A very good evening aspirants, I have an interesting announcement for you to know. 3. Exciting opportunities are knocking on your door to assist you in achieving your dream of clearing the UPSS Civil Services examination. We are here to help you in your prelims and main preparation. Those who have missed to enroll into the most reliable prelims test series offered by Shankaray's academy, your wait is over. Pre-storming batch 3 is starting on November 9th. The first test in this batch will commence on 20th November. Like the other batches, it will also have 66 tests. Now for mains examination, you are all aware about the mains boost 2023, right? Under this, you will be provided 40 mains-oriented tests in 90 days. The booster is a quick plan drafter for you to boost your mains score. It starts on October 31st and will include sectional test, half papers and civil services examination emulators. It is available in both offline and online modes for just 4,500 rupees. On top of that, we are launching a new initiative called SA Augmentor 2023. Under this initiative, you will be provided with four tests to enhance your essays. It is also available in both online and offline modes. You will get a different approach towards essay writing along with pre-essay and post-essay discussions. To further enhance the content of essays, you will be provided with a summarized easy material combined with mentorship. All these for just 6,000 rupees. So, grab these chances to kickstart your problems and mains exam preparation. So, this announcement, now let us move on to the Hindu newspaper analysis. Today's date is 28th October 2022. So, display it here or the list of news articles that we are going to discuss today. So, without much delay, let us get into the news article discussion. Now, take a look at this news article. It talks about the coinage of various kingdoms present in ancient and medieval India. It also deals with Kushan Empire and their coinage system of depicting their gods in their coins. So, let's take this as an opportunity to learn about Kushan Kingdom in detail. So, firstly coming to the question, who were Kushans? See, Kushans were a part of your sea tribe people who populated the present day northwest region of China. Okay? See, because of the expulsion of them from their homelands due to other wiring tribes, they came to occupy the parts of regions of Pakistan, Afghanistan, Uzbekistan and northwestern India. See, the extent of the territory is given in this map. You can have a look at it. And remember the time period within which they ruled this territory extends from 1st century CE to 4th century CE. The founder of the dynasty, Kujula, Khadpis, followed Greek religious ideas and iconography. He also followed traditions of Hinduism being a follower of Saivism while the later rulers adopted Buddhism. Now, coming to the most important king of the Kushan dynasty, I hope you would have all heard about the name Kanishka, right? He was the ruler under whom the Kushan Empire reached its zenith. He ascended the throne in 127 CE and ruled it till 150 CE. Here note that Kushans controlled the famous silk root which resulted in economic well-being. As a result of this, they minted a huge number of gold coins. Even though Indo-Greeks were the first to introduce gold coins in India, Kushans issued the most number of gold coins with high intrinsic value. So, these coins only had the images of gods on one side with the opposite side having the ruler's portrait. Here the most important fact to notice, the 4th Buddhist council was held under the patronage of Kanishka. This resulted in Buddhism spreading to far wider areas in central and eastern Asian regions. Other than this, the famous Gandhara art form was patronized by the Kushans. So, the scintillating sculptures of Buddha owes their existence to the Kushans. The famous fasting Buddha sculpture belongs to the Kushan Empire. So, now coming to the capital city of Kushans. See, Kushans ruled with two capitals. One at Purushapura, the present day Peshawar and the other capital at Madhura. So, now coming to the end of the Kushan Empire. See, the empire started disintegrating in the first half of the 4th century CE. The resultant political vacuum was filled by the emergence of the Guptan Empire in the central Indian region. So, this is all you have to know about Kushan Empire with respect to Prevenor examination. So, in this discussion, we saw in detail about the origin of Kushans. We saw that their time period extended from 1st century CE to 4th century CE. We saw that the founder of the dynasty is Kujula, cut pieces. Then we saw about Kanishka. We saw that 4th Buddhist council was held under the patronage of Kanishka. And then we saw some of the important details that we have to remember with respect to Kushan Empire. So, with these learnt points, let us move on to the next news article discussion. Now, take a look at this news article. It talks about the newly introduced Indian Telecom Bill 2022. The article reports that the provisions about Troy in the said bill is now being considered by the government for removal. Also, the article says that a new bill regarding Troy is in plans of the government, which is to be introduced within the next two years. So, this is the crux of the news article given here. In this context, let us learn about Troy in problems perspective. Firstly, let us see the abbreviation of Troy. See, Troy stands for Telecom Regulatory Authority of India. See, we know that the entry of private service providers in the telecom industry brought an inevitable need for independent regulation. So, Troy was established in 1997 by an act of parliament called the Telecom Regulatory Authority of India Act 1997. So, from this, what you can infer? Yes, Troy is a statutory body. It aims to regulate telecom services including fixation or revision of tariffs for telecom services which were earlier listed in the central government. Okay. Now, coming to the issue discussed in the news article, see to address spam calls and spam messages, the draft telecom bill provides that communications of commercial nature like advertisement and promotion shall be made only with the prior consent of a subscriber or the consumer. See, earlier this was dealt by the Telecom Regulatory Authority of India that is Troy, but the draft bill takes it out from the purview of Troy and gives the government the power to take stringent measures against such violators. The other issue is that the government would no longer be required to seek recommendations from the Troy before issuing licenses to telecommunication service providers. So, because of these provisions, the primary responsibility of Troy which is to regulate the telecommunication industry is watered down. This led to concerns from various sections of the civil society. Because of this outrage, the government now plans to do away with all the amended provisions relating to Troy presented in the set draft bill. So, this is all about the issue discussed in the news article. Additionally, we shall see some of the facts relating to Troy. See, the headquarters of Troy is located at New Delhi and the Troy's mission is to create and nurture conditions for growth of telecommunications in the country so that it will enable India to play a leading role in emerging global information society. One of the main objective of Troy is to provide a fair and transparent policy environment which promotes a level playing field and facilitates fair competition. So, in perseverance of this objective, Troy has issued from time to time a large number of regulations, orders and directives. These regulations or orders provided the required direction to the evolution of Indian telecom market. This has helped the Indian telecom market to move from a government owned monopoly to an open competitive market. These directions, orders and regulations issued cover a wide range of subjects including tariff, interconnection and quality of service as well as governance of the authority. So, now coming to the important point, note that the recommendations made by the Troy are not binding on the central government. So, that's all you have to know about Troy. See, in this news article discussion, we saw in detail about telecom regulatory authority of India. Make note of all these points very, very important from the preliminary point of view. So, these learnt points, now let us move on to the next news article discussion. Now, take a look at this news article. It talks about the issue of single use plastics. Now, suddenly it is in use because Tamil Nadu Pollution Control Board that is TNPCB has issued orders for the strict enforcement on the ban of single use plastics. The article reports that the Central Pollution Control Board has written to all state pollution control boards previously asking them to crack down on single use plastics through an intensive drive till December this year. So, this is the cracks of the news article given here. In this context, let us see what is single use plastic and also the effects of it on the environment. Now, coming to the term single use plastics. See, single use plastics or low utility, high littering potential plastics which are used and disposed by humans in a very short period of time. To be precise, they are used only once before they are disposed of. Some of the examples of a single use plastics are plastic carry bags, plastic flags, plastic sheets used for foot wrapping, spreading on dining table, plastic plates, plastic coated tea cups, plastic water packets, plastic straws, etc. Remember to uphold the spirit of Azadika Amrit Mahotsa, a defining step to curb pollution caused by littering and unmanaged plastic waste was taken by the Union government. What it did? It banned the manufacture, import, stocking, distribution, sale and use of identified single use plastic items all over the country from July 1, 2022. So, to implement the ban strictly across India, Central Pollution Control Board had been issuing directions towards state counterparts. So, this is what is given in today's news article.  