 Now, I'm going to feature of alcohol. Alcohol is a good choice. Okay, suffix for alcohol is what? All, right? And we prefer alcohol. You go and clean your alkali. Or any other substance you can see. Write down the name of these compounds. So, can you increase the temperature? Sir, here we should take the functional group with the logistic. Yes, not in the longest chain. But the carbon atom which contains the functional group, that should be present in the whole formation. We won't count here. This is not first act. We take this into the longest chain. So, we can take this one. The carbon which contains the functional group, that must be present in the parent chain. What happened? Why? Someone is the one who happened. Yeah, thank you. I don't know who is this. It's Bobby. Oh, Bobby. I'm taking the phone. Second thing, you count the oldest part of the chain. The carbon which contains the oldest should be the part of the chain. Are you in a position of the alkali? Yes. But, sir, if it's first, then we need a time. Or should we? Not that fast. But if you write also, it's not wrong. Okay. But, sir, in all of them here, we won't have to write because alcohol has the most priority. So, it will automatically become one. No, it's not one. We cannot start numbering from the carbon atom which has the functional group, which you cannot. You have to take the longest possible chain which contains the carbon atom which has the functional group. So, you have to start either from this end, this end, this end, or this end, this end. You can't. You cannot start. No, sir. I thought because you can't give a whole number because it's not a carbon. We cannot start from here. But we can start from this carbon. One, two, three. Second one. Okay. What is the name of this compound? Methanol. One carbon. Methanol. This one, we have third position, we have methyl. So, what do you write? Three methyl. Two tetanol or butane. Two all. Again, you see, this E. It starts. Because that is all. Okay. This one is what? Should we start from this carbon or this carbon? Numbering. We'll prefer this, right? If you start from here, okay. Sir, we have pentanol. Literally, Y and O. Y and O, M and E. Y and I. Sir, the name is pentanol. We don't need to put that here. No, we have O. Position of I and position of O. Sir, when you're writing I and O, in this case you can write pentanol. Sir, if you have a cycle and then you have, I'll go out of here. Sir, when there's been a bandage, you have to function with that carbon. This becomes substituent. Maybe cyclopentanol. Cyclopentanol. Cyclopentanol. Cyclopentanol. What is the name of this compound? We'll start numbering from this carbon. We'll start numbering from this side. One, two, three and four. Because if you start from this side, then this position will be third. We'll give priority to this alcohol, right? So it will be a buton of buton. Buton. Buton. Buton. Buton. Buton. Buton. Buton. Buton. Buton. Buton. Buton. Buton. Buton. Buton. Buton. So the name is what? Three bromo, butanol, butanol. Three bromo. Buton. Buton. Buton. Buton. Buton. Buton. Buton. Buton. Buton. Buton. Buton. Buton. Buton. Buton. Buton. This one is methanol. If you have CS3 here, if it is replaced by hydrogen, there is C L 3 O S methanol. So, this is substituents and this is a parent chain because it contains function group with it. Substituents is this what we can write cyclopentyl, where is 2 cyclopentyl? Cyclopentyl methanol, it is C S 2 right, C S 3 O S may 1 H if you replace by cyclopentyl group you will get this, ok. Diode, ethane diode, ethane 1 comma 2 diode. What is the name of this combo? The link is already of diode D, this is E V 1 different. The second last one is cyclopentyl or the double bond? What do you name first? Double bond is IN, this is cyclopropyl. So, I will write cyclopropyl first and then IN. Cycloterm we include in alphabetical sequence, cyclopropyl first and then IN, ok. We will see nominating what can we do. So, 5 cyclopropyl, X 2 IN 3 O, X 2 IN 3 O and the IN last one is E you have to omit, right. Cyclopropyl, 5 cyclopropyl, X 3 IN, sorry 2 IN, 2 IN 3 O, 2 IN 3 O. Otherwise there is no way you can bring the alkene name before the cyclopropyl. Yes, you got this cyclopropyl, if it is cyclical. 1, 2 and 3, right. So, at third position what we have? 3. 1, 2, 3, 4, 5, right. So, with the cyclopropyl, cyclopentyl. No, it is cyclopropyl. It is cyclopropyl 2 IN, because it is substituent, cyclopropyl will write as substituent. So, it is cyclopropyl 2 IN I, why IN I, because we have double bond here. And the position of double bond is what? Second carbon, because this one is first, this one is second. This is now complex substituent. Cyclopentyl, right. Because this is the primary chain, this is substituent. So, in substituent we will start the nominating from the carbon atom which attach with the parent chain. So, then that has the double bond. But this is substituent, this is not the parent chain. No, it is, but why can't you include it in the parent chain? Because OH will prefer OH over double bond. Yes, sir. OH attached with this ring. Right, OH attached with this ring. And this ring is attached with this ring over here. So, this is substituent and this is the parent chain. Cyclopentyl group with OH. Okay. So, this is cycloprop 2 IN I. Then cyclopentenol, right. Pentenol or pentenol. What? Okay. What if I place one bromine here then? 3-bromocycle. Not 3, 2-bromocycle. Yeah, all are. The first, the two should be second. 2-bromocyclochropionide. Okay. What if I place double bond here? Is this what you are thinking? Double bond here. So you don't want to say 2. Cyclopentenol. Pentenol. Pentenol. Pentenol. Pentenol. Pentenol. Not 4, it's 2. Because if it is 1, this should be 2 then. 1, 2, 3 here. Okay. I did not write. That's why you place double bond here. If I place it here, you won't realize this. Okay. That's right. So if you have double bond here, then you have to put second, 2 here, 3 here and 4. Right. So what is the name of this one if you write? It will be cycloprop 2 IN I. Cycloprop 2 IN I. Cyclopent 2 IN I. Because of the second carbon, the numbering of this should be 2, 3, 4 and 5. So this one is cyclopent 2 IN I. Okay. This one. How many carbon atoms are there? 3. 3? Yeah. 1, 2, 3, try off. It's a propane try off. Propane 1, 2, 3, try off. No. Because this carbon atom can have more than 1 over it also. Okay. So it is propane 1, 2, 3, try off. The name is propane 1, 2, 3, try off. This one? Cyclopent 2 IN I. Cyclopent 3 IN I. Cyclopent 2 IN I. Cyclopent 2 IN I. Cyclopent 3 IN I. Cyclopent 2 IN I. Cyclopent 3 IN I. Cyclopent 3 I. Cyclopent 3 IN I. Cyclopent I IN I. Just try it on. And you do it so fast. No problem. 1, 2, 3, 4, 5 and 6, so 3 bro, 4 cyclo, 4 cyclo, 2 enyles, 2 enyles, xn, 2 ball, xe. Ok, 1 by half, 3 cyclo, 3 bro more, 3 bro more, what? Understood, got it?