After a long wait and anxiety, Senator Madness Abe has been elected and imaged as the candidate of the Social Democratic Party, SDP, to contest the 2023 governorship election in the river state, following the voluntary withdrawal of Pernan Morris, former candidate. Now the primaries, which was conducted in Port Hacker, the river state, is supervised by the Independent National Electro-Commissioned Senior Officer Pastor Immanuel Issa Madness from security agencies was described as the fairest primaries ever held in the state. Chairman of the SDP Substitution Electro-Commite Mohamed Ibrahim, the Kled Abe winner of the election on a post haven't pulled 30 votes. The emergence of the former senator who represented the river south east senatorial district in the red chamber has split pay to speculation as to whether he will make good his promise of being in the ballot. In accordance with the Electoral Act of 2022 and annual guidelines, nearby certified that Senator Madness Abe has participated in the Substitution Primaries and has scored 30 valid votes in the SDP Organatorial Primaries for River State here today. He is hereby declared as the duly elected Organatorial candidate of SDP River State and is returned elected. While describing Abe as a credible candidate, Ibrahim advised members of the party to go and get the permanent voter cards in order to lead candidates of their choice. Meanwhile, a retired permanent secretary from the River State Civil Service, Princess Patricia Obunna, was elected as a running mate to Senator Abe and she appreciated the party for giving them the opportunity to be flag bearers come 2023. In a solidarity message, a former member of the House of Representatives and a chieftain of the SDP in the state, Chidi Nguyenhawka, said with what transpired in the recent governorship election, Nigerians could cast a vote and the confidence that it would count. The state chairman of the SDP, Sol Suama Osaro expressed confidence that the candidate of Abe would make the SDP a party to beat in 2023. Please do subscribe to our YouTube channel and don't forget to hit the notification button so you get notified about fresh news updates.


So, what is your favorite hidden spot in Oslo? I don't have exactly one hidden spot per se. However, there's one category of hidden spots that I really like in the city. There are a couple, maybe three or four, tiny museums plotted around the city that I really like going to. They're not as extravagant, not as big. You can't spend hours in there, obviously, as you could in Oslo or at Moscow Museum. But they have this really nice charm because they're usually curated by people who have a very strong passion for these almost niche subjects that the museums cover. One example would be the Viking Museum, which is sort of hidden away in the basement of a bank in the middle of the city. There's also the antiquity museum here at Ores University, which I really enjoy going to because despite it's sort of a hidden location also in the basement, it just has so much charm and it's so well done in every aspect that you can even imagine. And usually those museums are not as crowded as the others. Obviously, they don't attract the same tourist crowds. All museums I really like is the Occupation Museum, located right next to the center of the city. It is located underneath the Women's Museum, which is also a great museum, I gotta say. But underneath there, there is the Occupation Museum. It's very tiny. You could spend perhaps an afternoon down there, but what I really like about it, as I mentioned before, is the genuine passion of the people who work there. You might not even realize some of the stories that you can find from the Second World War, the occupation of Denmark. If you just read on Wikipedia, you wouldn't get the same impression as some of these very personal stories that the curators at the museum are able to deliver to you. It's also very nice, but it's a very unique setting down there, where you really feel connected to this very dark spot in our history and past in the history of Denmark and the history of Ores. Usually, when you think about the Second World War and the Occupation, you think of the Grand Aspen, the war, the fighting. But you can sometimes forget the little man, the little people, the people who had to carry on with their everyday and relatively tiny city as orders. How did they manage? What did they experience during this time? I was very surprised when I went there a few years ago, the hardships that even privileged people in a not so war-torn country as Denmark in a city as orders managed to experience. But it's really, really interesting and unfortunately hidden away, but that's part of the charm as well. My favorite hidden spot now is the Ores stadium, also called Serres Park, because that's where the local football team, AGF, is playing their home games. And there's a really great atmosphere there every time and they just had the best season for 22 years by winning the third place. So that's a really great place right now to be.


Served now to the backhand slice of Barty. Backhand now from Collins goes into the backhand slice of Barty. Collins with the backhand, down the line, Barty with the forward pass. It's a winner! Barty's the champion! Barty is the champion! The drought is over! The drought is over! The drought is over Australia. We have an Australian Open Champion for the first time in 44 years. Pat Cash couldn't do it. Layton Hewitt couldn't do it. Ash Barty can! There is no one better than Ash Barty right now. She's got two of the three slams in her pocket. How can you not love her? How can you not love her? How can you not love her? I reckon she can win the US Open and finish the... She can complete the set.


