COMEDK Chemistry PYQs with Solutions | Haloalkanes & Haloarenes Last 10 Years PYQs

Added: 2026-05-08

[00:00:00]Hello champions, welcome to the channel.

[00:00:02]In today's video, we are going to solve the last 20 years of comet K pyqs of a very very important organic chapter that is hello alkansen hello arens. So let's get started with the first question. The first question is the number of possible enansumeric pairs that can be produced during the monocchlorination of two methileb butin. Okay. Now remember the number of stereois isomers we calculate by the formula 2 ^ of n where n is the number of stereo centers okay or the number of chyal carbons we have in the particular compound. Now if I see two methile butane it is like this okay now if you see in this particular we have CH3 here we have CH3 we have CH then we have CH2 and CH3. Okay. Now you see that the only hydrogen if I replace this one with chlorine then this particular carbon will be optically active. Now when this is optically active it will be a stereo center. So we will have 2 ^ of 1 that is equal to two.

[00:01:03]So correct answer here will be option number a here. Next the grignard reagent a very useful starting compound for a number of organic reactions can be prepared by. So gignard reagent can be prepared when you have RX and you treat it with Mg and ether and from this we get R Mgx right. So here it will be reaction of alky helides with dry magnesium powder in the presence of dry ether option number B here. So now students before moving ahead I want to inform you something very very important. So KCT 2026 uh answer key is out. So I'm very sure that now all of you would have calculated your marks also. So now if you want to know what will be your expected rank in KCT 2026 examination then me and my team has prepared a tool for all of you. You can just search college cututoff.in okay and you'll be directed to this tool or you can even uh the link will be provided in the description section. You can click on the link given in the description and reach here. Then after that you have to just simply sign in with your Google or Gmail account and then you have to enter your PCM marks here and you have to enter your KCT marks here. After that just click on predict rank. Okay. Now here if you want you can select your category as well and here you will get an estimated rank as well as a rank range. Now I want to inform you that please do not compare it with the 2025 cutff because we all know that in 2025 the paper was very very easy and the cutff was really really high. But this year since the paper was somewhere between moderate to tough the cutff will definitely go down. So please don't compare with the 2025 cutff. Okay. Now if you scroll down here you can also select the branches in which you are interested and it will show you all the possible colleges that you can get based on your rank. Now these all cutffs and everything is calculated with proper research and all. So definitely this will give you a very very uh I can say close to the accurate results. So do give it a try. The link is there in the description section. Okay that's all and now let's move to our third question.

[00:02:59]Which one of the following compounds most readily under goes SN1 reaction? So for SN1 reaction you have to understand the order is 3° 2° and 1°. Okay. And next thing is what there is a formation of carboatine. So more stable the carboation is okay. The carboatine if it is more stable higher will be or faster will be the reaction. Now here what happens when bromine goes we'll get a carocetine here. Now O CH3 is an electron donating group. So it will donate the electrons and this carbon positive charge will be neutralized and that is why this carocatin is highly stable. Now here also we will get a benzile carocatin but NO2 group is electron withdrawing here. So it will pull the electrol this charge from the carbon and that will definitely make this particular compound highly unstable. Now here what happens here we will get a carocatine and it is resonance stabilized and here also we will get a carboatine and chlorine is having plus I effect. Okay. And here it is plus R effect. So we definitely know plus R effect is more dominant than plus I effect. Here there is no effect. So that is why A will be the one with the highest reactivity. Now if I just want to order to understand the whole concept. A will react fastest. Okay just a minute. Yeah A will react fastest.

[00:04:13]After that D will react. Why? Because chlorine is there. Oh sorry one second.

[00:04:16]One second. Sorry. Chlorine we will we know that chlorine is having minus I effect. Okay. So it will also destabilize. Okay. So this one will minus I. This is minus R. So both of them will destabilize the carocatine. So after a it will be C. Okay. Now comparing minus I and minus R. Minus R is more effective. So that means here the carocatine is more destabilized. So this will react slowest. So after that we will have D and then we will have C.

[00:04:42]Uh sorry. So A I got here then I have this one will be B the neutral one.

[00:04:49]Okay. B then you have D and then you have C. Okay. Next. Fourth question. In the following sequence the product C is.

[00:04:55]So first of all we have alcoholic KOH.

