Mathematician's Guide to SJChO / Chemist's Guide to SMO

Added: 2026-05-19

[00:00:03]Hello everyone. Uh, this is the surprise video that I teased at the end of the number theory video. At the time you are seeing this, the junior cam olympiate should be in a couple of days time and all the best to those of you who are taking it. Uh, this video is not clickbait. We are genuinely going to be talking about crossovers between math and chem. But I am not a chemist. I am not going to teach you new chemistry. Uh this is unlike the NOI video where I have actually qualifications to teach NOI. I am not qualified to teach uh SJCH but uh there's some fun stuff to say which is actually more of like focusing on the inspiration behind solutions for math being from chemistry. We'll see that in a little bit. Uh but first again this is the last reminder of one the hard SMO which is my mock SMO for 2026 uh this year with live reviews on the 30th of May which is uh next Saturday.

[00:01:04]Uh if you're watching this just as I release it. Uh we'll review all three sections uh in full. Uh but the question papers will also come with a answering form. You can submit and see how much you score as one final mock paper if you are looking for something more challenging.

[00:01:22]I've also confirmed the dates and timings of the actual SMO. So unless something really weird happens about the SMO, I will be doing the senior and junior on that day at 3:30 and 7 and open on the second day uh at 3:30. Uh which is going to be a bit of a squeeze, but last year I found that it was just enough time for me to get everything prepared. So, I'm going to try again and hopefully the smo doesn't make the um senior even harder because otherwise I'll be really short of time.

[00:01:52]So, we'll see what happens with that.

[00:01:53]But for now, on to the main video.

[00:01:57]And uh I have to make a confession about my history with the chem Olympia.

[00:02:02]Uh I took the junior CAM Olympiad in its first year in 2010.

[00:02:08]uh as a year 2 N US high student I found myself with basically N US high's 1 and a half years of chemistry which is definitely not enough for chem Olympia I didn't prepare for it but uh I mean I was um pretty well versed with school chemistry uh it turned out that I accidentally got a gold for that year with a combination of school chemistry and one might say um uh overuse of math have to solve the chem Olympia questions.

[00:02:41]I basically did not know what an organic compound was. I mean, that's a slight exaggeration, but only slight.

[00:02:48]It turns out that sometimes you can get away with a good result uh just by using math. And I'm going to show some of the similarities that I found.

[00:02:57]Now, the first and most obvious thing is equations.

[00:03:02]There were a lot of questions in chem Olympians that were about balancing equations. And they'll talk a bit about oh this is a acidbased reaction. This is a substitution reaction. This is a redux reaction. Like let's say this is the oxidization of ammonia. Now I did know what a redux reaction was. uh but whenever I saw something like this in my math brain it's an equation where I don't really care about anything other than just trying to find the coefficients which were what number of the questions asked about. So coefficients to me then I think of comparing coefficients which is that for let's say n a = to 2 c for o 2 b = 2 d for h 3 a = to 2d uh and then I'll just uh substitute it in and uh let's say that I make uh c equals to 1 a = to 2 then b = to 3 and then b = to 1.5 okay so I guess if it has to be integer uh then 4 3 2 and six now this is how I solved a fair number of questions and I will say that uh the chem Olympia has matured a lot since I was a student but there are many situations where uh you will still be comparing these right And it has the same spirit as comparing coefficients in math. Here's a question from the junior few years ago.

[00:04:45]Like the chemistry equation. If you don't really understand the intuition of the math equation, it turns out to be okay.

[00:04:59]It turns out to be fine. Even if let's say that you didn't realize what was going on here, which is that this thing can actually be factorized.

[00:05:14]And so you could do something with this to basically help to expedite the process.

[00:05:25]But if you don't know anything about this at all and you didn't even realize that it was factorizable, it is okay to just literally cross multiply and compare coefficients.

[00:05:45]So expand this gigantic mess.

[00:05:56]First I'll get the integer parts which are these. So you have 3 C + 10 B + 7 A.

[00:06:10]Then you have got the square root portions which for<unk> 15 is b + 2 c.

[00:06:24]For the<unk> 35 portions it's 2 a + b.

[00:06:34]And for the square<unk> 21 portions, it's just a plus c.

[00:06:51]Now I need to match it with this, which means I end up with the equations.

[00:06:56]Well, these are all going to be equal to two. And then there's this one on top which is equal to 10.

[00:07:06]Turns out to be pretty easy to solve and it also turns out to be pretty anticlimatic that a= to 1, b= to 0 and c = to 1 and your answer is two.

[00:07:23]Comparing coefficients is something pretty useful in a lot of areas of math because of the fact that your let's say polomial powers of x cannot mix square roots here they cannot really mix and that's actually how atoms in chemistry are like um the original uh expression for atoms in Greek is supposed to mean indivisible cannot be divided it means atoms and prime numbers actually supposed to mean the same thing. Now thankfully mathematicians didn't get prime numbers wrong. Prime numbers are still indivisible.

