Hello and welcome to this session. In this session we shall study the concept of samples and populations. Suppose in a city there are various ice cream parlours which sell variety of ice creams like vanilla, butterscotch, almond, chocolate etc. A journalist wants to know which the most famous ice cream parlour is so that he can write an article on it. What will we do? He will ask the people of the city which their favourite ice cream parlour is. But question arise will he ask each and every person in the city to know most famous ice cream parlour? If no, what will we do? The answer is he will select any number of people from the city and ask them about the parlour and will draw a conclusion about the most famous parlour in city. The name of the ice cream parlour which most of the people will take will be the most famous parlours. Suppose there are 300,000 people residing in the city. The journalist chooses any 1000 people to answer the question. Then 300,000 is the population and 1000 is the sample. Thus we can say population is the entire group of people or objects and sample is a part of the population. Remember it is not necessary that population should include people only. Population can be of any item or object, living or non-living. It depends on what do we want to know. For example a factory manufacturer of toys, the owner wants to know how many of them are defective. So he chooses any number of toys and asks how many are defective and draws the conclusion. In this example the toys manufactured is the population and the chosen toys from the sample. Also we know that the sample chosen is random. It means we can select any object or person without any preferences and this is called choosing the sample randomly. Now in the above example we chose 1000 persons from 300,000 people. It means size of the sample depends on the size of the population. It means if we are having 200 objects as population then we can choose sample of 15 objects. We can summarize the above as a sample is some portion of a larger group called the population. Since it is impossible to examine every item in the population a sample is selected to represent the population. After the sample is analyzed the conclusions can be drawn about the entire population. The larger the sample size the more closely it approximates to population. It means our conclusions regarding the entire population will be more accurate. Now we are going to discuss reasonable sample. Now question arises how can we select samples which gives us the most accurate conclusion or prediction. Chosen sample gives us the correct inferences about the population if it satisfies the following conditions and the first condition is sample should be selected as random, second it should be a representative of the population, thirdly it should be large enough to provide accurate data if the sample satisfies these three properties then it is the best sample chosen from the population. Let us take an example, identify the sample and the population and examine whether the sample chosen is reasonable. The school principal wants to organize a total cultural and technical first for the students. He randomly asks 100 students whether they would like to participate in the event or not. Population of the students in that particular school is 2000. Here all the 2000 students form the population and randomly selected 100 students make the sample. So here we have population which is given by 2000 students and the sample that is 100 students. To check whether the sample chosen is reasonable we must see that these three conditions are satisfied by the sample that is sample should be selected at random. It should be a representative of the population and it should be large enough to provide accurate data. Now we know that here the sample is randomly selected so the first condition is satisfied. Now 100 students of school can be of any grade and these students are chosen from the same school so it is a true representative of the population. So second condition is also satisfied and we are choosing 100 students from 2000 students so it is a large enough sample for the school population. Therefore the third condition is also satisfied since all the three conditions are satisfied thus we can say the sample chosen is reasonable. And this completes our session. Hope you enjoyed this session.


And stuff just went from bad to worse for the Baltimore Ravens You know what and a lot of y'all may disagree with this, but this is how I'm feeling right now For me, this honestly hurts worse than losing to the Chiefs It really does and I know it's a much different situation and the Chief that was obviously such a huge game But I think it's the way that we lost to the Chiefs too since we were losing like for so long And the Ravens were in it but never really in it in it and they weren't being themselves But this this hurts me more than the loss to the Chiefs did because You had a guy man You you you had a guy that was special you had a guy that was proven you had a guy that showed you literally showed you Showed you so much growth from his first year to his second year. He showed you so much consistency From his first year to his second year. He showed you multiple times like hey, I can make some stuff happen He had been with the Ravens previously But then he went over to Michigan made stuff happen over there boom like that instantly right away quick faster than a hurry Had a huge impact on them Then Ravens are like, you know what? Hey Mike McDonald come back. Come on back Come come to the NFL come be our defensive coordinator. How about it? come try it out for a little bit see how you like it and He not only got the job done with the Baltimore Ravens, but he excelled with the Baltimore Ravens And he was just amazing like he was somebody no matter what About nine point nine times out of ten. He was gonna be on point About nine point nine times out of ten He was going to not only do his job, but do a phenomenal job at his job now. That's gone Now that's done Now and congratulations to Mike McDonald first and foremost. Well, I guess it's kind of like second But first of all most congrats to Mike McDonald. We're happy for him We were definitely happy for him because he earned it. He earned his job and just reading the report Listen to what the report says from Ian Rappaport. He says sources The Seahawks are set to hire Ravens defensive coordinator Mike McDonald as their new head coach Seattle was willing to wait for him if Baltimore made the Super Bowl So you see Seattle knew they knew yeah, they had interviews with Aaron Glenn Yeah, they had interviews with Ben Johnson. Yeah, they had interviews with a Dan Quinn. Yeah, they had interviews with other people But that is significant. The Seahawks were willing to wait for him if Baltimore made the Super Bowl So they were willing to put the the hiring process on pause just for Mike McDonald because they knew they knew how special he was and Continuing with the report it says instead their season ended last week, of course and he takes a big-time job He's currently in Seattle on his second interview And it says the Seahawks brought McDonald in with the goal of hiring him the top target all Alone for a deal to expect it to be finalized So that's that man That's that And we got a lot to talk about right now.  We don't like this news at all like it for Mike McDonald personally But coming from the Ravens fans perspective We don't like it But you can still leave a like on the video leave like on the video because it helps out the channel a whole Lot and somebody that helped out the Ravens a whole lot was Mike McDonald But now he's gone something that we talked about with Mike McDonald And he's just short tenor with the Baltimore Ravens He had a huge impact his short tenor as defensive coordinator with the Baltimore Ravens He had a huge impact with the Baltimore Ravens and We saw growth. We saw accountability. We saw development personal development. We saw him develop Players we just saw so much from him in such a short time But again, it was a tremendous impact that he had on the Ravens But again, the thing we saw from him was consistency and he proved himself We know what to expect from Mike McDonald. He gave us every single reason To believe in him every single reason to believe in him and literally any single situation Because when the situations were the hardest when the situations were the toughest what did Mike McDonald do? He showed who he is. He stepped up. He showed out in the toughest of the tough games minus the Rams game besides that one in the toughest of the tough games Baltimore Ravens they played again, we'll go backwards The chiefs AFC championship biggest game of the season obviously because this is a game to go to the Super Bowl Patrick Mahon's in them cheese. They get 14 points on you quick on your defense on your number one defense quick It's like whoa had a lot of people freaking out a lot of people like oh my goodness. What's gonna happen next? Are we gonna get blown out my McDonald's and no no no watch this. I got you three points for the rest of the game to Patrick Mahon the Travis Kip, right? Isaiah Pacheco 17 points a whole game No points in the second half At all like really? No, that ain't possible. Yes, it is what Mike McDonald is possible certainly was The game before that and we talked about it earlier today Just how do you hold a team to not not even pat they can't even get past the 25 yard line? That's impossible. Especially in these days with all the rules and all that how everything favors. How do you do that? Then week 18, that was just a whatever game, but previously before that the games before the Dolphins Two in that dolphins offense with all that speed even without water with all that speed. Oh my goodness and They were running all over the Ravens. I think