ليلة الأمتحان الكيمياء

[00:03:55][Clears throat] If everything is okay, tell me, okay, okay.

[00:04:04]Peace, mercy, and blessings of God be upon you. How are you, third-year high school students? We'll make the live session about two hours long so that anyone who was missing anything from the two study guides, whether the organic or inorganic chemistry guide, can finish them. Now, let's start with "In the name of God" because you'll find me reviewing the entire curriculum in today's live session. So, calm down and let's go.

[00:04:20]First question: The question says that Markovnikov's rule cannot be applied.

[00:04:27]People should remember that Markovnikov's rule applied to either alkenes or alkynes, but only asymmetrical ones. If I were to talk to you about alkanes, an alkane doesn't have a bond to break. Now, if we talk about alkenes, let's say my alkene looks like this, sir, and let's say my alkene looks like this. If you're a smart kid, you'll ask me, "Sir, what's the difference between them? This is an alkene and this is an alkene." I'll tell you No, if you try to connect the H bonds, you'll find that in this compound has two H bonds, while this one has only one H bond. If you connect here, you'll find an H bond here and an H bond here.

[00:05:02]Markovnikov's rule applies to asymmetric compounds. In simpler terms, if I add HBr to compound d, I don't have a problem because there are H bonds here and there. So, break the bond and add one H bond and one BR bond; it doesn't matter to me.

[00:05:15]But Markovnikov's rule only applies to asymmetric alkenes. If you remember, if I have the compound in the middle and they tell you to add, say, HCl to it [clears throat], the rule tells you, guys, that the first thing you do is break the existing bond.

[00:05:32]The rule also tells you that the compound with more H bonds ( these two) takes the H bond, while the compound with less H bonds takes the Cl or BR bond. So, don't forget that Markovnikov's rule only applies to asymmetric alkenes and alkynes. So, the question here is, I can't... Who should I apply this to? If I were you, I'd tell him the answer is B, of course, because the answer is B. He'll tell you, "Where are 3 dimethyl 2 butene?" Let's draw it. Butene means four carbons, like this. 2 butene means the bond at number two. Here's one, here's two. But there are 2 and 3 dimethyl, meaning here there's methyl like this, and here there's methyl [clears throat] like this. Look at the carbons, you'll see that this carbon has four bonds, and here the carbon has four bonds. So, it's symmetrical, guys. When is it asymmetrical?

[00:06:28]The number of H bonds here is different from here. But they're the same. So, if I were you, the one that would n't apply is B. Why, for example, didn't he choose answer C? He says propene. If you look at propene, you'll see that whether you put its bond here or here, it has two H bonds, and this one has one. So, it would apply to it. Why didn't he choose C?

[00:06:50]For example, phenyl bromide, the word " vinyl diet" means it's an alkene, like ethylene, but remove the "h" from it, the word "vinyl." So, Mr. Bromide, add "br."

[00:07:05]If you pay attention, you'll tell me, "Well, this is asymmetric, so the two "h"s apply here, but here there's only one. So, the answer that doesn't apply is answer number B in the question before it. I'm going backwards, by the way. When ethene gas burns in the atmosphere, listen to this information: anything in organic chemistry that burns produces carbon dioxide and water vapor.

[00:07:33]So, what's the problem with ethene, which is an alkene?

[00:07:37]This alkene, guys, when it burns [clears throat], has two options: if it burns in an atmosphere of oxygen, it produces one thing, and if it burns in air, it produces another. What's the difference between them?" What is it, sir? Does the air contain oxygen? No, of course not. The ratio is different here. Here, it's concentrated. Here, guys, carbon dioxide, water vapor, and a flame called oxyacetylene are released. It's a very strong flame, the kind used by welders to weld metals. But here, guys, the oxygen content in the air is low, so what happens is that carbon dioxide and water vapor are released, but carbon is also released as smoke. When we light, for example, wood in the street, sometimes smoke comes out because not all of it burns. So, in this question, anything will release carbon dioxide and water vapor except for alkene and benzene. By the way, guys, for them to release smoke, they have to be in the air so that there isn't too much oxygen. So, if I were you, when they burn in the air, they release a smoky flame, which is the smoke that comes out. The previous question asks which of the following reagents is used to detect the double bond in alkene. You know that alkene It has two bonds like this, one sigma and one pi.

[00:09:03]How do I detect them? With bromine, of course, assuming, guys, that bromine is supposed to break the bond here, and I add br here and br there. So if you're paying attention, you'll say that this bromine was red when I broke the bond, and here's br here and br there, and the color disappears. Some people might ask, "But why did n't I choose acidified potassium permanganate, which was called the Bayer reaction?" If you remember, it wouldn't work because it needs to be in an alkaline medium for the potassium permanganate to become KMNO4. And what is the Bayer reaction? The Bayer reaction, which is basically called oxidation, where I break a bond and add OH or OH here, and ethylene glycol comes out. For this to work, it needs to be in an alkaline medium, which isn't there. So the bromine has to break the bond. The previous question asks which of the following equations represents the alkane substitution reaction.

[00:10:05]To understand the question In this state, it's like, guys, methane. Here's methane, CH4. In this state, what's wrong with it? When I try to activate it, I make it undergo halogenation. This is what halogenation means. What does halogenation mean? It means adding, for example, chlorine, chlorine, Cl2. What would happen then, guys?

[00:10:27]What happens is that if there is a Cl here, it enters here, that's why they called it substitution, and the HD goes and reacts with the Cl, so the akan comes out, yes, but with the Cl, which they call, if you understand, the akan plus a halogen, which is fluorochlorobromide. The correct answer will be C, by the way. What will come out then? The akan halogen will come out.

[00:10:53]Halogen, the alkane, here's the alkane with chlorine, which is the halogen, and the alkaline halide, hydrogen. Here, guys, is the hydrogen here with the chlorine, which is a halide, meaning hydrogen or a halide. Sorry, it's chlorine and hydrogen next to it. What happened is that I have methane 4H, as you see, I removed some of it and added Cl, so the alkane came out, but with Cl, which is a halogen.

[00:11:21]So what will come out here? I'll remove the s and replace it with H. In simple terms, it will be a halide, which is this with hydrogen.

[00:11:34]Regarding the previous question, if you remember, we were wondering how to know how to react. Let me tell you how to know how many times it reacts.