[00:07:58]Chemists later realized that was a bad name because uh of course atoms do contain electrons, protons, neutrons and uh they are not indivisible per se. But still the name stuck and there is some logic to considering each prime each atom of equations one at a time.

[00:08:19]Now continuing on the spirit of equations in chemistry we have the concept of limiting reagents.

[00:08:27]Now the idea of limiting reagents is that if let's say you are trying to synthesize some compound there is sort of a recipe needed.

[00:08:39]There's a recipe needed and the recipe sort of demands let's say for sulfur and iron to iron sulfide there is a 1 is to1 ratio.

[00:08:49]Now uh of course in 1 is to1 ratios this is referring to the stochometric ratio which is the molar ratio. It is not the mass ratio it is the number of moles. So because of the different molecular masses, you are not going to get the same number of moles of sulfur and iron here and so there's kind of a leftover sulfur because there is not enough iron.

[00:09:20]This is the limiting reagent.

[00:09:27]Now there are many examples in math where you can actually use this intuition of the limiting reagent. One example can be as simple as the GCD. If let's say that I have the GCD of a few things if I want to find the GCD I can look at the prime factorization and I can look at each of the primes take the smallest one. So in this case let's say for two there are powers of two here the limiting reagent in this case is this guy because it has the smallest power of two which means that your gcd will have uh two square inside and you will easily also check that there is nothing else that repeats throughout. So the only thing you get is two square.

[00:10:25]So that's one simple way that you can think about limiting reagents in math.

[00:10:29]But there is perhaps also a more familiar example which is the questions on how many zeros are there at the back of this or that weird thing.

[00:10:44]If you were to have the so-cal primary school version of the question of how many zeros are there at the back of let's say 100 factorial, right? A very traditional question.

[00:10:58]Essentially what your teacher for Olympiads would have told you is that oh the number of zeros 10 is 2 * 5 and obviously there are more multiples of two than multiples of five in 100 factorial. So therefore you count the number of fives.

[00:11:18]That's essentially the intuition that five is the limiting reagent.

[00:11:28]It is pretty obvious in that case that five is the limiting reagent.

[00:11:34]Similar to if let's say I asked you um you know that bottle of silver nitrate that you have in your chemistry lab. Do you think the silver is more expensive or the nitrate's more expensive? I think it's pretty obvious, right? Uh but there are situations where it is actually not that clear.

[00:11:49]So if I ask you how many zeros are there at the back of this 1,00 choose 48. I just put a random number there.

[00:11:58]I don't think it is that clear which is going to be the limiting reagent, the two or the five.

[00:12:04]And so you really would need to check both. So 1,00 48 if you were to write it like this.

[00:12:20]It's still not that clear uh whether there are more tools or fives.

[00:12:32]Uh I would say that it's probably easier to count it by not writing it in that form.

[00:12:40]just leave it as factorials and then we count the number of twos and fives uh in those factorials.

[00:12:51]So the fives we know that the way we count it is 1,00 factorial as 200 fives 40 more multiples of 25 divided by five again eight more multiples of 1 to5 and then divide by five again every time you divide you round up.

[00:13:19]So 249 on top and below we do the same thing. So let's say 952 uh divided by 5 round down 190ide by 5ide by 5 round down and then do the same for 48 divided by five round is 9 and then what?

[00:13:42]What is this?

[00:13:45]246.

[00:13:50]So there's three more fives in the top than the bottom.

[00:14:00]So you have a five cubed available.

[00:14:05]Uh you can then do the same for the tools, right? So the tools will sort of be mildly more annoying because there are well more calculations to do but uh you are just rounding now each time 952 divided by two so uh I'm writing them uh just like this because it is a bit easier if we just uh subtract them off like directly and Then for 48 you have 24 12 6 3 and 1.

[00:14:55]So most of these actually going to give you nothing. So uh there's nothing here, nothing here, nothing here, nothing here, one here, one here. So you actually only get two squared. There's only two more tools in the top than the bottom.

[00:15:12]So in this case two is the limiting reagent.

[00:15:16]You have got 2 squ and 5 cubed. And so therefore there are two zeros at the back and the extra five could not be used.

[00:15:27]Uh this is often a useful thing in math whenever you're like I need this and this and this in order to get something.

[00:15:35]Occasionally you'll be able to use this intuition of saying that okay it is limited by this one thing that I have not enough of whether it be abstract in terms of like I need more information about the variables or whether it be literally just counting something like in the prime factorization.

[00:15:58]Now the next idea is of a catalyst.

[00:16:03]We know what catalysts are and uh I am very familiar and probably many of you are familiar with having your uniform stained by KMn4.

[00:16:14]A catalyst of course in a chemist in a any reaction is supposed to be one it speeds up the reaction and two it is itself not consumed by the reaction. So of course uh that means that if you were to have some reactions that are catalyzed by platinum uh it's fine because platinum although very expensive if it's the catalyst it doesn't get consumed so we just use it.

