Chepter 10 halogen derivative exercise solution || part1 #12thchemistry @a2zpractical991

So, hello guys and welcome to my channel. So, friends, my name is Sachin Shukla and you are watching A to Z Practical. So, friends, today I'm launching a series in this practical, which is a question-and-answer series. And in this question-and-answer series, friends, what I'm going to do is solve all the questions and answers from every chapter of 12th-grade chemistry. So, you don't have to go anywhere. You must subscribe to this channel because it features numerous practicals for 11th and 12th grade chemistry and physics, which you can easily access and review. So, I'm starting this new series where I will provide you with everything related to all the exercises or question answers. Right now, I am starting the exercise, and you note down this exercise, observe that all the solutions are very easy, and I have explained them in a very simple way. So, let me begin. So, friends, the first chapter I am beginning here is the 10th chapter, which is halogen derivatives. In it, the first exercise is a multiple-choice question in which you have to select one option. So, let me show you. So, our first question is essentially to arrange the following compounds in increasing order of reactivity of the C–X bond toward nucleophiles. You have to explain that your CX bond means this carbon and this X bond, this carbon and this X bond, this carbon and this X bond, and this carbon and this X bond. So, which carbon and X bond is present here, which exhibits a very fast nucleophilic reaction? Nucleophilicity refers to the ability of a nucleophile to donate an electron pair to form a new covalent bond. For example, consider CS₃ and X. So, what is the nucleophile? If you take NaOH, then Na is positive and OH is negative. So, what is the problem here? This OH will come here and you will get CS3OH. What will you get? CS3OH plus NaX will be produced. Basically, this is a simple reaction. This reaction will be shared by everyone. With everyone means which nucleophile group is present here? Oh, because what is minus? Our nucleophile. So, what will come here after replacing X? OH, here after replacing X, OH, and here also the same possibility. But what he asked for was in increasing order. Increasing order refers to the sequence in which the reaction will be very fast and in which the reaction will be very slow. Will OH come very quickly here, or will OH come very quickly here and replace X? So, let's see. So, what did we see? Already, if you have read this chapter, then we have seen that the NO₂ group is applied here. And what is the NO₂ group? Electron-withdrawing group What happens? Electron First, I am performing compression in the ring. Thus, the NO2 group is applied here, which withdraws electrons from its side. If this electron withdraws on its side, then all the electrons of benzene will move here. And this X group will be easily freed. What will happen? It will be easily free. That means it will come out very quickly. It will come out much faster compared to this. So, our first option in this case will be the second one. Its reactivity will be extremely fast. Because here the NO2 group is not applied, and it will come out much better compared to this. Which is your first one? Now, let's compare which one is excellent among these two. Because these two are chains. See this carbon and what is attached here? CS3 and CS3, and what is X here? The second thing is what is here? Here is carbon, one hydrogen, and two CS₃ groups. So, this is basically a tertiary carbon and this is a secondary carbon. Now, a tertiary carbon is more reactive. Which is a halide? This bond will break very quickly. Why? Electrons will come from here, and electrons will come from here. All the bulkier groups will donate electrons. And here the electron density will increase. Because of this, the bond will break and come out. Then it will be your turn. Which one will be your choice? The third one. And then it will be your turn. Which one will come? The last one. So, these two have the same differences. There are many bulkier groups here. Which will pressurize this carbon? Because of this, the bond will break and come out. And here, there are only two bulkier CS3 groups. Which will comparatively pressurize this carbon less. So, it will take longer to come out. But what will remain to emerge? It will come out very soon. What will any nucleophilic OH group do with it? It will react. So, what is the first thing that is being produced in your increasing order? Second. First. Then third. Then what will happen? Fourth. So, friends. Now your second multiple-choice question. You have been given an alkene group here. Which is doubly bonded. And here HI is written. And there is hydrogen peroxide. Okay. So, you must have already carried out this reaction. Along with HBr. Okay. You have