And traditional rulers and members of the seven communities in Ikole West local council development have staged a protest expressing dissatisfaction with the location of the council headquarters in Ijesha, Isuikiti. The traditional rulers let their grieved communities to their ancestral renewal market square located at Isabaikiti to seek divine intervention on the matter. Ikole West local council development area in Iketi state has been the focus of arguments between communities and traditional rulers for some time now. They are not happy with the location of the council headquarters in Ijesha, Isuikiti. They want it at Usinikiti as agreed by the seven concerned communities. This protest is to forward their press home dear demands. All the oba's of the seven communities making up Ikole West have decided that we are not going to decide for anything. And finally we said until they give us a ride, until they return the headquarters to Usinikiti and temporary headquarters as we will not participate in any election. And finally we have decided we will not pay tax until our ride is restored to us. It is serving our grieved community. We've been agitating for this kind of equity but the governor did not give a pace to it. Coincidentally we don't know. The governor lived around here for donkey years. He knows the history and background of this place. So we're feeling that probably we have offended him and if that should be the case he shouldn't have thrown a personal vendetta which will not help the situation. It's a situation we could resolve. The state government urged the aggrieved communities to tread the path of peace and allow the issue to be resolved amicably. All the countries receiving attention from government. I speak with them. I know many of the people that wrote the petition and I give them the assurance that the government will look into it. It's not only when the government looks into it that they can come up with a decision on it. But we should not forget that the other community they are talking about is out of the equity state as well. It's not another community outside the equity state. Ijesha Issu Ekiti is 12 kilometers away from other communities and his protesters strongly hoarded that it cannot be the headquarters of the LCDA.


Welcome to 8% Club every Friday you guys know the time two o'clock central standard time Today's topic is good. I just got out of a sales meeting with an agency and one of the things that we talked about is Exactly what we're gonna cover on today's show.  I appreciate you being here today's always about today's about the number one ingredient 92% of interest agents fill and I want to talk about the number one ingredient To help you create Insurance success and this is probably the number one thing that agents struggle with and I want to come and Challenge you today to pick it up in this area Before I jump to the number one ingredient and you're gonna want to stay with me I'm gonna walk through the whiteboard is gonna be really good. I want to make sure that you guys are aware We've got a new call-to-close series that trains agents on everything from the phone call and all The steps all the way to the clothes. It's a live three-hour virtual training with me It's on March 7th Thursday March 7th at five o'clock central standard time again Call to close three-hour live training you get lifetime access to the recording the e-book Live Q&A with me three to four hours of live training The way I think about trainings and you can go to kodiaskins.com to look that up We're only taking 30 seats. We have some seats left one thing I want you to think about is the one thing I think about I invest in my personal brand I invest in retreats Conferences I go to events. I have coaches. I learn stuff. I buy courses all the time. I'm sitting true story I'm sitting It's about seven o'clock at six seven o'clock at night on new year Maybe 530 whatever on New Year's Eve. Okay about well almost two months ago And I'm sitting there and an ad comes up for a chatbot course. I buy it I spend a couple hundred bucks on the fly. I just commit and pay for it. I'm like babe Where's my card? I want this because I want to know more when it comes to sales Marketing insurance being great at what I do I want to know if I then the way I think about courses and trainings If I pick up one nugget that I apply into my business one idea that I apply into my business one Concept that I apply into my business then that course was worth it So when you start thinking about hey should I invest in myself or not? Will you get one idea concept or? Specific strategy that you can use to apply into your business so that new call-to-close series I can promise you you'll pick up you'll pick up one idea. You'll make about a hundred X Return off of whatever you pay for that call-to-close series. So make sure you go to codyaskins.com to check it out All right, so the number one thing the number one ingredient that is able to help create success in this business, and I've always been told it and It's true and it's act Activity Activity is a number one thing that is no one ingredient that if you can do this you can have success Because a lot of agents a lot of agents we want we want more cells, right? We want more money, right? So we want all of these things What happens is activity? Leads to appointments Appointments leads to sits sits Leads to cells Actually, yeah, it leads to cells policies applications whatever you want to call it and then that leads to what it leads to Money you can't have cells. You can't have money without cells You can't have cells out sits you can have sits without set appointments, and you can't have set appointments without activity I don't care if it's doorknocks or calls I would start tracking something I did a lot and I preached to insurance agents and agencies to track this I would track I Would track my calls My doorknocks text messages Emails all of it especially these two calls and doorknocks I would track these things because I and I want to know as a As an insurance agent I want to know at the end of every week. I want about to do something like this and be able to say okay my activity was 300 right I set 15 I Sat with 10 I made five cells and guess what I made about 3620 in commission like that's practically an Ideal week to me for an insurance agent now It's not a fantastic or crazy week and your average premiums should be a little higher, but I'm doing some math and averaging it out about 720 725 724 AP per cell you need to know your numbers so when I think about Activity this is calls and doorknocks. This is set appointments. This is sits This is cells and this is calls and I used to track these things every Single day so I would have a whiteboard and I'll go ahead and draw this out for you I would have a whiteboard again. Make sure you remember this Activity appointments it sells money. All right, I would have a whiteboard on my wall And it would look just like this. All right And I still do but as an agent. This is what you need to do. So you need you need The whole month right you need a whole month broke out and then you need different days, okay Monday Tuesday Wednesday Thursday Friday You can do Saturday or you can just do a total column, okay? And then each one of these blocks I want to have activity and Then I want to have sets sits cells and Commission or premium whichever you want to track But every day so like I eventually learned this was what's cool about this. I eventually I'm going deep today I'm giving you ridiculous value. I used to Figure out guess what day of the week was best for Cody askins every single week as an insurance agent I knew and you guys probably don't know what day of the week you do best I realized that I had a great day and a bad day. You want to know what you want to know what those were my best day Was Tuesdays? Tuesdays were my best day Because I spent Saturday Sundays and Mondays setting appointments for Tuesday, okay I realized that Wednesday was my worst day you say well that should never be your worst day Well, it shouldn't but I was putting so much focus on having success early in the week that I forgot to have appointments on Wednesday So when did Wednesday ended up being a office and call day and doorknock day for me And so I got to where I knew that okay if I do You know if I do 60 died I had 60 dials today. I set Four appointments. I sat with three of the previous sits I sold three, right because you very very well may and then I did about 2800 in in in premium or commission So when you start to think through and you start to track your activity Every single day and every single week and then you end up totaling it like we talked about. Hey my activity was 300 I set 15 and it should be more right. It should just it should just be more. I Sat with 10 I sold five and my total commission was about Let's just say 36 20 again, right? You need to know your activity I'm a strong believer in the number one key ingredient being activity most people fell in the activity area They don't track their activity enough. They don't know what they're doing on good They have good days and bad days and they have no clue They don't know what their activity their sets it sells and commission or premium are because they don't and a lot of times They have low numbers for instance. I was in a sales meeting not too long ago where one agent I'm gonna give you an example one agent had 185 calls for the week. Okay, and the other agent had 80 calls for the week Who do you think out of agent number one or agent number two and vote in comments below right now? Facebook YouTube Instagram Which agent number agent number one or agent number two had more success that week without even looking at it? What do you think agent number one or agent number two? Who do you think had more sets? Who do you think had more sits? Who do you think had more cells and who do you think made more money? Agent number one or agent number two It was agent number one and it doesn't always work out way ten out of ten times But typically it does the person that puts forth more activity in this business Typically has better what? results You typically have better results the more activity you do it's funny how that works the more activity you do The better and more success you have in this business. It is all about I'm big on tracking.  You won't have enough sets you ought to have a minimum a minimum of 15 plus set Appointments every single week you ought to have a minimum of 10 plus sits every I don't care what type of industry you're in I Don't care what type of insurance you're selling minimum of 10 sets every week you should have a minimum of five plus cells every single week and you should always do at least 3000 of annual premium Minimum per week that's immediately and you say well Cody.  That's a hundred and 50k in premium I bet that less I bet that less less than 3% of agents are doing that type of premium Volume annual premium. I guarantee it. I don't even need to guess. I just it may be lower like the company I was with when I did I did about 250k in premium 240k in premium in eight months at 19 I Was like a freaking solar eclipse like I was like the dude that landed on the moon like no one else was doing it Especially at my age and in my office or regional area. No one was doing it I was setting the standard So that's my challenge to you do so much activity that you're the destroyer in your office Because there's a lot of office that track activity You go into a sales meeting and then you lay out your activity and the information and what you did for that week Right, you need to work. You need to be accountable to somebody need to report to somebody You need a coach everybody's got a coach the greatest people in the world have coaches had coaches And you need a coach so find someone that you can look up to that can be your coach Ask them to be your coach ask them to be a part of their coaching programs because what will happen is You'll end up You'll end up being accountable and when I talk to agents and agencies all the time I say okay who leads an activity every week that oh that person okay Do you typically lead in sets it sells and money every week too?  