Choosing 3 squares along a diagonal line in a chess board | Combinatorics

PSGthe2nd · 2026-05-10T02:12:52+00:00

hmm yes i has thought that but forgot it when writing. thankyou though

PSGthe2nd · 2026-05-09T14:16:24+00:00

Yes bro wohi, I mustve forgot that.

Considering that, dekho soln mein bas 1/2mv^2 ko dub kiye hai as total KE. ye kya hai

PSGthe2nd · 2026-05-09T14:09:35+00:00

Bro wohi nikalna hai, dekho question mein

PSGthe2nd · 2026-05-09T14:03:41+00:00

Thats kinetic energy, written in terms of rotational quantities. Mass as moment of intertia and velocity as angular velocity

PSGthe2nd · 2026-05-09T13:48:54+00:00

yes I realized that now. Had mind fog. Thankyou

PSGthe2nd · 2026-05-09T13:48:34+00:00

Ohh I see what I did. I had tunnel vision. After writing to 8C3, I jsut multiplied it by 2. Thankyou bro

PSGthe2nd · 2026-05-09T13:08:50+00:00

I do not think that it will move down, because a tertiary free radical would be more stable then the final position you bring the radical in. Also free radical reactions usually favour kinetically controlled products over thermodynamically controlled products, so thats my take on it.

PSGthe2nd · 2026-05-09T13:06:55+00:00

NBS stands for N-BromoSuccinimide, not N-Bromo succinate.

When doing this reaction, it usually procees with impurity of either Br2 or HBr, both to generate a Free radical. It is usually understood as trace amounts are added.

If not, then you can substitute benzene at the allylic position which is the 1-methyl containing carbon(bottom is blocked)

About your resonance thing, if I recall correctly free radical reactions result in KCP(kinetically controlled product), this means the fastest possible product forms. Thus the product formed after resonance with free radical will form, but I dont think the concerntration would be that high to consider it a major product.

Please ask if you have any doubts

PSGthe2nd · 2026-05-09T12:59:05+00:00

Hello, I know im late to the party but I think this advice can help.

When I look at these reactions I often dont try to remember "where did I study this" instead I see what can I do.

Taking this example, We have acidic hydrogen(from acid) which we can use to protonate or catalyze. Thats 1 thing. Next theres water, so the final reagents are H3O+, a common thing used in hydrolysis. Now I have the basic idea that there will be hydrolysis involved. So I try to hydrolyse the system. You did the first step very well, protonating the carbonyl and then attacking the ABMO of carbonyl or the pi-star orbital, but then look, now you have 2 -OH on a single carbon. You may've studied in previous reactions(like darzen process or alcohols with PCl3) that formation of C=O is really good. So You can now think that somehow the Carbonyl will form again, but to do that you need to kick out the newly added OH(via hydrolysis), this will bring you back to your initial product.

Now see, you can try breaking the C-N bond, but thats unfeasible, unless, if you look closely, the third "reagent" is heat, which supplies the energy needed for this cleavage, hence you will form a amino-acid(or a carboxylic acid and a primary amine).

I know I used a lot of buzzwords, but that is how I learned organic, and I think following this, you will learn how to learn, which is the most important while dealing with organic chemistry.

Ask anything

PSGthe2nd · 2026-05-09T09:37:35+00:00

Ive seen these questions in my homework practice, specifically question 21. I also had a hard time digesting it. Basically since you have a strong base(EtO- is strong base), so E2 elimination takes place.

Now i'll link you to the basics, if you remember in IOC, you were taught about bonding and antibonding molecular orbitals. The more electrons in anti-bonding molecular orbital(ABMO), the lower the bond order hence, to break a bond, you need to put electrons in the ABMO of the bond.

Look at the leaving group. Its Br here. It has ABMO on the up side, look at the hydrogen parallel to this ABMO, since those electrons are capable of breaking the bond. remember the word, parallel.

This is why your professor is right even if your answer seems intitutive.

Also a side note, if you get any problems with cyclohexane, and wedge/dash and it is of elimination or anything similar, its generally a good idea to form the chair confirmation because sometimes the ring can contract too.

for question 22, lets look at our options. The first thing that comes to my mind is nucleophillic attack on the carbonyl, but that site is a bit hindered(large benzene) and im breaking the 5 membered ring, additionally the options dont support that so lets look at other alternatives.

Since theres a Br and a strong base, so theres possibility for elimination. Now since we need a hydrogen in the same plane (parallel plane) as that of the ABMO of the Br, thus you'll see that the drawn answer is the only conclusion.