Welcome to the series, Photographic Chemistry, presented by the Foundation of the American Institute for Conservation of Historic and Artistic Works. This program was made possible by grants from the National Endowment for the Humanities and the Andrew W. Mellon Foundation. Each program in this series is presented as a short video. Depending on your video viewer, you should be able to pause, return to a previous section, or skip ahead to a later section by using a scroll bar or on-screen icons. You will find an outline of the course and short quizzes to test your understanding on the course webpage. When we view high magnification micrographs of silver halide grains, we notice that they have very regular shapes. They will have a shape ranging anywhere from a cube to either a hexagonal shape or even an octahedral shape, but that's about all you normally see. It's very regular. The crystals that are formed in the silver halide families all follow a cubic pattern. They make cubic shapes where their edges are of various sizes. And this is what changes their dimension from being a perfect cube to maybe more of a hexagonal or octahedral type shape. And how can we control these shapes and being able to make and predict the exposure characteristics for these silver halide grains? Well, to be able to study crystal shapes or habits, we need to know an area of chemistry known as solid-state chemistry. What I'll do is introduce some of the terms and have you go back and fill some of the details in at your own reading. But essentially all we're going to be using is designations that allows us to move into a three-dimensional cubic space, a room, if you will. I need to have a map where if you and I are going to discuss where we are and where we're going or where a certain location is in the room, we need to be able to know how to translate from where we are to where we're going. And in these crystal habits and the grains, remember there are a regular array of positive silver ions and negative halide ions, and they alternate positive, negative, positive, negative, and so on. But within each plane, they're going to be similar. There'll be planes of positive silvers and planes of negative halide. Remember earlier I said this is an important concept and we'll come to know more about why this is as we move forward. Looking at the slide shown, you'll see two designations. Suppose we move in just two different directions from the center point of a crystal. We're standing in the middle of a cube and we want to move. We have two distinct ways we can move under what is known as the Miller indices or the index by which we map our motion. I can move in the B direction and the B direction under the Miller system is two, zero, zero. Or I can move in the A direction. In the A direction I'm moving along the one, one, one direction. Or these two faces once formed are normally referred to in solid state chemistry as the two, zero, zero face or the one, one, one face. Now you can imagine just in the way I drew it that I can generate two types of crystals. If I have my B direction or my two, zero, zero faces which are all at right angles to one another both up and down and left to right. If these grow out the fastest in relation to A what I'm going to end up with is a cube. If on the other hand I have my one, one, one face growing at about the same rate that my two, zero, zero face grows this is where I end up with a one, two, three, four, five, six, seven, eight sided at most an eight sided or octahedral shaped crystal. If there is not such an equal growth amongst the two faces I may end up with only a hexagonally shaped crystal or maybe it has growth that is faster on three sides but not one so I get preferential growth for example on the two, zero, zero faces but my one, one, one is it's just hanging in there giving me these clipped corners if you will and this is how I get that very characteristic hexagonal shape where there's three equal long sides and three equal short sides that make up the grain which you see for example in the logo for I, S, and T what are these conditions that changes the rate of growth along either the A direction or the B direction? Well what we know then is that just mathematically not even thinking about the chemistry of the emulsion making process but just mathematically if I can get my A to grow faster than my B but not so fast I mean it can't grow so that I end up with a cube again if it grows about three times faster then with my A growing faster than my B I'm going to end up with the octahedra as shown I'm going to end up with one, one, one faces if however my B grows at just a little bit faster rate than the A this has to be only the square root of three faster about 1.5 times or so but just a little bit faster then I'm going to end up with a very cubic characteristic shape that we associate with other cubic crystals such as table salt so now let's picturing this graphic in our minds let's now turn these faces in the silver halide chemistry to illustrate just how we get the various shapes that we see in silver halide emulsions you have completed this unit depending on your video viewer you should be able to scroll back to any point in the video as desired the short quiz found in the course materials on the website may help you confirm your understanding of the concepts introduced here many thanks to the instructor, production editor, coordinator and the collaborative workshops and photograph conservation committee for their work to make this program possible