I don't want to just beat him. I want to freakin destroy him. I'm telling you you guys need to Want to destroy your competition when when you're in an office setting you're in a sales meeting and Someone is putting up 300 dials a week, and they're baiting you and sets it sells and money every week. I Would walk in the next week with 600 I would plan on being the destroyer in your company if the best person does this that I'm doing more But we don't because we just think about being Average and about not doing enough and about not being all in and committing to succeeding So my challenge to you is activity the number one ingredient Dylan I'm about to come to you buddy in just a second with some with some Q&A So we got going on in questions and comments the number one Thing that you need to track and that you need to know and that you need to be aware of and that you need to be ready to freaking smash and be a destroyer on is Activity because activity is the one thing that you can control You can't control sets you can't control sits you can't control cells you can't control money because without activity You won't get any of it Activity is what I would be great at I would be known as a flat-out The person that outworks everyone else in this office is this person the person that leads an activity is this person person That's a flat out freaky destroyer is this person when I was a new agent any any any office have ever been in I was the hardest working agent in that office, and I was the quote-unquote destroyer for that operation I want you to be the destroyer in your local sales office We think dig anything going on a youtuber Facebook. Thank you guys for joining us Yeah, we got a question from Robo on YouTube Robo He asked what's the latest time to call back a real-time lead at 7 p.m. 8 p.m. Mic'd up. Yeah, latest time to call lead 7 p.m. Or 8 p.m. I Would call in Missouri. I can call all the way till 9 central and I've done it so I Just don't call past 9 like 8 o'clock dude 8 o'clock. They're probably in bed watching Big bang theory and they're like Somebody please call me so 8 o'clock like 7 8 o'clock Prime prime prime for calls That's the thing most people in that scenario they would think it they would overthink it well And I I need to call this person. Oh, it's 8 o'clock. I'm not gonna do it Come in first figure the rest out later. Don't even think about it Just freaking dial the number if you take them off you take them off just do it So let's just say the lead comes in at 9 p.m. Yeah, would you recommend calling them right then or in the morning? Always immediately now if it's 10 11 midnight. No, but if it's 9 o'clock 9 15 it just came in They they they ask I mean one thing to think about okay agents forget about this. They say okay a Lead came in am I the agent thinks am I allowed to do this when in reality?  They are asking me to call them right now And and then you just freakin pick it up and do it so that's a wonderful question I get I get that question actually quite a bit That's pretty much our only question. We have some people commenting and some people talking From the comments who looks like a destroyer Looks like they're gonna just outwork people Quentin Quinton Kelly said I called one back at 9 p.m. Last night and set an appointment come on man Quinton Kelly is dropping bombs man. I love it Dude, we're gonna get you up here. You're gonna drop the mic in the middle of the show. I love it. That's good Thank you, buddy. Hey tell somebody tell somebody you know that's an insurance agent that needs this I mean if you want your people that you know that are also insurance agents to be important the 8% club and to be successful And I know you do then make sure they're watching these things when we do make sure that make sure that they know about them Now it's up to it's up to them where they watch them But make sure that like when I've got people in my life that are helping me I tell others Because I want other people to be successful and I don't I like there's enough fish in the sea like whatever But I don't be the destroyer in my space like when it comes to content. I challenge you guys right now when it comes to content videos helping Does anyone in the insurance industry do more yes or no? No, because and I challenge you like dude I you show me someone that's doing more content for this space and I'll give you a hundred dollars cash right now It isn't happening because we are like you want to be with activity We're the destroyer for the activity For the content for the insurance industry from an activity level so because like just as activity made me a ton of money at an early age in the interest business Activity when I'm talking about marketing training lead generation all that Coaching all the everything we do for the insurance space now Activity wins. I want to be the activity king That's all that's all I gotta say We're good Yeah, I think we're good. You guys are awesome. Thanks for being part of 8% Club Hey, make sure you go to Cody ask ins calm. I've got a complete three hour live Training with you. It's called call to close everything in between from the call to the close if you pick up one nugget And you've immediately 101,000 X your investment if you pick up one thing go to Cody ask ins calm to check out the new call to close series You want to be a part of that it's a live training March 7th at 5 o'clock. I want you there.  Activity thanks for watching have an awesome weekend Dylan we got some videos dropping for him this weekend We will let's do it always man. When do we not like like that's a let's like a let's like a loaded question That's like man like when I when I run appointments. I'm like, I'm assuming you want to buy for me today I'm assuming Dylan's gonna be giving you guys some free stuff this weekend. So all right. Thanks for watching 8% Club appreciate you guys See you Monday


So welcome back, we are now nearing the last 2 weeks of this class and so far we were looking at experiments to determine reaction mechanisms. So in the last class we had looked at how we can trap reaction intermediates to understand the mechanism. So we had looked at traps for anions. So logically if you have anions, something which is electrophilic would be a good trap for the anion. For cations you would need something which is nucleophilic. A lot of times solvent is used because solvent is nucleophilic and it is in a large excess. So it ensures that your intermediate is actually trapped. Towards the end we had looked at traps for radicals and in radicals there are 2 types of traps. One type of trap quenches the radicals. So these are called as radical traps. The other type of trap retains the spin or the radical nature and these are called as spin traps. So in spin traps we had looked at nitroso and nitrone compounds which could be used to generate a new radical which is long lived because it is more stable. And in the radical traps what you use is you use a stable radical which will react with your reaction intermediate radical to form a quenched or neutral species. So we had looked at several examples of radical traps. Now in today's class what we will do is we will look at trapping of other intermediates that cannot be classified as carbocations, carbanions or radicals. One classic example is trapping of benzene. So we had seen the benzene intermediate earlier when we were looking at isotope labeling. So here is another method to generate a benzene and how this reaction works is quite interesting. So if you have this molecule, so when you do the diocetization reaction, so you have a aromatic amine here. What is the product that you will form? Hopefully you have gotten the answer right. So you will generate the disonium salt. Now once you generate this zwitterionic intermediate you can then imagine these electrons coming in and forming the benzene and once you form the benzene, you would need to figure out a way to trap the benzene. Now this reaction is very very facile because the other byproducts are gases and these gases will go away. So the reaction is highly favorable to the right where you generate the benzene and you have these gases going away. Now once you generate the benzene, it is a highly reactive intermediate. So you need a good reagent to trap the benzene and benzene we have studied is an intermediate when you look at nucleophilic aromatic substitution reactions. But benzene is also a very good dienophile for the Diels Alder reaction. So if you react this with a highly reactive diene, you will be able to form the Diels Alder adduct. And what has been used to trap the benzene is furan. So the adduct that you form would be, so this would be the adduct that you form. Now given the fact that this is a Diels Alder reaction, so you have your diene and your dienophile. Looking at the structure of the adduct, you know that the intermediate formed is a benzene. So this is a nice method for you to figure out that the benzene intermediate was actually formed. Otherwise it is so reactive that it is very difficult for you to isolate the benzene. And this example also shows you a very elegant method to generate the benzene. Now we will look at application in biology. So as usual we were looking for all experiments, how one can use these simple principles to even understand biology. So let us see how we can use this concept of trapping in biology. What is shown here is the structure of a polynucleic acid. So it is a repeating unit and this is a polymer, I have just shown you one monomeric unit of the polymer. So this is an RNA structure. So the mechanism for its hydrolysis has been proposed where you can draw the arrows from this neighboring hydroxy at the two position and it gives you this intermediate. I will just write this as B and here you have the rest of the. So here you see that the sugar polymer backbone is still intact. So I can then imagine these electron coming back in and kicking this out to give you another intermediate structure. So I am not showing the proton transfer steps. So I have shown the free OH here because it has hydrolyzed that linkage. So you have this plus the rest of the, a similar unit will be there on this side. So this intermediate would then react with water to give you your final hydrolyzed products. Now what is the proof that you have this intermediate forming, this sort of a cyclic intermediate. Now in order to prove that a clever experiment that was done was where a model system was used to understand nucleic acid hydrolysis. Now in the RNA you have this OH group at the two position which helps in formation of this cyclic intermediate. So to mimic that the model system that was chosen had an OH group right next to the phospho ester here. So you have this OH group right next to the ester linkage. So if you have this OH group one can imagine a similar mechanism where you can visualize, you can visualize this lone pair coming in to give you the intermediate shown here. Now what is seen is if you just use the model system you are not able to isolate this intermediate. So what was done was in order to isolate this intermediate the reaction was done in the presence of acetyl chloride. So in the presence of acetyl chloride what was found was that both of these intermediates interacted with the acetyl chloride. So the product that was formed would be the corresponding OSI related derivatives. Now once this SILated derivative is formed it is not very easy for this to go back to your starting compound. So what was seen was in both these cases the corresponding SILated products were isolated. So the fact that this product could be isolated I will call this as P1 indicated formation of the intermediate I. So this is a very simple method by which one can indirectly show formation of the intermediate I by trapping it as the acetyl adduct. So once you form the acetyl adduct you know that you have found this particular cyclic phosphorane intermediate in this mechanism and indirectly this was used to ascertain that during nucleic acid hydrolysis such an intermediate is actually formed. So these were examples of how you could use trapping agents to get some insight into the nature of the intermediate. Now there is another slight variant to this one can think of varying the reactant a little so that you can actually trap the intermediate in some form or the other. So here you are not adding a trapping agent but you are by clever design modifying the reactant so that indirectly you can say oh this intermediate was formed in the reaction. So one example is understanding the mechanism of Mandelate Resimes. So what Mandelate Resimes does is it converts S Mandelate to R Mandelate. So this is the structure of Mandelic acid and the chiral center is here. So what the enzyme does is it converts one enantiomer to the other enantiomer and what is proposed is that it goes via an carbanion intermediate. Now as we have seen this is not very stable for you to isolate. So how do you actually know that the enzyme goes via a carbanion intermediate. So one clever design was using a slightly modified reactant. So here the reactant used instead of Mandelic acid was a derivative where at the para position you have this CH2Br group. So now I want you to think if you have this CH2Br group and say you generate the intermediate carbanion here what would happen? So I repeat the question. So now you have chosen a clever model substrate. Now in the model substrate you have this CH2Br group.  