Firstly, human health. See, human health is harmed by single use plastics. When left to degrade, single use plastics leech many harmful chemicals into the environment like bisphenol A. They release endocrine disrupting chemicals that bio-chimulate in humans and can cause cancer and impact reproductive hormones. Secondly, single use plastics which are also micro plastics get in the mouth, stomachs and digestive systems of fishes, birds and animals. This is making it hard for them to breathe and eat. The non-decomposition of all these plastics becomes part of the food chain of aquatic and subsequently human life also. Now, thirdly, as you know, single use plastic is produced from fossil viewers. The process of extracting and creating these plastics emits huge amount of greenhouse gases. Therefore, single use plastics increases the amount of greenhouse gases in the atmosphere. So, this is all regarding the effects of single use plastics. So, through this discussion, we came to know about single use plastics and the ill effects of single use plastics. So, with these learned points, now let us move on to the next news article discussion. Now, have a look at this editorial article. See, this editorial article says that the number of government schemes are reduced and this is done by either slashing the schemes or by cutting funds to the scheme or by merging several schemes into one umbrella scheme. The reduction in the government schemes have done even by improper utilization of funds. So, in this news article discussion, let us discuss some of the important facts given in the news article discussion. You can use all these details as a value addition in your main examination. That is why we have chosen this news article. So, now before getting into the news article discussion, the syllabus relevant to this news article is highlighted here for your reference. You can go through it. So, to begin with, see the article says that over the past three years, over 50% of existing central government sponsored schemes have been discontinued, subsumed, revamped or rationalized into other schemes. So, here you have to know about what is this centrally sponsored scheme. See, centrally sponsored scheme or CSS is nothing but a certain percentage of the funding is borne by the states and the implementation is by the state government. While there is another type of scheme which is called as central sector scheme, in central sector scheme 100% of funding is done by union government and it is implemented by the central government missionary. Okay. So, just have this clarity. This article talks only about centrally sponsored scheme. So, now coming back, the author says that this reduction in schemes has various impacts across different ministries. To understand this better, let us see an example. Take union ministry of woman and child development. Currently, there is only three schemes under this ministry. They are the mission Shakti, mission Vatsalya and Poshar 2.0. Do you think it is sufficient? No, right. See, India is a country where women and children contribute 67.7% of its population. So, their empowerment, protection and their wholesome development in a safe and secure environment is crucial for sustainable and equitable development of the country, right? If this happens only, then we can achieve transformational, economic and social changes. Now, when all the schemes are subsumed under one or when certain schemes are discarded abruptly, how will the objectives of these schemes will be met? It is a question, right? Likewise, take the example of mission Shakti. This scheme alone replaced 14 schemes like the beti bachau, beti padau, scheme, woman helpline, WHL and many more. So, here what is happening is, see, under mission Shakti, there are two subschames. One is sambal and other is samathya. Under the sambal subscheme, which is for safety and security of women, the erstwhile schemes like one stop centre that is OAC, women helpline, WHL, beti bachau, beti padau, BB, BP or subsumed. And under the samathya subscheme, which is for empowerment of women, the erstwhile schemes like Ujwala, swadhaar, gray and working women hostel have been included with modifications. Then, even the existing national creche scheme for children of working mothers and Pradhan Mantri Matruvandrayojana, that is PM-M-VY, which were under the umbrella scheme of integrated child development scheme, ICDS, have now been included in samathya. So, here the author says that, this kind of reduction in scheme is not good. Also, to support his view, he quotes some more schemes as examples. Let us see them one by one. Take the example of Nirbaya fund. See, we all know that Nirbaya fund is a dedicated fund set up post 2012, that is after the Nirbaya gang rape case. It was set up with the focus on implementing the initiatives aimed at improving the security and safety of women in India. The fund was called Nirbaya fund and it included schemes like One Stop Center, Universalization of Women Helpline, Mahila Police Volunteer, then the Emergency Response Support System, Central Victim Compensation Fund and many more. So, what happened to this scheme is, see rupees 1000 crore was allotted to the fund annually from 2013, but it all remained largely unspent. For example, as of financial year 2021-22, approximately 6214 crore rupees was allotted to the fund since its launch. But the dispersed amount was only 4138 crore and among the dispersed amount, just 2922 crore was actually utilized and the remaining 660 crore rupees was dispersed to the Ministry of Women and Child Development. In that also, only 181 crore rupees was utilized as of July 2021. So, here from the schemes we have discussed so far, you can come to a conclusion that schemes are subsumed in various other umbrella schemes and hence the actual funding is reduced. But women, they still continue to face significant risk in public spaces. This is the sad reality. Now next, let us take the case of fertilizer subsidies. See, this is also declining over the last few years. See, the actual government spending on fertilizers in financial year 2022-2021 reached 127921 crore rupees. But in the following year, that is financial year 2021-22 budget, the allocation was 79,529 crore rupees only. But later it was revised to 140122 crore rupees. Now, again in financial year 2022-2023 budget, the allocation is reduced. In that, the allocation for NPK fertilizers, that is nitrogen, phosphorus and potassium fertilizers was 35% lower than revised estimates in financial year 2021-22. See, this kind of budgetary cuts have led to fertilizer shortages and former anguish. This is because of the rise in fertilizer price after the Ukraine war. So, here are the other questions. How will we incentivize farmers to continue agricultural operations? Now, not only this, take the Mahatma Gandhi National Rural Employment Guarantee Act, that is MGNREGA. Even for this, the fund allocation is reduced. Yes, the fund allocation went down by approximately 25% in the financial year 2022-23 budget. But when you take the demand for this scheme according to the economic survey 2022-23, the demand was higher than pre-pandemic levels. Also, it is said that the actual funding dispersal for this MGNREGA has often been delayed. So, this kind of fund cuts and improper disbursement leads to a decline in confidence over the scheme. Then, take the example of Garib Kalyar, Rojgal, Abhyan. See, it is a massive rural public works scheme to empower and provide livelihood opportunities to the returning migrant workers and rural citizens. It was launched on June 2020 and it was a 125-day Abhyan with a resource envelope of 50,000 crore rupees. But what happened is, approximately 50.78 crore person days of employment were provided at an expenditure of approximately 39,293 crore rupees. So here, you must note that this scheme subsumed 15 other schemes. The author regards this as a black mark because over 60 million to 100 million migrant workers who were seeking informal jobs were left out. And the fund allocation is further shrinked. The issue does not stop there. Even in healthcare too, there was reduction of schemes. For example, the agritated social health activist that is Asha, who are the first reporters, they have been delayed in salaries for up to six months. And even regularization of their jobs continues to be a struggle with wages and payments struck at minimum levels. So, Asha workers whose family is solely dependent on this salary suffers. So by highlighting all these examples, the author concludes by saying that rather than downsizing government schemes and cutting funds, one should right-size the government. That is, we should rise our aspirations towards better public service delivery. For example, we need to build capacity for an efficient civil service to meet today's challenges. The government can provide a corruption-free welfare system by running a modern economy and providing better public goods. So, this is all you have to know about this news article discussion. Now, I presented all the data for you to understand the current reality. You really don't have to memorize all the data. You can get an idea about what was happening before and what is going on now. I hope this discussion is very helpful for you. So, with these learnt points, let us move on to the next news article discussion. Have a look at this news article. See, this news article talks about the hate speech.  