[00:16:51]There are many math problems which sort of look very scary and what they really need is a catalyst.

[00:17:03]A good example is this equation down here.

[00:17:11]The left side looks like one ginormous mess.

[00:17:15]It does look like however not a random ginormous mess because all the exponents are powers of two. So it's not entirely random ginormous mess.

[00:17:25]You see a lot of uh threes and twos and then you see five and you're like oh five is actually 3 plus two as well.

[00:17:33]So that's not at random. But I still don't know how I can simplify this left hand side.

[00:17:40]But here's when the idea of a catalyst comes in. Now a catalyst in math uh as some uh professors uh sometimes joke.

[00:17:49]They say that everything in math can be solved by either multiplying by one or adding zero. Uh that's of course a slight exaggeration, but I would say that it has happened uh way more often than you would expect it to. Here our catalyst is to multiply by one that is one but uh I don't have to worry about one being consumed because in multiplication times one is actually doing nothing right so it's just sort of to overcome the activation energy of having to expand all of this because now I can just use my repeated difference of squares and go all the way 2 3 to the 2 to the n + 1 - 2 to the 2 to the n + 1 9 to the 256 is 3^ 2 to the 256 is 3 512 uh and I'll leave you to uh do a quick check for the fact that uh 512 is 2 ^ of 9 and so therefore you should let n be equals to Yeah, convince yourself that's correct.

[00:19:06]Uh this is the idea of a catalyst. Uh another very common uh instance of this in the SMO senior and open is this very famous cosine 20 cosine 40 cosine 80.

[00:19:22]Uh this is a very common combination because uh of the fact that it is nice angles with also no special angles among them and yet it simplifies nicely with the use of a catalyst and the catalyst is I am going to multiply by sin 20° at the top and bottom from There you can guess what's going to happen.

[00:19:55]Why did I put sign 20°?

[00:19:58]To use the sign double angle formula.

[00:20:02]So we combine them like this and then the catalyst kind of as in the definition of catalyst uh does not get consumed or created uh because sin 160° and sin 20° are the same. So it disappears at the end and you're just left with 1 /8.

[00:20:29]I guess uh once upon a time uh for any of these famous chemical reactions that have a high activation energy uh people did not know how to make them go faster.

[00:20:41]But we know that uh catalysts are something that are discovered over time sometimes uh systematically by understanding the mechanism. sometimes by accident. So likewise for a math expression, see if you're able to find some sort of catalyst to make it simplify. Uh we do not need to worry about whether that catalyst is consumed or not. We just want to make it simplify. And if it just cost us a tiny bit of extra trouble, uh I think that is still worth the effort.

[00:21:14]Now the next idea is a bit more left field. Uh it is this notion of well before notion uh it's about equilibria.

[00:21:26]Uh there are equations which in chemistry we write with the single arrow right and we write it with a single arrow because essentially it almost entirely will go to the other side. There's very little left of the things on the left. Uh unless of course the limiting reagent is causing some wastage. If assuming you put things in the right ratio, uh everything goes to the other side like the typical acid base reaction.

[00:21:56]Then you have some which are basically partial equilibria where it doesn't entirely go to the other side. You end up with a mixture of both.

[00:22:06]And we are told that there is this equilibrium constant for any such reaction.

[00:22:16]The equilibrium constant uh will be something in this form uh meaning that this is constant uh where your abd are concentrations.

[00:22:31]So of course there are the cases where you use some solids or you have some gaseous component uh and that will be for your chemistry teacher to talk you through the details. But let's just look at something like this.

[00:22:43]Essentially when you start off this here all of it is A and B. You have no C and D. So one could say that this K is actually going to be zero as in this expression and it is going to go to infinity if the reaction completes entirely such that the A and B there is zero of it left. So the concentration is zero.

[00:23:31]All right. So this is of course also going to be increasing as the reaction continues.

[00:23:43]Now the question is therefore uh often going to ask you okay given that this is the equilibrium constant how much of let's say compound C is going to be synthesized and that's basically an equation where you know that there must be an answer because it is between 0 to infinity also it is unique because there is no such thing as there are multiple different equilibria.

[00:24:18]Now we're not talking about uh the multiple different uh like places where the activation energy messes things up.

[00:24:25]We're just talking about with respect to the equilibrium constant.

[00:24:29]Uh I remember that uh as a student uh I decided uh you know we we are a little bit too clever for our own good, right?

[00:24:36]I decided in the middle of my chemistry group project presentation on equilibria to explain that actually do you know that uh the reason why an equilibrium always must happen is due to the intermediate value theorem and I don't think my chemistry teacher appreciated it very much.

[00:24:52]Well, the intermediate value theorem and this idea of it being a monotone function that is continuous is useful for solving a lot of SMO questions in a way that you often don't expect.

[00:25:11]Now, here's an example of that uh simple equation and if you worked hard enough to rearrange it, square it, rearrange, square it again and factoriize, you will get the answer. But I can just tell you a much simpler way which is just guess.