You can press the pause button and try to work out the answer by writing the mechanism. So let us see if you have this answer correct. So here let us assume that in the presence of the enzyme. So this reaction is carried out in the presence of the enzyme Mandelic racemase. So your substrate reacts with the enzyme and what you get is this intermediate carbanion. Now once you form this anion this is in conjugation with the aromatic ring. So this is a hint I am giving you if I if you did not get the answer earlier now think of drawing the resonance structures possible for this and trying to push the arrows starting from the carbanion into the aromatic ring. So what will you get? So since you have a very nice leaving group here what will happen is you will get the elimination product. So once you form this elimination product indirectly you know that you have generated the carbanion intermediate. So this is a very nice method where one can use model reactant to track a particular intermediate. So this shows you a very nice example of how the reactant was modified to give you insight into how an enzyme works in biology. Now there are other experiments that one can do to gain insight into the nature of the intermediate and one such experiment is the competition experiment. So competition experiment is a slight extension of what we have studied so far with trapping reagents. So what happens just as the name suggests competition means you have something competing with the other. So in this case you have a reactant which forms the intermediate you add two traps. So now you have a competition between track one and track two to form two products. Now depending on the ratio of product one and product two you can gain insights into the nature of i. So the trapping experiment that we had seen earlier the same concept is actually used here except you are putting two traps so that you can compete so these two traps can compete with each other. So just to give you an example now you have studied the mechanism of SN1 and E1. So shown here is a substrate I want you to write the products that you get if you do the SN1 reaction on the substrate or if you do the E1 reaction. So quickly press the pause button on your screen and go ahead and write these two mechanisms. So let us see if you are able to write the mechanism. So in the SN1 or the E1 the first step is you have the leaving group go. So what you generate is the carbocation plus the leaving group. Now after this it can undergo substitution or elimination. So if it undergoes substitution the product that you get would be in this case your nucleophile is ethanol so the product you get would be OET. Again I am not showing the proton transfer step. If it undergoes elimination so the E1 mechanism what you would end up getting is in this case you would have this product being formed. Because now what will happen is you will have a proton being abstracted from any one of these hydrogens to form the corresponding double bond. So these are the two products that are possible. Now if you look at it based on the standard textbook mechanism that you have studied so far the rate determining step is formation of the carbocation. So then what you would see is you should be generating the same intermediate irrespective of which leaving group you use because we have studied that this first step is the rate determining step. So irrespective of whichever leaving group you would end up generating the same carbocation. So once you generate the same carbocation if your nucleophile is the same which is ethanol in this case one would assume that you would have the same extent of substitution or elimination irrespective of whichever leaving group you choose.  Now once you generate the carbocation it can undergo either substitution or elimination. So what I am saying is if you have the same nucleophile the extent of substitution and elimination should be technically the same. So here is what was observed. So here the leaving group was varied and what you see is the ratio of the products actually changes. So in the case of CL and BR what you see is that you have quite a bit competition between the substitution as well as the elimination. Whereas when you have a charged leaving group what you see is that you have a greater extent of substitution. Now this would be confusing right because I just told you that once you generate the same carbocation so in all these cases you generate the same carbocation you have the same nucleophile then why do you see this difference. So what this gives you a hint is that the leaving group seems to be important in this particular reaction. So in this case what is seen is the leaving group plays an important role. This is not true for any SN1 or SN2 reaction it depends on the nucleophile you are using the solvent you are using and the leaving group. In this case you see similar behavior between CL and BR but with the charged leaving group you see a different behavior. So now how do you explain this mechanism? So how you can explain this is shown here I have shown you the carbocation intermediate and the counter iron for that separately. Now you can imagine a scenario where once you have the carbocation formed you can have the X- actually in close proximity to this carbocation. So it is not fully solvated as I have shown in the scenario above where it appears that both these species are independently in the reaction mixture and fully solvated. Here what I am saying is they are still in very good contact with each other so these are called as contact ion pair, intimate ion pair just as the name suggests it indicates that they are still very much close to each other. So if they are very much close to each other now when your nucleophile comes it will not be able to immediately access your electrophilic center. So what you see is in the case of CL and BR you see a competition between elimination and substitution that is because it still exists as this ion pair. Now as you can see CL would probably form a better ion pair as compared to BR because it is more electronegative. So what you can see is the competition is more in the case of CL. In BR you have a larger extent of substitution as compared to elimination. Now what happens in the third case? So when your leaving group is charged so what you have is you generate a neutral species once you form the carbocation. So once you form the carbocation you generate a neutral species unlike the other cases where you are generating a charged species. So once you generate this neutral species it does not have as much as an affinity for the carbocation as compared to the earlier cases. So you do not form an intimate ion pair and you see a greater extent of substitution. So a very simple competition experiment here was able to tell you what is the nature of the intermediate. So in the first two cases the intermediate is a contact ion pair or an intimate ion pair whereas in the third case it has more of a carbocation nature. So this shows you how you can use a competition experiment to figure out the nature of the intermediate. Now there is the reverse of this. So imagine that you take two reactants which can form the same intermediate. So initially you are taking the same intermediate and trapping it to give two products. Here you are starting with two reactants which will form the same intermediate. Now this can be used to figure out the nature of the intermediate because if you design the reactants such that they will give whatever is the same proposed intermediate you know that you will get the same product irrespective of the reactant because the intermediate is common. So this method is called checking for a common intermediate. So as you can see you have two reactants forming a common intermediate hence you get a common product. Now let us see two examples as to how this can be used to figure out the mechanism. The first example is a solvallises reaction. So as we have studied solvallises earlier it is a reaction where your solvent is also involved. So your solvent is acting as a nucleophile in this case. So now what was probed is in the substrate shown above are you actually forming a carbocation or do you have direct displacement of the solvent which is water. So can you write both mechanisms one where you form the carbocation and the other where you have a direct displacement. And the second question is when you form the carbocation can it rearrange to give you other carbocations. So again I repeat once you form the carbocation can it rearrange to give other species.  So let us see if you are able to write the mechanism. So if this generates a carbocation so you have loss of nitrogen. So this would be the carbocation that would be generated. So can you get other carbocations by rearrangement of this carbocation let us see if that is possible. So one way is if I push an arrow going to this carbon here. So what I generate would be a 4 membered ring and a new carbocation here. Another thing you can think of is you can think of pushing the arrow such that instead of pushing it at the carbon you form a new bond here. So this cyclopropane ring opens up to give you a new bond here and the product that you would get would be or rather the new intermediate you get would be. So this would open up and you would get possibly all of these leading to product. You can probably form this product, this product and this product. So if it goes via the carbocation one would see all of these products. Now does this actually take place? So what is seen is that you observe all these products. Now we were talking about checking a common intermediate. So now suppose I use these two reactants. So these two reactants now I am starting with taking cyclobutane with a disonium salt attached or we are taking the open tosylate. Now both of these would generate the corresponding carbocation. These carbocations are similar to what we had seen in the first example and they can again rearrange. So you can draw all of these resonance structures starting from any one of these correct. So I can still draw from the cyclobutane. I can get the cyclo-propyl with the exocyclic carbocation. I can also generate the open carbocation starting from cyclobutane. So I can generate all of these carbocations even if I start with the other two reactants. So now how to actually understand whether this is taking place? So if I do the reaction in the presence of water what is seen is that I get an identical distribution of products. So I get 48% of both of these cyclic products and the acyclic product in 4% irrespective of which reactant I use. I can use the cyclo-propane, cyclobutane or the open compound acyclic compound. In all cases I get the same product distribution. And what this tells you is that indeed the carbocation is formed which can rearrange to give you 3 different carbocations which lead to 3 different products. So this is a very nice example of how you can use 3 different reactants to check for common intermediates. So before I go in the next class what we would be doing is we will be looking at another example for checking the common intermediate. So I will just leave you with the reaction that we would be looking at. The reaction we would be looking at is a rhodium catalyzed allylic alkylation. So this is your allylic group and you are alkylating at this position. So the product distribution you get is 97 is to 3. So what we are going to see here is what is the nature of this intermediate and how you can use this method of checking for a common intermediate to figure out the nature of the intermediate. In the meanwhile you can try to work out how this reaction works. Thank you and see you in the next class.


