 Next write down, okay write down, nomenclature of alkene and alkyne, in this the write down to select the longest carbon chain, select the longest carbon chain which has maximum number of double or triple bond present, select the longest carbon chain, longest possible carbon chain which has the maximum number of double or triple bond, means the double or triple bond must present in the primary chain, the parent chain, double bond is preferred over triple bond. How is it wrong? What does it mean? In the molecule if you have double or triple bond and choose in the initial, yeah then we will try to give least possible to the double bond. Sir, double bond chain has two double bonds and another chain has two triple bonds, means if the parent chain takes the double bond. Sir, what is the double bond? This one alkyne, what does it mean? We show examples, what if you do double bond? Sir, there is no alkyne with double bond here, all alkyne has two double bonds, it is two pi bond not double bond, two pi bond. Alkyne can't have double bond here, because then you give priority to the double bond to become an alkyne small bond. Like an alkyne, double bond, but we give priority to a double bond. So for it to be an alkyne there should be zero double bond. No, how? Because if there was a double bond we give that chain priority and it will become an alkyne. Yes. Sir, how would you say double bond? I will give you the example, suppose my guys are like now C, F, D, B, C and single bond, C and single bond, C, double bond, C, F, this is possible. Sir, is there an alkyne or an alkyne? So that's why it's a normal feature of alkyne and alkyne, it is a mixture of both, and it's possible we have double bond and double bond, because balance is complex. So what is this thing called? This one, me. One by one minute, I said jump, you cannot jump. What I was discussing. Double bond preferred over double bond, right? You see this example. Suppose we have a molecule C, H, double bond, C, H, single bond, C, H, C, triple bond, C, H. This is the molecule. C, H, 2. C, H, 2. Okay. Now in this you see the position of triple bond and double bond. If you start from this side one, if you start from this side one, right? So we have the same position on both sides. So we'll get priority to the double bond. So we start numbering from here. One, two, three and four. Okay. So name of this compound will be butte. Butte, 1, E, 3, I, and this will write alphabetically. E we should write first and then Y. Butte, 1, E, 3, I. One more thing we have here. This E you have to drop. This you should not write. It should be E and not E and E. If you write E and E, you'll get 0 again. Sir, is there a double bond for triple bond? E, E and I. No. In numbering we'll prefer this. But this we'll write not the vertical sections. What? I know sir, I have to E. So why are we dropping this E? For that we have a reason. Okay. So for this write down one note here. Because if we don't it becomes E and Y and that's weird. With this we don't have any personal grads with this E. There are rules. That's what we have in dropping it. This E is there because again this is not called up by any other elements like A, I, Y, X and Y. A, I, O, something like that. Y also becomes Y. So when we have to drop this E for that we have a rule that we'll write down. The terminal E in the name should be dropped when it is followed by a suffix beginning with A, I, O, U and Y. E is not there. Okay. So when it is followed by these suffixes beginning with these letters like here we have Y. So this E is very important. It's very important. This note is very important. You must take care of this. Sometimes what happens in the option they'll give you this name and this E will be there and one of the options will be none of these. Okay. This is not dance. Understood. Another example you see. What is the name of this compound? CH2 double bond CH. Single bond CH, double bond CH2. Okay. So that will like be like, okay. You called me. You put a C. I put a D. Super. Ok. Super. Super. Super. Yeah, super. Super. Super. Super. Super. Super. Super. Super. Yeah. Super. Super. Super. Super. Super. Super. Super. It is 2 methyl or 4 methyl, double and triple bond will give priority over the substituents. See any substituents if you have, then will give priority to the double bond or triple bond over the substituents. So, the correct number will be from this side 1, 2, 3, 4 and 5. So, what is the name? 4 methyl. Spend 2 i. The name of this compound we have dienes right. So, we call it as hexa diene, hexa 1, 3 diene. Sir, can we write 1, 3 hexa diene? Yes. Here also, buta 1, 3 diene. Whenever we have diene, then we will write, we don't write buta. We will write buta, hexa, penta. Try it actually. Try it. But sir, it is followed by a prefix i, 2p, not 3p. It is followed by the prefix i. This i. Here we have, why that's why we are dropping this i. Oh right, sorry. Sir, fill it down first one. First one, this one. Sir, is it fine to write 1, 3 diene? 1, 3 diene. 1, 3 diene. Yeah. This is also right. 