[00:25:31]Why guess? Well, you can always guess.

[00:25:34]Sure. But the reason why I want to guess here is that this is an increasing function.

[00:25:45]Now it's an increasing function. It also clearly uh when I pick the minimum x possible which is x is 47 over 3 for it to be defined it's clearly less than six and as x goes to infinity it of course goes to infinity as well which means that there will be exactly one point in between where it hits six that means the graph is just going up uh not necessarily in a straight line.

[00:26:18]But if the graph goes up somewhere, it's going to hit.

[00:26:22]And you can find it by basically using a binary search. Now I'm elapsing into NOI. Uh by playing a higher or lower game where you're just guessing that okay, uh let's say that I want to try to make this happen. Uh let's try uh 16.

[00:26:41]And then if I try x= 16, I get 1 +<unk> 15. uh and then that is uh bit too small but it's already quite close to five so I go up again and if you went up again you just end up with 2 + 4 = to 6. So that's already your answer without actually too much trouble at all.

[00:27:13]So this idea can be used in general to approximate roots of a polomial. This includes what happened uh as a problem in the senior last year 2025. Uh but if let's say that you just have any uh equation like let's say x cub - 3x^2 uh + x + 1 = 0. Let's say okay no this has too obvious a root. This has a root x equals to one. Let's uh just change this to two.

[00:27:50]Now the idea is that you can actually estimate the roots just by saying okay when x equ= to 0 it is two uh when x equals to 1 uh this is going to be you know I think I used a pretty bad example uh let's change this to uh five so that it's not so obvious. Okay. So when x equals to 1, you get -1. Now this means that there's actually a root between 0 and one.

[00:28:25]You put in two, you get -8, you put in three, you still get something negative, which is 13.

[00:28:42]-10. You put in x goes to 5, you get seven. So there's another root between four and five.

[00:28:51]Uh and you backtrack to minus one, you get -5. So there's a third root between minus1 and zero.

[00:29:03]So this is actually what we can uh use to estimate roots for a cubic. Let's say where you know the ES3 roots. Uh this is the idea of last year's uh senior round question. Just putting in some values you know that somewhere in between there is this equilibrium.

[00:29:23]It doesn't help you to solve for it. If let's say that the values of x are all irrational then too bad. But it allows you to at least know it happens and have some range of when it happens.

[00:29:42]Now keeping to this physical chemistry world uh there is also the second law of thermodynamics. Now the second law of thermodynamics basically states that uh entropy keeps increasing uh and entropy is actually the amount of randomness uh where randomness has its own computation which I'm not going to get into.

[00:30:08]I found this to be a pretty useful analogy towards uh inequalities for quite a long time especially AMGM. So I'm going to uh use this as an example because it checks a few features.

[00:30:25]Now when you say entropy keeps increasing it means that even though let's say you satisfy the conservation of mass the conservation of energy not all reactions can occur because it needs to also go in this one direction.

[00:30:48]uh it needs to go in this direction where things get more random and so this is sort of like imagine a ball rolling downhill.

[00:30:57]Uh of course that would be physics that is converting your potential energy to kinetic energy but uh I am not going to go into physics. Too many different things are getting into my video at this point. Uh but this one time I'll allow it. Imagine that it's also like a ball rolling downhill. The ball cannot go back up the hill again without a force being applied to it and then the force also needs to come from something else's potential energy blah blah blah.

[00:31:25]Now think of it this way also when you are applying AMG. Now if you see this here and you see this here this looks like the most difficult question ever in the world.

[00:31:38]Uh I would say a couple of things. First is that if you're doing this in round one, you should probably just guess that ABCD equal uh to the whatever the fourth root of 20 26 over 4 is. And then that would give the potential answer since I have carefully chosen all of these to be just degree 4.

[00:32:07]Uh that would be correct.

[00:32:10]But if you wanted a proof or if you just wanted to know where did I come out with this, the answer is that I am often inspired by this notion of entropy which is that I basically just jumbled everything up. Uh if you notice there are a total of six A's, six B's uh and uh if you continue counting there are also uh six C's and six Ds.

[00:32:55]But here was all four. Well, never mind.

[00:32:58]Uh we know that we can adjust the correct number of moles of everything by multiplying by three halves. So that now we have got six of everything.

[00:33:10]And basically I'm saying that look uh AMGM says that if you want to go downhill, you want to get something smaller, it comes from when you mix it up and you will keep getting smaller as you mix it up. It keeps going downhill.

[00:33:26]Now this is definitely more mixed up and if you wanted to justify using AMGM how you work is just a squ bc is less than or equals to 14 of a a b c this is a b b c and so on.

[00:33:53]And if you use each of these, you are guaranteed that you will have six of each, right? Because uh I have a6 b6 c6 d6. So which means the sum is going to be that this weird thing is less than or equals to one quarter of six of each of them which is the 3039 as advertised.