Okay, let's start thank you all for coming to our presentation in this very hot day after lunch last day so thank you very much and thank you to the people online if there's anybody online as well. So our session is about integrating charge BT into tutorial sessions to enhance hopefully critical thinking in university students, we both work at Oxford University. So, let's get this started. So, today we're going to talk about the context of the tutorials at Oxford University small group teaching our experience using AI in psychology tutorials, and some recommendations, something that we're working on at Oxford, and maybe if there's time for any questions, we'll do that as well. And so I'm Sandra Morales, I work as a digital education advisor at the Center for Teaching and Learning. I also have an academic background in research on technology for language teaching I am a language teacher. And that's why I do some teaching as well at Oxford. So I'm Xavier, I'm a digital education advisor as well. My background is in science. I do the same work as Sandra and also do some teaching and tutorial. I've got a chance to do some tutorial at the university in psychology. So this is where the experience come from. Okay, so this is part of a discussion that we had at the Center for Teaching and Learning in terms of artificial intelligence and who is the authority. And that got us to think about who has the knowledge, who has the power to make recommendations and who should we listen to as this is so new. So we started thinking about how to apply it and how to come up with recommendations on how to use AI and how to support our students, mainly through this rapid and rapid evolving rapid evolution of AI and child GPT particularly.  So that's very difficult to find who is the authority in this field and even now with AI. Okay, so a little bit of context in terms of the tutorials at Oxford. They are small group teaching sessions with a tutor and two or three students. Yeah, two or three. One and three students. And basically the tutor's role is to encourage students to be more analytical to think critically and to deepen their knowledge about subject area using a different methodologies. And as we will see later on, there are different types of tutorial sessions that the tutor can prepare for these students according to their needs, obviously, and as Oxford is so diverse, different programs and divisions, they organize their tutorials sessions independently they they make their own decisions. So it depends on the program how they run the sessions but basically the role of the tutor is mainly to support and provide guidance to the students in this mall. So, so the, you know, we get to get, we get to see the students either maybe once or sometimes almost all the time we can follow them over a period of four to eight weeks. So we get to, we get the chance to build a rapport with the students, which is nice. And it's an opportunity that Oxford obviously allows and it's not maybe relevant to other university but that's a good, good thing for us. So for example during the pandemic, we've, we've had that we've kind of designed some learning pathway so that that comes from the pandemic where we use our main tools at different stages of the learning. So we've developed one for lectures, some for seminars and also for some small teaching tutorial. So you can see different stages and use the different tools for for each stages. And so this is before the rise of AI, and now some other some some other stuff I've used, I've kind of integrated in the learning path pathway, the use of AI. Yes, so this example is from one of the tutors in the language center so she applied the structure of the fit pathways for tutorials and small group teaching and integrating the different, the different tools that we have at Oxford canvas teams, etc. But she included chat GPT in the different stages so in terms of knowledge application, and the student and the tutor works together. So they have an instance to work with chat GPT. The same in terms of online or in class person, I mean online class and in person class to collaborate and to have a more personalized session in which they can both work with chat GPT. So here's one opinion or one, the voice of the student in terms of how to work with the tutor and with chat GPT. Basically, what they say is that they, they appreciate the tutor being there as well to support this process. Yeah. So our in terms of the work that we did in tutorials. We have the question of is critical thinking the answer to generative AI there's a link there to to an article that I read and well we thought, maybe, maybe that could be something that can support students to better interact with the tool. In terms of analysis, evaluation, decision making reflection, and basically to familiarize with the tool. So my experience using chat GPT psychology tutorial so I teach experimental psychology or teach a tutorial is experimental psychology. So I've used it already since January 2023. But recently we, me and Sandra, we tried to use it in one to one tutorial where we kind of introduce chat GPT to the students if they haven't done so yet. And we use them during tutorial. So we looked really at how they, they, they interact with the, with the tool, and especially about prompting engineering as we are looking at, you know, how did they talk to the, to the, to the, to the, to the chat GPT. So for example, we give them a task to do. But for example, design a curriculum on, for example, you know, we put them in a scenario, we said you are a tutor and then a professor ask you to design to design a one hour lectures, because the professor can't make it so you have to be in charge and and try to think about a curriculum or like a small, a small one hour curriculum for your teaching for an hour. So we asked them that and then we looked at what the students write on chat. So we just started to do that with three students and I mean, it's interesting to see the prompt how depending on the personality of the students how they interact with the, with the chat bot. So we have different types of students and we, you know, depending on their, on their, because of their personality, yes, they'll ask different, different ways of the machine. Yeah, what caught our attention was the language that they use to interact with the, with the machine if you, if you can see there. One is more imperative write me a university level. The other one says hello, can you please so that, you know, a different end to the third one is I want you to be an expert so it's giving orders directions to the to the machine so that was actually something that matched with our students personalities. So we started to so all the other people in our team and the Center for Teaching and Learning are also working on AI, and we started to think about, you know, the competencies that set that students or even stuff could could could have to to interact those tools so we can see that you can start from a novice level to an expert level. And how do you engage. Five modes of engagements. What tool do you need to choose. How do you prompt. How do you interpret the outcomes. How do you get them into your professional practice and, you know, if you're very can become an expert. You can start to make decisions on how to develop those tools. Do you want to say about where does it come from. Yeah, just to add that this framework is based on the common European framework for languages which has a similar organization for when you learn a language from basic level to experts so that we, we thought that that would be a good starting point to do the same, perhaps for AI. So, and we don't need to read all these texts but you know this is an example of Dominique kind of design this example how you can judge someone if they are competence on using different type of prompting. So this is from the beginners. This is to the advanced to the advanced users. Oh yes so also we've designed a tool to kind of how students can self evaluate themselves so before they start to use the, the tool, they could choose, they could look at this and then there are some questions they can ask. They can, they can answer to kind of find level they are at when they start to use the AI. So based on that all of that in our experience, we thought about a framework that we could use in the tutorials or any small group teaching. Well this is based on the Oxford organization of tutorials. And you can see here. These are the different types of sessions that tutors can can have with the students feedback session problem solving questioning collaborative discussions and content knowledge so these are the types of sessions in the tutorial session so what we suggest and recommend is that chat GPT is used as a companion in these sessions, basically starting with a self evaluation with a self evaluation by the students. So they can both the student and the tutor they can both know what's the start the what the starting point is in terms of using the tool and for the type of session. Then prompting practice in terms of the language that they use the interaction. I was reading the other day, an article that says that, you know, as you saw in the previous, in the previous slide, one of the students said to chat I want you to be an expert. And that is, it was the way to go basically. So, doing some prompting practice in this tutorial sessions, but as well, record the process of working with the AI tool through reflection and awareness so in this sense we recommend the students and the tutor work together say what worked. What didn't work. Let's write about it. Let's discuss this process. So it will allow us to have some self monitoring. And basically, the students have the control over the AI rather than the other way around so there is a process of reflection and in which we are aware on how we use the tool and what can be done to continue using it effectively depending on the students needs. So yeah, this is good because if you got your students for eight weeks you can see the progress. By the way, this framework is just, you know, what we started to think about this is not a framework that's been adopted by the university at the moment is just a recommendation so far, you know, we just needs to be discussed. So it's not something that is developed yet. Yes, to finish this is a call from one of our students, student take control of the air tools rather than the air taking over control over the taking control and we have also started to like so many of you maybe started to build canvas or we're using canvas courses online for academics or students to learn about AI and use it. At the moment is still under development. Thank you.