already solved this reaction. So, here hydrogen peroxide is added. That means you understand. Which rule is this? Antimarconic rule Which rule is this? Anti-Marconic rule Okay. Now what happens under the antimarconic off rule? If you have HBr. Then there is HI. So, there will be a break in H+. And there will be a break in I+. In the red. Antimarconic has stated that the negative aspect. See this negative part. Where will it go? Doubly bonded carbon carrier. Will go to a greater number of hydrogens. Where will it go? Will go to a greater number of hydrogens. Now tell me. What is here? This is a doubly bonded carbon atom. This has only two carbons. Doubly bonded carbon carrier. Who has more hydrogen? Friends. So, it has a greater number of hydrogen atoms. So, where will iodine come from? It will come here. As soon as iodine arrives here. So, see the balance of this carbon. How much will happen? Two Already three, four, five. So, I. So, antimarconic. You will know. If iodine comes here. So, there is a sigma bond in it. There is a pi bond. 괌 So, where does it shift? Here. Why does it shift? Because this carbon must provide stability. Now see, iodine has arrived. Two, three, four. Now this carbon is stable. It is stable on four. Now the negative charge has come here. Because two. See how the negative charge has appeared. There are two electrons here. This is the electron of carbon. And whose electron is the next one? This is the electron of carbon. But both electrons are coming from the same carbon. So, what will happen to this carbon? Minus Now who will this minus attract? It will attract the plus one. So, here was one H. So, how many H's will there be? Two So, what is being made here? See. CH3. Two Hs will be created here. Single bond. CH2 And what is being attached here? I. So, where are you getting such an option? You are seeing that you are getting such an option B. So, what will be your answer? What will happen? B. Now we have been asked this. Which of the following is undergoing racemization during alkaline hydrolysis? So, what is racemization? Do you know? In which racemic mixture is formed. A racemic mixture means a mixture in which both D and L forms are present. Dextro and levo. So, I. which is active in optics. Active in optics refers to a chiral carbon. What is a chiral carbon? All four functional groups of which are different. What are the atoms attached to the chiral carbon? Here H has been made. Here, CH3 has been formed. Here OH has been made. Here CL has been created. So, you're seeing this carbon. What are these four groups? It is different. Now we can obtain the dextro and levo forms of this. How? When will we see it in the mirror image? So, here the chlorine you are seeing on the right-hand side. Where will it come from? Left side. You are seeing. If you look in the mirror, this is what you'll see. Where will it come from? Left side. So this is Dexsto, and what is this, friends Leivo Alright, so now I'm going to tell you about this here. Which option is it that will do the resumption and so on? Later, we can subject it to alkaline hydrolysis. There is a rumor that the two brides are involved. When making chapati, what does this contain? What can be done so that our re-simulation can be done, then you can What do you want to eat, what to cook, tell me, Hood Then you will get a dextro and levo. And Dexsto and Leivo represent the same thing. We got this from Khem Maizeshan As you see, you're seeing, aren't you? What are these four groups, why are they separated here? The four groups are different, so if you create a mirror image of this One will go with the left hand, one will go with the right hand. Alright, so we'll get the reassembly done with this. So our answer will be the first one. First one, what will happen to ours here, answer Fronts, you must be seeing the next question being printed right in front of you Alright, what is being asked here is, which is the base method? You must have studied only one reaction in haloalkanes, the SN2 reaction. What happens in a sort reaction, tell me the details, I think it's important So I wrote here, one per bromo, then one per bromo, and what will happen to my bromo fun? Okay? What did Third say to you, one you dibromo buten, I make, bromo fun. These four carbons, where will they stay in you, bro? Dye bro So that's it, you're a bromine. And this is bromine on you. What should I do with this... I'll just balance it. And the last one is to bromo butane So what will this become? It will become bromine on one to two. And what will this become, butane? You can also number multiple ones at once. You just need to fix the bus position. Are you bathing or is it not the Guru's command? You've been practicing art since childhood, haven't you? How does Aodeno do it? That's a wildcat, isn't it? But it's dangerous to approach. Here too, these are the cards, Khayral Carddan. Because