[00:11:44]Here, H4 reacts four times. Each time, I remove the a and add the cl. Okay, question number 198. The question says I have three compounds, A, B, and C, which are hydrocarbons. I added hydrochloric acid to them. Then, when I added the hydrochloric acid, what happened?

[00:12:14]Let's go through it step by step. It says compound C doesn't react. If you understand, I added HCl to them. The part at the bottom didn't react either. So, without thinking, C is the alkane because an alkane doesn't have anything to break it with.

[00:12:30]So, if I were you, this has to be the alkane. Either the answer is A or the answer is D. Alkanes don't react; they don't have a bond to break. Here, I added... Chlorine, but at the number of anine, it's an alkene because an alkene only has one bond, so it has to be alkene. Pay attention, why does it have to be an alkene? To break its bond, I add Cl, which is ASL, one takes A and one takes Cl, so I lost one bond, so it has to be an alkene. Look, I chose either A or D. A and D is called propene, so the alkene is wrong, so I'll choose answer number D for it. This will prove to you that we're right.

[00:13:12]You'll see that here chlorine is added twice, meaning, in simple terms, how many anine bonds did I break? I broke two, so that's why I said that this is the alkene, propene. The previous question was a memorization question that asks which of the following compounds is used to clean electronic circuits. Which one is it? Freons.

[00:13:36]Freon, which is what damages the ozone layer, guys.

[00:13:38]Freon, you have two solutions.

[00:13:42]Its general formula is that carbon is surrounded by four atoms, but the most famous Freon in Egypt is... You meet EF and 2CL like this, so it tells you what cleans electrical circuits and refrigeration and air conditioning systems. That's it, guys!

[00:13:58]What's number A? Look at it carefully so I can come back to it later.

[00:14:01]CHBRCLCCF3, this compound, you all know it, it's called halothane.

[00:14:12]This is a safe anesthetic. Now, remember the curriculum, what was the unsafe anesthetic? It was called chloroform. It was CCLSLH.

[00:14:25]This was chloroform, an unsafe anesthetic. This is halothane, a safe anesthetic. And this is of course Freon. This is methane. The rest is easy. This is CENH, how many? This is ethylene.

[00:14:42][Snoring sounds] The next question asks you which of the following happens when 3 moles of pro-butene are added to 1 mole of 2-butene. Don't think, butene has four carbons. The important thing is 2-butene, meaning the bond here. If you're smart, we said that the bond is broken by 1 mole. Well, this has a bond with which one, so how much will it take to break it? It's one. He put How many bromines are there? The one that 's red, well, one of them will break, but out of the three [clears throat], two will remain. To put it simply, you entered the metro and found three empty seats, so you sat in one of them. So, one of those three will be used and break, leaving two that will still be red, but their color will diminish slightly. So, you chose answer number A.

[00:15:39]When does the red color disappear completely? Focus: if I put one here, it breaks one, but you entered the metro and found an empty seat, so you sat in it, that's it, nothing's left. In simple terms, there's nothing left.

[00:15:51]But here, out of the three, one will break, leaving two. The next one tells you about catalytic hydration.

[00:16:04]These questions in organic chemistry are very important, and we're not just solving questions; I'm reminding you of the curriculum.

[00:16:10]Catalytic hydration, of course, for people, means adding water to the gas produced from Adding drops of water to calcium carbide was called the dripping process.

[00:16:22]We'll cover it a lot at the end of the lesson.

[00:16:26]When I went to get calcium carbide ( CI S2), I'd add water and drip it, and it produced acetylene.

[00:16:38]Acetylene is called an alkyne. For those of you who attended my organic chemistry exam, I told you there's a very important point you need to know: when you dissolve an alkene in water, it gives an aldehyde. What's the aldehyde? Acetaldehyde. Or it ends with something like ethanalpropane. You have two options: either oxidize it ( this part is important for tomorrow) or reduce it.

[00:17:09]If you oxidize it, it gives an acid; if you reduce it, it gives an alcohol. So, again, when you add water to an alkene, you have two options: oxidation (clears throat) or reduction. It gives an aldehyde; oxidation gives an acid; reduction gives an alcohol.

[00:17:27]He also mentioned the catalytic hydration of the gas produced by adding drops of this... Okay, guys, when I added drops of water to this, it produced alkyne. It tells you to make a hydrate for it. I add water, and it produces an aldehyde. The aldehyde ends with that, so I tell it to produce ethanal. But isn't ethanal acetaldehyde? That's the aldehyde, guys! So choose answer B. And B and C, which is answer D, choose answer A: ethanol, alcohol, alcohol. If it tells you to reduce the aldehyde after you get it, and you do a reduction at the end (I wrote another reduction), it will give an alcohol.

[00:18:09]If it says to reduce it here, it doesn't say, then it's acetaldehyde or ethane. The next question is: Which of the following choices is correct? This part is a good idea. What's next? Tomorrow, God willing, you'll get a section like this. It will give you a compound and ask you to count the methyl and methylene groups.

[00:18:31]Methylene (CH2) is important. Tomorrow it's methyl ( CH3). Let's go.

[00:18:40]It asks you to count the sum of the methyl and methylene groups in these. Which one is correct? If you understand, let's go. From the bottom of the line, he says butane, so I'll try drawing butane. Here's butane: CH3 CH2 CH2 CH3. Draw this compound. It shows CH3 twice, meaning there should be two CH3s. In simple terms, where are they? And these two are important. Are these two and two? No, that's why this answer is wrong. Now look at the one before it: methylpropane. So I draw it: Here's propane, and the methyl group is CH3. And complete the rest with MHs, like this.

[00:19:29]Here's the compound. It looks like this.

[00:19:33]He's telling you it's methylpropane. He wants you to add the methyl group, CH3. If you understand, you'll tell me this answer is correct: CH3 CH3 CH3 CH3. No, 3, no, methyl. Is there CH2? No, that's why I'm zeroing it out. So I'll choose the correct answer, number C. Okay, excuse me, Mr., so people can follow along. What is cyclopropane? It means propane, it means a trivalent group, but in simple terms. Tie it together like this, a cycle ring, meaning a ring, then you complete the rest with H's, and it looks like this.

[00:20:11]You complete the rest with H's, and it looks like this.

[00:20:14]Here, how many methylenes are there? 3, lah, CH2, so it's CH followed by no, that's wrong, and so on.

