COMEDK Chemistry PYQs (Last 10 Years) | Organic Chemistry: Some Basic Principles

Ajouté : 2026-05-10

[00:00:00]Hello champions, welcome to the channel and in today's video we are going to solve the last 10 years of comet K PQs of a very very important chapter that is organic chemistry some basic principles and techniques from PU1. So now if you see the weightage of uh this chapter so from the last year it was 2025 four questions were asked and in 2023 and 2024 two questions were asked. So if you revise this chapter definitely you can expect two to four questions. So let's see the first question for F minus CL minus sorry yeah for F minus CL minus and Br minus the increasing order of the nucleophilicity okay nucleophilicity means the tendency to donate electrons.

[00:00:38]So bromine is less electro negative it can easily donate electrons option number B here. Next we have which of the following is correct regarding the minus I effect. So minus I effect depends on the electro negativity definitely florine will be highest and then oxygen and then nitrogen. So option number A here. Next uh this compound name is asked. So here you can see this is a compound like this. CH3 CH2 CH2 CH CH double bond CH2 CH2 CH2 CH3.

[00:01:13]Okay. So 1 2 3 4 5 6. So three ethile I have uh one sorry three propile and I have hexene. That is option number B here. Next arrange the following carboan. So carboanions we have first one is like more is the electro negative. So this is an sp hybridized carbon. So it will be more stable. So one will be highly stable. Okay. And the least stable will definitely be this one which is an sp3 carbon. So here also I have four. Now if you compare sp2 and sp2 definitely phenile is more stable.

[00:01:46]So after one it has to be two. So correct answer will be option number C here. Now students before moving ahead if you have written the KCT 2026 examination and want to predict your rank me and my team has prepared a very very useful tool for you. So in this tool you just have to enter your PCM marks and then KCT score score and then just click on predict rank and you will be seeing your estimated rank and also the range between which the rank will fall. Okay. Now apart from this the best part is you if you select the branches here or you can just select all branches you will be seeing all the possible colleges courses that are available according to your rank. Okay this is based on the round three cututoff of 2025. So this will give you a clear picture regarding what all uh I can say like uh what all options are available which colleges can give you seed. So you will have a clear picture about that. So definitely check this tool the link is there in the description section. Now let's move to the next question. Which of the following carocatines is least stable? Okay. So least stable we have to find NO2 group is there carocatine. NO2 is electron withdrawing group. So it will destabilize. Option number B here.

[00:02:54]Next in the carious method the estimation of halogen is given presence of 08 g of bromine. The correct structure is so if I if 172 g contains 08 then 172 g contains 80 g of bromine.

[00:03:08]Right? So that means only one bromine should be present in the compound. So I can eliminate this option. Now the reactant must be 172 g. So if you measure this one, option number B will be 172 g. So option B is the correct answer. Next the structures are given represent. So here the NH is on the second position and here it is present on the third position. So this is positional isomers. Next during lassine test N and S present in the organic compound changes to okay N changes to Na CN and S changes to Na2S option number A here. Which of the following is an isomer of ethanol? So dimethile ether is a functional functional isomer of ethanol. The order of the decreasing stability of the anion. So this is a tertiary, this is secondary, this is primary and this is so this is resonance stabilized. Obviously fourth one will be highly stable. After that you will have primary. So after four it has to be three. So option number B here. Next which type of intermediate A is formed during this type of reaction. So here what happens? Actually this bond breaks here. This bond breaks here and you get free radical formations. Option number C. Next. Which of the following is incorrect order of the stability of carocatine. Okay. So incorrect order.

[00:04:20]Right. So here if you see tertiary is highly stable but it is given in the middle. So this is the incorrect one.

[00:04:24]option number B here. Next, this shows dash type of geometry. So when you have two groups attached to the carbon, same groups are attached.

[00:04:33]Okay, same groups are attached means what will happen here? This one will be different. Like when the same groups are attached, you cannot have geometrical isome. But you have an optically active center or a stereo center here which will give you optical isome. Next we have in Zelda method ammonia formed 8 g of the food neutralizes 60 cm cube of 0.1 N acid the percentage of nitrogen is. So this one will be 1.4 into.1 into 60 divided by 8 into 100. Okay the formula is 1.4 into normality into volume divided by weight into 100. Now if I solve this you'll get 1.05. Now, which of the following statements is incorrect about the partition chromatography? The incorrect statement is the mobile phase can be a gas. No, mobile phase is always a solvent. Next, in Zelda method, ammonia from 5 g of food neutralizes 30 cm cube of.1 acid. The percentage of nitrogen is again we can solve this one. So, we have 1.4 into.1 into 30 divided by 5 g. So, if I solve this, I'll get around 0.84 option number A here. Next, a missisible mixture of these two can be separated by distillation. Missible liquids are always separated using distillation.

