Thanks for your hard work Sir. The exam was easy Alhamdulillah....just came back to express my gratefulness. Keep going Sir. I just have a request....please make a topic wise (Chapter wise) last some years questions solved for A2. BTW I am also from Bangladesh but I live in Saudi Arabia.
Hey! You’re welcome. I’m glad i could help. Best of luck for physics. And sure! Thanks for the recommendation. I’ll keep that in mind for A2. Keep on supporting the channel :)
@@CAIEPapersSolved Sir, I didn't know that you also solved physics . I just searched for a solved Specimen paper for Physics and gladly found your video. Now watching it :)
In question 13, why is the rate of the reaction decreasing? The volume of gas given off is increasing over time Also in question 25, why u considered sterioisomerism only without structural
Hey! Clarification for both of you For question 13, the volume of gas is increasing over time, but the rate at which it is increasing is actually decreasing. Like compare, in the first 30 seconds, we get 40 cm^3, in the next 30 seconds we get 30 cm ^3, in the next 30 seconds we get 18 cm^3. Maybe you guys misinterpreted the question. In the second row its written total volume given off, so to find out the exact volume given off in that particular period you need to subtract the old one from the new one. For example at 60 seconds total volume is 70, and in 30 seconds total volume was 40, so to find out total volume in the 2nd 30 seconds, we need to subtract 40 from 70, 70-40=30 in the second column. And so on. Hopefully you get it now. Over time the value decreases! For question 25, I did consider structural isomers. In questions where they mention the molecular formula you are supposed to consider ALL POSSIBLE structural isomers. But for example, in this question, they have already given you the compound. Like in B, you’re given butanone. So you have to progress from that. Ive considered the structural isomers. After reduction it turns into butan-2-ol. After dehydration the two structural isomers are but-2-ene and but-1-ene. Now, but-2-ene has 2 stereoisomers. So in total B has 3 isomers(both structural and stereo)!
Hey! Sure. Basically, options a and d are wrong anyway. Cold, conc and hot, dilute aren’t part of our syllabus. So its between B and C. Rather than eliminating C i just went for B. Because cold, dilute kmno4 produces diols. And when a diol is produced from an alkene, two oh groups are added. Each oh group has an mr of 17, so 17x2=34. That’s how i reached the answer. As for why C is wrong, hot, conc kmno4 would cause oxidative cleavage of the alkene. We aren’t even sure what the alkene is. For example if it’s ethene, we’d get ethanoic acid (ch3cooh) it has an Mr of 12+3+12+32+1= 60 while ethene had an Mr of ch2=ch2 (Mr 28). So the mr of the product is greater by 32. Basically in the video i said there would be too many variations. Like it couldve been propene as well yk. Then we’d get propanoic acid etc. so rather than eliminating other options here you should approach it by looking at the most probably answer!
so from now on, q 25 to 40 will basically be organic right? and everything before that will be from other chapters? kinda makes me glad. I barely lose marks in organic questions while inorganic on the other hand are so annoying. I always get stuck in balancing the equations because I cannot construct them properly lol
Yes! Basically q25-40 will be organic. I know, right? Organic is so much easier compared to inorganic. Make sure you use oxidation numbers to balance equations! good luck
@@CAIEPapersSolved haha yup. I don't know how my friends find organic harder. And yeah, thanks. Also, I just wanna thank you again lol, you have no idea on how many of my doubts were cleared thanks to your videos, if only I found your channel earlier. Hopefully in A2, life will be easier if AS goes well
@@aq3605 Honestly, yeah that is the one thing I hate about organic chemistry, but I think you get used to it when you practice it a lot. I would recommend, writing down the names of the different isomers you are drawing, so that you make sure that you have not drawn duplicate isomers even tho they look different.
Hey! The key point here is balancing the electrons transferred. At first i found out which element was reduced and which element was oxidised. Then i found out the oxidation number. Then i found out the change in oxidation number and tried to balance the number of electrons by multiplying an integer. Using that we found out the values. See if it makes sense now, or else i’ll give an elaborate explanation.
@@CAIEPapersSolved pretty understandable, the points are cleared, I must balance it myself now...and then see if it's ez to solve...Thnks...are u Malaysian teacher btw?