[00:20:23]This question is important for tomorrow, God willing, when you go into the exam. They might ask you to count how many methyl H's 2 or how many methyl H's 3. The next one, when comparing the chemical activity between butane and cyclobutane, is a different story. I chose these questions specifically because each question reminds you of something from organic chemistry. For example, butane is an alkane, and alkanes are relatively inert; they do n't react because there's no bond.

[00:20:56]Cyclobutane is supposed to be an alkane too, with four, but the ring is always more active. It needs to be active because the stable angle here should be 109.5, but here it's not. So, if they ask you which is more active, the question is... We'll solve a lot more of this in a bit. Which is more active? What's the right choice? You'll tell him that cyclic butane has this ring because the sides are close together, so the structure is messed up. It's more active and wants to react, so this is the most active and this is the least active. Choose answer C for him. The next question is the same story: count methylmethylene, but instead of saying it like that, he gave it like this: CH3, here it is, meaning it's called methyl.

[00:21:48]CH2, here it is [clears throat], meaning it's called methylene, but instead of saying it, he gave it as a diagram. And here's C with ASH, which is n't called that, we don't have that name.

[00:21:59]He told you that anine is methylbutene, butine means four carbons, and here's the bond, methyl, meaning it has an integer in CH. And if you want to continue with the steps, here's one, here, here.

[00:22:23]Let's go. If I were you, let's go through it together, one by one. Here's the compound, guys.

[00:22:33]Look at the first one. Here, C has two bonds, but it has an integer. He wants C to have only two bonds. If there's one H, does C have two bonds with one? No, so the number here is zero. The answer is A. Complete the sentence: C here has two H bonds, but only two regular bonds. Well, C here has two H bonds and two regular bonds. Well, there's only one here, which is this part.

[00:22:59]Well, there's H3, which is methyl. Well, there's one, guys, and the methyl is already here too, so that's why there are two. So I chose answer number A. I'll save the live stream for you so that if you don't understand any question, you can go back to it again. But each of these questions is important because, guys, it helps you understand a different idea than the next one. The next one tells you about the tertiary polymerization of the simplest acetylenes. Let's go one step at a time. The simplest acetylenes [laughs], which is acetylene. Here's what it looks like when I do tertiary polymerization. Now go with me one step at a time.

[00:23:38]So, the compound is formed three times. Here's one time, two, then three.

[00:23:49]Let's count how many bonds there are in these three.

[00:23:51]Sigma and how many sigmas are there? He's just asking about sigma and pi. Let's go. If I were you, I'd say sigma where 3 is, so how many sigmas does the compound have? 3 is here, 3 is here, sigma is important because the people are my friends, the animals. Here's 1 where 3 is, how many sigmas does it have? 3 over 3 here over 3 here. So how many sigmas does it have, guys? t sigma, what is here in 2 pi and 2 pi and 2 pi, so that's 6 pi, so it's inside the reaction like that.

[00:24:23]When I polymerize it, what did it give? Go back, he was giving me gasoline. Well, gasoline looks like this, guys. And here, hh... Why didn't you put it at the beginning? If I had put it at the beginning, it would have been a chain.

[00:26:04]Since he didn't mention a number here, if I were you, I'd be talking about one propene. Look, the bond is at one. But, sir, why didn't I put it here too? Because I don't understand. The methyl group is white like that, so it wouldn't be propene, it would be butene. Four of them are wrong.

[00:26:24]So he wants the number of methylene bonds, the methyl groups in this compound. He completes the rest with Hs. It comes out like this: here it's bonded, and here it is. How many CH2s are there? There's only this one. So I tell him the answer is number A. The next one is the number of sigma bonds. He has to ask about this tomorrow. Sigma-pi will ask you in the exam: methylene, methyl. This is in dimethylbenzene. Give benzene dimethyl, meaning here there's CH3, and here, for example, CH3A. He wants the number of sigma bonds. If you get this question tomorrow, God willing, and it's important, guys, don't write CH3 like that, write it as H. And here it is like H, and here it is like H.

[00:27:20]Unwrap it, and here it is the same. The story is HHH and it's completely messed up, don't do it like that. Come on, let's count the sigma bonds. We'll write in a different color. You know that each of its corners has an H, right? Each corner has an H.

[00:27:41]Count its bonds with me, it will be 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17, so it will be 17. There's something we almost forgot. Let's draw it properly. It should be 18. Let's draw it again.

[00:28:09]Where's the eraser, guys? Let's erase this. I don't know. No, anything will do.

[00:28:21]Let's try this. Here, here it is. Okay, let's draw it properly.

[00:28:31]Here, guys, is the benzene like this, in one color now, right? And here are its bonds.

[00:28:38]Each of its corners has an H, right?

[00:28:41]HH. And he said dimethyl, meaning C has HHH here, and C has HHH here, instead of the mess we were drawing in the diagram that Okay, let's count.

[00:28:57]When you count the sigma bonds, go ahead: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17. So, it's 17.

[00:29:22]We've counted everything. Now, let's continue. This question is at the end. The question before it asks you, based on the following diagram, what you would call the IUPAC.

[00:29:39]These are the tricky questions, you hate them so much, and you'll find questions like these tomorrow.

[00:29:44]Look, my friend, there are a few things that, if you know them, will make organic chemistry much easier for you. Regarding alkynes, I mentioned this sentence a little while ago, but remember, let's talk about it. Remember, here it says "catalytic hydrolysis."

[00:29:58]What does it say here? "Hydrogenation." I don't see what it says here, but the word "hydrogenation" is correct.

[00:30:04]Okay, alkynes, alkynes, guys, you have two options.

[00:30:09]I'm still... I told you a little while ago that when you hydrate it, meaning when you add water, it gives an aldehyde. Remember I wrote that a little while ago? When you oxidize it, it gives an acid. We wrote that. If you know a little about these things, you'll solve what's below. Now, I want you to know this, you absolutely have to know it. Here's the alkene.

