COMEDK Chemistry PYQs (Last 10 Years) | Hydrocarbon PYQs for COMEDK 2026

Added: 2026-05-10

[00:00:00]Hello champions, welcome to the channel and in today's video we are going to discuss the last 10 years of comet K PQs of a very very important chapter that is hydrocarbons from PU1. So before moving ahead first understand the weightage of this chapter. So if you see from last 3 years you can expect in 20 like in 2023 two questions were asked and in 2024 also two questions were asked and in 2025 three questions were asked. So in this year also you can easily expect two to three questions from this chapter. So now let's see the first chapter here. In the following reaction the most stable carboatine intermediate is. Okay. So most stable intermediate is. So basically here carboatine intermediate will form. So what happens this double bond breaks like this. You get a negative charge and you get a positive charge here. Okay. That is the first way because this is a primary carocatine and this is a secondary carocatine. Now what happens? We know that tertiary is more stable than secondary. So what happens?

[00:00:52]There's a methile shift. Sorry hydride shift here. So this hydrogen which is present on this carbon shifts to this one. So what we will get now? We will have CH3 CH3.

[00:01:03]Okay. And then we have this plus and then we have uh CH2 and then we have CH2 minus. Okay. Now this H+ is there that will be taken by this uh so it will get CH3 and you will get a carocetine like this. So the most stable one will be option number C here. Next one mole of 12 dybrropane on treatment with X moles of Na NH2 followed by treatment with ethile bromide give pentine the value of X is now if you see one two dybrropane you have like this okay now when you use the first mole of Na and H2 you will get a compound like this means you will have HBr will be eliminated here okay so HBr will be eliminated here so you will have CB Br and still one H will be remaining meaning double bond will be there. You will have CH okay and then you have another CH3. Now what will happen when you use another mole of Na and H2. Okay.

[00:01:59]Then again this HBr will be eliminated and you will get a compound like CH triple bond CH3.

[00:02:07]Okay. Now after that when you treat it with ethile bromide ethile bromide.

[00:02:15]Okay. When you treat this one with ethile bromide and you take one mole of Na NH2. So what it will do? It will abstract this proton here. This will become negatively charged and this one will attack on this carbon. So I will get here CH3 CH2 uh I have then uh C triple bond CH3 right? So this is spent 3 Y by NA. So now if you see for total I need three moles option number C here. Next the correct order of reactivity towards the electrophilic substitution reaction we have analine benzene and nitroenzene.

[00:02:46]Electrofile is reacting that means the ring should be nucleophilic in nature.

[00:02:50]In amalene you have NH2 group which is an electron donating group and it increases the electron density and makes the ring nucleophilic. So one will be the highest then benzene and then nitroen.in. Nitrobenzin will be the lowest because nitro is an electron withdrawing group. Next before moving ahead. So students if you have written KCT 2026 examination then me and my team has prepared a very very useful tool for you using which you can easily predict your KCT 2026 rank. So for this you have to simply enter your PCA marks here and you have to enter your expected KCT score here and then just click on predict rank and then you can see your estimated rank as well as you can see a range between which your rank is most likely to fall. Now the best part is you can also see all the branches and courses that is eligible all the colleges that that is suitable for your rank. Okay, if you scroll down you can see for example for 1934 RBCE electronics and telecommunication is possible or a suitable option because the cutff last year was 1422. Okay. So this is how you can use it and you will have an idea what all colleges which all branches which all courses you will be uh eligible based on your rank. Okay. So this is really going to be a very helpful tool for all of you. So the link of this tool is there in the description section. Definitely check this out.

[00:04:04]Okay. Now let's move to the next question. Which of the what is the correct sequence of reagents for the following preparation? So we are just basically converting it cis to trans.

[00:04:13]Okay. Now whenever we go from cis to trans you should go through an alkine formation. Okay. And for alkine formation first of all what you have to do? You have to add HBr to the given compound. Okay. If I add HBr what I will get here. Now see HBr will be added uh like this. Suppose here bromine is added. Okay. And here is added. Next you have done alcoholic KOH and NH2. Okay.

[00:04:39]Now when you have taken alcoholic KOH and NH2 uh that will give you here a triple bond.

[00:04:47]Okay. Here you will get a triple bond.

[00:04:51]I'll just write this one.

[00:04:54]Yeah. Now when you do this liquid ammonia, this will release this one and you will get a cis product. With liquid ammonia you will get here a sorry trans product. So option number A is the correct answer. Next we have which of the following is aromatic. So for aromatic we need to have two 4n + 2 pi electrons. So if you see 1 2 3 4 5 6 here we have 4 n + 2 pi electrons.