Diko This guru is different. And what is this "wala" thing in Grub anyway? It's different Because Koranas Your Majesty Did you go to Yafena? One Jaffa on the net So these two chemistries mix together into a poisonous one. Just as this carbon will match with it. This carbon will match perfectly with it. This tax will be matched along with it. But what is the D one of yours? Is it a chiral carbon? Front, you must be seeing the question printed right in front of you. that an alkyl chloride undergoes a Wurtz reaction 2,2,5,5-tetramethylhexane We are told which one among these is carbon. It will form 2,2,5,5,6,6-hexamethylhexane. So, after coupling, which one is this from among these? I can build a structure. Let me show you how to solve it. Just look at it for a moment, check its structure, then 1, 2, 3, 4, 5, One two three four five and these six and what else has been said 2,2,5,5, So one two two you okay three four five, 5,5,4 times, so tetramethyl And what is this, Hexen? So, which structure among the options given in your MCQ could possibly be formed? So it will not make your first one a bicycle How did I find out? Because look at the first option—what's given there What did you give me? CS3-C That's what's given, so that's what will be made. How did I know that I didn't put the word "reaction" in it? How does it look? Your R is done, and your X is done. Plus 2NA, what do you need to apply? R X, that's correct. What is R? This is your R, and what is X? This is your X. So I'm staying here at 2NA. Plus, what am I doing to him? CL-CH2-CS3-CS3 Alright, and that's CS3 What will happen now? In your reaction, what did you all read? It will be the same here too. What on earth is this? It will go outside. And what will happen between these two groups? That which becomes them, they will make it so. You see, this is the real deal; it will match right here. And this is the VLA structure; it will match here. And what do the auras do to each other here? They merge together. All right, the first answer number you gave is correct. If you do it together with the others, you won't get it. Takents, what has been said here, number seven of the Kausyo's Yes, your gaze is mesmerizing, Param, what did you say? Who is laughing at whom, meaning which one among the two? Reactant HOKA, which will produce butene nitrate If reading the stock exchange before the night of the nitrate Hey! Botan has already passed 4 carbon chains, ketones, underest, soreasat daron So Gonn Shagr is hot. He went to the botan garden. When you were counting nitraile count, it didn't become carbon count. I mean, that goes on your chain. Lake 1, 2, 3, 4 So this became butane, and what about this acrylonitrile? But if you're going to count the cyanite, you'll write the cyan. So, the numbering you have here is what goes through here. As Vanpish is a syringe, and this is what will happen with propane If you've performed a nitro reduction, your nitro group goes into the carbon chain. What exactly is going to happen to your butane? Alright So, let me now tell you with whom I can stay together. Look here Alright So what's this? Butane, and this will be nitrate Butane, nitrite What will happen to your right? It will become right. But if you're thinking about doing this with it, it won't work, it won't happen. Look, this is beryllium chloride. One, two, three, four, and this Chloride If you react this with Kesian Look here So what will you get? Look here. This cyanide will go here. One, two, three, four So, tell me, what is confusion? If you are beautiful, you will go along with beauty. But here, a nitrile should have been attached. Look, I'm counting. I initially told you If you're taking nitrile, What are you becoming, then? Primary Primary halide Valen. This half, which is counted from carbon, is never captured, as this is indeed a separate case. And the hermitage is used by him for meditation. Harankinl It is used in Harankal. Is it used? 3.1.0 Tasri - Upper Tréken It seems like Zune It is shining. And Dhrayt Yes, up to that point Prach What's the matter? Yes, that inch Come here too It seems like Hello everyone You take the broken car, make another one. And Ido proposal Hey, Dathan, and Ya'ar Ivar Din I create the third one, and the ID is Mifaij. || Butane Hey, two, three, four, beautiful ones, speak up And what was said at 1? Sorry... What did he say at 2? What did Mithai say at 2? Okay? 1 IDO Sorry You see the one with the 'B', what's written on it? IDo Propan So here is the residence of the Iduin. What I have made C. What did D say 2 I do 3 methylbutane What does butane mean? Four carbons should be two. What should be done? Two. And three should be methyl. Friends, I told you about what happens in the tertiary group in Sn1. This is used in a tertiary carbon. Let's see where iodine is bonded to the tertiary carbon. The carbon to which the iodine is attached It's