[00:05:53]Next, this one is neither an electrofile nor a nucleophile because so you know already it has donated its electron. It cannot further donate and nitrogen does not have any vacant d orbital to accept electron. So that is why option number B. Neither has electron pair available for donation nor can accommodate electron since all the cells of the nitrogen are fully occupied. Next you have to what is the number of primary, secondary and tertiary and quartonary carbons. Okay. So if you see the first one here only one carbon is attached so it is primary. Here this carbon is attached to two. So this is tertiary.

[00:06:24]This carbon is attached to four. So this is quartonary. This is primary. This is primary. This is primary and this is primary. Here this one is attached to two. This is secondary. And this one is attached to three. So it is tertiary. So I have five primary. That is option number B. Here next one minus I effect is shown by so minus N2 group because it is an electron group electrogative compound. Next question is the decreasing order of reactivity towards an electrofilic substitution means it should be mean the ring must be nucleophilic in nature. Right? So if you see here CH3 is a donating group it has plus I effect. So that will react with the electrofile. So fourth one will react faster. So I can eliminate other two options. Next after that one will come. So one is present in both. Now Cl is the one which pulls the electron with minus I effect and NO2 pulls by minus R effect. So obviously this is like most electrophilic. So this one will be the least reactivity. So option number C is the correct answer. Next an organic compound weighing 0.15 gives on carious estimation 0.12 g of AgBR.

[00:07:30]the percentage of Br in the compound is so the formula to calculate the percentage of Br will be 0.12 into 80 by 188 into.15 right so if I just solve this I'll get around you can see 188 is the total mass yeah you have to just solve this and into 100 also you need to calculate so if you just solve this you'll get 34.1% so option number B here next identify the correct UPSC name from the given 1 2 3 4 5. So it is pentane nitral that is option number C here. Next malic acid and fumeric acid.

[00:08:07]So if you see the structure they are geometrical isomers. One is this one is trans. Next up name of tb bututile chloride. So t butile chloride structure is like this. So this is two methile two chloropropane. So that is two chloro2thyl propane option number a here. Next arrange the following in the increasing order of stability. So here you can see that this is the stability depends on the number of hyper conjugation structures. So here I have 3 + 3 6 + 2 8 alpha hydrogen. Okay, here I have 9 alpha hydrogen. Here I have just six alpha hydrogen and here I have three alpha hydrogen and zero alpha hydrogen.

[00:08:48]So two will be highly stable. I'll eliminate this one. After that one will come. I'll eliminate this one. After that six will come. Uh sorry 213 okay just let's see here we have C H CH3 and CH3 okay so after that three will come right yeah three so 213 213 is there after that you can see fifth one will be the least one because it has no group to stabilize the carocatine so option number A here next which what is the hybridization state of the carbonial so CH2 plus is there this is sp2 hybrid ized option number A here. Nitrogen detection in an organic compound is carried out by lassine test. The blue color is formed due to the formation of F4 F whole CN6. Next, homolytic fusion of the following aloquins form free radical. So you have to find the stability of the free radical. So if you see this is a primary radical, this is secondary. This is 1 2 3. This is also tertiary. And like 1 2 3 this is also tertiary. But here you can see the number of alpha hydrogen's is less. Here the number of alpha hydrogen's is more. Option number B here. Next an organic compound which produces a bluish green flame when heated on a copper wire is actually chloromite. Next, the number of sigma and pi bonds in pent 2 y. This is very simple question but yeah you have to be very careful with the structure. Pent 2 y.

[00:10:15]Okay. So definitely I have two pi bonds.

[00:10:17]So I can eliminate this last option. Now 1 2 3 4 5 6 7 8 Okay. 1 2 3 4 5 6 7 8 9 10 11 12. So we have 12 I can say sigma bond. So option number A here. Now let's see next question. In lassine test for nitrogen in an organic compound the blue coloration is due to the formation of feriferyanide. That is option number B here. Next the reaction of ethine in presence of H minus ion can be an example of so when you have a reagent so this is due to minus E effect. It's option number C because it will pull electrons. Identify the electrofile among the following. Al3 is electron deficient. So that is why it is electrofile. Carbon atoms in ethine are your sp2 hybridized. Group with electron withdrawing resonance effect is nitro group. Next minus I is exhibited by minus I is exited by CN only. Next myth mythoxy methane and ethanol exhibit functional isomers. Okay. Yes. So students, these are the important questions from the organic chemistry chapter. Very basic and simple questions will be asked. So just revise the concepts. That's all in this video and do check out the predictor. The link is there in the description section. Thank you so much and stay tuned for the next chapter.