Now, let us not get deep into the issue as it is not very important.  And then we shall also see some of the laws related to hate speech in India. So, let's start with the question, what is hate speech? See, there is no specific definition of hate speech. But in the 267th report of the Law Commission of India, hate speech is defined as an incitement to hatred primarily against a group of persons.  So, in simple words, hate speech is any word written or spoken, sign or visible representations within the hearing or sight of a person with the intention to cause fear or allurem or incitement to violence. So, here what you have to notice, some of the provisions in law actually criminalize all actions like speeches, writings, signs and the presentations that create violence and spread this harmony between communities and groups. So, now let us see some of the legal provisions regarding these hate speech in India. See, provisions related to hate speech can be found both under Indian Penal Code that is IPC and under Representation of People's Act. See, under IPC there are three sections which you have to make note of. First are Sections 153A and 153B of the IPC. Both the section punishes acts that cause enmity and hatred between two groups. Then comes the section 295A of the IPC. This section deals with punishing acts which deliberately or with malicious intention outrage the religious feelings of a class of persons. Now, the third section that you have to remember are Sections 5051 and 5052. See, both make the publication and circulation of content which may cause ill will or hatred between different groups and offence. So, this is regarding IPC. Now, under Representation of People's Act in Sections 123 Clause 3A and 125 of the Representation of People Act. Both bars the promotion of animosity on the grounds of race, religion, community, caste or language in reference to elections and included under corrupt electoral practices. So, these are all some of the safeguards provided against hate speech in our Indian Penal Code and Representation of People's Act.  See, there are numerous reasons for that. First is internal security. I hope you all remember about Musafar Nahar Royce of 2013. It was triggered by a fake video that incited communal passions. So, to prevent such kind of spread of hatred among two communities and to maintain internal security, these laws are very important.  Now, thirdly, to prevent mob lynching and spread of misinformation and disinformation, curbing the hate speech is very, very important. I hope you all remember the Delhi Royce. The Delhi Royce is a very good example of misinformation and disinformation.  So, in this news article discussion, we saw in detail about what is the hate speech and we saw about some of the legal protection under IPC and RPA and then we saw why hate speech must be curbed. So, these learned points. Now, let us move on to the next part of the news article discussion which is the preliminary practice question discussion. Now, look at this question. Which of the following kingdoms ruled the northwestern part of Indian subcontinent during the post-Maurian period? First kingdom is Parthians, Satavahanas, Kushans, Sunga and Kanwa. So, you have to pick the correct pair here. See, the correct answer for the question is option B1 and 3 only. See, Parthians ruled the northwestern part of India. They were people of Iranian origin. So, statement one is actually correct. Statement two is wrong because Satavahanas, they ruled the present-day Deccan region. They did not rule northwestern part of Indian subcontinent. So, this statement is wrong and from our discussion itself, you could have known that Kushanas ruled over the northwestern part of India. So, this statement is correct. Now, coming to last two options, both these two options are incorrect because Sunga and Kanwa dynasty ruled over the central Indian region. So, the correct answer for the question is option B1 and 3 only. Now, moving on, this question is about Troy. Statement one, it is a statutory body and statement two, it is mandatory for the government to implement the recommendations of Troy. So, you have to choose the correct statement here. See, from our discussion itself, we know that statement one is actually correct. It was established as a statutory body in the year 1997 by Telecom Regulatory Authority of India Act. Statement two is incorrect because from the discussion itself, we saw that government does not have to listen to Troy's recommendations. Even though Troy functions as an independent regulator. So, here the correct answer for the question is option A1 only. Now, the two questions displayed here are the quiz question for you today. One question is a previous year question. Try to answer that in the comment section and the other is regarding hate speech. The question displayed here is the main practice question for you today. Just go through the question, write an answer and post the answer in the comment section. So, with this we came to the end of the news article discussion. If you like the video, hit like, do comment and don't forget to subscribe to Shankarae's Academy YouTube channel. Thank you for listening.


Good morning Hank, it's Tuesday. I'm at the beach and it's very pretty here, but my vacation has been completely consumed by this book, The Parable of the Solar by Octavia Butler, which is the best book I've read in- I don't know, in years. Like every now and again I'll read a book and literally feel like a spell has been cast upon me that allows me to escape the prison of myself. And for a little while I will not feel stuck inside my own consciousness, but instead I'll feel like I have magical access to worlds outside of me. And I find that feeling to be a tremendous consolation. And then when the book ends, I'm left with this weird wabi-sabi feeling, like I'm overwhelmed with the mysterium tremendous of having seen and having been seen. And I also feel this big sadness in my chest, maybe partly because I know I'll never get to read that book again for the first time, but also partly because I don't know how long it'll be until I come across a novel again that properly turns me inside out. Now of course Hank, the books that leave me, Florida, aren't always going to be the ones that leave you, Florida, but I do think you would love this particular book if you haven't read it already. The Parable of the Solar is a hard book to summarize. It's a coming-of-age story. It's also a dystopian novel about the collapse of the United States. It's about race and class and gender and wealth. And it's also about how religions form, and even though it's more than 20 years old, Hank, it is terrifyingly contemporary. Right, but so anyway, throughout this very page-turny, roaring, adventure-y novel, the narrator writes short poems about her understanding of God, which she defines as change. I usually dislike that kind of novelistic device, but Butler is such a beautiful writer that it succeeds brilliantly. Like, consider this magnificent short poem, Hank. In order to rise from its own ashes, a phoenix first must burn. I want to share just two more lines with you. First, like the Great Gatsby, the Parable of the Solar begins with a fictional epigraph, which concludes, without positive obsession, there is nothing at all. Hank, one of my favorite things about you is your positive obsession, a.k.a. your nerdiness. Like, you have this ability to be unironically enthusiastic and to find stuff that you can be strongly in favor of. And you aren't the kind of person who tears down without building up. Like, when we were in high school, I felt like in order to survive, I had to create ironic distance between myself and the world. You know, like, if you told me you liked a band, I would be like, yeah, their old stuff was good, but now they've sold out. Or if you told me you liked an author, I would tell you what was wrong with their work, etc. But Hank, you've encouraged me to let go of that, and throughout my entire adult life, you've really been a model for positive obsession, kind of even a guiding star at times. So, thank you. The second line is a simple one, as the narrator's life is falling apart, she writes, kindness eases change. Hank, I'm really sorry that Lemon is dying. I know that she's just a dog, but she's a great dog. And in the comments of your video on Friday, I saw so many people whose kindness is easing this awful change. It made me think about Lemon's life. Like, she was a racing greyhound, which often isn't the best dog life, but then when she retired, she came into your life. And just as you've made Lemon's life better, she's made your life better. We often imagine the world as a zero-sum game, as a place where good news for someone is necessarily bad news for someone else. But I believe that kindness can change that arithmetic, and as evidence, I would point to the love that Lemon's life has brought to you and to many. The internet can be a hard place to find kindness, maybe especially right now, but I believe it's worth fighting for because I believe it makes the world bigger and that it eases the hard but inevitable changes that define life. Nerdfighters, thank you for your kindness. Hank, I love you. I love Lemon. I'll see you on Friday.


