IIT-JEE Ultimate Challenge: 5 Questions, 30+ Important Concepts for exam & bonus questions #jee

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[00:00:02]Hello students, welcome back to organic mwords. Many of you have requested me to solve questions which will be helpful for your upcoming J advanc exam.

[00:00:11]Therefore, I have chosen these five questions which will cover at least 20 to 30 subtopics from organic chemistry.

[00:00:19]You can also try solving these questions on your own. Pause the video, solve it and then look for the solution. If you are stuck somewhere and if you still have question you can always ask your queries in the comment section. So let's get started. In the first question we have a halo arene attached to an acetal and it is treated with magnesium ether followed by ethanol and hydrayis. So what will be the final product? So whenever we have acetal upon hydrayis we know that it's going to form an alihide.

[00:00:52]So just look for the options where we have an alihide A and D. So the answer has to be either A or D because in C it is remained as it is not possible and here it is an acid. So there is no oxidizing agent. So you can rule out C and B. So how to decide between A and B?

[00:01:11]For that it's better to write down the sequence of reactions. So whenever we have hoarene or halo alkanes so when it is treated with magnesium the magnesium will get inserted between the carbon and halogen because that is the polar bond and it is forming mgbr and what is this rignar reagent.

[00:01:33]You can see that here the carbon is electrophilic. The moment magnesium comes here it will become nucleophilic.

[00:01:39]So what do you call this as oompolum nature? Oompolong refers to a German word wherein you are changing the polarity from positive to negative because carbon is negative here. Now it can act as either nucleophile or a base.

[00:01:55]How to decide that? Depending on what reaction condition we have here it is ethanol. Now ethanol is it an electrofile? No, it is a weak acid.

[00:02:08]because it is a weak acid it will react with the grignard reagent and here it is acting as a base. So the reaction is between acid and a base. So you can see the oxygen will gain negative charge and this will become positive charge. The negative will abstract the proton and hence you are going to get a proton over here. What about this group?

[00:02:34]No change here. this acetyl will remain as it is. What is the last step?

[00:02:40]Hydrayis. Whenever we have an acetyl, right? Upon hydrayis, what we are going to get here? As I told earlier, it's going to form an alihide. What about this one? It's going to form a diol.

[00:02:54]That is a byproduct. Don't worry about that. That's a byproduct. So, this is our main product. So, what is the final product? We have benzel deihide.

[00:03:03]Therefore the final answer is option D.

[00:03:07]Students just summarize what we have learned here. So we have a bromoenzalihide which is converted to benzaldihide.

[00:03:17]So for that what is done? First protection of the alihide. Why protection of the alihide is important here? Because when you do prepare the grignard reagent here there is a possibility that that grignard reagent will react with the other molecule having alihide. In order to avoid this we do protection [clears throat] the protection of the group and then carry out your favored reaction. What was our favored reaction? Removal of this bromine and we are substituting there by the hydrogen. So there is no direct step here. The hydide attacking here and making this broine leave the molecule will not happen in one step. So for that what you have to do make it a grigard reagent treated with either ethanol or simply by the water. So then what happens you will get the hydrogen here and then deprotected like this. So what all topics we have learned here protection deproction hydraysis of the acetal grigard reagent acting as a base or a nucleophile. So these many things we have learned just from one question. Now let's see the second question. For the second question, an aromatic compound A with the molecular formula C7 H6 CL2 gives AgCl on boiling with alcoholic AgNO3 solution which will yield C7 H7 O treatment with NaOH. That means what?

[00:04:49]AgNO3 followed by NaOH will give this organic compound and the byproduct is AgCl.

[00:04:56]So when this AgCl white precipitate will be formed. So remember whenever there is a halogen attached to the alifhatic chain. Okay. So now when this A upon oxidation gives monocchloroenzoic acid. So this oxidation means what?

[00:05:17]Maybe the reagent used is like Kmen4 in alkaline medium. It's forming monocchloroenzoic acid.

[00:05:26]But what is the position of chlorine? We don't know. And that is the question here. So in that case what you can do?

[00:05:32]You can simply write like this. Chlorine can be either on or metapar. This upon nitration will give only monitro product. This tells us that chlorine is preferably at the parap position. How so? Remember if it's forming monitro product that means what? The reactant must have more symmetry among ortho meta parah para is considered as a molecule with more symmetry. So with this much information you can tell that option A could be the right answer. But we can verify that by writing the reactions. If you consider option A as a reactant. So this is paracchloro benzy chloride. So when it is treated with AgNO3 followed by NaOH. This bond will get cleaved and it's going to form benzile alcohol like this. And its formula is what? C7 H7 O. And what happens to this chlorine? It will precipitate as a white solid like this.