Hi, I'm Daphne Genocopoulos. I want to thank the National Archives in Washington, DC for inviting me to speak today to you about my new book, The Pirate's Wife, the Remarkable True Story of Sarah Kidd. I'm especially honored to be here. And I'm especially delighted because I did some of my research for this book at the National Archives. I'm very grateful to the kind and knowledgeable staff who helped me find many 18th century documents held at their great repository. The Pirate's Wife is a book I've been working on for many years. It is the first ever direct account of the life of Sarah Kidd, the wife of the notorious pirate, Captain Kidd. Sarah lived a long and storied life, and I'm going to give you a glimpse into it. I'm going to show you some slides, and I'm going to do a reading for my book. I hope you all will find Sarah Kidd to be easily as interesting and remarkable as her famous husband. Generations of readers the world over have known the story of Captain William Kidd, the noble sea captain and privateer who turned ruthless pirate. The history of the golden age of piracy, in which Kidd played an outsized part, is replete with the exploits of pirates of many stripes, who hoisted the black flag and prowled the seas, attacking 17th and 18th century merchant vessels bound for the West Indies, West Africa, and North America. Kidd's story is particularly striking. In 1699, Captain Kidd became the subject of a deadly political scandal involving top officials on both sides of the Atlantic, including the King of England. There was a worldwide chase and eventual conviction and execution at the hands of colonial officials who may have been complicit in Kidd's darkest acts of piracy. It was a drama laced with lies, secrets, double dealing, betrayal, and buried treasure. The story of his journey from privateer to murderous pirate has been immortalized across the centuries in print, on stage and screen, in video games, and even in an epic 22-verse ballad. But our understanding of his legend is incomplete. What is largely unknown is that Kidd had a partner and accomplice, a behind-the-scenes player who enabled his plundering and helped him outpace his enemies. This accomplice was his wife, the English-born Sarah Kidd, a young, well-to-do New York socialite whose extraordinary life is a lesson in reinvention and resourcefulness. Sarah was running a thriving merchant business in the New York City seaport when she met Captain Kidd. The encounter set off a high octane romance that often operated outside the law. While Captain Kidd was plundering the high seas, Sarah Kidd was pirating in her own way within the confines of polite society, working to ensure her husband never got caught and that the location of the buried treasure stayed secret. With bold determination to survive and to protect her husband, Sarah secretly aided and abetted her husband, fighting alongside him against his accusers. The most impressive part is that even after Captain Kidd was put to death for his crimes, Sarah remained an incredibly beloved figure in the community. She secured a long and prosperous life for herself and for her children. Sarah knew how to persuade and she knew how to hustle. And in a time when women held little legal power, she found a way to ensure her future and to protect her family. But for more than 300 years, the story of Sarah Kidd has been all but erased from history. My journey into the world of pirates began in 2002 when I was commissioned by The New York Times to write a freelance piece for their museum's special section about the Witte Pirate Museum on Cape Cod. The museum contains the artifacts recovered from the Witte pirate ship that crashed during a fierce nor'easter off the coast of Wellfleet Cape Cod in 1717. Prior to writing this article, I didn't know much about pirates, except what I had read in books and what I had seen in movies. I thought all pirates had a pet parrot, like Long John Silver in Robert Louis Stevenson's Treasure Island, or that they looked like the handsome and fit Earl Flynn in the 1935 movie Captain Blood. Or more recently, walked, talked, and wore eyeshadow like Jack Sparrow in the movies Pirates of the Caribbean. But looking at the everyday artifacts recovered from the pirate ship, silver coins, cutlery, pewter plates, a teapot with the shoulder blade of a pirate, pistols, cannon, navigational instruments, medical supplies, and even a size five shoe, leg bone, and silk stocking, I realized that these mythological figures were mere ordinary men. The legend of the Captain Samuel Bellamy and his love interest, Maria Hallett, told me that pirates had connections on land. They had families and community links, and we don't know much about them. As I was researching my dissertation turned previous book, The Pirate Next Door, the untold story of 18th century pirates' wives, families, and communities, I kept encountering evidence of this mysterious woman who seemed oddly on the periphery of the story of the notorious Captain Kidd. Even the influential and authoritative early source book on pirates, a general history of the robberies and murders of the most notorious pirates by Captain Charles Johnson, did not include Sarah in the lengthy chapter on Captain Kidd. Sarah was alive at the time that book was published in London in 1724, and it would have been possible for the author to interview her or people who knew her. Finding Sarah's initials, SK, scratched on a colonial document at an archive in Boston, started me on a thrilling journey to learn more about her. I would soon learn that her initials marked the spot where love and law divide. To research Sarah's story, I had to use a variety of sources because women left little record of their affairs. Most women, including Sarah, could not write. Men owned all the property and exercised most of the legal rights of the time. I read the contextual history of the pirates and of the time period to learn the political, economic, and cultural events that shaped Sarah's life. I was able to find some important primary sources of Sarah's, such as her petition to regain her seized property from the governor of New York, Lord Belmont, and her last will and testament. Many of the ancient sources that I relied on were handwritten, and I transcribed over 250 of them to make them easier to work with. I visited archives in the places Sarah and Captain Kidd had been, including those in New York, Boston, and Rhode Island. I traced Sarah's steps in the surroundings she had visited, especially in Manhattan where she spent most of her life. I also found great resources in the Admiralty Papers at the National Archives in London. I found a rich cache of letters that were in a sea chest in Captain Samuel Burgess's pirate ship when it was captured by a British privateer in 1699. The letters were in the pirate's mailbag and contained a few letters from Captain Kidd's crewman aboard the adventure galley to their wives and the wives to Kidd's crewman. The letters are dated from 1695 to 1699, and they show that correspondence was conducted thousands of miles across the globe between Indian Ocean pirates and North American colonists, and that there was a successful and direct communications between New York and Madagascar and back again. No letters from Sarah or Captain Kidd were among these and none have survived, but there is every likelihood that they communicated during his three-year voyage through the Mariners Mail Service located on Ascension Island, a remote outpost in the Atlantic, New York merchant captains trading with the pirates in Madagascar stopped for fresh supplies of turtle meat and dropped off and collected the mail, left under a rock with a hole in it near the harbor. A letter from the wife of Kidd's crewman, Jacob Horne, was especially helpful and informative. Writing from their home in Flushing, New York on June 5th, 1698, Sarah Horne told her husband, quote, we hear abundance of flying news concerning you. This meant word had spread from port to port and ship to ship, and that there was trouble aboard the adventure galley. That trouble we would later learn was murder, mutiny, and piracy. To further understand the maritime world in which Captain Kidd was a part of, I attended a workshop at the National Archives in London and conducted research with scholars from the Prize Papers Project, a collaborative effort of the National Archives in London with the University of Oldenburg in Germany. To research and categorize, thousands of yet unopened documents captured by the British in wartime in the 17th and 18th century. I examined trial records, depositions, personal correspondence, ship slogs, cargo inventories, and even a mariners personal journal that was worn from wear in his front pocket. The manuscript's room at the Library of Congress was a terrific resource, and Captain Kidd's own recorded statements gave strong evidence of his devotion to Sarah. Archivists and other historians I met were enthusiastic and helpful during my research. As was my trusty research assistant, Higgins, a Cavalier King Charles Spaniel named for a runaway pirate, Jeremiah Higgins. Here he is inspecting the galleys of the pirate's wife. Unfortunately, there are no contemporary paintings of Sarah, Kidd. She can't be seen that way. But in 1911, the artist Jean-Leon Jérôme Faire created this painting of Captain Kidd in New York Harbor. The attractive, beautifully dressed woman with a lace fan holding Captain Kidd's hand on the deck of his ship is an interesting rendition of a woman who has captured Captain Kidd's attention. One contemporary of Sarah's described her as lovely and accomplished. And the woman in this painting is certainly lovely. Kidd's velvet knee length coat, a sword at his side and a tricornered hat are historically accurate depictions of the attire of the period and what he would have worn as a successful sea captain. His gesture towards his lovely visitor, a suave gentlemanly bow and a warm clasp of her hand fits with the charming demeanor he was known to display. While this painting is a figment of the artist's imagination and in 1911, we did not know the story of Captain Kidd and Sarah. There is coincidentally some historical semblance of truth in it. Born Sarah Bradley around 1670, the future pirate's wife arrived in New York from England with her widowed father and two brothers when she was 14 years old. At 15, she was married to one of the richest men in the colony, a much older wealthy New York merchant named William Cox. When Cox died tragically four years later in 1689, Sarah married a Dutch merchant and mariner named John Orte. It was then that she met Captain Kidd, a well-respected gentleman with whom she began a friendship. Captain Kidd was Sarah's third husband. Sarah was a twice widowed 21-year-old, considered one of the most eligible and sought after women in New York when she and Kidd married just two days after her second husband's death. While the circumstances might appear suspect for Sarah and Kidd to marry so soon after her second husband's untimely demise, Kidd and Sarah were stellar members of New York society. They were New York's power couple and no untoward behavior was ever proved. At the time of their marriage, the well-built, well-dressed 37-year-old sea captain who spoke with a hint of a Scottish accent was one of the most respected men in Manhattan. He was a celebrated war hero and sought after privateer. A privateer was hired by the government to legally plunder and seize enemy ships. He was a legal pirate. During wartime, the resources of warring countries were stretched to the limit and privateers were extra hired hands who own their own armed vessel and served as an auxiliary to England's Navy. A privateer's assignment was detailed in a document called a letter of mark and reprisal. He had investors and the captured prizes and cargo was shared with them. The captain and crew received a smaller proportion of the take. There was sometimes a fine line between a legal privateer and an outlaw pirate. Many a privateer once out at sea and beyond the reach of the authorities turned pirate to avoid having to share the loot. There were other reasons for turning pirate. Turning pirate was an attractive alternative for some men, especially those with wives and families because pirates lived in a highly civilized democratic society. They were paid well when the going was good and they were treated fairly. Pirates lived by a set of rules called articles and each pirate had to sign his name or if he could not write, he left his mark as an ex to join the crew of the pirate ship. With that commitment to be true to his fellow, brethren of the coast, as they were called, came benefits. Each pirate had an equal vote and most were given an equal share of the booty. There was a form of health insurance, life insurance and retirement benefits. The pirate community was designed to support and maintain the relationships on land while they were at sea. For example, if a pirate died in action, his share of the treasure was smuggled halfway around the world and given to his family. Turning pirate was a risky and dangerous choice, but for some men, a merry life in a short one was their motto. Sarah and Kids' Wedding took place in Manhattan on a rainy Saturday, May 16th, 1691. It was a day of high drama and grisly 17th century justice. England was at war with France, pirates were plundering the high seas and two traitors, the self-appointed Governor Jacob Lester and his son-in-law Jacob Milbourne were hanged for treason against King William and Queen Mary in the public square. Sarah and Kid attended the hanging after their wedding. A public hanging was