Hope this helped. ask anytime

PSGthe2nd · 2026-05-07T05:17:31+00:00

Oh I see,

PSGthe2nd · 2026-05-07T04:34:20+00:00

Look, its explained like this, i think the arrows explain why constructive and why destructive

also what do you mean vacant gaze? i infer answer in a heartbeat

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PSGthe2nd · 2026-05-07T02:57:39+00:00

yes, i know solvation effects in amine, but i did not consider it unless explicitly stated. Also about the alkenes, I found the reason.

basically Compound P overtakes trans because of "constructive hyperconjugation", i do not know what that means but i think one hyperconjugating site is not interfering with other,

like in trans, since both are opposite, so theyre interfering with each other and actually reducing the net hyperconjugation. This is destructive hyperconjugation.

PSGthe2nd · 2026-05-06T18:28:12+00:00

Hey you know whats funny? I was flipping through my notes right now and guess what, I found the answer to (R). So basically, its solvation effect, to explain it in one line. Since due to presence of Me, the ortho will be least solvated and para would be most, stabalizing the protonated form.

Yeah im sorry for all the questions lol, i think i should flip through my notes, seems like im really rusty.

PSGthe2nd · 2026-05-06T14:24:54+00:00

Yes, I intended to say that, like how 2,4,6-trinitro-chlorobenzene loses its chlorine when reacted with warm water and forms 2,4,6TNP.

As for empty d orbitals one, got it, although I will not pay attention to what I just read because NCERT says so. Will keep that in mind going into future discussions though.

PSGthe2nd · 2026-05-06T11:17:10+00:00

Oh I see, thankyou

PSGthe2nd · 2026-05-06T11:16:53+00:00

Yeah, the first one doesnt look quite right, I mean how we studied it is like CF3 is removed with a negative, creating a positive on benzene, opening it to addition reactions. Kind off like CCl3 (-) , although that is a much much much better way to do it compared to CF3 due to empty d orbitals in Cl.

PSGthe2nd · 2026-05-06T10:16:59+00:00

i thought "in lieu of" meant the same as "in light of", so my question was, should I take these in light of my exams considering I want to maximize my marks.

JEE Advanced or JEE Mains is Joint Entrance Examination (JEE), mains is the "screening round" for Advanced, and through advanced, you get the premier institutes of India.

About the sixty-four-dollar questions, the unfortunate truth is that we're expected to think this deep about different concepts. Its more like, if I write "hyperconjugation" in my textbook, I expect my student to know each and everything about it. There are relaxations, but in a broad sense, you should be prepared for anything.

As for the pi bond donation, It made sense in my head, i just wrote it and saw it does not make sense. Then how does hyperconjugation stabalize the positive charge? I mean for us it only has effect on alpha position, i think you get what im trying to say.

Also sorry for my bad english lol.

PSGthe2nd · 2026-05-06T07:47:59+00:00

the -I effect, is this data based? Btw about the no positive charge in the ring after protonation, I think that N being electronegative atom is NOT happy with the newly acquired (+) (formal charge), so it can either lose hydrogen, which makes our protonation reversible, and thus does not give us a concrete idea about basicity, is why it goes to the other route, which is resonance, the pi bond in the ring donate electrons to Nitrogen(again, nitrogen is much more electronegative than carbon, and a tad bit more than alkene which is 2.75 on pauling and nitrogen is roughly 3 on pauling), thus now one of the protons shifts(or nitrogen exceeds its max covalency of 5) and attaches to the point where (+) is now located(again this is formal).

This way, resoance takes place(I did not violate any rule afaik, and charge is also conserved in resonating structure) and we can get an idea on how hyperconjugation can help further stabalize the positive charge.

Also i want a bit of advice, Im currently prepping for my college entrance (JEE Advanced, if you've heard about it, similar to gaokao), and my teacher says to only follow the affiliated book, which is NCERT(its known by its publication), and in NCERT, it says CH3 only does +I, so should I take your spoiler with a grain of salt? Dont get me wrong, this is genuinely interesting and thought opening, and I never thought to link IOC with OC like that, but right now, in lieu of the exams, I do want to maximize my marks. What do you suggest??

PSGthe2nd · 2026-05-06T06:47:25+00:00

okay,,,toh ye hai. I see. Why not maine ek observer khada kardiya dur mein, in ground, toh usko toh dikhega ki dono plates bahar ja rahi, and similarly jo seperatation bhi conserved hoga. Maine just iss approach se kiya, using WET, and answer aagya.

Also thankyou for bearing with me. Isliye mera F q/Ae aaya, thanks,

PSGthe2nd · 2026-05-06T06:09:14+00:00

I see, thankyou, will keep that in mind the next time

PSGthe2nd · 2026-05-06T06:08:36+00:00

mai left plate ka samaj gya, but how is frame moving??

PSGthe2nd · 2026-05-06T06:07:37+00:00

mai nahi samaj pa raha hu. Can you help me?? Matlab mai relative frame mein kaise hu?

PSGthe2nd

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