I might be missing out on some key information, but how exactly did you solve question 7 so fast? A small explanation would really clear out my doubts.
first off the molecule has to have an electronegative element, so A is wrong. Then in C and D we can see that the electronegative element is followed by another electronegative element on the other side, which lowers the overall diapole movement, which makes B the best answer
You should practice section B. Compare this speciman paper with 9701/M/J/11/2017. These two papers are literally same. Cambridge has literally converted questions like statements into options from that 2017 paper in this speciman paper. So this is clue that section B questions can be asked as section A questions. So all the best for paper. Do pray for me as well. 🙂
@@iloveallah3991 Thank you. Your observation is great!! These papers are exactly same but one follows old pattern and other follows new pattern. This gave me some idea how Cambridge is going to give questions now. May Allah gives straight A's in AS and straight A*'s in A2.
Hey! Basically, at 50:33 , after reduction of citral using h2+ ni, we end up with the purple compound which doesn’t have any alkenes. (Alkenes reduced to form alkanes) so oxidative cleavage doesnt occur, only the aldehyde is converted to cooh
in q37, you said tertiary react slowly, but on google it says .... Why do tertiary halogenoalkanes react the fastest? Image result for which halogenoalkane reacts the fastest Tertiary haloakanes react via an sN1 mechanism that has a much lower activation energy than the sN2 mechanism with the high energy transition state. Hence tertiary haloalkanes react faster then secondary, which in turn react faster than primary.
Let me clarify. Sn2 is a 1 step reaction. Sn1 is a two step reaction. So typically sn1 reactions are USUALLY slower, there is no fixed rule though as it depends on a lot of other factors. Also, in 37, B was already wrong in the first place. They are saying that the Carbon atom bonded to the Cl atom is more positive in the molecule compared to the others, however that’s wrong. This is because due to the positive inductive effect of 3 methyl groups, the intensity of the positive charge on the carbon is actually less. So it should’ve been less positive. Hopefully you understand now!
Aoa can you make videos in which you explain mcq 25 to 40 again because this section is of organic chemistry and its really hard so i need a very clear explanation video😢
Hey! You’ll learn more about this in A2 Electrochem. Basically we do that when H+ is required as a catalyst, for example if we use KMnO4 as an oxidising agent(the example you are talking about). But here, we actually have K2MnO4, in which Mn has a charge of 6+. It doesn’t require H+ as a catalyst. So, we don’t use those. Hopefully it makes sense 🙏
I tried to explain as simply as possible but let's try again. A proton has a charge of +1 and a mass of 1 so the charge/mass ratio is +1/1 = +1 that gives us an angle of +15 degrees B is the correct answer because, it has 3 protons and 5 electrons so charge is -2 and it has 3 protons and 3 neutrons so mass is 6. Electrons have negligible mass. The ratio of charge/mass for B is -2/6 which is -1/3 . According to unitary method, since a ratio of +1 is equal to +15 degrees, a ratio of -1/3 will be equal to -5 degrees. Hopefully it makes sense now!
@@CAIEPapersSolved Also an easier way to save time is that B is the only one with a negative overall charge, therefore it is the only one that can give a negative angle.
@@CAIEPapersSolved oohhh haha didn't watch the full video just watched for a few questions that I got wrong and couldn't understand why. Also huge respect for replying so quickly, subbed!
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You are an absolute legend! This channel deserves so much more than it does! really!
Thankyou! ❤️
Thanks for your hard work Sir. The exam was easy Alhamdulillah....just came back to express my gratefulness. Keep going Sir. I just have a request....please make a topic wise (Chapter wise) last some years questions solved for A2. BTW I am also from Bangladesh but I live in Saudi Arabia.
Hey! You’re welcome. I’m glad i could help. Best of luck for physics.
And sure! Thanks for the recommendation. I’ll keep that in mind for A2. Keep on supporting the channel :)
@@CAIEPapersSolved Sir, I didn't know that you also solved physics . I just searched for a solved Specimen paper for Physics and gladly found your video. Now watching it :)
In question 13, why is the rate of the reaction decreasing? The volume of gas given off is increasing over time
Also in question 25, why u considered sterioisomerism only without structural
yeah i also didn't get that
Hey! Clarification for both of you
For question 13, the volume of gas is increasing over time, but the rate at which it is increasing is actually decreasing. Like compare, in the first 30 seconds, we get 40 cm^3, in the next 30 seconds we get 30 cm ^3, in the next 30 seconds we get 18 cm^3. Maybe you guys misinterpreted the question. In the second row its written total volume given off, so to find out the exact volume given off in that particular period you need to subtract the old one from the new one. For example at 60 seconds total volume is 70, and in 30 seconds total volume was 40, so to find out total volume in the 2nd 30 seconds, we need to subtract 40 from 70, 70-40=30 in the second column. And so on. Hopefully you get it now. Over time the value decreases!