[00:04:58]So that will remove this one that is double bond and you will get a compound like this CH CH2. Then this is product A. Now when you treat it with HBr you will get CH3 CHBR following maronicos rule and CH3. Now this is product B. Now if you treat it with aquis Q this Br will be treated with O. So you will get CH3 CHO CH3 okay so that is nothing but propane 2 O option number D here next when a hydrocarbon A okay is treated with excess of HCl a dihalogen derivative B is formed dihalogen derivative means two types of hogen is added now the compound B is treated with aquis Q okay to give compound C with two carbon atoms with two carbon atoms okay now compound C gives a silver mirror test.

[00:05:51]So, compound C is giving me silver mirror test. Okay. Now, first of all, when it has given that the compound C is having two carbons and we haven't done any decarboxilation reaction or any that type of reaction where the number of carbons will decrease, that means the reactant and the product will have same number of carbon. So, I can eliminate these options where there are more than two carbons. Now, if I see you have dihalogen derivative, right? Dihalogen derivative means this HCl is added two times and HCl can be added two times if you have a triple bond. So B has to be option number C. So if I show the reaction now A is ethine. Okay. Now when you add excess of HCl what will happen?

[00:06:32]You will get CH3 CL and then CL. And then when you treat it with aquis Q you will get CH3 CH O and O. Right? Now what happens on one carbon two O groups are not stabilized.

[00:06:50]So the removal of water takes place here and you will get CH3 C double bond O H.

[00:06:57]Okay this is ethanol and you know since it is an alihide it will give you positive silver mirror test. So that is why option number C here. Next which one of the following is not the correct order of boiling point of a or alkalhalides. So for boiling point we have two things that boiling point is directly proportional to mass. Okay. and it is inversely proportional to the uh branching right. So now if I see here which is not correctly matched here you can see that the mass is more. So definitely this one will be more boiling point. So that is correctly matched. Here if you see this is a primary and this in here we have two three carbons. Here we have four carbons. So definitely the first one will be more boiling point. That is also correct. Now here what I can see that this is a tertiary and this is secondary. If you see option number C, we have a structure like this.

[00:07:50]CH3, CH3, CH3 and Cl. And here we have CH3, CH3 and we have CH, then we have CH2 and then Cl, right? So definitely when we have branching, okay, more is the branching, less is the boiling point. So in this case, this one will have more boiling point because it has only one branching. But in the option it is given that this is having more boiling point which is incorrect. So option number C here. Next which one of the following is most reactive towards nucleophilic substitution reaction. So if your reactant is an electrofile it will be more reactive towards the nucleophilic reagent. Okay. So now you can see that here this carbon will not leave because this is positively charged which is unstable. And here also you will this bond will not break because of partial double bond character. Here again this will not break. But if you see this chlorine leaves here and you get a positive charge here and this carbon this carocatine is then what? Resonance stabilized.

[00:08:48]Okay. So if the carocatine is resonance stabilized then it will react faster with SN1 mechanism. So option number D here. Next elimination of bromine from two broin results in the formation of.

[00:08:59]So we have two bromoin and according to setif rule we will get the more substituted alken. So that will form an alken here that is but 2 ene next we have in the given zn and ether.

[00:09:13]So this is like your uh mg ether or sodium ether reaction. So what happens here in case of this one you get a woods reaction. So this one will eliminate and you will get two radicals here and this will join here. So you will get a compound that is a here. Next, an alky chloride produces a single alken on reaction with sodium ethoxide and ethanol. The alken further under goes hydrogenation to yield two methileb butin. Uh identify the alky chloride from amongst the following. So first of all what you have one alky chloride. It produces a single alken reaction with sodium hydroxide and ethanol. You get a single alken here. Okay, that means there is no saf product here. The alken further under goes hydrogenation to give you two methile butane. So if I have a structure like this yeah 1 2 3 If I have a structure like this and if I do hydroenation then what I will get here CH3 CH3 CH CH2 and CH3 right so this is two methilebutin I'll get in case when we add hydrogen to this compound so that means this particular compound can be obtained if I have chlorine or chlorine group here then with sage or with the elimination reaction. The only product that I'll get is the alken which is here that is two methile but one ENA. So I need chlorine on the first carbon and I think that is present in option number C here. If you see I think Cl is there I have CH2 then I have CH then I have CH3 then we have CH2 and CH3 right? So in this case what? HCl will be removed and you'll get a double bond here. So option number C is the correct answer. Next we have isopropyl chloride under goes hydrarolysis by so that is because it is a secondary halogen it can undergo by both SN1 and SN2 mechanism option number C here. Next we have PH CHB Br2 Br with alcoholic KOH gives you A. Then with Na and H2 and this one will give you B. So we have a compound like this pH we have CHBR CH2 Br okay now when you treat it with alcoholic KOH and Na NH2 so first one HBr will be removed you'll get a double bond here then again when you use NaNH H2 it is also a strong base one more HBR will be removed and you will get a compound like this see both the HBr will be removed and you will get a compound like this. Okay. So, A and B are okay.