Hello guys, we are here to discuss the J main paper which has been asked in this few days back. So this paper which I chose today, it is of 9th of January 2nd shift. Before starting the discussion here, let me tell you first thing. These questions are memory based questions, so you may see some difference from the original paper. But more or less the question paper will be same. So here in this session we are going to discuss the first 10 questions of chemistry. The first question that has been asked is which is the following form nitride. So you see the question here is the formation of nitride. We know the general form because all these elements belong to group 1. So the elements of group 1 we have here. And the general formula of nitride if I write, it will be M3N type. Suppose if lithium it is there, the lithium will form Li3N, K3N. Like this we can have the formula of nitride. Now the question is which will form nitride in all this. You see as we go, if you talk about lithium, potassium, rubidium and cesium, they are going actually down the group. So as we go down the group, we know size increases. The size of all these elements increases as we go down the group. So lithium has the smallest size. And hence the charge density on these ions. If you find the charge density on lithium here, that will be maximum. So what we can write, the charge density of lithium will be maximum. And that is why the bond between lithium and nitrogen is the strongest over here. So what happens here when this lithium and nitride form, so this reaction at room temperature of lithium and nitrogen, this evolves a huge amount of heat and this is an exothermic reaction. Because of high lattice energy of this molecule, high lattice energy. So what we can say because of the high lattice energy, the nitride of lithium is stable but that is not found to be the same case with all the elements as we go down the group. So all these elements will not form stable nitride except lithium. The answer will be option 1. Now the next question is in which of the following bond order increases and magnetic behavior changes from paramagnetic to diamagnetic. So first of all, this question is based on molecular orbital theory. The concept that involves in this is molecular orbital. So according to that we know there are bonding and anti-bonding molecular orbital. Bonding molecular orbital and anti-bonding molecular orbital. And the bond order formula we have is equals to the number of electrons in bonding molecular orbital minus the number of electrons in anti-bonding molecular orbital. This is divided by 2. This is the formula we have. So we can draw the electronic configuration here according to these orbital which is sigma 1s, sigma star 1s, sigma 2s, sigma star 2s and so on. Now we can fill the electron into this and we can find out the number of electrons present in bonding molecular orbital and anti-bonding molecular orbital. So you see the question now coming back to this question, these are all theory or concepts required to solve this question. Now if you are coming back to this question, O2 molecule we know it is a known fact that oxygen is a paramagnetic molecule. So first option that you have is O2 to O2 plus. Now paramagnetic is because of what? Unpaired electron. So this is paramagnetic we have and in this the number of unpaired electron is 2. This is also paramagnetic. Number of unpaired electron is 2 here also. Similarly we can find out for NO and NO plus. We can fill the electron into these orbital and we can find out this bond order actually. So for paramagnetic behavior we must have unpaired electron and diamagnetic all the electrons must be paired. So the correct answer in this, if you find out the bond order of NO, it is found to be 2.5 or here also we have 2.5. Number of unpaired electron or bond order is 2.5, here the bond order is 2. It is not the number of unpaired electron. And NO converts into what? NO plus. So we are removing one electron into this and hence the bond order here you will get is 3. So this is paramagnetic and this is diamagnetic. So this is what the answer we have. NO converts into NO plus. The magnetic behavior changes from para to dia and bond order is also increasing here. Correct? Option 2 is right. Question number 3. M3 plus ions show blue, green, red color with ligands L1, L2, L3 respectively. Arrange the ligand in their strength. So in this the concept that is required here that we know the strength of any ligand, strength of ligand is directly proportional to the absorption energy or delta of absorption. Now you see here the question is the color is blue, green and red. So the wavelength of emission because it shows blue, green and red color, the wavelength that emitted in this process is we can write lambda for blue, then lambda for green and then lambda for red. Correct? So like this the wavelength increases. Red color will have highest wavelength. So this is the wavelength emitted. So the wavelength emitted is this. So this wavelength has to observe to emit this. So del of absorption if you write down, del of absorption it will be again minimum for the blue light. Then we have green and then we have red. Emission if you write del of emission that will be reverse of this, which is nothing but we have del V maximum, del R sorry del G here, del G and not lambda G and lambda R. This is the emission we have. Strength of wavelength, strength of ligand is that proportional to the absorption right here. So for B it is maximum then green and then for B it is minimum and then G and then red. So maximum is for red and for red it is L3. L3 is maximum we have only one of two options. These two we can eliminate after L3 we have L2. So L3, L2 and L1 option D is correct here. So strength of any ligand that you have is that proportional to the absorption value that you have. And more wavelength, more will be the absorption. So that's the concept we need to have to solve this question. Question number three, answer is option C. Which of the following transition elements will have enthalpy of atomization? Okay, will have enthalpy of atomization. One thing is missing here will have least enthalpy of atomization. The question is will have least enthalpy of atomization. Okay, this is actually a factual question we have right. And the elements are vanadium, first of all we have vanadium in 3D series. Then we have iron, copper and zinc. So first of all if you see the energy here or enthalpy required for atomization. For vanadium it is 515 kilojoule per mole. These data I am giving you iron it is 416 kilojoule per mole. For copper it is 339 and for zinc it is 126 kilojoule per mole. Okay, so clearly if you see the answer for this question will be option three. Enthalpy of atomization is minimum for zinc according to this data. Another thing is what this actually this facts depends on the size which is you know decreases as we go from left to right to a certain pair to a certain element. And then it keeps on increasing also. First it decreases correct and then it becomes constant in between and for last two elements it increases. Right, so zinc is the last element of 3D series we have. Correct, that is why from this the electron the removal of electron is easier and hence its enthalpy of atomization is minimum. Answer is option C. Unless you see by which iron arsenic sulphide is maximum coagulated. Right, this you see arsenic sulphide is maximum coagulated. Right, now the soul that you have here is arsenic sulphide AS2S3 it is the soul we have. Now if you know the nature of the soul what charge we have here we can easily do this question. Right, arsenic sulphide is a negative soul actually negative soul. Okay, so if it is a negative soul so it requires positive charge for coagulation for coagulation. Right, and we know more positive charge more will be the coagulation. Right, so coagulating power coagulating power is directly proportional to the magnitude of the magnitude of positive charge. Why positive charge because this is a negative soul. Okay, this is a negative soul.  So hence we can easily say that this aluminium has maximum positive charge here which is AL3 plus. Right, this is AL3 plus and that's why the answer will be this because they are asking maximum coagulation. Right, if they ask you minimum coagulation then we can have any answer possible between 1, 2 and 4. So answer is option C. The only thing you should know to solve this question is the charge of this soul which is a negative soul. Then you can do this question. Next question, S2PO2 is a strong reducing agent. S2PO2 is a strong reducing agent. Okay, again I will do some correction into this. The question is S3PO4, S3PO2 sorry. S3PO2 is a strong reducing agent reason for this what we have. So for this question you should know the structure of S3PO2. Okay, so the structure will draw is P double bond O. We know in all these oxy acids of phosphorus one P double bond O is always present. Right, all other oxygen is attached with hydrogen if it is there. Right, so we have one oxygen only so one OH and other two hydrogen will be attached directly to this phosphorus. Right, now reducing agent is what? Reducing agent is because of, see if you take this OH bond here. Right, so this molecule has basicity, this molecule has basicity one. Why because we have only one OH bond? That's why basicity is one. Right, so oxidizing nature of this compound is because of this OH bond. Right, so reducing nature will be what? Because of this pH bond it will show reducing nature. Right, so this will oxidize other molecule because of this hydrogen present, but this will reduce others because of this hydrogen that you have here directly attached. Correct, so we have two pH bond and that's why it shows the reducing agent. So hence the answer will be option two. The only thing you require in this question is the structure of H3 CO2. Which is nothing but this. Next, copper crystallize in FCC lattice with unit cell edge length is given, calculate density. Right, so what is the formula of density? First of all I will write down the formula then we will see what is the data given here. So density D is equals to ZM divided by Avogadro number NA edge length Q. Right, where Z is the number of effective atoms, number of effective atoms. M is the atomic mass, if atom is there atomic mass. Do you see what all things are given here? Edge length is given X, FCC lattice we have. So FCC lattice we know we have eight corners and six phase centers. The value of Z from this if you calculate it is 8 into 1 by 8 plus 6 into 1 by 2, which is nothing but 4. Now this value we have to substitute here and we will get the answer. So density will be 4 into 63.5 6 into 10 to the power 23 into X angstrom it is given, X into 10 to the power minus 8 whole cube, right centimeter cube. So this will be equals to when you solve this you will get 421.71 by X cube gram per centimeter cube, okay. Option A is correct, right, direct formula we have to use here. Next question, okay, this reaction is given if at 300 Kelvin temperature E naught of the cell is also given 2 volt and find out the equilibrium constant, okay. So we know the relation of del G, del G is equals to del G naught plus RT ln cube, okay. Now it is given E naught of the cell is given and we know also know del G is equals to minus nfe this is the formula we have. If it is del G naught, then it is E naught is standard state, correct. And since we are talking about the equilibrium constant, so at equilibrium what we can write? Del G is equals to 0, right. At equilibrium del G is equals to 0 and Q is nothing but Kc equilibrium constant. These two formula if you apply here, so we will get del G naught is equals to minus RT ln Kc and del G naught is minus nfe naught is equals to minus RT ln Kc. The value of n is 2, right, one faraday it is given, right, this we have to assume. So n value is 2, right, 2 into f is 9600, E naught is also given 2, R is the value is given 8, temperature is 300 ln K, right. So when you solve this, you will get K is equals to e to the power 1, 6, 0. No theory, only calculation we have here. The formula is this and when we use this formula, the answer will be option 1, e to the power 1, 6, 0, right. Next question, the reaction is given 2A plus B gives C, if the initial rate of the reaction is 0.3 molar per second. On doubling the concentration of A and B, the rate becomes 2.4 molar per second. Doubling the concentration of A alone, it is 0.6 in order of the reaction. Okay, suppose the rate law we can write for this reaction is R is equals to K A to the power X and B to the power Y. This is the rate law I am taking. If you know this X and Y value, the order of this reaction will be what? X plus Y, we will add these two. So we have to find out this X and Y, correct. Now see, initial rate is this, correct. So first of all, when rate is 0.3 is equals to K, concentration of A to the power X, concentration of B to the power Y. When concentration of both increases twice, right, so the rate becomes 2.4. 2.4 is equals to K 2A to the power X and 2B to the power Y. When we double the concentration of A, rate becomes 0.6 is equals to K 2A concentration to the power X, B to the power Y. Okay, you see first and third equation if you compare, right. So rate becomes twice when we increase the concentration by 100%, right. Or when the concentration of A becomes twice, rate also becomes twice. You can easily solve these two equations and you will get the value of X as 1, right. Similarly, you can use these two equations and you will get the value of Y also that is 2. So basically, you need to solve this first, second and third equation. You will get X and Y value, the order will be with respect to A it is 1, with respect to B it is 2. The overall order, overall order is 3. What we can write, overall order X plus Y is equals to 3 with respect to B it is 2 with respect to A it is 1, right. It is 2, first one is wrong, second one is right. This is wrong, this is wrong. Option 2 is incorrect. And the mass of H2O formed, the combustion of this gamma, this takes place, right. So you see the molecule here, suppose if I take CX, write down the same thing. We will take the given molecule only. You see the molecule is C57H11046 and combustion of this. So we have O2, gives CO2. And if car number of carbon is 57, we will multiply here 57, carbon gets balanced. Hydrogen is 110, to balance hydrogen it should be 55 H2O. Now oxygen is 55, 114 plus 55, that is 169 approximately. You will have 163 by 263 plus 669. Number of moles of this molecule, moles of this will be equals to the mass taken divided by the molecular mass of this. You can calculate the molecular mass of this, 12 into 57, 1 into 110 plus 6 into 16. And when you solve this, you will get 445 divided by 890, which is 1 by 1 moles gives 55 moles of this. So half moles gives number of moles of H2O. So 1 gives 55, so half gives half of 55, that is 27.5, but you have to find out the mass of this. So mass of H2O will be half into 55 into 18, that will be 495 gram of H2O. Answer is option 2. These are the first 10 questions we have discussed. We will come back to another question very soon. Thank you.