[00:34:26]Um this also conforms with uh so-called Merhead intuition. If you uh are familiar with Merhead, uh the notion of majorization basically can also be thought of as this second law of thermodynamics turned into a more precise mathematical term for uh homogeneous inequalities.

[00:34:53]Next up is a more experimental thing which is actually more like your SJCH round two uh which I'm going to just dab into a bit of separation techniques. Now separation techniques uh you have all your filtration, you have all got your crystallization, your uh distillation etc. Uh and uh I'm just showing a filter paper because that was the easiest thing to Google.

[00:35:21]But basically what you want is that you have a mixture and you would like to separate a few different types of things.

[00:35:34]Now this is very commonly seen in questions on combi with uh coloring arguments which I have uh briefly mentioned in my combi video.

[00:35:47]So uh how many ways are there to delete a unit square from a 11 by11 square board such that the remaining board can be tiled by 3x1 and 1x3 rectangular tiles.

[00:35:59]Essentially uh what you would like to do is to distinguish the types of squares in the 11 by11 board. I'm not going to draw the 11 by11 board. Uh I am going to uh leverage technology.

[00:36:15]So, let's make an 11 by1 11 board. Uh, and let's make it look more squarish.

[00:36:22]Okay, that's quite squish.

[00:36:28]The idea is that I want to be able to separate my different types of squares so that I know which are legal and which are illegal to remove slashdelete.

[00:36:40]And I'm thinking that in each 3x one cell sorry 3x1 tile we can treat it as there are three types of tiles inside which I hope I can just number one two three.

[00:36:53]So if I can find what is the odd one out type of tile that will answer the question.

[00:37:01]So to that end I am just going to separate them like this in the most boring way imaginable which is basically to just uh ensure that every of my potential 3x 1 or 1x3 tiles is going to look like one two and three in some order. Now of course what's really happening is that these are just diagonals. So it's not very hard. I'm just going to fill this all up. Uh, delete the last extra row which is not needed.

[00:37:38]And so if we were to count this, we would see that okay, let's say for the ones 4 7 10 9 6 and 3 there are 40 of them.

[00:38:00]For the tools, there are 41 of them. And for the trees, likewise, there will be 40 of them. So therefore, you must remove a two.

[00:38:14]All good, right?

[00:38:17]But if you actually try to remove those tools, you realize that it doesn't work yet because well, I mean, you can try to just say remove this tool and then cover the rest.

[00:38:34]And if you try to do it, you'll realize that you do get stuck. Let's say for example uh at the bottom here now you're trying to do this. You will end up with uh well let's just say that it doesn't seem to work and you are not 100% sure why.

[00:38:51]The explanation is very simple which is that like separation techniques you have more than one separation technique or you can apply the same separation technique more than once like fractional distillation.

[00:39:04]So you can actually say that I have got extra tools. So the one to remove must be a two. But it must still be a tool if I were to just take this entire grid and reflect it. Is it not letting me right? Reflect it or rotate it in any direction.

[00:39:34]So if we look at let's say each of these rows we also realize that oh they're not actually rotational symmetry because these are the this is the line of symmetry.

[00:39:45]The only ones that actually obey this rotational and reflective symmetry are these guys.

[00:39:58]So now you're only left with nine of them. And you can go and uh check that actually all of these nine are going to work. It is not very difficult to find a way to do the tally.

[00:40:16]Now I just released a video on number theory a few days ago. So uh I did mention about mod filtering. Now mod filtering is also another sense in which you can think about separation techniques. So the very typical sort of examples of like okay if let's say that I have got x^2 uh plus y^2 is equal to uh 1 one one uh this is not going to work because if you take modulo 4 your squares are only zero or one but this one is 3 mod 4 so it's unable to combine. Now you can think of it that way which is basically that your uh modulos also serve as ways that you can separate the two sides of equation sometimes and confirm that they can never agree therefore get ending up with no solutions.

[00:41:11]And just like your uh separation techniques, you need to know what solvent dissolves certain things, right?

[00:41:16]It depends on whether this let's say is a metal or whether this is a polar or non-polar solvent. So likewise your mods, you will gain experience in knowing what are your options available that will be able to have the desired effect.

[00:41:35]Um, I'm sure many of you were pretty offended when I said at the start that uh I basically knew almost nothing about organic chemistry and accidentally got a SJCO gold. Uh, because right now, unfortunately for you guys, you needed to know a lot of organic chemistry, right? Okay. So uh organic chemistry is something that uh there is way too much to talk about and uh I will say that uh it is kind of annoying precisely because there are so many functional groups and there are so many specific properties of particular uh organic compounds like let's say a benzene has this whole long list of things that like one extra carbon changed everything.

[00:42:24]I'm not going to help you with that.

[00:42:25]Sorry. But I am going to say that this has actually a lot of resemblance to graph theory.

[00:42:33]Now, graph theory is also like that where you can have one specific type of graph. There are lots of names of specific graphs with specific properties.

[00:42:45]because there is so much of it. Uh you do need to know some theory but sometimes you just want to know some common sense and sometimes there are also some general properties that hold true.