[00:30:32]When you add hydrogen to it, it turns into an alkene. Add hydrogen again, and it turns into an akane. It only needs one bond to turn into an akane. So, it's telling you that you hydrogenated the alkene once, and it gives this compound, the alkene. What's new? Look, my friend, if you remove water from an alcohol, it gives an alkene. And if you add water to an alkene, they used to call that dehydration. But here, if you hydrate an alkene, meaning you add water, it gives an alcohol. So, the alkene, if you add water, it will turn into an alcohol. I'm sure you get it. There's something like that. We already mentioned this a little while ago. So, you know that an alkene gives an alkene, then an alkene, add water to it, and it gives alcohol. Remove the alcohol water, and it returns to the alkene. What are you trying to get at? It turns out that here's an acid and here's an alcohol. It's a very well-known principle in organic chemistry that an acid and an alcohol form an ester and water. So, what will come out is this because he's asking for the ester. If you understand, this ester is benzoate, which is the ester of these things. But he wants to call it IUPAC. So, if I were you, I'd say it's wrong, impossible. Why? Acetate isn't IUPAC.

[00:31:56]Acetate means two carbons, but it's called ethane.

[00:31:58]This is IUPAC, not IUPEC.

[00:32:01]So, sir, how do I know from the others? If you're smart, whether this side or that side comes from the same alkene, meaning the same carbon, then if I were you, I'd choose C. Why C? Because he said ethanoates, meaning two, and ethyl, meaning two. See the logic? They all come from the same alkene, so this will produce the same carbon and that will produce the same carbon. I'm only doing oxidation and reduction by adding water, meaning I'm not adding carbon. So, what happened is that here, it's the same as this. But why is this wrong?

[00:32:31]If I were you, he would have found that this is the same as this, and we excluded propanoates, which are Okay guys, to put it simply, it comes from a three-membered system. We don't have a conservation hydration system here; we only had hydration with two elements, which is ethene, which is acetylene. Okay, the previous question was the same. He gave a diagram, guys, and said, "Let's go."

[00:32:57][Clears throat] He gave a diagram and said, "I have a compound called CNH2N because it's not clear.

[00:33:07]CNH2N, NaCl, I think." Now look at this law, guys. By the way, you know it, but you're not paying attention. What is this? It's an alkene. You 'll say, "How come, teacher? Isn't an alkene CNH2N?" Why did he put a missing one here? He put chlorine in its place. Isn't organic chemistry carbon and hydrogen? What did he do? This is supposed to be an alkene. He removed one hydrogen atom from here but replaced it with chlorine. So this is an alkene. It's either CNH2N or it removes one hydrogen atom and puts Cl in, which removes two hydrogen atoms. And he put Cl, the important thing is to replace them here. He told you this compound is an alkene. The arrow is coming this way, right? The arrow is coming this way. The alkene, but I don't have the picture, it's not clear. The alkene.

[00:34:01]If I were you, this alkene, guys, he's telling you, I got a compound here, I got a compound here, and this compound, I put Cl in it, and it came out like this.

[00:34:13]If I were you, it came out as an alkene, so what was it?

[00:34:16]Look, and say " God is One" and pray to the Prophet. We agreed that it's an alkene. The alkene is CnA. It got CnA2n. Now, if you're someone who understands, he removed one of the H bonds. I'll repeat it again here and put Cl in its place. So he removed one H bond and put Cl in its place.

[00:34:41]If I were you, I'd choose answer number C. Now let's see. C tells you that the Y bond is the alkyne. Why? Because this alkene actually has three bonds, like this.

[00:34:57]He put HCl. Well, I'll break its bond right here and put one H bond and one Cl bond. So, what is the compound we were talking about? The Y molecule is indeed an alkyne, so I break its bond and add one H and one Cl.

[00:35:14]He said to add Cl, okay, so it's an alkene. But why did you say that this is considered a cycloalkan? What is a cycloalkan?

[00:35:25]If I were you, this cycloalkan is the same as an alkene, with the same formula. What did he do here in the problem? He added chlorine, so I break its bond and add this and this. You know that this and this have the same formula. Here you broke its bond and added one H and one Cl, so it looks like this. So this is a cycloalkan. Yes, cycloalkanes and alkenes are similar in the formula. What did he do? Let me tell you what he did. Can it be like this?

[00:35:54]For example, here was a cycloalkan, and I added Cl to it. So I'll remove... you know that here there is H and here H, and so on. H and H.

[00:36:02]I'll be with you until you understand. These have the same formula. What did he do? He removed and added chlorine, so I'll remove one of the The H that's here is S, so what is it? We agreed this is the same law, but I removed one S and put one and put it in. What is it? The question before it, the question before it tells you which of the following represents the use of the product of polymerization of compound X by addition in. Let's see what the compound said. Fill me like this: CnA minus A. So this is the alkyne.

[00:36:48]I added ACl to it. What will I do? So, I'll break one of their bonds and put one H and one CL here.

[00:37:00]This compound is in polymerization. Let's remember about polymerization; we've studied four compounds: either this compound called ethylene ( we'll explain polymerization now), or this compound that's used to make Teflon (the halal product), or this compound, which is propene, or our compound here (H and H), which is this one. Now, let's go one step at a time.

[00:37:39]This was called ethylene. What does polymerization mean? It means I take this compound with this compound with this compound and combine them. It's called addition polymerization. The two are similar. So, if I polymerize this compound, it gets polyethylene.

[00:37:51]What do I do? I break the bond and combine them.

[00:37:54]This is the compound they use to make the plastic bags they sell oil, sugar, flour, and things like that. But this one only has two carbons, so it's light. This one is similar, but it's called propene, so it's polypropylene. The compound with two carbons is a little heavier. This is what they make sacks from.

[00:38:13]Now, if I were you, I should This is what they use to make Teflon pans. On the other hand, if I were you, this is what they use to make the containers for water bottles.

[00:38:28]So let's see, he gave you a compound like this and said, "I have alkaline, I added HCl to it."

[00:38:38]He doesn't mean either of these, he means this one because it contains Cl. Okay, what is this compound used for? He used it to make water bottles, floor insulation, and sewage pipes. So I'll choose answer A for him: when? B for plastic bags (if he's asking about this, when?) C for cans and sacks (if he's asking about this, when?) and cooking pots (if he's asking about this).

[00:39:11]Polymerization is of two types: addition or condensation. Addition is similar. Condensation is different and produces water.

[00:39:23]Which of the following represents the compound resulting from the reaction of ethyl Propene, here's the propene, and ethyl, meaning C, take, here's the biotin, here's four, and ethyl, here's the ethyl. Okay, guys, I'll tell him here, put two here, and he put it at two, meaning give one, give two. Let's go.

