Hello, good day. Today we continue with our study of organic chemistry. The topic we will analyze today is the study of cycloalkanes and cycloalkanes. We begin by studying the nomenclature of cycloalkanes according to the IUPAC, checking in particular what must be done to name cycloalkanes with a double bond as a starting point. Point number one tells us that carbons 1 and 2 will be those that form the double bond. If one of them has a double bond, it is assigned the number one. This means that we will take the ring as our main chain or trunk. The numbering will begin from those carbons that have, in this case, the double bond. We will decide which of them to give the number 1 to. It clarifies that if one of these two carbons has a double bond, then the tiebreaker leans towards that carbon. The carbon with the double bond is assigned the number one. Okay, from there we will continue. We now have as number Point 2 tells us the following: the carbons of the ring are numbered towards the nearest double bond. In parentheses, the number obtained when writing the position indication must be the lowest possible. We close the parentheses. Point 3: the position and name of the essential elements are indicated, as is done with the names. And finally, point 4: if the ring has more than one double bond, it is indicated by the suffixes -d, -trip, -tetra before the ending -en. Well, now we have the four points that allow us to name the rings. We continue to the examples. From this topic onwards, we are going to work with the structure, in this case, linear. So let's go with our first example, which is point a. In point a, we can see that as a base, we are given a ring, since the double bond appears here. Subsequently, at one of its ends, a branch appears, and right at the opposite end, the other branch appears, and we are asked to give it a name. Well, as we recall, one of the points in this case consists of, or told us, that we should... Selecting carbon number 1 and carbon number 2 from those carbons that shared the double bond, that is, those located here between this one and this one, also mentioned that a tie-breaking criterion for numbering was that if one of these two carbons had a branch, then it was called carbon number one. As you can see, the carbon with a branch is the one at the top, so it will be number one. The next carbon will be number two, and the numbering continues to the carbon here, which is number three, and finally to the carbon at the top left, number four. Once I have numbered the carbons, I proceed to identify what are the 'tree essences' of adolescence. The one on carbon number one is a useful being. Adolescence exists on carbon number 3; it is a metal. Again, as I mentioned before, to avoid complications with alternative nomenclatures in this case, where the branch is named based on complexity, we will name them according to the UPAC system, in alphabetical order, using the letters we are going to use. The contrast in this case is the methyl group versus the main group, and the important thing to remember is that the prefix tert- in primary branches, which are those that are directly connected to the main chain, doesn't count in terms of alphabetical order. Okay, so the name of this structure is one that will be useful: 3-methyl-cyclobutane. It's not cyclobutane because in this case it's a cyclic structure of 4 carbons that has an unsaturated bond, which is the double bond. And there you have it, we have our first structure. Let's go now to our second example. We see that as the base structure or main chain, they are now presenting us with a proper cycle. Again, it's not a cycle because it's a carbon structure of three carbons that has a ring shape. That is to say, this cyclobutane is not a proper cyclic because it has a double bond. Again, as I mentioned at some point, if we are already working with cycles with a linear structure, then everything must be a linear structure. Okay, from there it's time to number the important parts, and we remember that these are the carbons between which number 1 and 2 will be, or the carbons that It will share the numbers 1 and 2, as none of them have adolescence in this case, so the designation can be given arbitrarily. I can say 1 and 2, or I can say 1 and 2. Okay, so it's carbon 1, carbon 2, and obviously the next one is carbon 3. The name... well, once the main chain has been numbered, I was forgetting, we have to identify the branch of carbon 3. It's a carbon of its own. We wo n't stop to explain it now. When we studied branches in this case, and carbon trees, we studied this type of branching or carbon trees in depth. We studied these radicals of the carbon. Okay, so we begin. Its name will be, in this case, 3, and its own number... We have that first the name is given to the branch. Since there are no more branches, now we name the main chain, which is a cyclotroph, not a cycloprol. We continue now with our next example, which is section c. They are going to present us with an example, and a more