1, 3 hexa diene. Okay. What is the name of this compound? Cyclosautic. Sir, will you start from this end of the compound of that end of the compound? No way, sir. Can we use… Tell me the position of this study. You… Sir, can we start from this end of the compound? Sir, can we start from this end of the compound? Sir, can we start from this end of the compound? Sir, can we start… Okay, stop. See whenever the ring contains double bond then we… When the ring contains double bond, that means always is first and second position with that double bonded carbon atom in it, right? When the ring has double bond right down, in this right down note, when the ring contains double bond, then we always give first and second position to the double bonded carbon So, now you see here, this can be first or this can be first, both possibility. If this is first, one, then it is true, we cannot take this as two, because first and second position will have to give to this carbon atom. So, one, two, three, the position of this will be six. But if you go this way, one, two, three, so here we have the least possible number, correct numbering is this. Here also the rule is what? We always give first and second position to the double bonded carbon atom, so you don't have to write x1e, simply hexene. Cyclohexene symbol, because it is understood that it is the first atom, the rule we have, first carbon atom will have the double bond. If you write x1e, it is not wrong but not required, because the rule is what? So, let me finish this. So, if you have two consecutive terms, consecutive is, you can't have a ring, in that ring it is not possible, because high angle is stable, not stable. If you have double bond here also, it is sp hybridized. So, the angle is still very large. Generally, the ring is two consecutive double bonded molecules. Sir, it is a 3-bromo, 1,4-cyclo-battery. Or a q3-bromo. 5-bromo. 5-bromo. 5-bromo. 5-bromo. Say, what are the things we can do, you see? One thing is what? We can take 1, 2, 3, 4 and 5. But then, that double bond is? Okay. Another possibility is what? Oh, priority has to be double bond. Yeah, so, that's a 5-bromo. 1, 2, 3, 4, 5-bromo. Okay. Now, in this way, you see, the position of double bond is first and third. Yeah. Here, it is first and fourth. This double bond is preferred over the locans, bromine and all, whatever locans we have. So, this numbering is not correct, because the double bond has higher number. Sir, but this we won't see that. So, fine. We won't see that. No, we will prefer this. Double bond over the locans. Okay. This one is correct. Double bond over the locans. Double bond over the locans. Double bond over the locans. Double bond over the locans. So, the sum is the same for both. So, the super has one and double bond. So, what is the difference? Take 2 as 1 and 2 as 1 and 2 as 1. So, here we have, you see? No, sir. Take 2 as 1 and 3 as 2. Size? How do you think? That's not possible. One is normal. No, that's not possible. No, sir. No, no, sir. One. One. One. Ten. One on the right. One on the right. How do you think? First and second to the left. First and second to the right. Again emotional error. First and second, okay, sir. We can reduce it, but according to the rules. Okay. So, this is the correct bond. Bipromo. Bipromo. Bipromo. Bipromo. Bipromo. Bipromo. Bipromo. Bipromo. Bipromo. Bipromo. I don't know what order. Pat者. Penta. I. One on the left. One on the right. One on my 3. Ten. Nine. Ten. Ten. Ten. Nine Nine. Wait, wait. Let me finish this after this video. Not here. C-H-2 double bond C-H-2 double bond C-H-2 So, do you count the diazons as C-H-2 double bond C-H-2 double bond C-H-2 double bond C-H-2 double bond must be taken to the parentage The smaller ring will be the parentage, the higher the number of fiber atoms One comma two is not required because when you have one comma two is not required because when you have props, this is the only position you are left with. There is no other way where you can place the double bond, propa diene. See in this one, when you have double bond in the ring and any double bond carbon atom, if it contains any substituents, then we are bound to give first position to this carbon atom. In the sixth group, half sixth group, some of blocans we will not follow. Whenever the substituents present onto the double bonded carbon atom, that carbon atom will be the first position. So, if it is first, we cannot take this as second because second will be this. Substituents first position. If it is present onto the double bonded carbon atom, another rule is what? If any substituents present onto the double bonded carbon atom in the ring, then that carbon will be the first carbon atom. Question will be the first carbon atom. Then we will see alphabetically which one is the first one. Or if it is same, then we will see the sum of blocans. So, in that we will see. So, that is sum of blocans. It depends what position it is, how you are going. So, when you see the molecule, if you follow this rule, then you will understand what is the correct numbering of that molecule. What is the name of this compound? Six-chromo-1-clotocyclohexene. This one, this is one because bromine comes first alphabetically. This is one, this is two. One-chromo-2-clotocyclohexene. One-chromo-2-clotocyclohexene. In this one, the only difference is what? Instead of this bromine, we have cyclo-propyl. So, name will be 5-cyclo-propyl, cyclo-penta-13-tion. Okay? Now, in this one you see, this ring has double bond. So, this is the parent chain of this compound. However, the number of carbons it will give priority to the double bond. Now, in this one, this carbon is one, this carbon is two, this is three. Because double bonded carbon at a must have first and second position. Three-cyclo-penta-3-cyclo-propyl. Okay? Got it? Why don't you do some practice so you can do it? Now, the best thing is what you can do. What is the double bond? What is the double bond? How do you name the other one? Propenyl. Cyclo-propyl becomes cyclo-propenyl, cyclo-propyl. If it is double bond, then it is cyclo-propenyl. If double bond is there, then simply cyclo-propyl. Right? If it is placed at the substituents, then it is cyclo-propenyl.