[00:42:56]For example, uh for your alkanes and alkenes, you know that for your alkanes, the general formula is CN H2N plus2 and for alkenes it is CN H2N.

[00:43:17]And this also applies even if let's say it goes beyond whatever uh this again the first search on Google gave me because if let's say that you had uh butane uh this is also going to be uh C4 H10 if you had uh what's this called again uh isopropane right uh isop propane is also C4H10.

[00:43:52]Now you might be thinking that uh this is like a chemistry property uh and that's half correct. It's because of the four electron pairs that you want your eight electrons uh for each of your carbons. Therefore, it needs four bonds.

[00:44:09]But another way of thinking about it is actually as a graph.

[00:44:14]Because when you have got n carbon atoms at first they basically will have four n electron pairs.

[00:44:30]Now each bond basically is going to uh well uh reduce the need for one hydrogen well one less hydrogen on each.

[00:44:52]And in order to get let's say uh n carbon atoms in this general form you are going to have n minus one bonds because in graph theory terms it's a tree.

[00:45:18]Therefore, you will have uh 4 n -2 * n -1 which is 2 n + 2 hydrogen atoms.

[00:45:28]However, if let's say that you have uh cyloalk like let's say if you have uh cyclopropane now you know that of course in cyclopropane uh which I'm sure is not very stable and therefore uh there is a whole lot of other chemistry to say uh but we just know that this cyclopropane is going to only have uh six hydrogens's and that's because every extra bond that is formed is basically an extra cycle in a graphical sense which then is an extra edge that removes two more hydrogens.

[00:46:19]So graph theory is a very useful way to think about some of these things. I don't think it's going to help you very much for chem Olympia but it's just an interesting thought and uh I will leave this on screen to let you think about for a bit. This is a junior round two question which uh does not have much to do with uh alkanes and alkenes uh but it does philosophically have a similar like mini graph theory kind of flavor which you might be interested to think about.

[00:46:57]I'm not done yet. Uh surprisingly uh I'm not going to talk about isomers specifically. Uh there are two types of isomers which there are the geometric isomers and there are stereo isomers.

[00:47:11]Let's start with the geometric isomers which often times we'll be focusing on cy trans isomers.

[00:47:20]Um the reason why uh these are isomers meaning that they are considered as not the same is that uh chemically speaking uh they can have different properties.

[00:47:35]Just imagine that this chlorine is just very big compared to the rest and therefore in terms of let's say whether a reaction is possible uh it can block things. Um there's also of course the fact that it will become potentially slightly polar because of the fact that your uh is your chlorine electrofilic or electrophobic? I do not to remember but uh whichever direction it is, it can become polar versus non-polar.

[00:48:05]Uh I'm also not sure if chlorine is the right idea. Sorry. But uh the behavior is similar and yet a bit different.

[00:48:15]Now this reminds me a lot about uh config issues in job specifically uh external bis sectors versus internal bis sectors.

[00:48:33]Let's try to draw this Now the external bis sector of angle A on CA produced uh produced means extended as well.

[00:49:00]So what does that look like?

[00:49:04]It looks like if you extend it and then biseect the exterior angle something like this and then uh yeah the question is drop another perpendicular find af* fb The properties of external bis sectors are probably are not that well known. Uh we have mentioned the external angle bis sector theorem is the same as the internal angle bis sector theorem.

[00:49:47]But there's also more.

[00:49:53]Essentially whenever you have internal and external they come in pairs.

[00:49:59]And so it is sort of that whatever properties hold for one you can see what properties hold for the other and also what things are sort of the opposite of what held for the other so-cal isomeic construction.

[00:50:21]This would be the normal internal angle bis sector.

[00:50:26]It is well known that if you have the internal angle bis sector, what is going to happen is that you have hit the arc midpoint because if you were to just uh use your angles in the same segment, these angles are going to run to here and here.

[00:50:44]Meaning that this is isoclesles, meaning that these two arcs are the same.

[00:50:50]Okay, I'll erase that because it's not especially important right now.

[00:50:56]But did you also know what it means for the external bis sector?

[00:51:04]Well, the external bis sector is perpendicular to the internal bis sector.

[00:51:12]So if the external bis sector is perpendicular to the internal bis sector, that would mean that this is actually going to be the diameter And if this is going to be the diameter, it means that you also will have that E is the midpoint of major arc BC instead of minor arc BC.

[00:51:46]So that becomes the main characterization that you actually need.

[00:51:51]The actual line's not very important.

[00:51:54]The more important part of it is that oh so because of this I know that my uh be is equal to C. Perhaps I did not draw that very well.

[00:52:14]And uh in fact I can draw it better now that I know where it's supposed to be.

[00:52:19]Uh, it looks like it should be even closer to A based on my drawing.

[00:52:33]That seems a bit better.

[00:52:42]And if you have got this, it is going to give you the idea of finding some congrent triangles because you have now got equal stuff.