[00:39:52]He told you to do this compound with HBR. Let's see one by one.

[00:39:59]First, complete the rest: Hs, give ash here, and ash here, and that's how the four are.

[00:40:05]Marknikov's rule, remember it? It told you that whoever has more ash takes A. So this one has more A, so it will take A after, of course, breaking the bond.

[00:40:16]So it's supposed to be all of them. The organism has already broken the bond. I put the one with more H, so it took H, and this one took BR. Then choose the longest chain from the beginning. It comes out like this.

[00:40:31]How many five carbons does this have? It's called pentane. Which one has pentane? There's only this one. So, again, guys, again, because this question is also famous, by the way. It brings a compound that you'll apply. Marknikov broke the chain. The one with more H takes H, the one with less H takes BR. He chose the longest chain, which came out to five, meaning it's called pentane. The only one with pentane is the answer A. Which of the following are the steps necessary to obtain water gas from heptane? If you're smart, where did the water gas come from? Remember, it came from the reaction of carbon dioxide. Look up! It comes from the reaction of carbon dioxide with methane. What should I do? Well, when I did that, it produced CO + hydrogen. That's water gas combined. So, he's telling you how to get this water gas from heptane. What should I do?

[00:41:34]Burn it. Al-Baroudi just said that anything that burns produces carbon dioxide.

[00:41:40]So, I'll take this compound and burn it. That's wrong. It has to burn first to get carbon dioxide. Then I react it with methane. What does dry distillation mean?

[00:41:50]Dry distillation means heating, and dry distillation happened twice.

[00:41:59]First, sodium benzoate reacts to produce benzene.

[00:42:05]Second, sodium acetate reacts to produce equilibrium.

[00:42:14]So, guys, I'll react it directly with methane to produce water gas. The next question is: what's the correct IUPAC name for the compound called isopropyl butane? Let's take it one step at a time. Butane means four. Some people still have trouble with the word " iso." Of course, you know that "propyl" means three, so you'd go and find a branch for three. But the three look like this... no, sorry, no, that's not how the three look. Erase it. Okay, the three look like this.

[00:42:59]You'll feel them forming a 90-degree angle.

[00:43:03]If I were you, he'd say isopropyl. Here's butane, four, and isobutyl. Isopropyl means... C, C, here, C, here. In simple terms, the three aren't like that. They're not one after the other. If they're one after the other, it's called propyl. If they're at this angle, it's called... The important thing is to choose the longest [clears throat] chain. So when I go to choose the longest chain, count with me: 1, 2, 3, 4, 5. It's called pentane. Which one has pentane in it? Okay, your answer is correct.

[00:43:38]The question is, which compound won't have Marknikov's rule applied to? It works on the asymmetric. He tells you not to work on which one? On the homologous one. Why? Biotin, meaning four, but he said two, two, guys. So, should I put the bond here? Well, this one is missing an a and an ash. Well, this one is homologous, just like the others.

[00:44:04]The rule says five words: whoever has more ash takes the h. So, like the right of Wasel or what? The important question for tomorrow is: Which of the following compounds produces this compound? Let's take it step by step.

[00:44:27]Is one mole of bromine for propene? I'll test it to show you what's right and wrong. Here's the propene.

[00:44:36]If I add one mole of bromine, no, that won't work. Why?

[00:44:40]Because then I'll break its bond and add BR here and BR there. So, BR would be added once at 1 and once at 2. He wants them. BR is only at 2. So the first is wrong, of course. The second one is illogical. I mean, adding bromine to bromide?

[00:44:56]What does that mean? Chlorine? Next is HBR for propene. So, propene also has three Hs. Here they are. And the rest are Hs.

[00:45:11]When I add HBR, Markovnikov's rule says that breaking the bond breaks the bond. The one with more Hs takes the H, and the one with less H takes the BR.

[00:45:20]What's this? I put BR in the middle, but only one BR. He wants which BR? I'll choose answer D. Why D? Because D says that's how it looks.

[00:45:31]Read this: Bromine propene, that's what it looks like.

[00:45:36]So when I go, guys, I add the HBr that it wants. Well, I'll actually break its bond, and whoever has more H will take the H, and whoever has less H will take the BR. So the BR is now at the same carbon atom in the middle. The one before it tells you what the wrong statement about urea is. Remember that was the very first lesson?

[00:46:01]Berzelius developed the theory of vital forces and told you that organic compounds in the lab are in the body of a real organism, but they aren't formed in a lab until Fuller came along and prepared urea, which is organic from inorganic. Okay, read this: Is it the first organic compound that can be prepared from inorganic compounds? Right, but he wants the wrong one.

[00:46:26]Does the urea molecule contain subsigma and one pi? Draw it, teacher, draw it, teacher. Here's the urea molecule.

[00:46:36]Count with me, my friend: 1, 2, 3, 4, 5, 6, and one from here, seven, seven sigma, and one pi, which is So, this answer is actually correct. He wants the wrong one.

[00:46:52]Okay, look at number D. Is it an organic compound that dissolves in water?

[00:46:55]Yes, it's urea. In a living organism, this is urine, or rather, it's one of the components of urine, so it dissolves in water. So, the wrong answer is number C. Let me tell you why. C is used as a fertilizer. Yes, this answer is correct. Continue, but does it give the soil nitrogen and phosphorus? Yes, it gives it that, but does it actually have phosphorus? The wrong answer is that it doesn't give phosphorus; it doesn't actually have phosphorus. Next: Which of the following compounds is an isomer of an open-chain hydrocarbon containing four carbon atoms and two pi bonds? If you're smart, go ahead. The question is silly. Two pi bonds means alkyne.

[00:47:38]But that's alkanes. Wrong. But that's butene. Alkyne only has one pi bond. But that's propene.

[00:47:45]Alkyne only has one pi bond. But that's butyne.

[00:47:49]Alkyne has two pi bonds. The previous one wants the number of carbon atoms in an alkane. Alkenes and alkynes, excuse me, what's this? Alkanes and cyclic alkanes, and the alkenes where this conformation begins, which are isomers. Let's go one step at a time. The simplest organic alkane is methane. Focus, the simplest chain is 3C. Listen to what I'm saying: the simplest organic compound is methane, the simplest chain is 3C. So, what does branching start from? From four. So, my answer is this. Why? Because the four can either be drawn side by side or three. One branching isn't possible. I can't make a ring. This ring, if the alkene is similar to the cyclic alkane, we solved it in a problem a little while ago, so it starts from four. The cyclic alkane also starts from four. Why? Because when I draw the ring, does it have another drawing? No, it doesn't have another drawing.