interesting one, in some ways a little more complex, because now it has two double bonds, and not only that, it also has two carbon trees in those... Double bonds, okay, we have this structure and they ask us to name it, so how are we going to do it? How are we going to start numbering? Well, quite simple, actually. We remember that they are telling us that the numbering starts from the double bond, and this is where the first problem arises: how can I start from the double bonds? And in this case, there are two double bonds inside. So, it goes in the simple way, it goes like this: in the case that it only had those two double bonds, and we'll explain it in parts, it starts with the bonds at the top, so it would be carbon 1, 2, 3, 4, and 5. In the case that it only had these double bonds and had a second bond at the bottom, then it starts with the carbon that obviously has the double bond and the second bond, so it would be carbon 1, 2, 3, 4, and 5. In this case, it's a rather complex example in the first instance because we ca n't apply the first tie-breaking criterion in this case since both carbons have two double bonds, they have a double bond in both double bonds, in all three. So what is it? The tie-breaking criterion in this case consists of observing the bond at the top. If this bond at the top has a branch point, then it will be carbon number 1. We observe that there is indeed a double bond on the top carbon, and this double bond has a branch point. So, this one here will be carbon number one. This one here will be carbon number two, and the numbering continues: carbon number 3, carbon number 4, and carbon number 5. We erase the previous examples of the situations that, in this case, are primary situations that can be carried out before arriving at the solution of this exercise. Okay, so we have identified them. Let's name them. As always, first we must mention the branches. These are on carbons 1 and 3. Both are methyl, so it's 13-dimethyl. Again, the branch at number 13 is a methyl, the branch at number 1 is a methyl, so its name is 13-dimethyl. Since we now have double bonds in this case, we have to name the carbon from which each one originates. These double bonds, the first one goes from carbon 1 to 2, so it's said to originate at carbon 1. The second one goes from carbon 3 to 4, so it's said to originate at carbon 3. So, if you look at these numbers I'm using, they only indicate where the double bonds originate. So, it's 13- cyclopentacyclohexanthine. If you count it, it refers to the fact that the structure has 5 carbons in a ring. Since I have two double bonds, I have to add the ending. Okay, so its name is 13-dimethyl-13- cyclohexanthine. That's good, pretty good. Now that we've explained it, it's a little easy, and we move on to our last exercise, our last example of what rings are. Now that we've reached this point, I want you to analyze this figure. Pause the video and see if you can name it, mainly so that you can discover what you haven't understood yet. If you manage to name it by applying the criteria we've seen, then congratulations, you'll have understood the topic very well. If you don't... If you manage to name it, don't worry. My advice is to review these topics again. The advantage of this class being in video format is that you can repeat it as many times as you want; you can even pause it directly to the parts you didn't understand the first time. The advantage is that here in video format, the teacher can explain the topic to you about 100,000 times, as many times as you want to repeat the video. Many times, people think this is helpful, but on the contrary, the first time we don't grasp the topic properly because it's unfamiliar. So that doesn't mean that if they explain it again, you won't understand it, because the next time they explain it, you already have some background knowledge. You already heard it the first time, so you already have a more or less clear understanding of the concepts. The second, third, or fourth time you hear it, you already have 60 points of reference, which is excellent and will help you understand it even better. In any case, if any of you, in this case, after watching the video, after doing the activity, and after repeating the video, still have doubts, you know that you can tell me. You can contact the subject's WhatsApp group or my personal WhatsApp number, which I'll provide if needed. So, I'll tell you this is the last one, and this is the last exercise of the cycle. I ask you to please pause the video, try to name it, and restart it. Okay, we're going to start naming it. The first thing I want you to notice is the presence of these two double bonds. Therefore, the numbering should start with either of these two, either 12 or 12. Now, which one is correct? In this case, the correct one is the one on the left, the incorrect one is the one on the right, because the correct one is the one on the left. If you observe, the difference here lies in the fact that the double bond on the