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Hello, so welcome back and we continue from where we left off. So, we derive this celebrated functional equation of Bernard Riemann. Riemann wrote his 8 page memoir on the distribution of primes and the zeta function and in that memoir he gives 2 different proofs of this functional equation and we have given a proof which is Fourier analysis based which is the theme of this particular course. Now, we have derived this in the strip 0 less than real part of z less than 1. Now, if we believe that the zeta function can be continued analytically throughout c minus 1, then by the principle of analytic continuation or the identity theorem as it is known this equation 1.42 will hold for all values of s in the plane minus 0 1. But we have not proved the analytic continuation of the zeta function on c minus 1 on the complex plane minus 1, we only establish the analytic continuation in real part of s bigger than 0 that is what we wanted. Now, I shall indicate in a series of exercises how to do the analytic continuation of the zeta function on the complex plane punctured at 1. Again we start with a very simple observation. It is a Laplace transform or a gamma function formula integral 0 to infinity e to the power minus n t t to the power s minus 1 dt equals gamma s into n to the power minus s equation 1.43 that you see in the display. Again I am going to put n equal to 1 2 3 etcetera and I am going to add. Again you have the task of justifying carefully this exchange of integration and infinite sums that I am leading it to you to do and on the right hand side what am I going to get? I am going to get gamma s into zeta s. So, I have proved that gamma s into zeta of s equal to integral 0 to infinity t to the power s minus 1 dt upon e to the power t minus 1. The summation that you are going to get when you put n equal to 1 2 3 etcetera and add up is going to simplify to 1 upon e to the power t minus 1. Now, what I do is that I write this t to the power s minus 1 s t to the power s minus 2 times t and the other factor t by e to the power t minus 1 I am going to call it phi of t. So, now the next exercise is look at this function phi of t equal to t upon e to the power t minus 1 for t not equal to 0 phi of 0 equal to 1. With this prescription at the origin this function phi of t is actually an infinitely differentiable function of the real line. You can think of this function phi as a holomorphic function of t. Note that 1 upon e to the power t minus 1 has a simple pole at t equal to 0 and the numerator has a simple 0. So, this function phi of t will be holomorphic in a strip containing the real line. Of course, when t equal to 2 pi i it will still have a pole. So, this phi of t is not some entire function or anything like that this phi of t has a pole at t equal to 2 pi i etc. But we are not interested in that we are interested in the behavior on the real line. In fact, we are interested in the behavior in the positive real line and if you have holomorphy in a neighborhood well and good in particular it is infinitely differentiable. So, let us go further and see now what you do is you perform an integration by parts. You take this equation gamma s zeta s equal to integral 0 to infinity t to the power s minus 2 phi t. Write t to the power s minus 2 as d dt of t to the power s minus 1 divided by s minus 1. This is derivative. Of course, all this drama holds when real part of s is bigger than 1 because the series that we obtained by summing with respect to n is valid only for real part of s bigger than 1. So, this equation 1.44 is valid for real part of s bigger than 1 alright. So, perform an integration by parts perform an integration by parts and show that for real part of s bigger than 1, we still have zeta of s equal to minus 1 upon s minus 1 gamma s integral 0 to 1 t to the power s minus 1 phi prime t dt. So, this 1 upon s minus 1 we picked up because of integration by parts. Now, observe that this right hand side of 1.45 this right hand side of 1.45 is going to be holomorphic on real part of s bigger than 0 except when s equal to 1 this right hand side of 1.45 is holomorphic as a function of s. Of course, you will ask a very crucial question. How do I know that the gamma function does not have any zeros in the right half plane when s is real positive the gamma function is a integral of a positive function. So, the gamma function does not vanish on the positive real line, but here the s is complex real part of s is bigger than 1 s is complex. So, how do I know that this gamma function does not have any zeros there. Further I am going to look at the right hand side of 1.45 not only on real part of s bigger than 1, I am in fact going to look at it on real part of s bigger than 0. So, I need to be sure that this gamma function that appears in the denominator does not vanish on real part of s bigger than 0. How do I know that? There are several ways of looking at it. One very simple way to see that is to remember the Euler's formula gamma s into gamma 1 minus s, gamma s into gamma 1 minus s is pi cosecant pi s, gamma s into gamma 1 minus s is pi cosecant pi s that is called the Euler's reflection formula. But looking at this Euler's reflection formula for the gamma function you know that the gamma function has no zeros the gamma function cannot have any zeros. So, the right hand side of 1.45 as far as these factors are concerned minus 1 upon s minus 1 gamma s they are no problem except that s equal to 1 is forbidden. What about this integral? Looking at this integral you need to look at the derivative of phi t phi the derivative of phi t is going to decay very rapidly as t goes to infinity. So, near infinity there is no problem near the origin also you have to show that there is no problem that this particular object here will be a holomorphic function on real part of s bigger than 0. So, the equation 1.45 immediately gives you an analytic continuation of the zeta function on the larger domain real part of s bigger than 0. Now, what you can do is you can perform further integration by parts you can do d dt of t to the power s d dt of t to the power s is t to the power s minus 1 s times t to the power s minus 1.  Of course, when I finish integration by parts the derivative will fall on phi prime and you will get a phi double prime. So, when I repeat this process k times when I repeat this process k times I will get a total of k plus 1 derivatives on phi and every time these factors that you get on the the denominator s s plus 1 s plus 2 they can be combined with the gamma function and I will get gamma s plus k. This s minus 1 will remain it will persist and every time you perform an integration by parts you will become minus sign.  The phi can be written in terms of the hyperbolic cotangent the phi can be written in terms of hyperbolic cotangent after you do that when k is bigger than 1 when k is bigger than 1 there will be a t part which will go away and you will get this equation 1.46. So, 1.46 is 2 for k equal to 1 2 3 etcetera. So, further integration by parts gives you this formula, but now the right hand side makes perfect sense it is holomorphic in this larger domain real part of s bigger than minus k. So, we have now extended the zeta function from real part of s bigger than 1 to real part of s bigger than 0 and further on to real part of s bigger than minus k with the one single exception that we got this factor of s minus 1 in the denominator. So, s equal to 1 has to be excluded. So, the zeta function is meromorphic on c minus this singleton 1 and 1 is a simple pole 1 is a simple pole and I will leave it to you to calculate the residue at s equal to 1 using equation 1.45. So, we have completed the proof of the analytic continuation of the zeta function and thus we have completed the proof of the functional equation not only on the strip 0 less than real part of z less than 1 where indeed we have completed it on the complex plane punctured at 2.0 and 1. So, that completes this section on the Riemann zeta function. So, we proved the basic convergence theorem for Fourier series and we use that to study two special functions the Jacobi theta function and its functional equation and from there on to the functional equation for the Riemann zeta function. Now, we move to a different special function a very different special function the Bessel's functions or the first kind. You have no doubt encountered the Bessel's functions in your undergraduate education. Where did you encounter the Bessel's function? If you look at Simmons GF Simmons differential equations with applications in historical notes there is a beautiful section on Bessel's functions and with a nice historical commentaries about the various mathematicians who studied the Bessel's functions. So, the Bessel's functions appears in connection with problems on wave propagation that is how you encounter them in undergraduate courses. You will encounter them when you study modulation problems in wave phenomenon. When you study the vibrations of a circular membrane standing waves. So, what do you do? You take the wave equation del 2 u by