[00:05:18]Option number C here. Here 1 2 3 4 5 6 but this is also part of the ring. So we have total eight electrons. So this will not be there. Here 2 sp3 hybridized carbon is there. So that is not the aromatic compound. In aromatic compound all the carbons must be sp2 hybridized.

[00:05:33]Here also we have a lot of sp3 hybridized carbon. Next which of the following has the minimum boiling point.

[00:05:39]So the order of boiling point is alkanes, alkans and then alkyes. Okay.

[00:05:42]Alkyes have the highest boiling point.

[00:05:44]And in alkanes if the branching increases then the boiling point decreases. So if you see this one will be the highest then but one then butin and isobutin will be the last because it is branching. Next we have the alken which on ozonolyis yields acetone only. So for ozonolyis of acetone like you need to have product like this right. Uh so now what you have to do you have to just remove this double bond and you have like remove this oxygen and put it in the form of double bond. So you will get here 1 2 3 4 2 3 dimethile but 2. So that is your option number C. Next in the reaction given. So if you see the reaction is first there is an O group.

[00:06:29]Okay. Now this O group first gets protonated and you will get a compound like this O plus H2. Now this goes out and you will get a compound like this.

[00:06:42]Okay. Now we have carbon here one here two here three here four and here five.

[00:06:48]Now this bond comes here and you will get a five membered ring.

[00:06:56]Okay. Now again uh if you see this second carbon will have a positive charge here. Right? Now you can see that this is a secondary carocatin and tertiary carocatons are more stable. So this bond comes here and you will get a structure like this where you have something uh yeah now this carocatin is formed here. This is H here. So this bond will come here and you will get a double bond like this.

[00:07:26]Okay. So that will give me option number D here. Next, the mechanism of the reaction in terms of the intermediates or type of reaction is given below. Mark the incorrect option. So when you have addition of HBr, this is through carocatin intermediate. So this is a correct statement. Now for this one, peroxide effect, this is free radical.

[00:07:46]This is also correct. And when you have X2, this is electrphilic addition and not substitution. So option number C is the incorrect one. The most acidic hydrogen atoms are present in ethine.

[00:07:57]Next we have complete the following reactions and select the correct option.

[00:08:01]So phenol with dash will give you benzene. So phenol on reduction with zinc will give you benzene. So I can eliminate these two option. Next benzin sulfonic acid with steam will give you benzene. So I can eliminate this one. So option number B. Chlorobenzine if you treat it with hydrogen and nickel alloy that will give you benzene. Next the reaction of the given compound with HBr gives. So now you have this compound.

[00:08:22]You can see it is like CH3 CH double bond CH and then we have this C6 H4 and O.

[00:08:33]Right? Now see always this benzile carocatin is more stable. So the carocatin will form here. So that means Br will attack here and H will attack here. So I'll get CH3 CH2 CH Br C6 H4 whole O. So option number B here. Next in the given reaction you have Na NH2 first. So what it will do? It will abstract one hydrogen. So what is X? Now X is a compound with the negative charge. Then if you treat it with CH3 I then what I will get CH3 I this one will attack on this one. So you will get CH triple bond CH3. Okay this is Y. Now when you treat it with mercury that is H2O and Hg2+ what will that do? That will just add water following maronic rule. So what will happen here? CH double bond CO and here we have CH3 and this one will be H. Now there will be tomization. So this bond will come here. This bond will come here. So this one will give me CH3 C double bond O and then we'll have CH3.

[00:09:37]So this is option number C here. Next from the following compounds choose the one which is not aromatic. So here we have 1 2 3 4 5 6 7 8 four pi electrons.

[00:09:49]So for aromatic we need 4 n + 2 pi electrons. Option number B here. Similar to alken and alkyes benzene also undergo ozonolyis in the sequence x and y. So when benzene will undergo this one ozonolyis. So first of all you can see the structure of benzene is like this right? Now when ozonolyis happens first of all ozonide is formed in that what happens if you remember. So you here on each of them you will get one one ozonide formed here. Okay that is structure of ozonide is like this I'll just and this uh metal bond is removed. So like this three ozide will form trioide will be the first one. So I can eliminate these two option. Now when benzene will undergo this one ozenis if you break the bond you will have compound like this double bond.

[00:10:38]Okay. So this is a gioxil compound option number a here. Next, which of the following reaction is expected to readily give a hydrocarbon product in good? So, this one never gives because we get a mixture of products. So, electrolyis is the best method to give or to get hydrocarbon in the highest.

[00:10:54]Next, which of the among the following is the most strained cyloalkan?