left has a branch, and that's where the numbering begins. So, we have carbon 1 and carbon 2. Again, it tells us that if there are two, in this case, unsaturations or even an insaccharide, and if it has a branch, then the numbering should begin with that branch. That's the tie- breaking criterion that we're going to use. To apply then, it's carbon 1, carbon 2, and there, obviously, the sequence of numbers is given. The next would be carbon 3, carbon 4, carbon 5, and we finish with carbon 6. Once we have numbered the ring that today serves as our base structure or our trunk or main chain, we now identify the branches. Both the branch at carbon 1 and the one at carbon 6 are the same branch. If you observe, it has two carbons, so it's a pair of carbons. Each of these branches is a methyl group. So we begin to name our compound. The name you should have obtained is 16, useful because there are two. From there, we name the carbons where the double bond originates. The double bond originates at carbon 1 and carbon 4. 14 cyclohexam, good, 14 cyclohexam 1. And that's it, we now have the name of our structure. Again, I hope you have successfully completed our structure for today. With that, we finish what is the topic of the cycle, which we were going to move on to now. Half of the video with the nomenclature of carbon rings according to IUPAC. Again, the exercise remains the same: the nomenclature of carbon rings according to IUPAC also specifies that this nomenclature is for naming carbon rings with carbon atoms. In point number 1, carbons 1 and 2 will be those that form the triple bond. Number 2, the carbons of the ring are numbered towards the carbon closest to the triple bond. Number 3, the position is indicated in the name of the carbon atoms as is done with alkanes. Number 4, if the ring has more than one triple bond, it is indicated by the suffixes -tetra before the ending. And we are going to indicate the first series of examples. We have our triple bond, and therefore it is going to be a continuous ring, and at the end we have a carbon atom. Well, then obviously the numbering will start from any of these carbons, which are the carbons that are involved in the bonding of the triple bond. In this case, what numbering is correct? 1 2 3 4 and 5, or 1 2 3 4 and 5. In this case, if you observe, you'll notice it's a gunslinger. Both numberings end at carbon number 4. So either of these numberings is correct, and we'll leave the numbering in blue. So my structure in this case would be 4-membered cycle 21. That's the name of our structure: 4-membered cycle 21. Now I'm going to modify it a little to explain what I wanted to explain. Initially, when generating this structure, I didn't consider that it was an isomer. But now, with this new structure, the one from section b, we can study the point I wanted to address, which is where the numbering should begin, and if this number is correct and incorrect. We have the structure now; it's still a cycle, only now we've slightly moved the essential tree. So, which numbering is correct in this case? Numbering 1 2 3 4 and 5 or numbering 1 2 3 4 and 5. We analyze both numberings again. We could say that we can freely choose which of these two numberings to take; however, the rule tells us that the numbering must correspond to the dvd numbering. In this case, generate a low numbering with respect to the carbon atom, which we have to try to get as close as possible to the essential tree. If we observe the numbering in blue, we quickly find the essential tree at carbon 3. If we observe the numbering in red, which in this case would be incorrect, we find the essential tree at the end of the ring, at carbon number 5. So, having explained this, the correct numbering is the numbering in blue because we quickly find the essential tree and because we take into account the carbons that involve the triple bond. So we erase the numbering in red, which is incorrect, and give it a name. In this case, its name is 3-methyl- cyclo21, 3-methyl-cyclo22. Okay, let's continue. Now let's go with our example, which will be carbon 22. Let's now make a slightly more ambitious structure, as you know. As I mentioned before, these structures resemble geometric figures. We have an eight-sided figure here. We have our eight-sided figure with two triple bonds and a. Again, I want you to analyze this, since this topic is quite similar to the one we just discussed. I think you should try to name this structure, especially regarding the numbering. So, let's begin analyzing the possible numberings. The first numbering would be carbon 1, 2, 3, 4, 5, 6, 7, and 8, giving us the name 6-methyl- 14- cyclo(J)-dyn. Now we'll look at the other numbering: carbon 1, 2, 3, 4, 5, 6, 7, and 8. The other numbering will be 8-methyl- 14- cyclo( J) -dyn. So, which is correct and which is incorrect? In this case, remember that the correct one is the one that allows me to first identify the branching when we talk about the presence of two or more saturations within the ring. As