For question 25, I did consider structural isomers. In questions where they mention the molecular formula you are supposed to consider ALL POSSIBLE structural isomers. But for example, in this question, they have already given you the compound. Like in B, you’re given butanone. So you have to progress from that. Ive considered the structural isomers. After reduction it turns into butan-2-ol. After dehydration the two structural isomers are but-2-ene and but-1-ene. Now, but-2-ene has 2 stereoisomers. So in total B has 3 isomers(both structural and stereo)!
@@CAIEPapersSolved got it thank's 😃
@@CAIEPapersSolved alright thanks alot!
hey can you please explain 27 again? i didnt quite understand how you ruled out all options and chose cold dilute
Hey! Sure.
Basically, options a and d are wrong anyway. Cold, conc and hot, dilute aren’t part of our syllabus. So its between B and C. Rather than eliminating C i just went for B. Because cold, dilute kmno4 produces diols. And when a diol is produced from an alkene, two oh groups are added. Each oh group has an mr of 17, so 17x2=34. That’s how i reached the answer.
As for why C is wrong, hot, conc kmno4 would cause oxidative cleavage of the alkene. We aren’t even sure what the alkene is. For example if it’s ethene, we’d get ethanoic acid (ch3cooh) it has an Mr of 12+3+12+32+1= 60 while ethene had an Mr of ch2=ch2 (Mr 28). So the mr of the product is greater by 32. Basically in the video i said there would be too many variations. Like it couldve been propene as well yk. Then we’d get propanoic acid etc. so rather than eliminating other options here you should approach it by looking at the most probably answer!
@@CAIEPapersSolved ohh okay i understand now! thank you for your work🙌🙏
so from now on, q 25 to 40 will basically be organic right? and everything before that will be from other chapters? kinda makes me glad. I barely lose marks in organic questions while inorganic on the other hand are so annoying. I always get stuck in balancing the equations because I cannot construct them properly lol
Yes! Basically q25-40 will be organic. I know, right? Organic is so much easier compared to inorganic. Make sure you use oxidation numbers to balance equations! good luck
@@CAIEPapersSolved haha yup. I don't know how my friends find organic harder. And yeah, thanks. Also, I just wanna thank you again lol, you have no idea on how many of my doubts were cleared thanks to your videos, if only I found your channel earlier. Hopefully in A2, life will be easier if AS goes well
really, figuring out structural isomers and chiral carbons is so hard
@@aq3605 Honestly, yeah that is the one thing I hate about organic chemistry, but I think you get used to it when you practice it a lot. I would recommend, writing down the names of the different isomers you are drawing, so that you make sure that you have not drawn duplicate isomers even tho they look different.
Hi again, can u plz explain question 11, like whats the key point to do in it?
Hey! The key point here is balancing the electrons transferred. At first i found out which element was reduced and which element was oxidised. Then i found out the oxidation number. Then i found out the change in oxidation number and tried to balance the number of electrons by multiplying an integer. Using that we found out the values. See if it makes sense now, or else i’ll give an elaborate explanation.
@@CAIEPapersSolved Thank you so much sir , this question had me confused 🙏
@@CAIEPapersSolved pretty understandable, the points are cleared, I must balance it myself now...and then see if it's ez to solve...Thnks...are u Malaysian teacher btw?
No. Im from Bangladesh! :)
@@CAIEPapersSolved okay thanks hopefully todays ppr would be easy :]
Which software do you use?
OBS for recording, OneNote for solving papers
I might be missing out on some key information, but how exactly did you solve question 7 so fast? A small explanation would really clear out my doubts.
first off the molecule has to have an electronegative element, so A is wrong. Then in C and D we can see that the electronegative element is followed by another electronegative element on the other side, which lowers the overall diapole movement, which makes B the best answer
Hi, why isn’t the 40th answer D?
should we be practicing section B as its now removed
Nope. Skip
You should practice section B.
Compare this speciman paper with 9701/M/J/11/2017.
These two papers are literally same. Cambridge has literally converted questions like statements into options from that 2017 paper in this speciman paper.
So this is clue that section B questions can be asked as section A questions. So all the best for paper.
Do pray for me as well. 🙂
@@iloveallah3991 Thank you.
Your observation is great!!
These papers are exactly same but one follows old pattern and other follows new pattern. This gave me some idea how Cambridge is going to give questions now.
May Allah gives straight A's in AS and straight A*'s in A2.