I think they're running back Not where he most of but uh a chain devin a chance that he had over 100 yards that game Dolphin score what 15 points or 19 points something like that something crazy low But they first two drives they went right down the field boom like that. What happened after that nothing nothing nothing Because Mike McDonald again in the biggest games against it then the previous week before that against the 49ers You know what happened there. He made those two MVP candidates look like nothing Brock Purdy what five picks right five in a set Now I know obviously it's about the Jimmy's and Joe's more than X's and O's but Mike McDonald put those Jimmy's and Joe's in the proper places to have success and he would do that every week Just about every week That's why this hurts bad and another reason this hurts bad. We talked about people being proven We talked about him being proven. We talked about knowing what to expect from him now once again the focus is on John Harbaugh and A lot of focus is on John Harbaugh reason being and we're gonna talk about this more later on throughout the weekend and really in the coming months to but the focus is on John Harbaugh because It's like hold up John Harbaugh You you've proven to us who you are in the toughest of the tough moments last week against the Texans I was thinking oh, okay. Hey Ravens done turn the corner. They got it now. Yes. We have another demo for 2019 We 2019 the Baltimore Ravens were a historically historically Great franchise when it came to the running game. They were historic They made they broke so many records in the running game And what did that do that made it easier in the passing game the March accent have like he went This dude would have like 13 pass attempts like four or five touchdowns in a lot of games Why because they were running so well that made life easy everybody's like oh, man We got to stop them from running then they oh, oh, they threw the ball over the top of our heads Oh Lamar Jackson through another touchdown. Oh, there goes again through another touchdown and They would do that like almost every single game But then what happened in the playoffs this historically great rushing team in the playoffs. They decide. Hey, you know what? Mark Ingram.   He got hurt So then they didn't make the what they made they didn't make the playoffs in 2021 They had it and then Lamar got hurt had a number one seed in the AFC north or number one seed in the AFC Then Lamar got hurt everything fell apart In 2022 Same thing at number one seed at one point a FC number one seed and FC North all that good stuff Lamar and MVP race He got hurt But the Ravens still because of everything that they had already done they had ended up making a playoffs But of course, you know how that went the whole Tyler Huntley fumble at the one that are a they fought hard though But then this year this year it's like all right Lamar is healthy. Oh, this is gonna be different Oh, Baltimore Ravens was steamrolling teams. They were doing that thing. It's like, oh, yeah This is not a fluke. This team is not fluky at all. They are the real deal Get to the playoffs take care of business in a major way against the Texans again That's what had a lot of us food. It sure had me. I'm like, oh Ravens got it I saw the adjustments from Todd Munkin. I see adjustments from Mike McDonald I'm like hardball got these boys, right? Let's go John. Come on, baby And we've been giving him his credit all year for sure because he's deserved it for sure because he has had this team Right, you don't win 13 games by fluke. Uh-huh. You don't do that Especially the way they've been taking care of some of these teams But then the FC championship game rolled around Against John hardballs his the guy he used to work under Andy Reed and everything went out the window again Again Again Same exact scenario as 2019 Everything's right there in front of you You know what you got to do and what made the game worse What made the game worse It's like, all right Chiefs got one of the worst run defenses in the league one of the worst So a lot of us we have in conversations leading up to the game. We're doing our game previews and all that We're talking about oh, rave is about to run wild on these cheese. Let's go I can't wait this game might even be a blowout, but it's still patreon homes in them So got to respect them because they done been here done that but right. Oh ravens. Oh, they could take off Run all over them. Chiefs can't stop the run. The bills just showed you the week before That the chiefs they ain't stopping nobody from running Oh, yeah, we about to have a field day on these boys. Let's go. Nope Raven said we are going to run with our running backs six times Six Now, I know a lot of the blame has been going the top mucking a lot of people been like monkey What kind of game was that? What kind of game did you call and that is true? What kind of game did he call? Then a lot of the blame goes to Lamar Jackson. Lamar Jackson. What were you doing?  They wasn't blocking like that. What is going on with the offensive line? Why are you not blocking for the mother so MVP quarterback got to protect them? They get some blame too but the common denominator Common denominator because this is literally the had the ravens lost that game And they've put their best foot forward because that's what I always say I always wanted to be to where Everything should fall on the players all the blame should fall on the players. It should not fall on the coaches That's how I I think it should be If there if the rate that that's the best type of I know there's not a good type of loss But if a team fails for something I think that the coaches should put all the players in a position to have success But then if the players don't execute right then it should all be on them But I can't say that in this case. I couldn't say that in 2019 either and these were literally the exact same things that happen Baltimore Ravens historically great running team this year too Historically were great running team has historically great defense And they decided, you know what? We're gonna go directly against everything that works They did it in 2019 in 2019. Todd monkey was nowhere to be found. I mean, he was with Cleveland He was not with the Baltimore Raven. He was the offensive coordinator. Greg Roman was Everybody was all of us was so angry at Greg Roman. Greg Roman is Greg Roman that man. Greg Roman. Why'd you do that? What's wrong with you Greg Roman? What what what kind of game plan is that? In this year He brought the top monkey. What kind of game plan? What's that? What are you doing? What's going on? Who oversees those guys? Who's in charge of those guys? Who can step in with those guys and not just with the coordinator with the players too? Who's that coach? So many times last year We we talked about especially when um before they even let Greg Roman go We said this both before and after they let Greg Roman go both times or both scenarios We said that the issue is much deeper than just Greg Roman It's much deeper than Greg Roman much deeper Because we talked about how the Baltimore Ravens they've been through so many different offensive coordinators So many different ones And it's been the same issue Same issue And it's like now the Baltimore Ravens. It's like they graduated to a new level of the same issue Because before the issue would be the offense in the regular season We like man the offense in the regular season. They just can't get it done. They can't score They can't get no y'all. They can't get no points. What's going on?  Oh doing regular season audience They put enough points. They put enough points even in games that they struggled that they put enough points It's like, okay, let's go. But they didn't got to post season and it'd be like, uh What's happening So then go from Greg Roman top monkey regular season Oh, yeah doing that thing even in games where they not all the way on point. They put enough points for enough points for enough points and then boom postseason First half they put on they put up 10 but then the second half they took off. It's like, okay. Yeah, let's go But then in that chiefs game, it's like, oh same issues John Hobo, we looking at you my friend We looking at you my friend We talked about it throughout the season how um, because a lot of people they knew we knew that Mike McDonald was out of here We knew one that he was going to be out of here And I said the only way that Mike McDonald was was was going to stay the only way that I felt like he could stay That he would stay would be if the Baltimore Ravens won the Super Bowl and John Hobo decided to move to a front office role and Mike McDonald got promoted the head coach That was the only way I saw him staying. That was it That obviously didn't happen. He sure ain't winning the Super Bowl um And now Yeah, we're here Mike McDonald is gone He's gone um And I know we got some guys in the building who they could promote they could promote Zachary Or who used to play for the Baltimore Ravens used to be a nice inside linebacker for the Ravens But then he had that unfortunate neck injury Uh, so he could be a defensive coordinator. You got Anthony Weaver who was a defensive line coach He used to also play for the Baltimore Ravens was a defensive lineman for the Baltimore Ravens You could promote him to defensive coordinator and I'm sure both of those guys could do a phenomenal job But the thing where Ravens fans will be hesitant now and how what they've been hesitant about before Will still be the head coach It will still be the ultimate leader on the team John Hobo And I can't fault any Ravens fans for being hesitant and fully believing in what the Ravens can accomplish Because they have the same head coach because he's been the common denominator and so many of the Baltimore Ravens different things Now there have been other factors for sure There have been other factors for sure There have been times when Lamar might have a bad game There might be times when the tackling is bad. There might be times when this, that and a third But there is one common denominator And that is John Hobo So with Mike McDonald leaving Is we'll see what happens. I'm sure that they're gonna hire somebody and they're gonna do a good job as defensive coordinator Will they do as good as Mike McDonald? We'll see. We gotta hope so. Of course. We hope so Of course, we hope so We don't want to see the Baltimore Ravens fail But I know many Ravens fans are scared that the Baltimore Ravens will continue to fail If they keep doing the same things over and over and over if in the biggest moments in the biggest stages They continue to do this malpractice And they continue to fail at this just ultimate level Then that won't be a good thing team. Keep it clean. I We'll see We'll see how this whole thing works itself out. Uh, that just made the press conference on Friday that much more interesting