an event everyone turned out for. It was a carnival-like source of entertainment. The stark contrast of the day, a love match and an execution foreshadowed the dark drama that would be their life together. The kids lived in a mansion, Sarah inherited from her first husband, William Cox. It was located on the corner of Pearl and Hanover Street in the Hanover Square neighborhood of Manhattan. The waterfront property faced a wall now called Wall Street in the financial district of Manhattan. Coincidentally, their home was just a few blocks from the offices of my publisher, Hanover Square Press, an imprint of Harper Collins. I like to think Sarah found my publisher for me. Around 1692, Sarah gave birth to a daughter, Elizabeth. And in 1694, she had another daughter, little Sarah. During the colonial period, giving birth was a social and bonding event where women looked after one another. Sarah's babies would have been born at home in a separate area away from the main living quarters. Men were not involved in the birthing process, so Captain Kid would have been busy somewhere else. It was the job of women, relatives, neighbors, friends, and elders in the colony to act as midwives to assist in the delivery. Special birthing linens were prepared and laid out. Sarah would have used a birthing stool or perhaps a loved one held her upright and supported her as she progressed through labor. In the early stages, Sarah would have acted as the hostess of the festive occasion. It was an old English tradition for new mothers to serve groaning cakes, a sweet nutritious baked good made of spices, molasses, rum, apples, and carrots. And to go with it, groaning beer. As they waited for the blessed moment, the women would entertain themselves with gossip, jokes, and stories. In 1696, Captain Kid was given a dream job, two privateering commissions from the King, William III. His investors were some of the most important men in England, including the newly appointed governor of New York and Massachusetts, Richard Coot, known as Lord Belmont. Belmont's complicated relationship with Kid would lead to Kid's tragic downfall. In his brand new ship, one that looked like this, the Adventure Galley was a 287 ton warship with three tall masts, square rigged sails, and 34 big guns. It was a hybrid ship called a galley, meaning that it could be rowed as well as sailed. His job was to hunt French enemy ships and to rid the seas of pirates who disrupted international trade. His commissions were for one year. His initially unsuccessful voyage led to frustration, mutiny, and eventually Kid's turn to piracy. Sarah was key to Kid's fight for his life against the men who accused him of turning from privateer to pirate. She and their daughters spent time with Kid on his pirate ship and Sarah helped Kid hide his stolen treasure. As an accomplice to an outlaw, she was arrested and imprisoned with him in Boston. Once released, Sarah helped construct the narrative Captain Kid presented in his defense. She ensured he was taken care of in jail. She worked tirelessly to help secure a pardon, and she even tried to help plot an escape. After Kid was executed in 1701, Sarah lived another 40 years. She reinvented herself and she managed to go from one half of a criminal outlaw couple back to a high society socialite. She secured her family's inheritance, remarried, had more children, and lived the rest of her life as a prominent and respectable citizen. She even learned how to write her full name. Sarah Kid's extraordinary life is a rare example of the kind of life that pirate's wives lived during the golden age of piracy. Hers is a tale about love, marriage, motherhood, and survival, Sarah's life, and particularly her transformation from a New York socialite to a stateless outlaw sheds new light on the political, economic, and cultural events of the late 17th and 18th centuries. From her, we learn about the economic hardships of widowhood, the political repercussions of piracy, the effects of war on the new and emerging colonial economies, the business of slavery. Her life tells a broader story about how certain women were able to assert their will and reclaim their agency within the oppressive strictures of colonial America. During her 74 years, Sarah lived through seven British monarchs, 21 New York governors, and she experienced firsthand the golden age of piracy. She survived four husbands and three of her five adult children. The pirate's wife, the remarkable true story of Sarah Kid, recasts the image of Captain Kid from a diabolical pirate to a flawed but decent man who tried to please his investors and protect his wife and family. He was a pirate, but not one with a black heart. Sarah's initials, SK, scratched on a few colonial documents gave clues to her existence. Those bold pen strokes reveal a history we have only imagined, the dangerous, difficult, and thrilling life of a pirate's wife. They shed unexpected light on a young colonial woman caught up in a world of politics, passion, and grisly 18th century justice. The unknown SK is finally identified. She is Sarah Kid, the first lady of pirates. I'd like to close by reading a passage from the pirate's wife. This is from the prologue. Sarah Kid laid in a weakened state in the bedroom of her Manhattan mansion. A highly contagious lethal disease raged through the colony, striking young and old, richer poor, black or white. It was September 12th, 1744. And the 74-year-old Sarah had first taken to her bed to get warm under the soft quilts and to rest her head on the goose-down pillows. Then the chills, fever, and fatigue set in. She was nearly certain she had contracted the deadly disease everyone called diphtheria. As a precaution, she asked her family and friends to stay at a safe distance. She arranged for soft foods and a soothing drink made from the medicinal herbs in her garden to be left outside her bedroom door. Her mind wandered in a fever-induced haze. She closed her eyes and remembered herself in another time and place. She was a young woman with her husband, Captain William Kid, on his pirate ship, the St. Antonio, a vessel laden with gold, silver, and jewels. As his closest confidant, she learned that he'd buried some of the stolen treasure for safekeeping, and he described to her where it was hidden. She was not to tell a soul. For more than 40 years, since his death in 1701, Sarah, the pirate's wife, kept his secrets safe. Not even her five children knew. She alluded to it in her will, noting that she had assets in the city of New York and elsewhere. She did not identify elsewhere. Sarah worried about the consequences if her children were caught with stolen pirate loot. Her strong instincts told her it was best to leave well enough alone. As she thought back over her life, not all of her memories were fond ones, especially the time when she was a pirate's wife. But now the memory of the hardships and the heartbreak had softened, and Sarah wouldn't have traded it for anything. She felt proud, very proud, to have been a pirate's wife. And she bore the title as a badge of honor. Sarah repeated a prayer as her condition worsened. Almighty God, have mercy on my soul, and pardon and forgive me all my sins and offenses so that I may, after this miserable life, arise with our Savior, Jesus Christ. She became delirious from the fever and shook uncontrollably. The sheets were soaked with her perspiration. Still, the thought of that secret weighed on her, as well-kept secrets often do. As she prayed for forgiveness, she may have thought it was time to identify elsewhere, to her three children who paced downstairs in the sitting room. It wasn't long before Sarah developed a sore throat that felt like a razor when she swallowed. She tried to speak, but it hurt so much she could only whisper. Her daughter, Elizabeth Kid Troupe, peeked through the keyhole to check on Sarah. The once vigorous woman now appeared very small among the many furnishings and tasseled curtains.  Elizabeth saw her mother's lips moving, mouthing words, but she could not hear her. She strained through the keyhole to hear what she might be whispering. Elizabeth called for her brothers, William and Henry, who had stepped outside on the front stoop that faced the harbor. The cry of the seagull seemed to signal the alarm. Elizabeth told them to hurry. Each took a turn at the keyhole looking and listening. Sarah's breathing was loud and strained as she gasped for air. The three of them looked at each other with tears in their eyes when the room fell quiet. There was not a sound, not even a whisper. For over 300 years, treasure hunters have scoured the North American Eastern seaboard trying to find where elsewhere is. That secret is with Sarah, buried in the churchyard of Trinity Church Wall Street in Manhattan. Thank you.


This lecture is part of an online course on Galois theory and will be about Hilbert's theorem 90. So I should first just say a bit about why it has such a funny name. The reason it's called Hilbert's theorem 90 is that it is Hilbert's theorem 90. More precisely, Hilbert wrote a famous report on the theory of algebraic numbers back in the 19th century. And one of the theorems in his report, you can see this is theorem 90, is exactly the theorem we're going to talk about. I don't know why this name has stuck, but anyway. So, incidentally, this theorem wasn't first proved by Hilbert. You can find more or less the same theorem in one of Comer's papers from many years before. So the original form of Hilbert's theorem says the following. And say the original form because there's another form due to eminertia, which is more general that we'll talk about later. Anyway, the original theorem says suppose K contained in M is cyclic. That means the Galois group is cyclic. Then, if the norm of alpha is equal to 1, this implies alpha is equal to beta over sigma beta for some beta. So here sigma is a generator of the Galois group. So sigma to the n is equal to 1, where n is equal to the degree of m over K. And alpha is any element of m, and n is just the norm. Well, so how are we going to find such a beta given that norm of alpha is equal to 1? Well, what we can do is we can take this equation and we can rewrite it a little bit. So this says that alpha times sigma of beta is equal to beta. And now what you do is you look at alpha times sigma and you think of this as being a linear transformation from m to m, where you think of m as being a vector space over K, of course. So sigma is an element of the Galois group acting, and alpha just means multiplication by alpha, but we're just composing them and thinking of them as being a single linear transformation. And what this says is that beta is a fixed point of alpha times sigma and also another key condition. We need beta is not equal to zero, otherwise you can't divide by sigma of beta. So we want to find a non-zero fixed point of this linear transformation, in other words, something with eigenvalue 1. Well, if we've got a group action, it's easy to find fixed points because we can just average over the group. So let's try and find a group action. Well, the thing to do is to work out the order of this linear transformation alpha times sigma. And in fact, it turns out to have order n, because you can see this as follows. If we take alpha sigma to the n, this is just equal to alpha, sigma, alpha, sigma, alpha, sigma, and so on. Now here, this sigma is acting on all these alphas and this sigma is acting on all the alphas to the right and so on. So if you work this out, it's alpha times sigma of alpha times sigma squared of alpha and so on, times sigma to the n minus 1 alpha times sigma to the n.  