[00:52:57]You have also got this uh I should probably draw it again but never mind. uh you have also got this right angle here.

[00:53:09]Fine, I will draw it again.

[00:53:16]So since I know that I actually want uh E to be the midpoint of major arc BC, why not I just um draw that first so that uh that is never going to be messed up.

[00:53:36]And then uh AB is larger. So I can draw it something like this. So now I know that uh it is going to be much more accurate.

[00:54:22]to finish off the location of the actual uh congrent triangles.

[00:54:33]I guess what we can use is that we can go back to that external angle bis sector and use that as a way of getting a reflection because a reflection is how we are able to get congrent triangles. So, I'm just going to reflect the f over to there.

[00:54:59]And we're going to find that uh by doing the reflection uh bf is congruent to cf prime.

[00:55:10]Sorry, CF prime e Right. So the implication is that because we have done that uh BF is equal to CF prime.

[00:55:35]So uh a which is 75 minus a f is equal to a c which is 53 plus afr prime. These are equal. So af and afr prime are half the average and your answer is 64 * 11.

[00:56:13]Now after geometric isomers we have uh stereoism or enantus.

[00:56:21]I thought of this example because I've been running a call lately and then uh I took some Zerek and then I remembered vaguely uh something I learned about uh how you could have stereoisomers uh which have different properties uh and in the case of Zerek there's actually these two isomers uh and one of them is the thing that gives it its property of uh helping with a cold and the other one is actually kind of not the active in entrema. It's not the one that uh is doing anything uh for my code. Uh a lot of times it's just there because it is really expensive to separate the two and one is useful, the other is harmless. So let's just have both and double the dosage.

[00:57:05]That is legitimately uh something that is true also of quite a number of our medications. uh when we are saying drowsy and non-drowsy it could be that one of the inential meals is a drowsy inducing one it doesn't actually uh treat you and then the other is the one that treats you so therefore to make it non-drowsy they just remove the other one that could be sometimes how it's done now the thing about this stereo isomer is that basically you have this uh carbon atom and then you have got these who basically swap places, right? One's going in, one's going out. And it turns out that uh this leads them to have some different properties just like the geometric isomers just now.

[00:57:56]But obviously uh if we are familiar with uh carbons, we know that there are often only going to be two uh stereoism uh for each of these uh carbons where there are these uh four different um groups.

[00:58:28]Which leads me to a combi question.

[00:58:35]How many ways are there to color the vertices of a tetrahedrin using five colors where rotations are treated as identical?

[00:58:52]In other words, it's not just 5 to the^ of four.

[00:58:59]How many rotations are there though?

[00:59:04]So there are five choices for each vertex.

[00:59:10]Obviously, it's not 5 to the^ of 4 divided by some factorial because that's not even integer.

[00:59:22]Then some of you uh may think oh yeah actually this is just a stars and bars question in disguise because it just depends on how many of each of them there are. All right. So uh if let's say that there are a total of uh four vertices going into five boxes right because of the five types this is just the same as you arrange four objects and four partitions and so you have 8 choose 4 which is equal to 70.

[00:59:56]Now unfortunately uh this is also not correct and legitimately the easiest way that I could explain this symmetry is by talking about this uh stereo isomers which is that if there are four different colors there are two different permutations.

[01:00:34]Now why are there two different permutations? A simple way to imagine this is that if let's say that you have uh this is your tetrahedrin you have for any permutation you can choose the base in four ways and then once you've chosen the base you can rotate it in three ways.

[01:01:07]And so in order to get uh 4 factorial is 24, you can drop it into two groups of 12.

[01:01:17]Now if you're familiar with group theory, we can refer to this as actually orbits and then something called Burns size lema is going to come in uh which is beyond the scope of this video for sure. But we can say that for four different colors there's two different permutations. But with repeats uh any repeats you no longer have this right because if you have any repeats then you have got just a total of 12 or fewer permutations and you know that one of them you can already rotate in that many ways. which is why we do not normally have stereo isomers unless there are four different groups coming out of your carbon.

[01:02:05]So that's just one permutation.

[01:02:08]And now we can count how many ways are there to have four different colors.

[01:02:15]Well, to have four different colors, it means that you must have 21 1 0.

[01:02:25]Sorry, not 211. uh 1 1 1 0 which means of course that there are just five outcomes.

[01:02:39]And so this 5 * 2 is 10 different permutations with four different colors.

[01:02:51]The ones you repeat however can be is it uh all four are the same? Is it three and one? Is it two and two? Well, we don't need to worry about that because uh with the previously counted uh 8 through 4 is 70.

[01:03:10]We can say that uh the remaining 65 ways are going to all just have a repeat somewhere.

[01:03:20]Right? So our answer in this case is 10 + 65.

[01:03:30]Incidentally the reverse question of uh how many different let's say uh possible chemical structures are there for C6 H12 or C6 H12 or C6 H10O. uh any of these turns out to be a pretty tricky question and the aforementioned Burns size lema is actually uh used by chemists to determine the number of different uh either geometric or stereo isomers or just different uh chemically named compounds corresponding to a chemical formula.