Friends, it is a good time to summarize what we have learnt in the last couple of lectures. So we are looking at the fluid solid heterogeneous, non-catalytic heterogeneous reactions. And then we identified two possible cases, two possible modes. One is the situation where the particle which is actually participating in the reaction, it does not shrink with as the reaction progresses. And then another situation where the particle which participates in the reaction actually shrinks in size as the reaction progresses. And then we looked at two possible models, progressive conversion model and the shrinking core model. And it was observed that the shrinking core model is more common in most of the situations. And so we developed a model for the shrinking core model for both the particle size, unchanging particle size and also for those particles which were shrinking in size. And that for situations where the overall reaction, overall conversion is actually controlled by different rate controlling steps. For example, the diffusion of gas phase from the bulk through the gas film to the surface of the core or the diffusion of the species through the ash layer or it is the reaction controlling. So the expressions that we get for the time that is taken for the unreacted core to reach a certain size which is directly correlated with the conversion of the solid species can be summarized as below. So for spherical particles, so it can be summarized in this table here. So for a spherical particle suppose we take the constant size and varying or changing size. These are two possible modes. Here changing size refers to the special case of no ash layer present. As noted in the earlier lecture, the size can change even if the ash layer is present if the densities of solid product and the reactant are different. Now suppose if it is gas film diffusion control, if the overall reaction is controlled by the film diffusion control then we found that T by tau where T is the time taken for the unreacted core to reach a certain radius because of the heterogeneous reaction that should be equal to the conversion of the solid itself which is directly related to the amount of solid that is reacted. And then in changing size the expression for the same would be T by tau is 1-1-xp to the power of 2 by 3. Here tau is the time taken for the complete conversion that is all of the solids which is present in the core has gone for complete reaction. Now next possibility is ash layer control, ash layer diffusion control. So if ash layer diffusion is actually controlling the overall conversion in that case T by tau is given by 1-3 into 1-xp to the power of 2 by 3 plus 2 into 1-xp. That is the expression for T by tau and of course in changing size the ash layer does not exist and therefore it does not offer any resistance to the overall conversion. Then the third case is where the reaction is controlling the overall conversion, it is the reaction control then T by tau is given by 1-1-xb to the power of 1 by 3 and here T by tau is given by 1-1-xb to the power of 1 by 3. In both cases the time taken for the unreacted core to reach a certain radius depends on the conversion in exactly the same way because the presence or absence of the film or the ash layer does not contribute to the overall conversion if it is in the reaction control regime. Now all this we looked for spherical particles. What happens here? What happens if it is other geometries? For example if there is plating of metals then normally it is not plated on a spherical particle it is plated on other kinds of geometries, it could be a cylindrical geometry or it could be a flat plate geometry or it could be any other curvature. So let us look at two specific other geometries which are fairly common. One is the flat plate, suppose if it is a flat plate and where the solid is now present in the flat plate and there is a fluid which is now coming on to the surface of the flat plate and the heterogeneous reaction is occurring at the surface of the flat plate. So in that case if we consider the constant particle size, if we consider the case of constant particle size then if the for a flat plate one can actually define one can define conversion as 1-1 by L where L is essentially the half thickness, L is the half thickness of the plate that is actually being used for used as a solid reactant and if it is a film diffusion controlling, if it is a film diffusion controlling regime, in that case T by tau is given by xp where tau is given by tau is the time taken for the total conversion that is for all the solids to undergo reaction that will be given by density of the particle divided into the half thickness length scale divided by mass transport coefficient into concentration of the species in the gas phase. If it is ash layer diffusion control, then T by tau is essentially given by xp square where tau is given by rho B L square divided by 2 diffusivity into concentration of the species in the gas phase. If it is reaction controlled, then T by tau is simply given by xb and tau is given by rho B L divided by k double prime which is the intrinsic rate constant multiplied by CAg. K double prime is the intrinsic specific rate and CAg is the corresponding gas phase concentration. Now similar expressions have been obtained for cylinder geometry and they are as follows in this table. So for cylinder, for a cylindrical geometry the xp conversion of the solid is defined as 1 minus R by R0T whole square. Now here R0 is the initial radius of the core, R0 is the initial radius, initial radius of the cylinder and R is the instantaneous radius, instantaneous radius of the cylinder. So with this definition, if it is gas phase diffusion control, then T by tau is essentially given as xb and tau is equal to rho B into R0 which is the initial radius divided by 2 times kg into CAg. So that is the time taken by the core for complete conversion and if it is ash layer diffusion control then T by tau is given by xb plus 1 minus xb multiplied by the natural logarithm of 1 minus xb where tau is tau B, tau is the time for total conversion that is given by rho B into R0 square divided by 4 de into CAg. So that is the concentration of the species at the gas phase and if it is reaction control, if it is reaction control then T by tau is essentially given by 1 minus 1 minus xb to the power of half and tau is equal to rho B into R0 divided by k double prime that is the specific reaction rate multiplied by CAg. So this sort of summarizes the various time taken for the core to reach a certain radius if the overall reaction is controlled by different resistances which is actually available in the system. So all these are for basically constant size system and similar expressions can actually be worked out for varying size as well. We so far looked at the cases where the overall reaction is actually controlled by the resistance of one of these three resistances which are available that is either the diffusion through the gas film or the diffusion through the ash layer in the case of constant particle size or if it is a reaction controlled. So in reality this is not the situation, in reality what happens is that all three resistances actually contribute. So as reaction proceeds and in fact that is true because as reaction proceeds, as reaction proceeds even in the constant size case, even the constant size case what is observed is that the of course the unreacted core radius is going to decrease, unreacted core radius decreases as the reaction proceeds and therefore as a result the relative importance of various resistances is going to change. So that is very important because there are three resistances as the size of the unreacted core changes then the relative importance of these different resistances towards their effect on the overall conversion also is going to change. So which means that not one, so no one resistance controls overall conversion all times, all time t. So therefore it is important to consider simultaneous action of all the resistances, so it is important to consider simultaneous action of all three resistances. So if you want to incorporate all three resistances then the model for the radius of the unreacted core as a function of time that the differential equation which basically tells what is the rate of change, relates the rate of change of the radius of unreacted core with respect to time to all the other properties and concentrations of the system can actually be rewritten using a combination of all the resistances and the rewritten model will be dr by dt that will be equal to minus CAg divided by rho b, CAg is the gas phase concentration rho b is the density of the solid that is used divided by r square by r naught square into kg plus