del t squared minus Laplacian of u equal to 0. What are standing waves? You look for a special solution u of x, y, z, t it is a function of four variables x, y, z are the spatial variable and t is a time variable. So, use separation of variables you write it as e to the power i k t e to the power i k t times another function v of x, y, z substitute it to the wave equation and you will see that this second factor v of x, y, z must satisfy the Hemmholz's equation and when you write the Hemmholz's equation in polar coordinates you are going to get the Bessel's differential equation. So, the vibrations of a circular membrane and when you study standing waves you will get the Bessel's functions. When you study problems involving cylindrical symmetries such as diffraction of light through a circular aperture you know about Newton's rings the diffraction patterns are circular rings of light and dark rings in alternating. So, that those are called the Newton's rings the radii of these Newton rings the radii of these Newton's rings they are expressible in terms of the zeros of the Bessel's functions. Why is it that the Bessel's functions appears in connection with cylindrical symmetries when you write the Laplace's equation in cylindrical coordinates you will see why the Bessel's differential equation would appear. Apart from wave phenomenon Bessel's function also has a tendency to appear in other places unexpected places problems in analytic number theory. The book of Ivanish and Kolevsky analytic number theory published by the American Mathematical Society in 2004 is highly recommended for how Bessel's function makes its appearance in problems in analytic number theory. We shall be looking at a very special problem that appears in celestial mechanics the inversion of the Kepler equation how Bessel's function appears when you look at the problem of two bodies and when you want to invert the Kepler equation then how the Bessel's functions can be used. So, let us begin with the Bessel's differential equation which is an ordinary differential equation and that ordinary differential equation is displayed in the slide x squared y double prime plus xy prime plus x squared minus p squared y equal to 0. Here I am going to assume that p is a real parameter I can also allow the p to be complex but we will not need the complex values of p we will stick to real values of p and we can assume that p is non-negative because p is real and only the square appears in the differential equation. Let us rewrite this differential equation 1.47 as x squared y double prime plus xy prime minus p squared y equals minus x squared y I push the x squared y on the right hand side and I can regard the x squared y as a forcing function or I can regard the minus x squared y as a perturbation term in the differential equation. So, when x is small so if x is the radial coordinate or something and near the center of the circular membrane I can think of this minus x squared y as a small perturbation. So, suppose I ignore this small perturbation for the time being minus x squared y what happens I get a Cauchy Euler equation but everybody knows how to solve the Cauchy Euler equation we teach it in the bachelor's courses at the sophomore level. The solutions are x to the power p and x to the power minus p I am going to ignore the x to the power minus p because I am looking at the vibrations of a circular membrane and the vibrations are finite it is not shooting off to infinity or anything like that. So, take x to the power p with non-negative p and so if x to the power p is a solution of the Cauchy Euler equation we would believe that for the perturbed equation the solution will be x to the power p times something nice and the nice part is a power series a naught plus a 1 x plus a 2 x squared plus dot dot dot and I am going to assume that the a naught is non-zero because if a naught is 0 then essentially it goes out and x comes out as a factor it becomes x to the power p plus 1 and we know that that is not the solution of the Cauchy Euler equation it is x to the power p and x to the power minus p. So, it makes sense to put the condition a naught is not equal to 0. So, with this condition a naught not equal to 0 we can we can compute all the coefficients a 1 a 2 a 3 successively and you have probably done that in your undergraduate courses and you got a power series solution of the Bessel's function except that there is a factor x to the power p in front of it. This series x to the power p times a naught plus a 1 x plus a 2 x squared plus dot dot dot is called the Frobenius series solution of 1.47. So, with a suitable normalization notice that that is a linear differential equation if you have one solution I can multiply the solution by nonzero constant I get another solution and so the choice of a naught I must I must fix the value of a naught to normalize this. So, with suitable normalization we get what is called as the Bessel's function or the first kind. Now, we are going to be interested in the values of p where p is an integer p is an integer. So, I am going to call it k I am not going to use the letter p I am going to use a letter k where k is an integer and k is a nonnegative integer it is a natural number 0 1 2 3 etcetera and the series that I got will be denoted by j k z there is summation k from 0 to infinity minus 1 to the power n upon n factorial into n plus k factorial z upon 2 to the power k plus 2 n. That is equation 1.49 that you see on the slide. The first exercise of course is to study the convergence of this series that is a power series why is it a power series because the k is an integer and k is nonnegative. So, it is a straight away power series. Now, you can use the ratio test the D'Alembert's ratio test and you can prove that this power series converges for all values of z all complex values of z and so the sum is an entire function. Sum is an entire function and you know from complex analysis that the power series can be differentiated term by term within the disc of convergence. So, term by term differentiation is valid. So, what I am suggesting is you multiply this equation 1.49 by z to the power minus k when I multiply by z to the power minus k the z to the power k goes away you simply get z to the power 2 n that is a power series and then you differentiate.  The other exercise is multiply by z to the power k and then do the differentiation you get one more identity z to the power k j k z prime is z to the power k j k minus 1 z. So, these are the two identities that you need to prove.  So, these are some simple algebraic manipulations and I would like you to carry out these exercises. Now, once we do these then we are ready to play with this sequence of Bessel's functions. So, what you have is the following you got for k equal to 0 1 2 3 you have got a bunch of functions j 0 z j 1 z j 2 z this entire sequence once you have a sequence of objects you can cook up the generating function for that sequence. So, the generating function for the Bessel's sequence of Bessel's functions is g z t which is summation k from minus infinity to infinity t to the power k j k z. But now you will ask k is going from minus infinity to infinity whereas, we have defined the Bessel's function only for positive only for non-negative integers k we are defined only for non-negative integers k here negative integers also show up. So, for convenience when k is less than 0 what is the definition of j k z the definition of j k z for k less than 0 is given in 1.50 it is by definition minus 1 to the power k j minus k z. Remember that this is a convention and this definition is only given for negative integers k if k is not an integer the story is completely different and we are not going to discuss that. So, with this convention 1.50 I write down the generating function as an infinite series going from minus infinity to infinity t to the power k j k z. Now the first problem that we have is we need to understand how does this series converge the convergence properties of 1.51 has to be settled before we proceed further. The infinite series 1.51 is a series in t involving both negative and positive powers and the coefficients are holomorphic functions of z. So, we should be looking at this right hand side as a function of t in the punctured plane c minus 0 and z varies over c. So, on these sub-domains how does this series converge we need to prove some kind of convergence uniform convergence on compact subsets. So, the term by term differentiation will be valid we want to differentiate this equation term by term and for that we need to estimate the coefficients. We need some convenient estimate on the size of j k z mod j k z we need to estimate the size of mod j k z. A very coarse estimate will be quite enough for our purpose we do not need a very very fine estimate. There are lots of things known about these Bessel's functions. In fact, there is a very big and massive volume by G N Watson. What we need here is very very elementary estimates on j k z. We shall stop here and we shall derive these estimates in the next capsule. So, this would be a good place to pause. Thank you very much.