@@faizan2647 Thank you brother. ❤️
May you also get straight A's and A*.
@@iloveallah3991 thanks for pointing that out! I’ll try to solve Section B in my next videos :)
Also good luck to you all
in q 30 how did u know options b and c are not correct i dont get it
Can you please share the link of the bonus video ASAP as I have my exam day after
Thanks
ruclips.net/video/S75xYUO4CJg/видео.html
Here you go, I was talking about this video for the Organic Bonus Content. The timestamp is 28:35!
honestly while solving these questions you do realise how easy was o levels
Right?
this was helpful though, do you take online classes?
You're welcome! I do actually, but only in Bangladesh
@@CAIEPapersSolved heyyy YOU TAKE ONLINE CLASSES IN BD? i wanna take your classes for A2
Hi, are you going to do 2020 papers also ? please do them.
Hey! I'll try, but I have exams coming up so it's unlikely :(
Okk my bad, also I just wanted to ask, is it ok to do only 2021 and 2022 MCQ papers? If not, till which year do you recommend?
2020-2021, FM 2022 and specimen paper 2022 should be enough!
in question 35, why is the compound not broken?
Hey! Basically, at 50:33 , after reduction of citral using h2+ ni, we end up with the purple compound which doesn’t have any alkenes. (Alkenes reduced to form alkanes) so oxidative cleavage doesnt occur, only the aldehyde is converted to cooh
thank you so much you’ve helped with tomorrow’s external
Thanks man we appreciate it
❤️
in q37, you said tertiary react slowly, but on google it says .... Why do tertiary halogenoalkanes react the fastest?
Image result for which halogenoalkane reacts the fastest
Tertiary haloakanes react via an sN1 mechanism that has a much lower activation energy than the sN2 mechanism with the high energy transition state. Hence tertiary haloalkanes react faster then secondary, which in turn react faster than primary.
Let me clarify. Sn2 is a 1 step reaction. Sn1 is a two step reaction. So typically sn1 reactions are USUALLY slower, there is no fixed rule though as it depends on a lot of other factors.
Also, in 37, B was already wrong in the first place. They are saying that the Carbon atom bonded to the Cl atom is more positive in the molecule compared to the others, however that’s wrong. This is because due to the positive inductive effect of 3 methyl groups, the intensity of the positive charge on the carbon is actually less. So it should’ve been less positive. Hopefully you understand now!
Do u know if oct nov 2022 As chem P3 will be exempted or not?
I'm not sure about that!
Aoa can you make videos in which you explain mcq 25 to 40 again because this section is of organic chemistry and its really hard so i need a very clear explanation video😢
Sir Q11, why dont we balance by adding water and h+ ions as usual... i have exam tomorrow
Hey! You’ll learn more about this in A2 Electrochem. Basically we do that when H+ is required as a catalyst, for example if we use KMnO4 as an oxidising agent(the example you are talking about). But here, we actually have K2MnO4, in which Mn has a charge of 6+. It doesn’t require H+ as a catalyst. So, we don’t use those. Hopefully it makes sense 🙏
absolute chad
Thankyou
9:37 the question i ask myself all the time💀
😂😂😂 everytime
Do you take online classes in bd?
Yes. Both online and offline
How can I contact you? for classes?
@@nazmun9d834 You can find my email in the about section of my channel
ruclips.net/user/CAIEPapersSolvedabout
You can email me there!
also explain Q2 again, its not clear
I tried to explain as simply as possible but let's try again.
A proton has a charge of +1 and a mass of 1 so the charge/mass ratio is +1/1 = +1
that gives us an angle of +15 degrees
B is the correct answer because, it has 3 protons and 5 electrons so charge is -2
and it has 3 protons and 3 neutrons so mass is 6. Electrons have negligible mass.
The ratio of charge/mass for B is -2/6 which is -1/3 . According to unitary method, since a ratio of +1 is equal to +15 degrees, a ratio of -1/3 will be equal to -5 degrees. Hopefully it makes sense now!
@@CAIEPapersSolved Also an easier way to save time is that B is the only one with a negative overall charge, therefore it is the only one that can give a negative angle.
@@1tubax yup! Exactly. I mentioned that in the video, but just gave an elaborate explanation if you’re given a case with 2 negative particles :)
@@CAIEPapersSolved oohhh haha didn't watch the full video just watched for a few questions that I got wrong and couldn't understand why.
Also huge respect for replying so quickly, subbed!
@@1tubax thankyou! Good luck if you’re giving the exam tomorrow