r naught minus r divided by the diffusivity into r by r naught plus 1 by k double prime that is the specific reaction rate. Now here it is assumed that the volume fraction of the solid in the unreacted core is approximately equal to 1 and so there are three terms here which correspond to three different resistances. So the first one here this corresponds to the gas film, this corresponds to the gas film resistance and this corresponds to the ash layer, this corresponds to the diffusion in ash layer and the last term corresponds to the reaction resistance, resistance due to the heterogeneous reaction that is occurring in these on the surface of the unreacted core. So now suppose this is for a constant size system, so if the particle size does not change, this is for the constant size system. Suppose if the same thing can be written for a shrinking size type of particles where there is no ash layer, so if there is no ash layer then the resistance due to ash layer does not exist. So simply we can obtain the expression by removing the resistance due to the ash layer which is present in this expression here. So for shrinking particles there is no ash layer, so therefore we can write this expression as dr by dt is minus Cag by rho B divided by 1 by kg plus 1 by k double prime where k double prime is the specific rate and kg is the corresponding mass transport coefficient. So now let us take an example problem and see how we can predict the dissolution of a particle of a certain type which is undergoing such kind of a fluid solid non-catalytic heterogeneous reaction. An excellent example of that is the drugs which are actually being administered. So the solid drugs which are administered for curing a certain disease, when they go into the body they have to dissolve and it is important to understand how much time does it take for the drug which is a solid particle to dissolve and that the dosage of the drug actually strongly depends on the time taken for complete dissolution. So therefore it is important to model this from the pharmacokinetic standpoint of you. So let us look at this problem. So dissolution of solid particles, suppose if we assume that there are all these drugs which are given inside, they are all mono dispersed particles that is all the particles that are actually in the drug are of same size. This is not true always but let us assume to start with that all particles are actually same size that is it is a mono dispersed system and then so this is, so we need to, the objective is to find out what is the dissolution time which actually if we place an important role in the pharmacokinetics. Now suppose if there is species A which is basically in the fluid stream that reacts with certain solid which is essentially the drug particle and that leads to formation of products. Now if we assume that the fluid A actually reacts with a core which contains the solid material and if we assume that it is first order with respect to the fluid, the reaction rate is first order with respect to fluid and zero order with respect to the solids which is present. Then the question is what is the dissolution time? So we need to estimate the objective of the problem is to find the dissolution time, the time that is taken for dissolution of these particles. So now we can observe, we can actually observe that the rate of mass transport, the rate of mass transfer to the surface is actually equal to the rate of surface reaction. Why is this because the particles are expected to as the property of the drugs are always such that the particles are expected to dissolve and so the particles are, this is actually a process where the particles are actually shrinking in size. So there is no ash layer which is actually present. So therefore the rate of mass transfer to the surface should be equal to the rate at which it is being consumed for the surface reaction. So therefore we can write the WAR if that is the flux at which the species is actually traveling from the bulk gas phase to the surface of the solid that is equal to kg into Ca minus CaS where Ca is the bulk concentration and Ca is the surface concentration. And that should be equal to the reaction rate Ra double prime and that is equal to if Kr is the specific reaction constant multiplied by CaS. So that is the surface concentration and this is the net reaction rate. So this is the reaction rate and this is the mass transport. This corresponds to the mass transport. So from here we can actually eliminate CaS. So note that CaS is a quantity which cannot be actually measured experimentally but Ca is the bulk concentration and that can actually be measured. So we need to eliminate CaS which is not a measurable property, measurable quantity. So kg into Ca divided by kg plus Kr. So plugging in this back into the expression for WAR we find that double prime that is equal to Kr into kg Ca divided by Kr and that is equal to Ca by 1 by kg plus 1 by Kr. So therefore if we know the mass transport coefficient and the reaction in specific reaction constant reaction rate then we will be able to actually calculate the required quantities just the dissolution time. So therefore now there is a need to find out what is this mass transport coefficient. So we can use correlations in order to find the mass transport coefficient kg. So suppose if we assume that the particles are small. So under small dp and if there is no shear stress at the, so if there is no shear stress at the boundary of the fluid and the solid then we can actually using the Frosling correlation we can find out that from Frosling correlation we can find out that Sherwood number which is equal to the mass transport coefficient into diameter of the particle divided by the corresponding fuselage that is approximately equal to 2. So from here we can find out that the mass transport coefficient is given by 2 times diffusivity divided by the corresponding particle diameter of the particle which is actually being dissolved. So plugging in the back into the expression for flux we find that WAR equal to Kr into C A 1 plus Kr by kg and that is equal to Kr into C A divided by 1 plus Kr into dp divided by 2 times de and with this can be rewritten as Kr into C A divided by 1 plus dp by d star. So where d star is nothing but the ratio of 2 times de into Kr so that is what is d star. So d star the meaning of d star is that it basically tells you the diameter at which the mass transport and the reaction rates are actually equal. So if we look at what is the expression for d star, d star is 2 times de divided by Kr and that is essentially the diameter at which the mass transport rate and the reaction rate are equal. So therefore if the diameter of the particle if this is larger than d star then it can be expected that it is mass transport controlling while if the particle diameter is less than d star then it is actually expected to be a reaction controlling scheme. The overall conversion is expected to be reaction controlling. So now in order to find the radius of the particle as a function of time we can now write a mole balance on the solid particle and the mole balance goes as here, mole balance on solid particle. So where whatever rate at which things are coming inside minus whatever rate at which the particles are leaving out plus generation should be equal to accumulation that is the balance. Now nothing is coming inside because it is the solid which is actually participating so there is no flow so nothing is coming inside and nothing is actually leaving but some of it is actually being reacted so that is a counter to the generation term. So if rbs is the rate at which the solid is being consumed at the surface of the particle multiplied by pi into dp square that is the surface area of the solid particle and that should be equal to d by dt into density into volume of the spherical particle. So suppose if we assume equimolar counter diffusion, suppose if we assume equimolar counter diffusion then the minus rAs double prime that is the rate at which the species A is actually being consumed because of the heterogeneous reaction that should be equal to minus rbs the rate at which the solid is being consumed this is because of the equimolar counter diffusion and the stoichiometry which is associated with the particular reaction. So now this equation can further be simplified as rho into 3 pi by 6 into dp square into that is equal to rAs into pi into dp square. So this can be simplified as d by dt of dp that is equal to minus 2 minus rAs double prime divided by rho. So that is the expression for the rate of change of the diameter of the drug particle as a function of time and how is it related to the surface reaction rate and the density of the particle. So now this can actually be we know the what is the rate so that can be written as minus 2 into kr into Ca divided by rho into 1 by 1 plus dp by d star and this can be written as minus alpha divided by 1 plus dp by d star. So that is the expression for rate of change of the diameter of the particle as a function of time. So if you assume that this whole thing is a constant like alpha so we can rewrite this expression as d by dt of dt equal to minus alpha by 1 plus dp by d star where alpha captures the rate of the reaction divided by the density of the corresponding particle. So now we can integrate this with the following initial condition that at time t equal to 0 the initial size of the particle was dpo and based on this initial condition the equation can be integrated and integrated expression would be dpo minus dp plus 1 by d star 1 by 2 d star into dpo square minus dp square that should be equal to alpha t. So this provides a relationship between the properties of the system and the diameter of the particle and time.  