[00:48:50]Draw it like this. Give me another shape for the three. So, he's telling you, you'll say, "But teacher, it's possible to use three." Look at the brain, look. The brain, what is it? It is possible for the three to be drawn, but the alkene is drawn.

[00:49:06]You just said that they are drawn instead of a ring, so they are drawn as three. I will tell you that is correct, but he is not specific, meaning if he wants the alkene to be ring-shaped or the alkene, but okay, even if it is this, let's look at the next one, alkynes. Alkynes cannot be three, and why is it isomer? Tell me, bring me isomers of this, here, bring me isomers of this. There is no isomer. Isomers when there are four so that the four are drawn like this.

[00:49:40]Either the four are drawn like this, the four will be drawn in the line shape. No, this is also four, and here there are three, and one, so I will choose the net from the four, so the answer is number B. The next one is dead. Are you focused or what? Let's not waste our effort, guys!

[00:50:15]This is a really tough question. One of the following compounds needs six hydrogen atoms to become saturated. So, how many pi bonds are there? Pay attention! The pi bond is broken and replaced with H and A, meaning H and A.

[00:50:33]Look, guys, here's the bond: pi needs a hydrogen or bromine molecule, any molecule. Or, the pi bond needs two atoms.

[00:50:49]So, this molecule, which is A, plus the two atoms, is H and A. Now, it says it needs six atoms, and each one needs two atoms. So, how many pi bonds are there? There must be three pi bonds. Pay attention! Let's draw it as 3 pi so it looks like this. I'll give you an example just so you understand.

[00:51:11]I'll break a bond here and add H and A, break a bond here and add H and A, and here I'll add H and A. So, I broke three pi bonds and added six.

[00:51:20]If I add six, then it's 3 pi. Okay, the question is asking you... Who has 3 pi bonds?

[00:51:27]Let's go through it one step at a time. The name of the resulting compound is: Who has 3 pi bonds?

[00:51:32]Let's look at the word diphenyl.

[00:51:35]Phenyl means a benzene ring, and here's another benzene ring.

[00:51:43]Does this one have 3 pi bonds? No, this one has 3 pi bonds, and this one has 3 6 pi bonds. So the first one doesn't work, and neither does the second one. The second one tells you: Where is phenylpropene? Here's the propene, and here's the bond. Phenyl means a benzene ring. It wants 3 pi bonds. If this is 3 pi bonds, and this is one 4, then it doesn't work. Now, pentene. Pentene means an alkyne. It has two pi bonds, but not 3 pi bonds. Now, chlorobenzene. Here's the answer: chlorobenzene. Be careful, these answers are wrong. The one that's correct is misspelled in the printing. The answer is chlorobenzene. Let me tell you, here's benzene.

[00:52:31]This circle is for chlorine, and indeed, this is what contains chlorine. So I'll tell him that the chlorobenzene before it, the molecular formula, did the same thing, but it's a bit tricky. The molecular formula of an unsaturated hydrocarbon, two of which react with four molecules, one at a time.

[00:52:50]I don't deal with two moles; I want the mole. What does mole mean? Look, it means all of these were before it, and I don't take anything; I deal with the mole. So when he tells you that two moles reacted with four, it means that one mole reacts with whichever molecule, molecule, meaning H2.

[00:53:14]Focus first; he said atom or molecule. Atom means H, molecule means H2. The two reacted with four, so one will react with the most important.

[00:53:25]Well, that's how it all turned out to be 4 H. So he's telling you that it was unstable, and after it stabilized, its symbol became CXHY. After it stabilized, it became like this. So before it stabilized, what was it like? It was CXHY, but I moved next to the carbon, towards the H.

[00:53:46]He went Take two molecules, meaning take 4H.

[00:53:52]Before he took the 4H, he was missing A.

[00:53:56]So if I were you, I would tell him he was missing A. So when did he choose C? We chose A. If it was an atom, meaning H, but here it's a molecule. A molecule means A. Focus again. He told you that two moles react with four, so one reacts with two. Two are important in B.

[00:54:13]4H. He is after it stabilized. So before it stabilized, it was missing A.

[00:54:18]I hope you understood. The aromatic hydrocarbon that gives an explosive substance when nitrated. What is the explosive substance we have?

[00:54:34]Of course, it's a TNT bomb. So he is telling you that the aromatic hydrocarbon that gives an explosive substance when nitrated, before I add the nitrate, before I add NO2, the one in black that I will mark, what was it called? His name was Talween, who was Talween?

[00:55:03]Next, by adding 1 mole of hydrogen and then 1 mole of chlorine to this compound, let's write it like this. He wrote CH, then three bonds, then CH, then CH and CH. Come on, my dear, he tells you to put 1 mole of hydrogen, so I will break its bond here and add a, and I will break another bond too and add a, because he put 1 mole.

[00:55:38]Then he told you to put chlorine, so I will break this bond too and add Cl here and Cl here.

[00:55:46]Let's name it.

[00:55:48]What is the longest chain? 1 2 3 4 4 means its name is butane, they're all butane, okay, but this one is at number one and this one is at number two. Which one has one and two? The answer is number C. I got it. I broke here, two H, and here I put anneale chloride. The next one is at halogenation [clears throat]. Halothane with bromine. I said it at the beginning of the lesson, which was basically a safe anesthetic.

[00:56:23]I'll halogenate it. What does halogenation mean? We agreed then. I'll remove the H here and put BR. We'll call this bromine. And this one, of course, they're all called ethane, not two carbons.

[00:56:38]This bromine, and this one is at number one because this is the first alphabet. So which one has one and one?

[00:56:44]Dibromo. Of course, this is wrong because I don't write the fluorine first. And this one and two is wrong, and one and two is wrong too, because this one is at number one and this one is at number one.

[00:56:55]Which one has one and one?

[00:56:58]The answer is number B. Oh, my joy! I can't see, Esther. Stay with me, Zakaria. Esther, I put anneale or oh. Okay. Listen up, my friends.