[00:10:58]Cyclopropen is the most strained one because its angle of strain is more. A reagent used for the unsaturation in alkenes is broine with CCL force. Option number C here. Methane can be converted into ethane by the reaction. So when you methane you do chlorination so that will give you CH3 CL and then with reaction or with woods reaction you will get this product that is ethane. So option number C here which of the following is the most stable alken. So in alken you have two carbons and next stability depends on these R groups. Okay. If these groups are more carbon then this will be highly stable. So if you see this one R2 R2 is there. So this will be highly stable.

[00:11:36]Here we have two R groupoups. Here we have only one R group and here we have zero R group. Next when chlorine is passed through the boiling tolene we get so if you have towine okay and you when you're allowing chlorine with boiling means with high energy so this hydrogen will be replaced and you will get a compound like this that is CH2 CL or that is benzile chloride option number D here. Next one mole of symmetrical alken on ozonalis gives two moles of an aldihide having a molecular mass of 44 U. the alken is okay now if you see but 2 ene I'll just you have to check all the options okay but 44 means you can easily expect that it will have two carbons so and it should have a symmetrical alken because you're getting same product so propin will not give you ethine will give you symmetrical but this is also not symmetrical but ethine is having only one like if it breaks like if ethine breaks like this only one product you will get HC double bond O and its mass cannot be 44. So that is why but 2 EN has to be the correct answer. And if you see the product here you will get CH3 CO H and mass will be here 24 16 uh and then you have four. So that gives me 44. Okay. Next we have iodine value is used to determine the degree of est unsaturation. So option number B here.

[00:13:00]Next question. Which one of the following is the most energetic confirmation of cycllohexane? So if you see the confirmation of cycllohexane the half chair you can see here the half chair is the one with highest energy and chair is the one with lowest energy.

[00:13:14]Okay. So you have to remember these two terms. Now I'll show you the options. So yeah so half chair will be the answer option number D here. Okay. Next identify the main product in the reaction. So CH3. So you have a tertiary alcohols and so CH3 CH3 CH3 and then you have O right and how can I uh this one suppose if I write like this it will be 1 second. Yeah it will be C H2 and H. Now this O and H will be removed and you will get a double bond here. So option number A.

[00:13:51]Next the compound which on ozonolyis produce a mixture of propanon and ethanol. So propanon is CH3 C double bond O CH3 and ethanol is something like this.

[00:14:05]Okay. So now this one will be removed and you will get a structure like this.

[00:14:08]CH3 CH3 okay and double bond CH3 H. Okay. So 1 2 3 4. So we have two methile but 2 ene option number D here.

[00:14:23]Next when two broen is healed with alcoholic solution of potassium hydroxide. The major product obtained is when you have two brooentane okay and it is mixed with alcoholic potassium hydroxide. So there will be dehydrohalogenation and you'll get a double bond here that is pent to EN.

[00:14:44]Next, which of the following chemical system is not aromatic? So if you see option number D here we have sp3 carbon which is not aromatic. Next in the given reaction we have this one the product P will be so if you see HC triple bond CH is there. Now if I add here H and O so it will be like this.

[00:15:04]Okay. Now this one will come here. This one will shift here. So you will get here CH3 C double bond O that is an alihide. So tolerance reagent will react. Bread reagent will react.

[00:15:15]Iodophor test will also be there because CH3 CO group is there. Victor test is actually for alcohol. So this one will not react. Next the organic compound obtained during the addition of HBr propane in the presence of peroxide. So antimaronic rule will be followed and one broen will be the major product.

[00:15:32]Maronic rule is applicable to unsymmetrical alkanes. Okay. So C3, H6 and HBr. Yes, this is not applicable because it is an what I say symmetrical alken. Okay. So same case here and with Cl2 we cannot add we need HX for that.

[00:15:47]So option number A here CH3 CN with dilute HCl will give you CH3 CO A will be your acetic acid. Then lithium aluminium hydride will reduce this one.

[00:15:58]So B will be CH3 CH2O and then with PCL5 you will get CH3 CH2 CL and then alcoholic KO will give you CH2 double bond CH2. Okay. So the product D in the above sequence of reaction is ethine. That is option number B here. Next, which of the following is aromatic?

[00:16:20]So tyrosine is the one which is an aromatic amino acid. Next, isopropyl chloride and isobutile chloride reacts with sodium ether to give you isopropile chloride. Okay. So this is isopropile chloride and we have isobutile chloride.

[00:16:37]Okay. Now when these two will react what is the major product that will form you will get a product like this.

[00:16:46]Okay 1 2 3 4 5. So 24 dimethile pentine penten which is option number d here. So students these are the last 10 years of pqs from the chapter hydrocarbons. Just revise the basic reactions. You will be able to do this chapter and stay tuned for the next chapters. And the case rank predictor link is there in the description section. Do check it out.

[00:17:08]Thank you so much for watching and all the