So that is the relationship between the time taken for complete conversion and the diameter of the particle and other properties of the system. Now this is the expression this is for a mono dispersed particles but normally the drugs which are actually administrator they are all poly dispersed that is particles are actually of different sizes. There is a population of particles and each of these particles can actually be of different sizes and all of them are actually simultaneously undergoing dissolution leading to the shrinking of the particles. Now the only issue here is that different particles will actually do different things because the sizes different although the reaction is same their dissolution rates are expected to be different because their sizes are completely different. So let us look at dissolution of let us see how to capture this behavior dissolution of poly dispersed particles. So what is poly dispersed particles? Suppose you have collection of particles which are of different sizes for example the this big one could be of size d1, it could be d2, maybe it could be d3 and this could be d4 and so on and so forth. So there will be a collection of particles and each of these are different in size and so there are particles of different sizes. Now not just that the initial stage that the particles will be of different size there will be a distribution of sizes of particles all through the time and the dissolution is actually occurring. So therefore clearly there is a distribution there is a size distribution and the question is to find out what is the dissolution time, the objective is to find the dissolution time and in order to find the dissolution time for particles who are actually placed in a certain distribution one needs to actually follow the distribution dynamics. So the distribution dynamics have to be followed. So let us look at what is the distribution. So suppose if I look at the plane of number of particles versus the fraction of the number of particles that are actually of a particular size. So then we expect that it is a certain histogram where the location between let us say dp and dp plus delta dp which is a small increase in the particle, small difference in the particle size. So in that case the area and under the curve in this location it signifies the number of particles which are actually present in the system whose diameter is between these two numbers. So therefore f dp multiplied by delta dp that is basically the number of particles between dp and dp plus delta dp. So that is the number of particles and moreover if we integrate this expression from 0 to infinity f dp into dp into the differential of diameter of the particle that should be equal to n naught which is the total that is the total initial particles which are actually present. Now if we assume that it follows a log normal distribution let us assume that the initial distribution of the particles they follow a log normal distribution. Let us assume that they follow log normal distribution. So now the fraction of the species, fraction of the particles whose size is dp at time t equal to 0 and that normalized by n naught is actually given by 1 by dp square root of 2 pi into ln of sigma 2 multiplied by exponential of minus ln of dp by dg and square of that divided by 2 times ln sigma 2 square. So that is the expression for the log normal distribution where dg and sigma 2 are essentially the distribution parameters these are the distribution parameters. So now the question is in order to find out what is the dissolution time of the particles which are actually in this polydispers system is to actually find out what is the how the distribution itself changes instead of monitoring every drug particle it is better to monitor simply the population itself. So the question is how does the how does distribution change? So that is what we are going to look at. Now suppose if we look at the distribution at any time and if it is some general curve like this now we can write a simple balance on the population of these drug particles and suppose if we take a small element between dp and dp plus delta dp. So therefore the thickness of this element is delta dp if we assume the thickness is delta dp and by definition of this histogram the integral of 0 to infinity f of dp, t into ddp should be equal to n0 that should be the total number of particles that were actually initially present and because there are no new particles are added it is only the particles actually dissolve and disappear. So therefore the integral under the curve up to infinity if we assume that the particles are actually of some finite size at all times then that should be equal to the total number of particles itself. Now if that is not the case then this should be a function of time. So the number of particles will change because some of these particles will dissolve and disappear and when that happens then this integral between 0 to infinity should actually be a function of time. Now one other quantity which is required in order to model the system is the growth rate. Suppose rdp is the growth rate of every particle and of course it depends upon the size of the particle itself. So that is the growth rate, growth rate of particle whose diameter is dp then one can actually write a population balance equation in order to capture the dynamics of the whole distribution itself as the reaction actually proceeds. So the population balance equation can be written as the balance between the number of particles which are actually growing and reaching the diameter whose sizes between dp and delta dp and also we need to account for the particles which are already present in this small element dp and delta dp and they grow and they actually become bigger than dp plus delta dp. So which means they leave this small element delta dp and there is no addition of new particles because once the drug is being fed it is just being dissolved and so there is no other mechanism by which the particles are actually being added into that small element delta dp and then the other term is the accumulation term. So therefore putting them all together it will be number of particles growing into region between dp and dp plus delta dp minus the number of particles growing out of region between dp and dp plus delta dp that should be equal to the accumulation of particles in delta dp in that small element. Now suppose if r is the growth rate which is what we have defined a short while ago then the number of particles growing into the region between dp and dp plus delta dp is actually given by r into dp. So that is the growth rate when the particle is right whose size is exactly dp and then that multiplied by the corresponding fraction will till will provide a clue as to what is the number of particles are actually growing into the region in this small interval of diameter and that evaluated dp minus delta dp and that should be equal to d by dt so that is the accumulation of the particles in delta dp equal to f of dp comma t multiplied by delta dp. So that is the rate of change that is the rate of which the particles are actually being accumulated in that small element. So now rearranging this equation and setting delta dp equal to 0 so set delta dp to be 0 then can be written as minus d by d of the particle diameter into f of dp comma t that should be equal to dou f by dou comma t. So that is the expression for that is the population balance. So now what do we do with this population balance? So we can solve this equation in order to find the radius of the particle in order to find the how the distribution the radius distribution actually changes. So now this can actually be rewritten as df by dt plus r into df by dp plus f into dr by dp that is equal to 0. So now where rdp is essentially the growth rate of the particle of size dp and if suppose in the earlier case if the particles were mono dispersed we actually found out what is the growth rate. So therefore for every particle of a certain size that expression can be used as a growth rate of the particle of that particular size. So therefore from here it will be ddp by dt that should be equal to minus alpha by 1 plus dp by d star. So that is the rate at which the mono dispersed particles are actually growing. So plugging that into the expression plugging that into the population balance we will find that so this is for the mono dispersed case. So plugging that into the expression we will find that the rate of rdp that is equal to minus alpha by 1 plus dp by d star. And so the population balance will be df by dt plus minus alpha by 1 plus dp by d star into df by dp plus f into alpha by d star into 1 by 1 plus dp by d star the whole square that is equal to 0. So that is the expression for the that is the population balance that actually captures how the dynamics of this distribution actually changes with time, how the distribution changes with time. So now if we introduce a few dimensionless quantities which is actually useful in solving this problem so psi equal to f into d star divided by n naught where f is the distribution of the particles based on the size and d star is the ratio of the diameter when the mass transport rate and the reaction rate are equal to each other and n naught is the initial total number of particles. Then epsilon is defined as 1 plus dp by d star and theta is defined as alpha into t by d star. So now if we introduce these dimensionless quantities we can rewrite the population balance as d psi by d theta that is minus 1 by epsilon into d psi by d epsilon minus that is equal to minus psi by epsilon square. So that is the expression. So this equation is actually of a very, very familiar differential equation form which is of the form p of x, y into dz by dx plus q of x, y into dz by dy that is equal to r of x, y, z. So from here we can see that if p of x, y is nothing but 1 and q of x, y is like 1 by epsilon and the r of x, y is like