Now let's look at what's going on. Here, this expression here is just norm of alpha, which is equal to 1. On the other hand, this is also equal to 1 because sigma to the n is equal to 1, sigma has ordered n. So alpha sigma to the n is just equal to 1. So alpha sigma generates a cyclic group of order, well, dividing n. And now we can find a fixed point. So to find a fixed point, all we do is we take beta to be theta plus alpha sigma theta plus alpha sigma squared theta and so on plus alpha sigma to the n minus 1 theta. And here we've just summed over all elements of the cyclic group generated by alpha sigma and applied them all to theta. And obviously, alpha sigma beta is equal to beta. Well, we haven't quite finished because there's a slight problem is beta none zero. So all we need to do is to find some value of theta such that this is none zero. And it seems kind of trivial, but if you try it a bit, it's not at all easy to show that some theta exists. It might be that this big linear transformation is identically zero. So we need to find some theta so that the sum is none zero. And how can we do this? Well, there's a very useful lemma called Arten's lemma, which can be used to show this. So this says suppose G is any group. And suppose sigma 1, sigma 2 and so on are characters of G. Well, what do I mean by character of G? Well, this means a homomorphism from group G to the multiplicative group of the field M. This is more or less the same as a character in representation theory, if you're wondering. Then Arten's lemma says that sigma 1, sigma 2 and so on are linearly independent over M. So what does linearly independent mean? Well, this means that if a1 times sigma 1c plus a2 times sigma 2c and so on is equal to naught for all c and some finite collection of these characters, then a1 equals a2 equals and so on equals zero. So there's no linear relation between distinct characters. So I should say these have to be distinct characters. Obviously, if sigma 1 is equal to sigma 2, we could just take a1 equals minus a2, which would be kind of rather silly. So let's show that there is no linear relation. Suppose there is a linear relation, choose the shortest possible relation if one exists. So we might have some relation a1, sigma 1 plus and so on plus ak sigma k equals zero. And since it's the shortest possible all ai's are none zero, so we can divide by a1, so we can assume a1 is equal to 1. And secondly, we can divide all the characters by the character sigma 1, so we can assume sigma 1 equals 1. So this just makes the notation a little bit simpler. So 1 just means the trivial character, so it's the character such that sigma 1g is equal to 1 for all g, which is obviously a character. And now what this relation means is that for all c, we have c plus a2 sigma 2c plus a3 sigma 3c and so on is equal to zero. So this is just a1 sigma 1c, which is equal to c by our assumptions. And now we just multiply it by b. Well what's b? Well we pick b with sigma 2b not equal to b, which we can do because sigma 2 is not equal to sigma 1. So we just multiply this by b, we get bc plus b a2 sigma 2c and so on is equal to zero. And then we can also change c to b times c, so we get bc plus a2 sigma 2bc plus and so on is equal to zero. And now we use the fact that all the sigmas are characters, so we get bc plus a2 sigma 2b sigma 2c plus a3 sigma 3b sigma 3c, so on is equal to zero. And now what we do is we just subtract this relation and this relation. And the factors of bc cancel and we find b minus sigma 2b times a2 sigma 2c plus various other things is equal to zero. And now we notice that this term here is none zero because a2 is not equal to zero because we had a shortest possible relation. And b minus sigma 2b is none zero because sigma 2 is not equal to 1 and we picked it. And what we have here is a shorter relation. And this contradicts our assumption that we'd already chose the shorter relation so no relation exists. So in other words characters are linearly independent. Now let's go back and use this to show that we can find the theta such that beta doesn't exist. Actually before that we should just have a sort of warning. We're going to take g to be the multiplicative group of m, so a character sigma i is just a homomorphism from m star to m star. And let's put a warning in fluorescent pinks that everybody notices it. Warning, if sigma i is in the Galois group then sigma i is a character. But there are two possible ways of multiplying elements sigma i. We can take a product of characters, so if we've got any two characters which are homomorphisms from m to m we can obviously just multiply them together and we get sigma i, sigma j of x. This is the product of characters is sigma i of x times sigma j of x. So this is a sort of point wise multiplication. And the product of character is not equal to the composition of elements of the Galois group. So composition of Galois group, instead of doing point wise multiplication we actually compose two maps here. So this would say sigma i, sigma j, x is equal to sigma i, sigma j of x. So you see that this notation sigma i, sigma j is actually ambiguous. We can either be talking about point wise multiplication or about composition and these are completely different operations. For example, this thing here is abelian and this can be non-abelian because the Galois group might not be abelian but this multiplication is also abelian. And you also notice that this product here might not be in the Galois group. If you take two elements of the Galois group and multiply them by this operation it's a character but there's absolutely no reason why it should be an element of the Galois group. So that's a rather confusing thing about characters. It's sort of slightly ambiguous what you mean if you're multiplying two characters because there are two ways of multiplying them. Well anyway, let's go back to finish the proof of Hilbert's theorem 90. So we have characters 1, sigma, sigma squared, up to sigma to the n minus 1. And these are linearly independent by Artein.  So if we apply this to theta it means theta plus alpha sigma theta plus alpha squared sigma squared theta and so on is not equal to naught for sum theta in M. And this is what we wanted to prove because this is the element beta. So we found an element beta that's fixed by alpha sigma and is none zero. So this finishes the proof of Hilbert's theorem 90. Funny thing that the hardest part of this proof is proving the apparently trivial fact that some rather general element isn't always zero. So let's do an example to show that Hilbert's theorem 90 really proves a result about cummer extensions as a special case. So I suppose K contained in M has Galois group, say Z over PZ. And suppose K contains a primitive p-th root of 1 denoted by zeta. And let's see what Hilbert's theorem 90 says about it. Well we notice that norm of zeta is equal to zeta to the P which is equal to 1 because it's fixed by sigma. So by Hilbert 90 we know that zeta is equal to beta over sigma beta for sum element beta. Well this means that sigma of beta is equal to zeta minus 1 of beta which is exactly the formula we use to show that beta is a p-th root of unity. So beta not equal to zero. So sigma of beta to the P is equal to beta to the P. So beta to the P is in K. Beta is not in K because of this condition here. So M is equal to K with the p-th root of beta adjoined. So we recover the theorem about Kummer extensions in that if K contains a p-th root of unity, then M is generated by p-th root of something. So Hilbert's theorem 90 is a sort of generalization of this fact to the case when the field doesn't contain a p-th root of unity. We can't show that it's generated by the p-th root of something but we can still show that it has this property that if something is norm 1 then it's given by whatever that formula was. Just before going on to Nerta's generalization of Hilbert's theorem 90, let's just have another application of Arting's lemma. We can show that the trace is not identically zero for Galois extensions K contained in M. Well we've already shown this for arbitrary separable extensions so proving it for Galois extensions is just a special case of that. But if you remember the proof for arbitrary extensions used a lot of rather messy looking calculations with determinants and so on. Using Arting's lemma it's completely trivial because we notice that trace of theta is equal to sigma 1 of theta plus sigma 2 of theta and so on plus sigma N of theta. And this is not not for all theta because if it was it would be an obvious linear relation between the elements sigma 1 up to sigma N of the Galois group. So trace of theta is none zero for some theta. Notice incidentally this immediately implies the bilinear form is none degenerate because if A is not equal to zero then the inner product of A and theta over A is the trace of theta which is none zero. So for Galois extensions Arting's lemma gives a very short proof of the fact that the bilinear form is is none degenerate. So now we're going to go on to Nerter's form of Hilbert's theorem 90 and life is really unfair because not only was Hilbert's original theorem 90 not really due to Hilbert but Nerter's version of Hilbert's theorem 90 is always called Hilbert's theorem 90 and everybody forgets about Nerter but whatever. So what Nerter's theorem says is suppose K contained in M is any finite Galois extension. So instead of taking just a cyclic extension we're going to allow any Galois extension. And now it says that any one co-cycle is a co-boundary. So what is a one co-cycle? Well a one co-cycle is a map from the Galois group to M star. Strictly speaking this is a one co-cycle with coefficients in M star. And it's usually noted by sigma maps to an element alpha subscript sigma and I sometimes accidentally write my sigma so it's hard to tell their subscript so watch out for that. And it has to satisfy the following rather bizarre looking condition. Alpha sigma tau is equal to alpha sigma times sigma of alpha tau. And I'll explain where this weird looking condition comes from a little bit later. By the way this is the condition when you're taking a one co-cycle with values in a multiplicative group. If this were an additive group then you'd have alpha sigma tau is equal to alpha sigma plus sigma alpha tau which people sometimes use. And what is a one co-boundary? Well a one co-boundary is something where alpha sigma is equal to beta minus sigma beta or beta divided by sigma beta depending on whether you're an additive or multiplicative. Here beta is a fixed element of I guess it would be M star in this multiplicative case. Let me put this in brackets because we're not really going to use it. And the first co-homology group of the Galois group with coefficients in M star is just a group of one co-cycles divided by one co-boundaries. So this is just like an algebraic topology where a first co-homology group is some collection of co-cycles over co-boundaries except usually co-cycles are a little bit more intuitive than this weird expression here. So Nertis theorem says that this group is the trivial group. In case you're wondering, yes there are co-homology groups for higher dimensions. For instance the second co-homology group of the Galois group with coefficients in M star turns out to be the Brouwer group that I mentioned earlier. But we won't be talking more about this now. So we should try and explain. Well I've given you the definition of a one co-cycle but I haven't really explained what it is. So I should say what on earth is a one co-cycle? So I mean what is the motivation for it? Not what is this rather incomprehensible definition? Well let's look at the following expression. I take alpha sigma times sigma and I'm going to multiply it by alpha tau times tau. And I'm going to think of both of these as being linear transformations on M. Sigma and alpha sigma are both linear transformations because this is an automorphism of M and we can think of this as multiplication by the element alpha sigma. And if I work at what this is, it's alpha sigma times sigma alpha tau times sigma tau applied to something in M. And by the one co-cycle condition this is equal to alpha sigma tau times sigma tau. So in other words the map taking sigma goes to alpha sigma dot sigma is a sort of action of the Galois group on M star. Except it's a bit of a funny action because it's an action on M star as a set. It doesn't necessarily preserve the group action. So the action of the Galois group on M preserves the group action. But this is a sort of twisted action which doesn't preserve the multiplication. However it still has an action of the Galois group on the set M. So that's what a co-cycle is. You can think of