[00:06:43]Now when it undergo oxidation this one right so say suppose a strong oxidizing agent like kminophor in basic medium first it will convert it to benzil alcohol which upon oxidation will give the benzoic acid like this and it is paracchlorobenzoic acid. This upon nitration what we are going to get? See co is a meta directing. What about this? This is orthopara. Par is already blocked. So where it will go to ortho. So with respect to this meta is this with respect to this or therefore you are going to get only one product like this.

[00:07:28]Therefore this statement will hold good for option A. Hence this is the final answer. Say suppose if you would have taken this when it undergo oxidation here it will form caroxyic acid with respect to this the metaposition is this but with respect to this oro and para so what happens here you will get multiple products not possible similarly here you will get SIBO H with respect to this meta is here and here for this orthoine paras you will get two products at least so not possible for option D this is not obeying the first and second statement itself. So not possible. So you can even eliminate this option in the first step itself. So this is how you have to solve this question using the reactions of ho alkanes, halo aren electrophilic substitution of benzene ring and oxidation. So so many concepts in just one question. For the third question, there are series of reactions given. In this scenario, it's better to identify the structures of A B C and then choose the correct option rather than eliminating the options. What is the first reaction? It is a reaction between a ketone and hydrogen cyanide. So cyanide acts as a nucleophile and the ketone will undergo nucleophilic addition reaction. Where did you learn this? Alihides, ketones and caroxyic acid. So what will be the structure of the product? It is a cyanohydrine.

[00:09:00]Therefore, you can write directly like this.

[00:09:04]So this is product A. When this A is treated with lithium aluminium hydide, a strong reducing agent which will furnish four moles of the hydide ions. First it will react with this alcohol and then it will reduce this nitral to primary amine where you have learned this preparation of amines from the nitrals and upon hydrayis what we are going to get here the final B product here is you can see this O will remain as it is upon hydrayis what about here a primary amine complete reduction of nitral to amine time this methile will remain as it is.

[00:09:48]So what is this product? B.

[00:09:51]Now the last step here is it's a diazotization.

[00:09:56]Diasotization because NO2 with H2SO4 will give us HNO2. So it will convert this NH2 to dasonium salt. But since it is alifhatic amine these alifhatic dasonium salts are not stable and what it will form a carboation over here. If it is aromatic amine it would have been stabilized by benzene but here it is not possible. Hence we are getting a reaction intermediate carboation.

[00:10:29]Now what happens because we have an alcohol here and there is a carboation this will undergo drink expansion. How do you write that ring expansion? You see when the bonding electrons are given to this carbon this will become pentavalent. It cannot withstand it has to break the bond where it will break the bond here.

[00:10:51]Since it is breaking the bond here this carbon will go and attack this CH2.

[00:10:57]Hence there is a ring expansion from five member to six member. So how do you write the product? Very simple. You write down this as it is. You write down a six member ring and look at this carbon attached to benzene benzylic carbon that is the one which is forming a ketone here. And what happens to this carbon? It is attacking here. That means CH2 CH2. So this will be our final product. And where do we have it in the options? Check. Now look at this option B. This is five membered. You can eliminate that. You can see B and C are very close. Both of them are in fact isomers. The position of the ketone is different. Here it is benzylic. Here it is non-benzylic. But we got this as a final product. So what all topics we have covered here. The reactions of ketones, conversion of nitrals to amines. So this is similar to pinol pinolone rearrangement. You have to remember that it's a pinnacle pinolone kind of rearrangement.

[00:12:03]you have an alcohol which is converted to a ketone. Hence the final answer is option B. So in the fourth question, you have to arrange the given molecules in the decreasing order of their nucleophilic substitution reaction.

[00:12:20]What is nucleophilic substitution reaction? Here since all the molecules are aromatic in nature, we are discussing SNR mechanism. Say suppose on benzene ring if there is a h hallogen like chlorine when the nucleophile attacks like hydroxide methoxide ethoxide when it attacks this carbon the chlorine will not act as a good living group. Why so? Because the lone pair of electrons present on the chlorine will exhibit plusa effect which makes this carbon chlorine bond as a partial double bond. Therefore when the nucleophile attacks this bond is difficult to break.