psi by minus epsilon square. So the way to solve this equation is basically to use the method of characteristics is to use the method of characteristics where this problem is posed in a slightly different way. So dx by p sorry d theta by 1 is equal to d epsilon by minus 1 by epsilon and that should be equal to d psi by minus psi by epsilon square. So that is the way in which the population balance equation which is actually written here can actually be posed in terms of the method of characteristics. So now the first two terms in the in the repose problem basically it looks like d theta by 1 is equal to minus epsilon into d epsilon. So from here we can find out that epsilon square plus 2 theta is equal to a constant c 1 and that should be of the form of the form h of epsilon square plus 2 theta. So that should be the functional form functional dependence of this particular expression and then next let us look at the other case d epsilon by 1 by epsilon that should be equal to d psi by psi by epsilon square. So this can actually be rewritten as d epsilon by epsilon is equal to d psi by psi which can be solved to obtain ln epsilon equal to ln psi and that is equal to some constant k. So that is a constant. So from here we can find that psi by epsilon is nothing but some constant c 2 and that can be equal to h of epsilon square plus 2 theta. So that is the functional dependence. So next we can actually see how to take this forward and find the distribution profiles. So suppose if we can rewrite the relationship between the psi and epsilon as psi equal to square root of epsilon square into h of epsilon square plus 2 theta. So in order to obtain the final distribution which is actually present in the definition of psi. So psi is almost like a non-dimensional distribution of the various size of the particles which are of different sizes. So if you look at the initial distribution which is actually given by log normal distribution. So the log normal distribution is as it goes here root 2 pi into ln sigma 2 into exponential of minus ln dp by d star dp by dg the whole square divided by 2 ln sigma 2 and square of that. So that is the initial distribution. Now if you know the initial distributions then we can actually introduce the transform variables. So basically we introduce epsilon psi and theta into this initial distribution and if we assume that epsilon equal to 1 plus dp by d star. So this comes from the non-dimensional form of epsilon. And from here we can get that dp is equal to d star into epsilon minus 1. Actually this is the non-dimensional form of epsilon and from here we can find out the relationship between dp and the other parameters. So if you assume that dr, if you assume a new variable dr is equal to dg divided by d star where dg is some parameter of the distribution and d star is the diameter at which the mass transfer rate and the reaction rates are equal. Then we can find out that psi is equal to some f of epsilon, theta divided by n0 into d star that should be equal to square root of epsilon square into h of epsilon square plus 2 theta. So that is the functional form of this variable psi. So now if you want to find out what is the final distribution we need to find this expression f of epsilon, theta. This expression needs to be found in order to find out what is the instantaneous distribution of the sizes. So how can we do this? So it can be done by actually using the distribution at the initial state where the reaction has not started then f of epsilon, theta. So this can be obtained by simply replacing epsilon square with plus 2 theta in f of epsilon, 0. So note that this is the initial distribution and the final distribution can simply be obtained by replacing epsilon square in this initial distribution with epsilon square plus 2 theta. So that serves as the solution methodology. So now we can rewrite this as psi of epsilon, theta. So that is equal to f of epsilon, theta. So we can introduce, we can now use the initial distribution in order to find out what is the value of psi, what is the expression for psi as a function of epsilon, theta and that would be equal to epsilon divided by square root of 2 pi into ln sigma 2 into 1 by square root of epsilon square plus 2 theta into 1 by square root of epsilon square plus 2 theta minus 1 multiplied by exponential of minus ln of square root of epsilon square plus 2 theta minus 1 by dr and square of that divided by 2 into ln sigma 2 to the power of square of that. So that is the distribution at any time in the non-dimensional form. So once we know this distribution, we can look at the distribution profiles as a function of time. So suppose if this is dp in the particle diameter and this is f of dp, time then if suppose this is the initial distribution, so that is time t equal to 0, that is the initial distribution, then as time progresses all the particles are actually undergoing the dissolution because of the heterogeneous reaction and so as time goes by the distribution changes and so this is t1 which is greater than 0 and then as time further goes by more dissolution will happen and so this will be t2 which will be greater than t1. Then further time elapses and then the distribution will be this will be t3 which is greater than t2 and then eventually when the conversion is going to be almost complete, then the distribution will look like this where this is t4 which is greater than t3. So that is the kind of distribution that one can actually obtain as a function of time for this polydispersed particles which are undergoing dissolution. So let us summarize what we have learnt in the last 3 lectures. So we have looked at the fluid solid non-catalytic heterogeneous reactions and this can occur through 2 modes. One is where the particle size does not change and other one is the case where the particle size actually changes as the reaction progresses and there are 3 different possible resistances which are actually existing in this kind of a system. One is the gas film resistance for diffusion resistance and the other one is the ash layer diffusion resistance and the third one is the reaction controlling resistance offered because of the surface reaction. So these 3 under these 3 regimes the time that is taken for the time that is taken by the particle to reach a certain radius has been modeled and calculated and that for different geometries have been looked into for both the shrinking or the changing size particle case and for the constant size particle case. And then we looked at an example of dissolution of monodispersed particles and extended for a polydispersed system. Thank you.


In the 14th century, Scotland produced one of the most inspired political documents in medieval history. The fact that the Declaration of Her Growth was created during a historical period of darkness makes it all the more remarkable. Addressed to Pope John XXII, the Declaration of Her Growth was a letter signed by over 30 nobles, barons and freeholders on behalf of the community of the realm of Scotland. Written on 6 April 1320, this date is now recognised as Tartan Day in various countries around the world. The Declaration of Her Growth emphasised Scotland's independence, the nobility's support for Robert the Bruce to be the King of Scotland and called on the Pope to help halt the attempted domination of Scotland by Edward II, King of England. In part, the Declaration of Her Growth was sent as a response to the Pope's decision to excommunicate Robert the Bruce, who had been crowned King of Scotland in 1306, after Scotland disobeyed papal efforts to try and agree a peace treaty with England. Originally written in Latin, the Declaration of Her Growth was produced during the First Scottish War of Independence, which had started with the English invasion of Scotland in 1296 and ended in 1328 with the Treaty of Edinburgh, Northampton, a treaty where the independence of the Kingdom of Scotland was recognised, yet this was reneged upon in 1332. This remarkable document is thought to have been written in our broth Abbey by Bernard Abbot of our broth, an administrator in Bishop, one inspiration for the writing style and language used in the document was a book written by the Roman historian and politician, Salist, called The Conspiracy of Catalina. One of the most revolutionary aspects of the Declaration of Her Growth pertain to the perspective of the Scottish nobility towards Robert the Bruce. Even though they endorse the legendary Scottish figure as the rightful King on one hand, they did so with a caveat that can be considered truly revolutionary in an age of the divine right of kings. The writers state that they are bound to Robert the Bruce, both by his right and by his merits that are freedom may still be maintained, and by him come what may we mean to stand. However, the writers add an inspired caveat, yet if he should give up what he has begun, seeking to make us or our kingdom subject to the King of England or the English, we should exert ourselves at once to drive him out as our enemy and a subverter of his own right and ours, and make some other man who is well able to defend us, our King. It truly cannot be underestimated how politically enlightened this clarification was, given the dismal lack of individual rights and political freedoms in the 14th century. The pronouncement that the community of the realm of Scotland were willing to support their King, only if he served their interests, was centuries ahead of its time, as kings were akin to gods in most other regions of the world. Daring to even whisper what the writers pronounced in practically any other region of the world, 700 years ago, would have been met with persecution and the sword. Yet for the writers of the declaration of our broth, absolute freedom was far more valued than absolute monarchy. In other words, liberty was king.