[00:57:14]Esterification involves three reactions, and this is the last lesson in the curriculum, so pay close attention, please.

[00:57:18]People seem to have overlooked it. Esterification undergoes three processes, and this part is very important.

[00:57:25]Either it's an acidic hydrolysis reaction, which produces an acid and an alcohol; or it's a basic hydrolysis reaction, which produces a salt of the acid.

[00:57:40]This acid salt could be sodium benzoate, potassium benzoate, or sodium acetate, plus alcohol. Or it's ammonia hydrolysis, which produces alcohol plus an amide. This amide is CO₂, NH₂, acetamide, or benzamide. The important thing is that here, he added an "n" (meaning a base), so it will produce an acid salt and an alcohol.

[00:58:08]This is definitely the alcohol, if you're paying attention, because he told you he dehydrated it. We said it a little while ago: dehydration of an alcohol gives an alkene. Now, remember, you dehydrated an alkene to give an alkane. Remember, you hydrogenated an alkene to give an alkane?

[00:58:17]Remember, or have you forgotten? Alkyne, then alkene, then an alkane. So, this is definitely the acid salt.

[00:58:30]This acid salt undergoes dry distillation. Remember what I said? In dry distillation, let me remind you that if I have benzoates, I get benzene. If I have acetate, I get methane, which is an alkane. So we're on the right track. Which one of them is the answer that works for all these choices? Let's look at option B. It tells you that X is sodium ethanoate. X is sodium ethanoate, right? That works. Z is an acid and Z is ethene. That's the answer, actually. Amazing! So why not option C?

[00:59:05]C tells you to wait a minute.

[00:59:12]No, wait, let me tell you why it's wrong.

[00:59:17]If I were you, I'd say, " Look, D, X will be propanoate." Pay attention, X will be propanoate. It's similar to this. Why did I say this at the beginning? Because it's similar to this.

[00:59:31]But here it says that Z is ethene. But Z is actually an alkene.

[00:59:38]What's the difference between ethanoates and acetates? Look, let me tell you, alcohol gives water. I'll remove water from alcohol. Yes. The origin of this is an ester, so the ester will have the largest number from here, which are the three propanoates, which is the three that break the previous one.

[01:00:05]God is One. What is this? If you know that the CN is two in this compound, what is the compound? CN means CI2N, which means 4 + 2, which is approximately six, right?

[01:00:23]Or, well, that's why this is an alcohol, this is ethyl alcohol. So when I oxidize it, it will give an acid, right? Well, the one below told you that the CN is three in this compound. So, apply 3D2N, which means 6, or 2D2, which means you made it an acid. What is it, which is COOH? What acid? When I reduce it, it gives an alcohol. He brought two compounds, one acid and one alcohol. The alcohol is oxidized against an acid, and the acid is reduced against an alcohol, and so on. Let's see which answer is correct from all of this. Let's look at number B. Is the boiling point of C, which is the alcohol, higher than that of D, which is the alcohol? Do you remember, or have you forgotten, which is higher? An acid makes two bonds, so is the boiling point of C, which is alcohol, higher? No, look at answer number D. Is the compound an isomer of this compound? Look, here's B and here's D. This is an acid, yes, and this is an acid, yes, but this one has one carbon atom and this one has two carbons. So this answer is also wrong. So this one is wrong, and this one is wrong. The next one asks if, when compound C combines with this compound, it produces a pentone, which is a ketone, or an isomer. No, an acid with an alcohol gives an ester, so this one won't work. So I'll choose answer number C. When compound C combines with this compound, it produces an acid and an alcohol, which gives an ester. And the ester has an isomer, which is why it can be an acid. Even the pentone has five because here it has two and here it has three, so it actually has five.

[01:02:18]I hope you understand which of the following steps are correct to obtain a compound used as a vasodilator from the compound called lah- bromo or lah-chlorobrombin.

[01:02:42]First of all, why did you choose to nitrate it at the end? Which one is used as a vasodilator? In short, you have to remember. Here's where this compound came from. I'll explain everything to you.

[01:03:02]This compound came from glycerol. I nitrated the glycerol, so this is glycerol.

[01:03:11]Let's go through it step by step. First, chlorobrombe.

[01:03:17]What did you do, guys? First, I did addition halogenation. The compound he's talking about is called chloropropene. Here's the first thing I'll do: addition. Now, when I do the addition halogenation, here's what the compound looks like. Let's go through it step by step. Here's the compound he mentioned. When I do addition halogenation, what does halogenation mean? It means I'll break a bond. Look, I'll break a bond and add Cl and Cl.

[01:03:50]Now, all three contain Cl. Focused? What do I do?

[01:03:53]I add 3 Cl. Then I add a base to remove the Cl and add 3 OH, which is alkaline hydrolysis. Then I do the nitration at the end. It's a very interesting topic.

[01:04:07]Which of the following choices represents the additive or necessary processes for a crop? It's a substance added to textiles to give them softness and suppleness. It's three bromopropenes. The question is the same story, by the way. Let's go with it.

[01:04:31]He told you three bromopropenes, so propene, then three bromopropenes, so propenes. So, here's H, here's H, and here's H. I'm writing the compound he mentioned first, the one added to textiles and all that stuff, so you know.

[01:04:51]Also, glycerol. You know, if you understand, why I marked the number C directly like that? Because he told you it has oxidation. You know what I'm going to do, right? Well, the first oxidation means I'm going to break the bond here, but I'll add OH.

[01:05:10]I did that. I broke the bond, I added OH, then a basic hydrolysis, meaning I remove the BRD and add a base, which is also OH. There you have it, the compound. Okay, two minutes, guys. Let's take a break and we'll come back again. See you later. Now, bring the compounds. Let's take a break, guys. Anyone for question 74? We'll go in. There's still the story about benzene and all that stuff, and the disc, water, and para, and all that stuff is important for tomorrow, God willing. So, question 74, come on, quickly with me.

[01:09:17]The question asks you about the aromatic compound with the formula C7H8. What is that, guys? You need to know that the compound called talc is called talc. What's that, sir? Talc used to be prepared in two ways: either by using heptane to perform a catalytic reforming, or by using benzene and then performing a free-craft reaction or a chelation reaction. So, the question asks you how to get talc. I have two options: did you find benzene and perform a chelation or free-craft reaction? No, but I have heptane. Heptane is made up of seven Cs, but next to each other. Here's how they look next to each other.