[00:11:50]Uh A will be alken like only one CH HBr has been removed. Okay. That will be A.

[00:11:57]So, A will be 1 second.

[00:12:01]Okay. So here what happens? First when I add Na and H2 I get this product. Okay.

[00:12:05]After that there is one more that is CH3. CH2 CL is also added. Now see here this H is highly acidic. So this H goes out and here we get a negative charge and you have now CH3 CH2 CL. Okay. So this is delta minus this is delta plus.

[00:12:19]So this negative charge will attack on this carbon and your final product here will be uh C triple bond C H2 CH3.

[00:12:31]Okay. So we have this one is option number B. Right. First I'm getting B U product A is this one. Okay. And after that when you use again this one will give you this final product is this one.

[00:12:43]Okay. Now when you use alcoholic KO one HBr will be removed. Again when you use Na and H2 another HBr will be removed.

[00:12:50]So we will get a triple bond in product A. Now when I use this Na NH2 it abstracts this proton. Okay. So you will get a negative charge on this carbon.

[00:12:59]This NH2 that we are using in the second case will remove this acidic hydrogen.

[00:13:04]Now we'll get a negative charge here.

[00:13:05]Now this one can attack on this particular RX and you will get a compound like this. Okay, I hope this is clear. Next, which of the following is an example of SN2 reaction? So primary alkans react here. So CH3 Br is primary.

[00:13:17]So option number A here. Next we have this reaction. The major product in the following reaction. We have sodium dither. So there will be woods reaction.

[00:13:24]So 1 2 3 4 and five. So five carbons will combine here and you will get a product like this.

[00:13:33]Okay. Now when you are treating it with Br2 H new what will happen in this case you will get a more stable bromine will form here right? So this hydrogen is highly stable. So here the bromine will be attached. Now with alcoholic ethanol or alcoholic Q what will happen? It will remove this Br. So there will be removal of HBr. Alcoholic Q will be removed and you will get a double bond here. So that will be product number B here. Next.

[00:14:01]Which of the following is not true for SN1? Favored by polar solvent, that is correct. Tertiary reacts faster. That is correct. Rate of the reaction does not depend on the molar concentration of the nucleophile. That is also correct because in the starting of the reaction, we just have the reagent. Next, primary alkalhalides generally react by SN1. No, tertiary alkalhalides react. So that is why option number D here. Next, arrange the given compounds in the decreasing order of their boiling points. Okay. So this is completely primary decreasing order. Right? So one will be the highest because it is primary. This is secondary and this is tertiary. So 132 which is option number A here. Next gamin is benzene hexa chloride. A compound that under goes bromination easily.

[00:14:43]Bromination is an electrofilic substitution reaction. So your reactant must be electronrich. And if you see in phenol we have O group which donates electron. So that is why it is electronrich. Option number C here.

[00:14:55]Which of the following compounds is responsible for the depletion of the ozone layer? that is fre. Here the reaction is given with aquis Q we get is classified as so this is a substitution reaction because you can see I is replaced with O group and this one will be a nucleophilic substitution uh reaction right O minus here is a nucleophile it will attack on this I so this is nucleophilic substitution reaction for SN1 reaction the order of reactivity of halo alkans is tertiary secondary and then primary which is option number C.

[00:15:31]The major product obtained when chlorobenzine is heated with concentrated HNO3 and H2SO4. So this is nitration and we will get para as the major product. So you will get one chloro and four nitroenzin. Okay. So option number A here.

[00:15:47]Next we have given two structures one and two R. So if you see that the last part okay this carbon configuration is exactly same and here they have slightly rotated it okay so these when you rotate across a single bond these are called as confirmers that is option number B here an organic compound is optically active if it is super if it is non-s superimposable on its mirror image this is the most important uh condition okay not even the chyal center sometimes it is kyal center but if its mirror image is super imposable then it is nonoptical active. So students I think yeah this is the last question. So that's all in this video. Very less number of questions are there. In the next video we are going to do the next chapter. So please stay tuned and subscribe the channel and the link of the KCT rank predictor that is your college cututoff.in is given in the description section. You can definitely check and see your expected rank in case 2026. Thank you so much for watching and if you have any doubts do let me know in the comment section. I'll definitely address it and all the