Good morning and welcome back to the course on Classics in Total Synthesis Part 1. And we have been discussing total synthesis of various natural products. In the last lecture, we talked about the total synthesis of a marine natural product called elitrobin and where we talked about the total synthesis reported by K.C. Nicolau. So today we will talk about another total synthesis but this time reported by Danyevsky and in the case of Nicolau's total synthesis, he started with the carbon, the monotterpin and here Danyevsky started with another monotterpin called alpha philandry and his work required you know lot of stereochemical studies particularly with respect to carbohydrate so that he could prove the correct isomer of the final natural product that is elitrobin. And his total synthesis involved 3 important key reactions. One is the Nosakhi, Kishi, Hayama, Rinklos reaction to form the five umbered ring and the second key reaction which is also very interesting rearrangement that is pyranose to furanose and furan to pyranose okay. So these 2 key reactions he had used to first convert the furan ring into pyranose and then later the pyranose ring was converted into five umbered furanose ring. The third key reaction was the stillly coupling between the carbocycle and the carbohydrate mighty to form the core structure of elitrobin. So from retrosynthetic point of view as you can see here it is very clear you have a complete carbocycle and a sugar unit okay. So the easiest way to disconnect is just you can disconnect either here or here. So if you are using stillly coupling then the disconnection of bond A is better than disconnection at bond B. So the stillly coupling disconnection gives Ranshevsky these 2 fragments and the left hand side you have the vinyl triplet and the right hand side you have the trivital tin derivative which is required for the stillly coupling. So now for the synthesis of the carbocyclic derivative that is the tricyclic compound the triplet can be obtained from this protected alcohol by removal of the pyrolyte group oxidation to ketone and followed by enol triplet formation. One can easily convert this into the required triplet in 3 steps after protection of this hydroxyl group. Now this five umber ring as I said this is one of the key reactions where the pyranose form okay where the pyranose form was converted into furanose form using acidic condition okay. This pyranose form again was obtained from furan okay. So here using this hydroxyl group an epoxidation of one of the double bonds of furan followed by the rearrangement one can easily get this 6 umber pyranose ring okay that was the second key reaction and the third key reaction is the intramolecular Nosakhi you see ring closer reaction between this the bromofuran and this aldehyde to form the 10 umber ring. Here now if you look at this it is very easy if you have this aldehyde and you can use the lithiobromo furan addition to this aldehyde to generate this chiral center with OH and for the southern hemisphere you need CH2CHO that can be obtained from this ester by reduction and homologation okay. So now these two substituted ones these two substituted ones on this 6 umber ring can be easily introduced from Philandrine a commercially available mono terpen using photochemical reaction as the key reaction okay. So as I said there are two key reactions which he has used at least I will talk about one key reaction that is Nosakhi Kishi reaction Nosakhi Kishi reaction is nothing but if you have a vinyl triplet or vinyl halide and this on treatment with aldehydes in the presence of chromus chloride you will get an allylic alcohol. This reaction can be done intermolecularly as well as intramolecularly basically if you want an allylic alcohol so this is one of the very important transformations and one can also get homolylic alcohol using similar conditions here instead of vinyl bromide what you need is an allylic bromide. So allylic bromide on treatment with aldehyde in the presence of chromus chloride it gives homolylic alcohol and if you have an ester at appropriate place then the homolylic alcohol which is formed can also intramolecularly attack the carminate group of the ester forming a lacto okay. So this Nosakhi Kishi reaction has been widely used for making allylic and homoallylic alcohols and also many pachyocycles. So now let us see how Dainishevsky's group synthesized eletrobin and as I said the synthesis of eletrobin requires synthesis of two different fragments one the carbocyclic ABC in fragment the other one is sugar fragment. So now let us start with the synthesis of sugar fragment for which he started with the commercially available D-arabinose and the first step was the peracetylation that is all the hydroxyl groups including the lactol hydroxyl group where acetylated with acetic anhydride and DMAP then this particular anomeric acetoxy group okay can be easily displaced with ethane thiol under Lewis acidic condition and here you can see the alpha ion number is the major isomer okay. So once you have that then the three acetates okay three secondary acetates can be easily hydrolyzed by treatment with sodium ethoxy methanol. Now if you look at this triol these two hydroxyl groups are cis to each other and these two are trans to each other. So one can easily protect the syn diol to get the corresponding astronaut under standard conditions. Now what needs to be done is you have to introduce a CH2O SNBU3 and also protect this hydroxyl group. So these are the two things left for the synthesis of sugar fragment with tributyltine group which is required for stelaic coupling. So you take this free hydroxyl group now protect the hydroxyl group as TBS ether protect the hydroxyl group as TBS ether. Then treat this anomeric thiol ether with tributyltine methanol tributyltine methanol in the presence of methyl triplet. So that gives you the required and see CH2 SNBU3. So here beta isomer is the isomer which is required for the stelaic coupling.  So now you have synthesized the sugar fragment successfully in few steps starting from D-Arabino's. Now let us look at the synthesis of ABC ring of elitrobin starting from a chiral monoturpin called phylandrine. So the phylandrine, so this is the structure of phylandrine. Now if you look at this structure of phylandrine there are two double bonds. One is a trisubstituted double bond other one is a disubstituted double bond. And as I mentioned when I talked about the retro synthesis it involves a photochemical reaction. It has two double bonds and selectively one of them has to undergo 2 plus 2 cycloaddition reaction. So as I said one is trisubstituted other one is disubstituted and between these two for stelaic reasons one can selectively carry out photochemical 2 plus 2 cycloaddition reaction with the disubstituted alkene. And the other alkene was the dichlorochetene. The dichlorochetene can be easily obtained by treatment of zinc with trichloroacetyl chloride or one can also get it from dichloroacetyl chloride with mild bases like triethylamine. So now once you do this reaction what you get is a mixture of two isomers and once the dichlorochetene is formed because of the presence of the bulky isopropyl group in beta position this dichlorochetene will come only from the alpha side. So that is how these two hydrogens you can see when it approaches from the alpha side these two hydrogens will be beta. So this is the major product once you have this the two chlorines can be easily exchanged with hydrogen by treatment with zinc methanol. So zinc methanol removes the two chlorines. Now we have to open the cyclobutanol drink. So one way is to introduce a functional group here okay you can introduce a functional group and open this cyclobutanol and if you use Bayer-Riegel oxidation. So then it will open up here which is not required you need to open the cyclobutanol at this place. So for that it is easy to introduce an aldehyde equivalent okay if you look at this. So this is an aldehyde equivalent if you hydrolyze this enamine okay this is an enamine basically it is an enamine derived from dimethyl formaldehyde protected dimethyl formaldehyde okay. So now this enamine upon hydrolysis with para toluene sulfonic acid. So what will happen? So you will get first you will get an aldehyde like this okay then under acidic condition and the acidic condition that is with methanol what will happen? The methanol will attack here and open up. So that will give you the corresponding ester that is you will have this aldehyde and ester you will have this corresponding aldehyde in ester okay this is the product you get is it clear? So now what you need to do one you have to add the furan ring to this aldehyde now it is protected aldehyde other one you have to homologate this ester is not it what you need is a CH2CHO already you have CH2CHO here but here you have only CO2MB. So that should be homologous okay. So first you remove this acetone okay para toluene sulfonic acid and water you remove the acetone you get the aldehyde. Now once you have the aldehyde take this 2-5-dibromo furan this on treatment with butyl lithium. So one of the bromines will be exchanged with lithium to get the corresponding lithiospecies this furan lithium that is bromofuran lithium will add to this aldehyde. So you have aldehyde and ester and as you know aldehyde is more reactive than ester. So once it adds to the aldehyde so what you get is a mixture of this alcohol which is as a result of direct addition of this