you can see, both end almost the same; both are 14-cyclo(J)-dyn. However, the difference lies in which one allows me to find my branch or adolescence more quickly, which in this case is the methyl group. If I follow the blue numbering, I find the methyl group on carbon 6. If I follow the red numbering, I find the methyl group on carbon number 8. So, the one that works for me is the blue one, the one that doesn't. The correct one is the blue one, the incorrect one is the red one. Correct, incorrect. So, it's very important for those who wonder, in this case, how do I know these two numbering systems are generated? Well, we'll explain it in detail. First, let's look at the blue numbering. How is the blue numbering generated? Why wasn't the blue numbering 1, 2, 3, 4, 5, 6, 7, and 8? Why wasn't it like that? The answer is simple: we established that the numbering starts from these carbons involved in the upper part, including carbon 1 and carbon 2, and it has to go, in this case, to the tree. The closest saturation to the nearest triple bond, if there is one, would be the numbering. If there is n't one, then the numbering would have to go closer to the tree. It is the closest, but as if someone were saturated, then the numbering has to go to the tree. It is the closest. If you observe and we did the numbering in this way 1 2 3 4 5 6 and 7 8, obviously we find the unsaturated carbon up to carbons 6 and 7, and before reaching saturation we find my essential tree, which cannot be in this case because first my saturation has to try to reach the other as quickly as possible, and if in the way a tree gets in the way, exceptional problem. But the idea is that an unsaturated carbon has to quickly seek to join the other in saturation. Regarding the numbering, that is why the numbering starts from the top part, which is 12 3 4 5 6 7 and 8, in the case that we consider The correct numbering is the blue numbering, the numbering that starts from the top. Now, if we consider the numbering to be the one at the bottom, then it would be carbon 12, and it's numbered this way because I'm trying to reach the other one in saturation as quickly as possible. That's why it starts in this direction and not the opposite, because the same thing would happen as I explained: it takes longer to reach the installation. So the correct one is 6000 14. This should be separated. or by commas 14 cyclo dyn well and we finish with our last exercise of today, once again we generate a cycle and we are going to put these two trees science again take your time and let's see then what the name of said structure is well after analyzing the structure we found two possible names the first is 6 be useful to transmit cyclohexane of 1 how do we get that name I will explain now the numbering starts from the carbons that are at the top because they in this case apart from being at the top they have the insaccharide so the numbering is 1 2 3 4 5 looking in this case to get as quickly as possible to what is the insaccharide so there is this routine 3 methyl 15 cyclohexane are in this case okay the 15 I was missing 15 cyclohexane the other numbering in this way 1 2 3 4 5 and 6 as we can see both numberings are found on carbon 4 to what comes the other in saturation so as Since they are equidistant, the tiebreaker in this case is which is closer to the branch. Unfortunately, both meet at the first branch at carbon number 3. So, if both unsaturated carbons are equidistant from the next unsaturated carbon, this is the first tiebreaker. If both are saturated, and both numbers now reach the nearest branch at the same distance, which is the carbon following the double carbon in the triple salt, which is carbon 3, how do I know which of these two names is correct? The second name is 33-methyl-6-ethyl, which is also 114- cyclohexanedyn. So, as I was saying, which one do I know is correct? The answer here is that the more complex branch, in this case, and the numbering in blue meet the more complex branch, which is the ethyl carbon at carbon 3, while the numbering in red meets a fairly simple branch at carbon 3. So, that is the tiebreaker that allows us to know that the correct name is 6- methyl- 14-cyclohexanedyn. He examined well, so with that we have finished our exercise for today, and with that we have finished today's video. I thank you for paying attention to this class, and we will see you soon in the next session where we will see students on loan, and in the last station where we will see carboxylic acids. After these two videos, we will have already completed the syllabus for the third partial exam, and from there, next week we will have a short series of videos, two or at most three videos, where we will cover the integration of what we have learned. In this case, we will see sample exam exercises so that you are more familiar with the type of exercises that will appear on the exam. So, that's all from me. Goodbye, thank you very much for watching the video, and see you in the next installment. Goodbye.