we also eliminate him from here remember that we have already eliminated fluorine and we have also already eliminated chlorine now simply in the last halogen that we have left in this case to name is iodine which at some point we will do so we continue a b c d e f g h i j k l m there is the Lincoln M if the three are in so it's 2 44 I promised I listed my methyl too they are already out of there I only have two branches left which are the cne open deal and the iodine m n so the next six neo gentil and very important in the case of iodine referring its name is iodine and not iodo so I'm going to take it by first letter is y so practically it would be the last name in most cases so finally it's 88 of iodine and finally we are going to name in this case the trunk or main path naming first in the carbons where it is The hydroxyl group would be 11 or 12, where cano friol 1 112 or decane, but okay, and with this we have finished naming our structure. Today I thank you for having been attentive since this class. Remember that we still have one last class, which is carboxylic acids, for which the video will be sent to you tomorrow. Over the following weeks, a maximum of 23 videos will be generated to integrate what you have learned, where we will be working with exercises quite similar to this one. In Globe, we will do everything we have seen, and these exercises will be very similar to those that will be on the exam. So, that's all from me. I say goodbye, take care, see you later, and have an excellent weekend and good luck on your upcoming exams.