[01:10:02]So, these seven Cs, guys, perform a reforming reaction. So, choose answer number C, the next one.

[01:10:15]75 asks you about an aromatic compound with the general formula C. How can you fill me in like this? CN, CN, CN, HN, O 3. Look, my dear friend, to convert this compound into the simplest aromatic compound, meaning benzene, I want to convert this compound to benzene. Benzene doesn't have any oxygen units. Here, what was I doing? I was getting zinc to remove the oxygen. But here, there are 3. Or you could get 3 moles of zinc. So choose answer number B, the one after it, to get the simplest hydrate or aromatic acid. He wants the simplest aromatic acid, which is benzene, guys. But benzoic acid, you do it like this. He wants this from an aromatic compound in O.

[01:11:16]What do I do to get it? We just said the first way is to do reduction. Go put zinc, guys, to get benzene. Who started with reduction? This is wrong, and this is wrong. The important thing is, who started with reduction? Either B or D. But after I do reduction, I will do [clears throat]. Remember alkylation? We said a little while ago that it produces color. So, the next thing is alkylation. We said B and D, so it's B. Well, after that... I'll bring up the topic of acid, and what does acid come from? Oxidation, because he told you that oxidation occurs during the hydrolysis of a monohalobenzene compound.

[01:12:04]What does that mean? So, monohalogenated benzene means it only has one chlorine atom.

[01:12:10]When I hydrolyze it in an alkali, you've probably noticed what I'm trying to do. I'm in an alkaline environment, so I'll remove the LDL and replace it with phenol. Then, when I add alkane—this aldehyde used to be formaldehyde— what does it do? Phenol with formaldehyde produces a compound called bacliad, which was used in cigarette lighters. It was the plastic used in air conditioner fans, the kind of plastic used in the main body. So, they use it in cigarette lighters instead of putting out the cigarette under your potatoes tomorrow.

[01:12:54]Next question: Arrange the following compounds in ascending order of sigma bonds. This question has been repeated many times: Which has the highest number of sigma bonds? Then which? Then which?

[01:13:07]But unfortunately, when a question like this comes up, you have to test it. Here's phenol, here's benzene, and here's benzene. We solved it from... A few questions, a lot, it's okay. This is diphenyl, count and... and of course, don't forget that here's H, and here's H, and here's H, and here's H, and here's H. Okay, I'm not going to count anything, but you try counting like this. You'll find that this one has the most sigma bonds. Here it is: 1 2 3 4 5 6 7 8 9 10 11 12 and so on. Then move on to the next one: 13 14 15 16 17 18 19 20 21 22.

[01:13:56]So, you'll find this one has the most sigma bonds.

[01:13:59]I'll draw it for you. Now, take six of them, and each one has an H.

[01:14:09]Count the bonds, and you'll find less than 22.

[01:14:18]Then, methylbenzene, count the sigma bonds again, and you'll find this one has the fewest H H H H, and so on. So, this one has the most sigma bonds, then this one, then this one. Good luck with the question: Which of the following methods is NOT used to prepare a compound? Aromatic, its formula is C8, right?

[01:14:48]C8H10. Does he want to prepare a compound with the symbol C8 or what?

[01:14:57]Yes, you're saying yes, Mahmoud Maneh, okay, he's telling you what method won't get me the compound called C8.

[01:15:07]If I were you, the answer would be number B. Why? Let me tell you, talc looks like this: six benzene, and here's one. Count how many carbons this is, guys. Here it is: 1, 2, 3, 4, 5, 6, and seven.

[01:15:28]That's seven carbons. This is talc.

[01:15:31]If you add ethyl here, the ethyl will be two on top of the seven here. So that makes nine, not eight. Why didn't you choose, for example, C? C? C? C? It says benzene. Focus now.

[01:15:45]Here's benzene.

[01:15:46]When I add ethyl, which is two, it becomes eight. The one above is talc, which is seven. So when I add one methyl, how many eights does it make? The one that wo n't work is BC7A18. We said what sub-C produces. If you remember, in the molding process, it produced talc.

[01:16:17]The coloring came from either Farid al-Kraft or Heptin. Here's talc.

[01:16:28]Do hydrogenation on talc. Break all the bonds inside.

[01:16:36]I broke all the bonds inside. Here's H, and here's H. Now it's a completely normal compound. So it's asking you how this compound reacts, right? Yes, it's asking about A and B. If I were you, this is the talc that contained benzene. And if you're smart, just give CH3. CH3 contains benzene. Benzene reacts with both, by addition and substitution.

[01:17:11]So, is there an answer that says A reacts by addition to substitution? That's wrong. And it does n't have a bond to break, so by substitution there's no addition, it won't add anything.

[01:17:24]Both react by substitution. The one who forgot this part, for it to be an addition, a bond has to be broken. So here there's no addition, there's a bond to break, so there's an addition. I want to get to Gamoxan. Gamoxan, guys, is an insecticide. It was here, CL1L, SL1L, from calcium carbide. Remember calcium carbide? We said it at the beginning. Now I have to work on it. I add water to it to give acetylene.

[01:17:57]Calcium carbide, I'll do a dripping process to give acetylene.

[01:18:04]Acetylene, I do a triple polymerization process, we also learned that, to give benzene. Benzene, I'll halogenate it, meaning I'll add chlorine to give gammaxane. So here it will be distillation or dripping, then a dripping polymerization, then a dripping polymerization.

[01:18:21]The answer is number one: dripping, then polymerization, then halogenation.

[01:18:29]The answer is number one.

[01:18:36][Sighs] What's the method? [Clears throat] To get an explosive substance, which is TNT, I want to get TNT from an aliphatic compound. What's the aliphatic compound? Didn't we say that seven gives seven, which is sulfur? So, guys, six will give six, which is benzene. So the first method is to get a compound containing 6C and reform it, and it will get benzene. I'll do a chlorination process and add CH3 to it, and it will get this. Then at the end, I'll add nitration, and it will be reformed.

[01:19:29][Sounds] [Snoring] Then Nitra, keep going like this, guys, and we'll continue together, but please, don't waste my time. Sit down and sort out the world. The one who has a share of something outside of membership, keep going like this and see the world. See you later.