lithiospecies to the aldehyde and the second product is nothing but this alcohol attacking the ester carbonyl and forming the 6-1-butlacto okay. So these are the 2 products and the major product is this alcohol okay. So now you have the alcohol next step is to protect the alcohol. So the alcohol could be protected as TBDPS ether and the standard condition then now you need to homologate this ester okay. How do you do okay you have to first reduce then that CH2OH should be homologated. So the reduction is normally done with the dye ball if you use excess dye ball you will get the corresponding alcohol. So here they use excess dye ball so that you get the primary alcohol. Now the primary alcohol can be homologated through mesylation. So first you mesylate the primary alcohol to get the mesylated compound this upon treatment with potassium cyanide okay when you this is a good leaving group mesylate. So now if you treat with potassium cyanide in the presence of 18 crown 6 ether you get the corresponding cyanide CH2Cm okay. So what we need now this cyanide should be reduced to aldehyde then the intramolecular cyclization should takes place okay. So dye ball reduction of cyanide will give you aldehyde so that was a very clean reaction. Then the second key reaction okay so the second key reaction in the total sense of Tanishib ski is the Noshaki-Kissi reaction. So this upon intramolecular Noshaki-Kissi reaction gave the corresponding 10 membered ring wherein the alpha isomer was the major product. The hydroxyl group which is alpha is the preferred product which you can see they got 15 is to 1 ratio of alpha and beta. So now if you look at this particular structure you have the 6 membered ring in place the B ring that is the 10 membered ring is in place. Now what you need to do is you have to convert this furan convert this furan one you have to get a double bond here you have to introduce a methoxy group here and you have to introduce a methyl group. So 3 things one has to do one introduction of methyl group introduction of double bond and introduction of methoxy group here. But at the same time you should know this 2 double bonds should not be there in furan okay. So what did they do? Before you do this you need to protect the newly formed hydroxyl group okay. So that was protected as pyrolytester by treating with pyrolyl chloride. Then you can remove the TBDPS group okay. So standard Tbuff tetrabutylaminium fluoride treatment will remove the TBDPS to give the corresponding alcohol. So once you have this alcohol if you look at the relationship between this double bond and this hydroxyl group it is an allylic alcohol okay that hydroxyl group will direct the incoming peroxidizing agent so that you will get alpha epoxide. So what happens once you treat with dimethyl dioxide DMD you get the corresponding pyronos ring. So not only epoxide is formed it undergoes a rearrangement okay it is a well known rearrangement to form this hydroxy pyrono. How does it happen? It is very simple it is well known in the literature. The first as I said it form the epoxide of the double bond which is close to the hydroxyl group it is like allylic alcohol epoxidation. Once it is formed then you can see the arrows the opening of this phymober ring gives rise to this enedio okay. It is basically nothing but if you have 1,4 diketone on treatment with acid you will get furan okay 1,4 diketone if you treat with acid you will get furan. The same thing it is a reverse reaction but since you are using oxidation you get extra double bond okay it is just the reversal of acid treatment of 1,4 diketone to furan. Here what you are doing is you are using an oxidizing agent so that is why you introduce a double bond as well. Once this enedione is formed then intramolecularly the hydroxyl group here will attack the ketone to form the 6-mombard ring okay. So what you have done now you have oxidized the furan ring and while doing that it underwent a ring expansion reaction to form a hydroxy pyromo okay and this is also good in one sense that if you look at the natural product, if you look at the natural product you need to introduce a methyl group here, you need to introduce a methyl group here. So now this has become easy because you have an enone okay you have an enone then it should be easy to introduce the methyl group here. So how did they do? First you protect this lactone okay so protect it as TMS ether then you add methyl lithium okay. So methyl lithium addition to the enone comes from alpha side that methyl group comes from alpha side to get the beta alcohol then you treat with acid okay. Here when you treat with acid this pyronos ring rearranges back to the furanos ring. Here the pyronos ring rearranges back to the furanos ring so how does it happen? It is a very interesting again a rearrangement it is like shuttle trip rearrangement. Once it was 5-mombard ring that is furan, furan to pyronos now pyronos to furanos okay it is like shuttle trip rearrangement how did this happen? The moment you protonate what happens this oxygen gets protonated then this lone pair on the O TMS opens this so what you get? This oxonium ion okay. Now if this hydroxyl group attacks back, if this hydroxyl group attacks back you get the same pyronos ring whereas if this tertiary alcohol attacks this carbonyl then you will get furanos ring is not it? So that is what happens here the tertiary alcohol attacks the oxonium ion carbon to get the 5-mombard ring. So now this is very easy to see all the functional groups required for the synthesis of elitrobin is in place. You have a hydroxyl group here okay then the OME is there methyl also and what you need to do is you have to convert this into enol triflate and also the northern hemisphere hydroxyl group you have to esterify okay these are the 2 things he needs to do to complete the total synthesis of elitrobin okay. So what you need to do first is you protected the hydroxyl group okay you protected the hydroxyl group as TBS ether okay so that you can remove the pyrolyl group and make it as vinyl triflate and then couple with tributyltin derivative of the sugar to attach the sugar fragment okay. So you protected the hydroxyl group then you reductively remove the pyrolyte ester so pyrolyte ester can be reductively removed because it is an ester is not it. So if you reduce with the di-ball the pyrolyte ester gets cleaved and then you will get the hydroxyl group hydroxyl group was then oxidized with tetra n-propyl ammonium perruthanate T-BAP and NMO to get the ketone. So once you have the ketone the next step is to make the enol triflate. So this was successfully done by making the enolate with lithium exomethyl diselicide and the triflate was introduced using well known comminceration okay. So now the enol triflate is made so that means the whole carbocyclic core structure of elitrobin is ready and already we have discussed the synthesis of sugar fragment. So what we need to do is we have to couple this triflate with the tin derivative using steely coupling and the steely coupling worked very well with tetra case palladium to give the complete structure of the elitrobin. So what is left now is you have to remove the TBS, attach the side chain then remove the astronaut and acetyl group. So these are the 3 things left for the total synthesis of elitrobin. So first easiest is to remove the TBS group okay treatment with T-BAP you remove the TBS next you have to attach this side chain okay. So this alpha-vita and such an acid is a known compound and now using DCC you can couple this carboxylic acid with the alcohol to introduce the side chain on the northern hemisphere okay. Now what is left you have to remove the acetate and you have to remove the acetone right okay. So both are done in one step actually if you look at the elitrobin this acetate is intact acetate is required in elitrobin only you need to remove this astromate okay you have to remove only the astronite and that was done easily by treatment with PPTS methanol you remove the astronite and that gives elitrobin so that is how you could complete the total synthesis of elitrobin and if you look at the total synthesis of elitrobin reported by Danieczewski so he started with two commercially available as well as naturally occurring compound for the carbocycle he started with a monotirpin called alpha phylandrine and for the sugar fragment he started with D-arabinose commercially available sugar okay monotirin. Then there are three key reactions which he used one the Nosaki Kishi reaction to form the 10 membered ring and then second the ring expansion of furan ring to pyranose and the third one is the pyranose to furanose that is 6 membered to furanose 5 membered ring was done under acidic condition that is the fourth key reaction that is that still like coupling between the carbocyclic triphlate and then tributyldin tin derivative of sugar fragment okay so these are the four key reactions which Danieczewski used to synthesize elitrobin overall this total synthesis was accomplished in 26 longest linear steps and with an yield of 0.25 percent so considering the molecule considering the complexity of the natural product 0.25 percent overall yield is really a healthy one okay so this way if you look at the two total synthesis which we have discussed on elitrobin, Nikolaou Nikolaou's total synthesis and Danieczewski's total synthesis actually involved some key reactions and also new chemistry were developed during these two total synthesis okay thank you.