[01:04:13]So finally we will bring up uh bond angles which is associated with your VPR.

[01:04:23]Uh this is just an excuse for me to talk about vectors. Uh this is one of the just obvious crossovers. It's not very deep. It's just that uh whenever you have any 3D geome uh vectors is a good way to handle it because not always are you going to have anything nice about them. They are just going to be in some sort of let's say uh geometric structure and I want to find some angles some lengths and I do not expect that uh there's any good reason to use like uh normal geometry. So uh factors can come into play.

[01:05:02]Uh all those of you who have uh seen this 109.5° many times I don't know whether you find it familiar or not. Uh but 109.5° actually apart from being your bond angle in a tetrahedrin it's it does have a simple derivation and a simple mathematical expression for it which we going to see.

[01:05:28]small issue is how am I going to actually model a tetrahedrin using coordinates and uh small trick to deal with a regular tetrahedrin at least is that start with a cube.

[01:05:49]So start with a cube and then just take the alternating points soal and you get a regular tetrahedrin. All the distances between them are exactly square <unk>2.

[01:06:09]In this case here we are saying that the carbon atom is going to be at equidistance to all of them. So that should just be at the center of the cube your carbon atom and we can give them all coordinates very easily. So if I just uh have this is the origin then as one would normally label them uh this is what they would look like and the carbon atom would be just half half half being at the middle of the cube.

[01:06:47]Sorry, what happened to my label there?

[01:06:48]101.

[01:06:56]Now supposing that I want to find what this bond angle is with vectors. That is actually not very hard to do. Now we can just take these as the two vectors. I'll call this A and B.

[01:07:09]And as we know to get a bond angle is just the dot product divided by their magnitudes. So CA cross CB sorry dot CB over the product of their legs.

[01:07:32]CA is equal to half half half and CB is equal to uh negative half you do the dot product it would begative - half * half * negative half half * half that adds up to minus one quarter and then the magnitudes of both of these are definitely the same which is uh root of 3/4 they multiply to 3/4s and conclusion is that actually uh your mysterious 109.5° is the cosine inverse of - 1/3.

[01:08:19]Sorry if everyone already knows this but I think I went through the whole of school life without anyone telling me that 109.5° actually is uh from vectors.

[01:08:27]It is certainly not that the hydrogen atoms just magically decided to form a 109.5° angle. It is just because of the electron pair repulsion pushes them to be as far apart from each other as possible. And it just so happens that uh that angle is cosine inverse of - 1/3 which is to one decimal place 109.5°.

[01:08:52]Now SMO doesn't have too much 3D geometry. There's a lot of like vector questions where they just in your face tell you here are vectors or let I jk be this and that. Uh I just want to uh remind you that 3D geometry can be tackled by vectors even if vectors are not mentioned in the problem.

[01:09:13]So this is a relatively simple question uh but just a simple illustration as well where I have a tetrahedrin OA is perpendicular to the plane ABC. Uh these are my lengths.

[01:09:25]I can set it up with coordinates starting with uh triangle ABC.

[01:09:31]As usual don't trust their diagram.

[01:09:32]Their diagram is really bad for the SMO problems. Uh it has not improved much since 1997.

[01:09:43]This is supposed to be isocive right triangle.

[01:09:50]Uh, OA just comes out of it up.

[01:09:55]So, we can assign all of them coordinates very easily.

[01:10:01]Um the nice thing about doing uh something with coordinates is that you don't have to worry about noticing things. Uh so if there was nothing to notice then you will save type. Uh the flip side of course is that if there was some nice features you may miss it because you're just writing down a bunch of numbers.

[01:10:21]So uh M and N are the midpoints of OB and OC.

[01:10:26]So uh m is going to be just one and n midpoint oc is 011.

[01:10:41]If you had uh done some geometry, you may have noticed that this means uh amn is equilateral, which would make the question very easy.

[01:10:54]But let's say that we did not notice that then we can use our cross product all the same in order to get the area.

[01:11:08]So let's just do that as a finale for video.

[01:11:16]The cross product of course uh is formula that I assume you know how to do. So just for each of them just ignore that row and then take the determinant uh so the cross difference of the other two and then negate the middle one you will get -1 -11 giving you an area of <unk>3 /2 consistent with this being an equilateral triangle with length <unk>2.

[01:11:44]So these are my few ideas uh of overlaps between uh Chem Olympia as well as the SMO. Uh for all the chemistry mains out there, please tell me if I have said anything too wrong about any aspect of chemistry. Uh I am very far removed from this. But uh on the best part, I am fairly certain I've said all the right things. Nonetheless, uh if you have any other suggestions of what else could be a crossover, let me know. And of course, all the best to anyone who's taking the chem Olympia in a couple of days or taking the SMO in a couple of weeks. So, thanks for watching. Bye-bye. And see you soon for the hard smo.