Good morning and I will come back to our NPTEL lecture series on classics in total synthesis part 1. So we have been discussing total synthesis of trichunanes and in the last week, we talked about total synthesis of trichunanes using photochemical reaction, for example pattern of bookie reaction as a key reaction to form trichunanes reported by Viresha Ravels group. So we will continue our discussion on total synthesis of trichunanes where photochemical reactions have been used as the key reaction, okay. So today we will talk about maybe 4 or 5 total synthesis and the first one which we will discuss is the total synthesis of silphi, perfol, 16, 5 ohm and this synthesis, particular synthesis was reported by Demuth and Hinskan and interestingly what they have used is oxodipy-methane rearrangement induced by photochemical condition and overall if you look at the whole synthesis though the number of steps are little bit more, the strategy of using oxodipy-methane rearrangement to get the trichunane is a unique one and the starting material that is the diene which is required for the asymmetric Dielsall reaction was started from chiral you know which we will discuss. From the retrosynthetic point of view, this natural product can be obtained from this particular compound, you can see what one has to do is 2 methyl groups as well as the double bond should be introduced that can be done by first introducing a double bond here then you add a methyl group followed by quench with methyl iodide then introduce a double bond. So one can do it in 3 to 4 steps and this ketone can be obtained from another ketone and why this ketone was introduced you will know that is because for the photochemical oxodipy-methane rearrangement this ketone is required and before that this ketone should be reduced and protected and this particular ketone can be obtained from this tetracyclic compound. See now you can see a cyclopropane so this is formed through the oxodipy-methane rearrangement and reductive opening of this cyclopropane followed by quenching with methyl iodide you will get this intermediate okay and the key reaction here is the oxodipy-methane rearrangement of this tricyclic compound and when you look at this tricyclic compound as you know it can be obtained from 4 plus 2 cycloaddition reaction wherein this is the diene okay. This diene can be obtained from commercially available compound as well as one can make this in large quantity from Hages-Paris ketone and that is nothing but if you have this instead of alcohol if you have ketone that is Hages-Paris ketone and that can be obtained from 2 methyl cyclopentane 1,3-dione using L-polline as the catalyst one can do a Michael addition followed by Alder reaction which we normally call it as a Robinson annihilation sequence okay. So the starting material for this is 2 methyl cyclopentane 1,3-dione which is commercially available. So before we go into the details of the total synthesis reported by Marty Debout we will discuss the key reaction which is oxodipy-methane rearrangement. So what is that? So if you have a system like this that is you know you can call it as allyl ketone. Now under photochemical condition first this carbonyl will form a di radical same thing will happen with the double bond. So that can lead to the formation of this type of cyclopropane assume that this is CH2 okay. So this right di radical what will happen next it will open up. So how it will open up? So they will open up like this and it will lead to the formation of original ketone. Ketone we started with you will get the ketone back as well the resultant di radical now you can see here this is a radical this is a radical they will combine to form the cyclopropane assume that this is CH2 okay. So basically if you do an oxodipy-methane rearrangement you can see the formation of cyclopropane okay. And this one can do on bicyclic system, tricyclic system and here is an example where you have bicyclic system so that will give like this tetraradical and then what will happen this tetraradical immediately you can see it will form a cyclopropane okay. Then this radical on the oxygen will come back and then open this cyclopropane to give this bicyclic di radical this bicyclic di radical one can easily redraw like this okay you can see here this is 1, this is 2, this is 4, this is 5. So that is the 5 m board ring and the remaining is 6 m board ring. Now these 2 di radical will combine to form the tricyclic compound in that the third ring is a cyclopropane so this is the oxodipy-methane rearrangement product okay this can be easily obtained from any bicyclic compound having a carbonyl and double bond at appropriate place. When you do that one can also expect another product which is normally minor product that is arising out of 1, 3 acyl shift that is when you do a photochemical reaction on this ketone first NaRis type 1 cleavage will take place where you have you generate this di radical followed by migration of the double bond okay the double bond will migrate and you will get the corresponding ally radical okay what will happen now this will migrate or here you can write like this and these di radical will combine to form cyclobutanol okay. So one can see using oxodipy-methane rearrangement you can get a cyclopropane or if you use 1, 3 acyl shift then you will get a cyclobutanol okay 6 ombre ring fused with cyclobutanol. Now let us see how De Booth synthesized this natural product he started with 2 methyl cyclopentane 1, 3 diome which we saw during the retrosynthetic analysis then using the Robinson annihilation sequence first Michael addition with methyl vinyl ketone followed by treatment with S proline and dehydration you get a hadges-parish ketone okay. You have an enone and a ketone and the ketone can be selectively reduced with 0.25 equivalence of sodium borough hydride in ethanol at 0 degrees to get this aleric or this alcohol and once you have this alcohol then you can protect this alcohol as mometer okay and if you treat with LDA TMS chloride if you treat with LDA TMS chloride so it will generate anion here and then it will form an enolate that enolate will become enol TMS and this is the diene now he did deal solve reaction with mellic anhydride to form this tetracyclic comp okay. So this is the intermediate and this intermediate you can see this is the transient state okay this is the transient state and this gives this tetracyclic comp. Now this anhydride if you look at this anhydride should be converted into double bond so he used a electrolysis method like hydrolysis followed by di-ticarboxylation to give the double bond and this is the key precursor for the oxodipy-methane rearrangement. So taking this compound and then shine light it undergoes oxodipy-methane rearrangement and as I mentioned earlier about how oxodipy-methane rearrangement takes place first it will form this tetraradical and followed by formation of the cyclo propane and opening of this three membered ring you will get another di-radical and whereas this will be a carbonyl group and once you have this di-radical and you can number it you know for better understanding always better you number the starting material okay give numbering. So the six membered ring you can give numbering and then see where the radicals are in the starting material and when you redraw the structure for better understanding you can give the number and then you can easily make out where these two radicals are. Once these two radicals you can write and that will form the corresponding cyclo propane as I said there is a possibility of forming 1, 3 acyl shift also of course the yield is only 5% and the main product is only the required one that is the cyclo propane formation. So this tetracyclic compound next he reductively open okay so lithium tri di isopropyl amine and TMS chloride. So what happens this will be cleaved this cyclo propane will be cleaved and then it will form it will form the corresponding enol TMS okay this will form the corresponding enol TMS because you are using lithium diisopropyl amine and conging with TMS chloride because that enolate once it is generated you trap that enolate with the corresponding TMS chloride. Once you have that that is the first step and the second step the enol TMS ether can be cleaved with any fluoride source. So that is what they have done they have done with this benzyl trimethyl ammonium fluoride to cleave the enol TMS to generate the O- and the O- you conge with methyl iodide okay you conge with methyl iodide you stereo selectively introduce the methyl group okay. So now if you count the number of carbon atoms you will see here in this ring you have 6 here you have 2, 8 and then here you have 3 and then here you have 1 okay. Next step is to remove the carbonyl group so you do not want the carbonyl group okay this can be done in many ways but what they have done is they have protected the ketone group as diethiol derivative okay. Then you remove the man group okay you remove the man group is using titanium tetrachloride you get the hydroxyl group at this point you can remove the titanium using a combination of titanium tetrachloride Lewis acid and reducing agent so they use titanium tetrachloride and lithium aluminum hydride to get the corresponding CH2. So the deoxygenation of carbonyl group was done in 2 step first you protected a diethion then remove that with titanium tetrachloride and LAH. Now what you need to do you have to oxidize the alcohol to ketone okay then you have to introduce the 2 methyl groups and then double bond.  