[00:12:59]That means the order of reactivity for this molecule is less. To increase the order of reaction for such hoarines you can introduce electron withdrawing groups like alihide acid nitro groups.

[00:13:15]Look at all the options. We have nitro in three of the options. Molecule two and three have benzene as an aromatic molecule. Whereas in option one and four we have pyodine. Now does it make any difference with respect to nucleophilic substitution? Yes, it does. How so?

[00:13:37]Pyodine will undergo nucleophilic substitution better than benzene. Is it so? Because of the hetroatom nitrogen.

[00:13:46]Nitrogen is an electron withdrawing atom. because of its electro negativity.

[00:13:53]So because it withdraws electron through minus I effect it decreases the electron density. If it is decreasing electron density then it is activating the molecule towards the nucleophilic substitution reaction. Remember if there is any electron withdrawing group so there is electron withdrawing group within the molecule. So it decreases electron density. Therefore it increases the SNR order of reaction. So from this you can say that molecule 1 and molecule 4 are already activated towards the SNR reaction. Compare one and four. Now there is a nitro group. That means what? it is still more increasing the order of reaction towards SNR. Therefore, four should be the highest. Where do we have four? Option C and D. And you can see in both the cases three is the least. Why three is the least? Because whenever we have electron withdrawing group, it has to be either on ortho and paraposition only then it will make this chlorine to leave as a good living group. Since it is at meta so you can see that the third molecule is at the end. Now the competition is between one and two. So how do you decide between one and two? As I told you the perodin is always more reactive than benzene irrespective of the position of the chlorine over here.

[00:15:28]Therefore one will be more reactive. So where do we have one as more reactive than two? It is in option C. Therefore, the final answer is C. For the last question, you have to identify the correct product with their reactions.

[00:15:47]The first reaction is a simple nucleophilic substitution reaction. See, although we have three chlorine here, there is a neighboring group effect here. The oxygen will help this chlorine to leave the molecule and then what happens? The eth oxide will attack the same carbon. Therefore the nucleophilic substitution happens only on this carbon. Don't worry about the stereochemistry of the product here. The question is will it be a monos substituted product or die or tri substituted. So it's only monos substituted because this oxygen is helping this chlorine to leave the molecule. Hence this is a correct statement.

[00:16:28]Let's check the second reaction. Here again it is a nucleophilic substitution.

[00:16:33]Where will it happen? On the benzylic carbon or this carbon? So whenever there is a benzylic hello compound it can leave the molecule easily and form a stable benzilic carboation which will be stabilized by the benzene ring resonance. Hence the nucleophilic substitution happens on the benzilic carbon not on this carbon. Hence this is a wrong option. Now let's see for option C. So we have a benzene attached to an alifhatic cyclic chain. The reaction is not happening on the benzene ring. It is happening on the side chain wherein the bromine is substituted by ethoxide and there is a rearrangement. So when the rearrangement happens the methile is supposed to come over here. Why so?

[00:17:19]Because when bromin leaves it's going to form a carocation over here and there is a methile shift one two methile shift and the carboatine formed here will be tertiary and ethoxide should attack over here. So this part is correct but the methile is missing from the product.

[00:17:38]Hence this is also a wrong option. Now let's check for the last one. So whenever there is a bicyclic compound just keep in mind about the Brett's rule. So the Bret's rule says that the double bond is not possible on the bridge head carbon. So whenever you have bridge head compound like this with eight carbon or less than eight carbon the double bond cannot form over here right or the double bond cannot form over here or here. So anywhere there is a bridge head carbon. So it's not possible. Why? So because whenever there is a double bond that should be planer in bicyclic compounds like this the double bond cannot be pler when it is on the bridge head carbon. But is there double bond possible on these carbons like this or here or here? Yes, this is very much possible. Hence since it is alcoholic Koh, it is undergoing elimination but not substitution and the double bond is formed over here. So this is acceptable based on bre's rule. Hence this is a correct statement. Among the given options A and D are correct. So what all topics we have here? There is a nucleophilic substitution and there is a neighboring group participation in the nucleophilic substitution and then we have benzylic carocatine its stability over here. And here we have a nucleophilic substitution with rearrangement like methile shift. And then last one is bicyclic component breath rule elimination. So this is how you have to solve comprehensive question so that you will be ready for your upcoming exam. If you still have any questions on this, you can comment your queries in the comment section. In the next video, I'll come up with more and more questions and don't forget to subscribe the channel. Thank you.

[00:19:37]D.

[00:19:49]D.