attached to the same carbon, so what is this? This is called Primary The carbon to which the iodine is bonded It has a single-carbon bond and is also primary. This iodine is attached to a carbon, and that carbon is connected to three other carbons. That's it, when will Tarsidar become a hansom? That's the answer number, dead. Even D can make a video. Here, the carbon to which this iodine is attached has only two carbon connections. So, what will happen to the goose? C So, frontiers, this is the nine-related question, you're looking at it here which is connected to MG So this is your question, and you need to provide the answer to it, right? So you made a Grignard reagent; there, magnesium is present. Whenever you see magnesium, you'll understand that it's a Grignard reagent. The charge of magnesium is +2, so here, the magnesium that exists means these two. In between, what happens if we get an app like Roadster, wouldn't something like that appear here? Will it become C here? Did you say this magnesium oxide is not suitable to be used alone? Yes ỉ Let's play after the month of August. So both electrons stay right here, making it minus minus, which attracts the positive one. Okay, so here's what you get when I first give you the left-hand side of the equation. What gets attached to you when you wear this ring? Here, you can also attach the letter H, or you can choose not to. But if H is there, then your answer will be answer number D. Your last AMSI query was that which of the following is used as a source of a Dichlorocarbene is formed from phosgene. Your answer number is B. In this, what is formed by bisically dichloro? How is carbene formed? Also Let me tell you, this is what happens with your chloroform, CHCl₃. 3 Cl3 How to make this carbene Tychloro-carbene case What happens here? This will be in the plus form here. Matlab are being removed Yeah, plus, what happens here in this form? These electronic devices just keep going on like this. If the carbon above the agar has a negative charge, then We will surely do it for you. If carbon has a negative charge, what do people call it? Carbene Negative charge with an electron contained therein It should have electrons. such a beautiful day, the sky is so clear, the air is fresh, the birds are singing, the flowers are blooming, everything is so peaceful, it's like a dream, I feel so happy, I just want to stay here forever, it's perfect Look here, watch closely, you watch closely, one, two, three, four, you must not stop here, you must take the logesh chain, so I'm counting, one, two, three, four, and this five, go all the way down, so what will this become for me? Logesh No, let me tell you that compared to halogens, If you go by priorities order, you'll find the priorities of the double bond. First, you need to meet to see, so what do you have to do, You must assign priority to the double bond first—how do you do it first? Here, a double bond is coming at position 2, on carbon 1,2. And from here, double bonds are coming out on carbons 1, 2, and 3. So where will your beloved Priyautis go? This is your correct order, 1,2 Tell me, what is on the 2nd floor? Your E and E double bond So what's up, Jagga, PENT, 2, E and E Okay, there's only a double bond at position 2, right? And look, there is an attachment at position 3 and an attachment at position 4. So, the attachment at position 3 is methyl, and at position 4 is bromo. So, shall we go with alphabetical order? Because whatever attachments you have, they're all just five kinds of wrath. Write from Vipothal with love So, the attachment is always in my possession. So huh, the one who knows the secret and Dai Aaya had to go to the hospital because he was feeling unwell. And what will this become? Faded. What do you need to do now, friends? Here you can see chlorine, and here you can see your C. Naderi. If you go to Priorities Order, what will you see there? Then, the chlorine will attack! The methyl will attack! So I know the Ensarath is correct, and the light that flies in the sky is called Chloro. So, what will happen next? Chloro. And what if it were methyl? After that, the system that is your main system is the main ring's system. It will be written later. So, tell me, what is on Attachment 1? Chloro is there, and at position 4, what is it? Ethyl is there. So, 1 Chloro, 4 Ethyl, and what else? What will happen to benzene? Benzene Above this, if I were to specifically mention its name, So, what do we call CS3 and the benzene ring, Tall-In? If you take this as the main chain, it will become your main chain, and the main group. And from where will the numbering start, from here Then do 1, 2, you can't do it here, 1, 2, 3 Because the attachment available here, where is it available, at 3? Here you have two attachments. That's why numbering will start here, why give numbering to this? Because CS3 connects to benzene and this is a tall-in configuration. Because you can also write its other name Look here, it's CS3 and chlorine, so you know If its priority is higher, you will assign it a higher number. So what will happen? 