Then for introducing the double bond you use GDQ and this exotic reagent which looks like exotic reagent but it is not it is a reagent derived from trifluoroacetamide okay take trifluoroacetamide and then you treat with trimethylsilyl chloride you get this. So this is the reagent you know you can get a double bond or you can introduce a double bond next to the ketone. Next you have to introduce 2 more methyl groups one here and one here but at the same time the double bond should be kept intact. So how it can be done? First you add the 1, 4 addition on the double bond with lithium dimethylcuprate and quench the enolate with methyl iodide to get the 2 methyl groups and then repeat the same process repeat the same process. So that is you introduce a double bond via enol TMS followed by oxidation to get the natural product okay. So that way if you look at this synthesis the key step in the synthesis was the oxa diphyomethane rearrangement and they started from commercially available 2 methyl cyclopentane 1, 3 dione and then they used Robinson enolation as a key step to make the next starting material that is agisperage ketone and overall this total synthesis was accomplished in 11 longest linear steps with an overall yield of 1.6 percent yield. So now we will move to another very short total synthesis reported by Wehera okay. Here again as I said we will be discussing only the photochemical reaction which has been used as the key reaction in the synthesis of trichonates and what he has used in this particular synthesis is photochemical 1, 3 acetylcide. The earlier synthesis which we saw where oxa diphyomethane rearrangement as the key reaction and there also he got 1, 3 acetylcide product as a minor product. In this particular case Wehera has used photochemical 1, 3 acetylcide as the key step and that is a main reaction to get the key starting material or key intermediate. So let us see his retro synthesis and when you look at this capnoline as I said whenever you have a functional group or if you do not have many functional groups you can introduce a functional group okay. So here you have a functional group a double bond but that may not be sufficient sometimes when you do a proper retrosynthetic analysis. So it is better to introduce another functional group which can facilitate the retro synthesis to get a simpler starting material. So that is how what Wehera has done he has introduced a carbonyl group at the middle ring. The idea is if you introduce a carbonyl group then one can do an intramolecular alkylation okay. So that is what he planned. So you can see you can if you make this as a good leaving group okay. If you make this as a good leaving group then you can generate anion and you can form the third fiber boundary is not it? So that was the idea. So for introducing the carbonyl group then you can introduce a double bond here. The reason for introducing the double bond is this four carbon unit can be added stereo selectively using a 1, 4 addition reaction. And now if you look at this you reduce this double bond and introduce a double bond here what for that is how he can use the photochemical reaction that is 1, 3 acyl shift to get this compound. Let us see how he made this compound in the DL synthesis and how from there he used the 1, 3 acyl shift to get this bicyclic compound. The total synthesis started with metacrysal methyl ether, metal ammonia reduction gave the diene and you treat with sodomyte when you treat with sodomyte then it will generate anion and the migration of the double bond will take place to give this type of diene or you can this will give this type of diene but this is the most stable diene. So you get this diene and this can undergo Diels-Aul reaction with alpha-chloroclonitrile alpha-chloroclonitrile is a synthetic equivalent for ketene, okay so ketene can normally cannot undergo 4 plus 2 cycloaddition reaction. So that is why indirectly you have to use some equivalent which can give ketene in the product. So alpha-chloroclonitrile is one of the best synthetic equivalents for ketene. So you do the Diels-Aul reaction with alpha-chloroclonitrile. Now if you treat with potassium hydroxide and DMSO that will hydrolyze this to ketone, okay try to write mechanism for this, okay so mechanism for the hydrolysis of chloronitrile adduct to the ketone it is a very interesting mechanism. Take this bicyclic ketone and then treat with potassium tertiary butoxide excess potassium tertiary butoxide and methyl iodide you can introduce 2 methyl groups next to the ket, okay. Then when you reduce this with sodium borohydride or lithium aluminium hydride you get a mixture of endo and exo alcohol. So this is endo and this is exo alcohol so almost you get 1 is to 1 ratio. These 2 will give 2 different products upon treatment with acids for example if you take this endo alcohol so first it will protonate the hydroxyl group and that will be a leaving group. Now the bond which is anti to the leaving group so this is the bond which is anti to the leaving group that is the OH bond and this is the bond which is anti to that, okay that will migrate and what will you get if you use Lewis acid is okay you will get this product. So this is nothing but if you look at you can write like this that is because the C this C C bond which is anti to the leaving group migrates. So if you do the same thing with exo alcohol now which bond is anti to that so this is the bond and this bond is anti to this. So that will migrate if you treat with acid and this will migrate so that will give you this bicycling ketone okay and this can be redrawn like this okay it is a 5 umber ring here and 6 umber ring here. So what you have to notice in this is if you use endo alcohol you get the corresponding alpha beta unsaturated ketone but if you use exo alcohol you get ketone and the double bond moves to the 5 umber ring other ring okay no problem and you can also write it like this and then now you shine light. So what it will happen first the Norris type 1 cleavage to generate the diradical followed by migration of the double bond. Now this C C bond can rotate is not it this C C bond can rotate and when you rotate you can see this diradical this diradical will combine to form the corresponding 5 umber ring and if you rotate this by 180 degree okay you can go through this and rotate it by 180 degree and you will get this couple okay hydrogenate the double bond as I said you have to shift the double bond here you do not want the double bond here you want the double bond here. So first you reduce the double bond then introduce the double bond. So this is done using a Suji trust method where you generate the enol carbonate enol allyl carbonate and followed by treatment with palladium acetate you get the double bond. So once you have the double bond next is to add the 4 carbon unit in a 1 4 fashion. So take this enol and add this 4 carbon unit you will get this product. Now as I said next you have to make this as a good leaving group generate anion and then form the third ring. So for that you have to remove the TBDPS get the alcohol converted to good leaving group. So TBDPS can be easily removed if you treat with acetic acid get the alcohol convert that into tosylate. So tosylate is a good leaving group. Now you take this compound treat with lithium hexamethyl disolxide okay so it will generate anion and intermolecularly attack the carbon bearing O tosylate in a sent to fashion and you introduce the third ring. So what is left now is to remove the ketone. So reduce the ketone to alcohol convert that into zandate sodium iodide carbon disulfide methyl iodide OH will be converted to zandate and the zandate can be reductively cleaved using tributatin iodide and AABN. So to summarize so if you look at the total synthesis of Wehera and Shinari Mamoto of Capnallin they started with a simple commercially available starting material that is Metacrysa methyl ether. So they did birch reduction conjugation real solid reaction to get the bicyclic 2 to 2 octinone then the key reaction that is 1 3 acyl shift was done under photochemical condition to get a 5 ombre ring and 6 ombre ring and later they did functional group transformation to achieve successfully the total synthesis of Capnallin. Overall they took about 13 longest linear steps however the overall yield is quite good about 3.7 percent yield. So with this we have discussed few more total synthesis based on photochemical reaction as the key reaction. We will continue our discussion on synthesis of few more trichunanes again using photochemical reaction in the next lecture. Thank you.


During the COVID-19 pandemic, it is critical that the Army continues to train and stay ready. Over the last two weeks, soldiers from the 101st Combat Aviation Brigade conducted an aerial gunnery, qualifying their air crews on multiple methods of weapon system engagements as they prepare for their upcoming deployments. Using rockets, 30 millimeters, and our Hellfire dummy missiles in order to shoot some targets and let down range, just to make sure we stay current and qualified and maintain our combat effectiveness for force combat. The exercise has also allowed for the unit to increase proficiency in FARP operations. Yeah, so the FARP is a forward arming and refuel point. So this is where they get all their ammo, get their hot fuel before they take the range and engage targets. Our FARP is pretty much our lifeline over there because they're the ones supplying us fuel and the armaments.  With the practice of social distancing and the wear of personal protective equipment, soldiers can stay capable and mission ready. No, we're stable to operate. Like I said, we're a little under manned just because we're trying to limit the amount of people that are out here that would possibly get exposed. But we're still working at a good pace, keeping everything smooth while still staying safe. So I think we're still able to accomplish everything we need to accomplish while working under these conditions. The gunnery ended with the entire battalion qualifying over a two-week period. Being out here hands-on, getting interact with the pilots, loading ammo, it doesn't get much better than that. Reporting from Fort Camo, Kentucky, I'm Sergeant Jeremy Lewis and I'll see you on your next rendezvous with Destiny. Air Assault.