it as being a way of twisting the action of the Galois group on M so it's still an action although no longer preserving the group operation on M. And now we can ask what is the condition for being a co-boundary? Well if alpha sigma is equal to beta over sigma beta this just means that alpha sigma times sigma beta is equal to beta. In other words beta is fixed by this funny twisted action. So if the action preserved the group action of M star then the fact that beta was fixed would not be terribly interesting. You could just take beta to be the identity element. But what this says is that if this action is a co-boundary then this funny twisted action still has a fixed point even though it might not be the identity element of M. Anyway with this interpretation of a one co-cycle Nurtis theorem becomes almost trivial to prove. If you don't think of this as being a linear transformation it's really hard to prove Nurtis theorem. So here's a proof of Nurtis version of Hilbert's theorem 90. Maybe it should be called Nurtis theorem 90 or something. So suppose alpha sigma is a one co-cycle. Then we get action of the Galois group on the set M star. This follows just by mapping sigma to the element alpha sigma times sigma. And we want a fixed point. And to get a fixed point we just take the average over the group. So we take an element beta which is sum over all elements of sigma of alpha sigma times sigma of some theta. So beta is always fixed by the Galois group. Well wait a moment we just said that this funny twisted action didn't preserve the group operation. And suddenly we're taking an average over the Galois group which implies it does preserve the group operation. So what's going on? Well this twisted action sigma goes to alpha sigma dot sigma does not preserve the multiplication on M but does preserve the addition. In fact these elements are just linear transformations so that they preserve the additive structure of M so it's perfectly okay to take an average over the group using the additive structure. So we found an element fixed by the Galois group and as before we need beta is not equal to zero and we can choose theta so beta is not equal to zero by Artin's lemma that we had before. So just as in the cyclic case there must be some theta such that beta is none zero otherwise we would have a linear dependence between all the elements of the Galois group. So next we'll just check that the Noter's non-Abelian form of theorem 90. So let's show that Noter's theorem 90 implies Hilbert's theorem 90. So let's suppose K contains an M is cyclic and suppose that norm of alpha is equal to one so we want to show that alpha is equal to beta over sigma beta where sigma generates the Galois group. In order to apply notice then we need a one-coast cycle where we get a one-coast cycle as follows we put alpha one equals one remember we've got to have an element alpha something for any something in the Galois group and let's put alpha sigma equals alpha. Now the co-cycle condition implies that alpha sigma squared must be alpha times sigma of alpha and it implies that alpha sigma cubed must be alpha times sigma of alpha times sigma squared of alpha and similarly you can go on and you can check alpha of sigma to the N must be alpha times sigma alpha times sigma squared alpha up to all the way times sigma to the N minus one of alpha times I guess that's it and now sigma to the N is equal to one so these two elements have to be equal in order for us to get a one-coast cycle well they're equal as norm of alpha equals one so norm of alpha implies that this is a one-coast cycle so saying alpha is norm one is really just the one-coast cycle condition so this means that alpha sigma to the I is equal to beta over sigma to the I beta for some beta not equal to zero and if we take I equals one we just find alpha is equal to beta over sigma beta which is the conclusion of Hilbert's theorem 90 so Hilbert's theorem really is a special case of Nurtur's theorem next we can just say a little bit more about what a one-coast cycle is first of all one-coast cycles sort of classify things that become same over M so for example suppose we take R contained in C and we want to classify Lie algebras over the reals so it doesn't really matter whether you're classifying Lie algebras or some other algebraic structure I'm just saying Lie algebras because that's a common thing people like to classify now simple Lie algebras over the complex numbers have been classified by Carton and Killing and you might want to classify simple Lie algebras over the reals well if you've got a simple Lie algebra over the reals you can tensor it with the complex numbers and you get a simple real Lie algebra over the complex numbers so the question is if you've got a Lie algebra over the complex numbers how many Lie algebras over the reals give rise to it? for example if you've done a Lie algebra course you know you can take the Lie algebras SL2 and SU2 and these are different over the reals but if you tensor them with the complex numbers they both become isomorphic so the general problem is suppose we've got some two algebraic structures they might be Lie algebras they might be something else and suppose if you tensor them over M they become isomorphic so suppose there's an isomorphism I between them then we want to sort of find you know if we've got one possible V we want to find all other possible Ws that become isomorphic to it over M so that we can classify whatever these structures are and now we've got an action of the Galois group on both of these because it acts on M so we can put a sigma equals i sigma i to the minus one sigma to the minus one so a sigma is going to act on this M see we'll apply sigma to the minus one then i to the minus one then sigma then i and similarly we're going to get a sigma tau which is going to be i sigma tau i to the minus one tau to the minus one sigma to the minus one and we get sigma a tau which is going to be sigma and then we put a tau i tau i to the minus one tau to the minus one sigma to the minus one so the action of sigma on one of these things is just going to be given by conjugation and then we can check that a sigma tau is in fact equal to a sigma times sigma a tau which is the one co-cycle condition. So in other words, an isomorphism between if we take two different things over a field, an isomorphism between them over the extension field is just a one co-cycle. And it's not terribly difficult to guess that a one it's going to be two one co-cycles are going to differ by one co-boundary if and only if these two things were isomorphic over the field k you first thought of. So in other words, the first co-homology group of one co-cycles modulo co-boundaries classifies things over k if you know the classification over this bigger field m. So one co-cycles tend to turn up quite a lot when you're trying to classify things over fields that are not algebraically closed. For instance, if you're trying to classify elliptic curves, elliptic curves are easy to classify over the complex numbers, but then number theorists want to classify elliptic curves over say the rational numbers. And this starts producing lots of one co-cycles with values in the automorphism group and elliptic curve and so on. Okay, so that's all about Hilbert's theorem 90. Next lecture we'll probably be doing something about infinite Galois extensions.


Hey everybody, Dr. O here. So we just got on talking about ionic bonds, which are caused by or created by a transfer of electrons. Now we're going to talk about covalent bonds. They're created by a sharing of electrons. So I mean, it's a mutually beneficial relationship. These two covalently bonded atoms are intimately connected, which is why these bonds are going to be stronger than ionic bonds. Really good analogy from the book talks about next door neighbors whose kids go to one house and play and then come back to the other house and play. They're not, I'm not giving up Oliver when he goes to the neighbor's house to play and they're not giving up their kids when they come here to play because they're going to go back and forth. So the sharing of electrons, the cycling back and forth is what makes covalent bonds so strong. Now, why do we care about them in human anatomy and physiology? They're very important in a lot of your organic compounds like your carbohydrates. For example, peptide bonds, which holds amino acids together in proteins. Those are going to be a type of covalent bonds. So very, very important. All right. So here you see you can have single, double, you can even have triple covalent bonds. So if two atoms are sharing one pair of electrons, they have a single covalent bond. If two atoms are sharing two pairs of electrons, they would have double. Here we see on the bottom that carbon is actually has double covalent bonds with two different oxygens there to form CO2, carbon dioxide. But you can even have triple covalent bonds where two atoms are sharing three pairs of electrons. But the key there is the sharing. So how many electrons are they sharing, that's what determines the type of covalent bond you have. Another very, very important thing to keep in mind when it comes to covalent bonds is we have nonpolar and polar covalent bonds. So whatever analogy you want to use with a nonpolar covalent bond, the sharing of electrons is equal. So think about maybe like a teeter tot or maybe you call it a seesaw. Two people of the exact same weight, they're sharing the load equally so it's balanced. But if one person is much heavier than the other person, there's going to be an imbalance there. My favorite example with polar covalent bonds would be tug of war. So let's talk about polar covalent bonds. It's when the sharing is unequal. So you see here at the bottom methane would be nonpolar covalent bonds because their, the sharing of electrons is equal. Water is going to have polar covalent bonds because the sharing is unequal. So if you look at, so like the oxygen has that partial, that's a little, that little symbol is delta. That means partial negative. So oxygen has a partial negative charge, meaning the electrons are spending more time near oxygen than they are hydrogen, which is why hydrogen has a partial positive charge. So if you want to use, so oxygen has eight positive protons pulling these negative electrons towards it. Hydrogen only has one positive proton, weakly pulling the electrons towards it. That's why the electrons, they stir, they're still shared, but they spend more time with oxygen. So if you want to go back to your teeter totter example, oxygen eight times heavier than hydrogen. It's very unevenly balanced or a tug of war. If I'm eight times the size of my kid and we're in a tug of war, I'm going to win easily. I'm going to pull and pull and pull and the electrons are going to travel my direction. So that's why I like that tug of war analogy. You have two people exact same size mass strength in a tug of war. It's going to be a battle back and forth. That would be a nonpolar covalent bond. Me having a tug of war with my five year old Oliver, that's going to be a polar covalent bond. I'm going to win almost all the time. So I'm going to get the electrons more often than not. Another example here would be hydrogen chloride. So chlorine has 17 protons, hydrogen has one. So the negative electrons are going to be way more attracted to the chlorine side of this bond. So yes, they're still sharing, but hydrogen is going to have the partial positive end because the negative electrons are being pulled away from it more often than not. Chlorine's going to have the partial negative end. So this matters. This is going to matter very much in our next video where we talk about hydrogen bonds because these partial positive and negative charges will determine how a compound interacts with its neighbors. And just so you know, the reason they're called partial positive and negative charges is because these forces, they're not as strong as an actual ionic bond. So that's why they're called partial positive or partial negative. Okay. So that's going to be your covalent bonds, both nonpolar and polar. I hope this helps. Have a wonderful day. Be blessed.