1, 2, 3, 4, so let me write down its name, okay? Now that you have considered everything as attachment, the attachment will be written. At the beginning, and this ring system here, where would you write "last"? So, look. If there is chlorine at position 1 and chlorine at position 4, what will 1,4 become? Dichloro, chloro, and 2-methyl, methyl, 2-methyl, okay, and ring System, what should the system name be, your benzene, so 2,4-Dichloro, 2 Methyl, Benzene, OK, here is your second question, you need to answer this I need to write the product, and I need to write the name, so look, it's written as NAI. And it says "Acetone," and whenever you see this, you should immediately know which one it is. Reaction is present, Finclastine reaction, what is it, Finclastine reaction, what happens in it, Halogen exchange occurs, meaning iodine here will move out, and NaCl will go outside. Do you know which reaction fluorine undergoes in the entire group of halogen derivatives, and with which substance it reacts? With Sorts Reaction, given with whom, Sorts, it contains, What is it, look, even Sembu will be there, halogen exchange will take place, Fluorine will go here, and this SBA will come out. Here there is fluorine at the 3rd floor. And the unpleasant free Fluorine name will just be revealed. So, what exactly do you have to do with this? The total collection of Sandhala is two dena. Those who are called 3 Then 3 felrines will go there. Mental free Flourine for 3 Chapter 3 Chapter 2 I'm not going anymore; I'm fluorinating this. So the three fluorines are coming forward here Because on the third day of pregnancy, what happened to me? Three fluorines And how many bromines come out from here? Three bromines. So what will they make, SB, SB, BR, three, three Now let's look ahead—acid has been provided, along with peroxide. So you'll understand that Marconic Sifree will go in time. What happened today at Marconic Siree? Life has been tough, everything's been stressful, and we're just going to take a bond—just make it better. So yes, I want it, it's good, is this also acceptable with it? The weak bond will be here, shifted to this site. Both electrons will be here, so they will be here. Who will attract the minus minus? The one with the plus So I'll write Prodrk here CS3 CH CS3 And here was CS Vaccine You will see it right here Now here S O C L 2 You know that S O C L 2 very much It is the best reagent. It often appears in MCQs. The best reagent is used. To tighten the hex grip Hey, look here Chlorine will come here. And what will happen to this O-H group here? He will be liberated. Look, what has come upon them? Chlorine, and this O-H came from here. So, Dikom, what do these SO2 and O molecules form? So, two sulfur dioxide plus two H, and here one chlorine remains, right? So, what will these two together make? A mixture of both gases from both. It is produced in the state, so we get this pure product. So you'll see how EPHI will break down the EPHI DARP condition chlorine, how it will break down, one There will be a break in chlorine plus, the second chlorine minus, because it's like that. They will both move together heterolytically bonded. It's a fall one. So both will go together, both will become one, minus one will become one, plus So, you need to remember that above the benzene ring, we have an acro-foil group. Your nuclear electric electric So, what will this electrophile group do? Your ash will rise above this ring. Let's see what will be created here. There's a toll here, but what should be done here? CL, and what will the H that comes out from here form together with the other CL? SCL You will see the Koshian print in front of you; Koshian Number Three is here. So, let's see what a chiral molecule is? If a carbon atom has four different groups attached to it, then it is a chiral molecule. So, friends, do you know which one is your answer in this? A is the one, because A is the one? Here it says that you are given four carbons, and in between, OH is given. OK? Here we have CS3, CS2, CS2, and H. So, look, this is not a chiral carbon; how did I know it's a chiral carbon? This is our chiral carbon, our chiral carbon. I'll take another example from it. What have you been given? Five carbons are given. One, two, three, four, five OK? And one woke, one Birumen OK? If you're the same, then carbon isn't yours anymore. So, basically, front side, I'm letting you know that I'm going to upload this in two parts. The first part is this; right now, in a moment, you are seeing this first part. And the second part is in the second video, so please watch this part. If you need the complete solution, please also watch the second part. Thank you.