What's going on? Everyone's saying I'm here with late in sports cards. You're up in 22, 23. Penini Court Kings, basketball hobby, 16 box, full case break, number eight, tiered, random team style. There's a look again at your random results. As always, you can find these on the website under the results tab, as well as a link in chat. Well, let's hit you a big LeBron case hit, Michael. Then you could sell it and buy more rookies. There you go. Ooh, modern problems require modern solutions. Aren't we Court Kings? What are we, what are we, five years old here? What are we, three? Nice. Pulled the trout and otani. Oh my God. Good Lord, Manny. We have to play at some point, Manny. Lamello ball to 149. I think they're too hard, dude. I stink at the showdowns. I'm terrible at it. 20 out of 75, that is Sean Kemp. What's up, Raiden? Sean Kemp going to the Thunder, Impressionist Inc. We've got Palo, Rookie Otto. Nice one there for the magic. Kawhi Watercombs. Palo, Rookie. Would I say Otto? Oh, just Jersey, Jersey, I'm Palo. Thank you, good call. Terry Eason, Rookie, and level three is Jalen Williams. Nice. See, my issue with the showdowns is like, I didn't play last year. I only played back in 21s. I don't know if they made this change last year. But with the showdowns, even if you lost, if you were doing the missions up to the final showdown, if you lost, you would just go onto the next one and try to get more runs.   So I'll get to like the fifth or sixth mission and then I'll lose. And I'll be like, well, that was a waste of my time. Like that's a little like obnoxious having to do that. It's made for people that don't do anything else besides playing the game. True, true. Yeah, you get to the last mission and you lose. Like it's ridiculous, dude. Pink Fred Van Vleet 2.99. Nice one here for the Pistons. The 1.99, Jayden Ivy, Fresh Paint Rookie Auto. Nice one there, Detroit. Christian Coloco, Rookie Expression. Jersey for the Rappers. Cade Cunningham, Watercolors. Jalen Durian, Works in Progress, Rookie. Jalen Williams, Level One. Blake Wesley, Level Two. Like dude, it's just like, it's so obnoxious. And like they make all the programs like you have to do the showdown if you want to get to the end. Like you can't just like avoid the showdown because usually it's like 25 points or whatever it is.   Watercolors, Red, Kevin Garnett, 2.149. 2.99, Jeremy Sohan, Fresh Paint Rookie Auto for the Spurs. Here we go, San Antonio. Got Al Horford for the Celtics, Art Nouveau. Here we go, Boston. Jaden Hardy, Works in Progress. First steps, Jabari Smith, Rookie. Level One, Jaden Hardy. Level Two, Blake Wesley. Next box. Yeah dude, like we were just talking about dude, there's one thing in the show. So basically what you do is it's kind of like, have you ever played like the showdowns and like Ultimate Team for Madden?  Yeah, that's like, you used to be able to do that in MLB too. So like you'd have to, so basically like your team is down like 16-nothing to start in every mission you do, you get like runs and then you have to go against the final boss, you have to score like, you have to win basically.  Now you get two runs, two runs, two runs, losing, you're done. It's like dude. Yeah, absolutely. So obnoxious. Hopefully. LePron to 149. There you go, Michael. Nice one there. Just following how to stroke. Redemption, Jalen Williams. Fresh paint going to the Thunder. Here we go. Okay, see, congrats on that one. Very nice. Very nice. Funda. Jalen Williams. Ty Ty Washington. Rookie expression jersey. I can't beat any showdown, dude. Like I said, I get to like the fifth mission I'm done. I just got absolutely cooked. Kairi, Oseme and Jain. Level one of Wendell Moore and level three in Junemhardt for the Pacers.   I've had it. I really have had it with the showdowns.   Pink is a Devon Bissell to 99 for these spurs. David Roddy to 199. Rookie fresh paint on it for the Grizz. Nice penmanship on that one. There you go, Memphis. AJ Griffith, Rookie expression jersey for the Hawks. Watercolor Scotty Barnes. Dyson Daniels. Ochi, Bahi, level one and level four. Wendell Moore, Jr. for the Timberwolves. Nice level four Rookie there. Okay. Yeah, true. Give some confidence back to Conquest. He's got to smoke him. Walked off Randy Johnson with a grand slam with home run Derby Soto. Oh, there we go. Good God, dude. Conquest are pretty fun. I still don't really understand it to be honest. I haven't taken the time to like figure it out. Devon Booker to 49. Hey, nice one here. Michael to 199, Rookie Auto. Scotty Pippin, Jr. for the Lakers. There you go, Michael. There's a Rookie Auto for you, my friend. Scotty Pippin, Jr. Carl Anthony Towns, Art Nouveau, Game Use Jersey for the Timberwolves. Donovan, Terry Eason, Works in Progress Roof. Level one, David Roddy. Level two, Patrick Baldwin, Jr. For the Warriors. It is, yeah, in a hall of fame, it's possible to see a ball, dude. It's ridiculous. So annoying. So good, dude. Just paint corners, like. The game is so fun, though. I'm addicted to the game, dude. It's so much fun. Yeah, famous. Kind of a famous name, yes, to some degree. There's a Tatum for you as well to 49. Watercolor. There you go, Michael. Second year, Josh Christopher. Master Strokes to 75 for the Rockets. Jeremy Sohan, Rookie Expression Jersey for the Spurs. Dude, Madden Stinks. Yannis, Kevin Garnett, Patrick Baldwin, and Johnny Davis. Madden is, we've gone on our rants about Madden in this chat, dude, it is, like, it's horrible. I haven't played it in months.  Like, I would just buy every Madden every year, and I'd play it for like a couple months, and I'd love it. And then I just played it less and less and less, and I'd stop buying it, and I didn't buy 21 or 22. Oh, yeah, it's absolutely a slot machine to pay to win.  Trey Young, Purple, Sue 49 for the Hawks. That is 249 Rick Fox on the Master Strokes. The Show, Madden, another Show, Chell, 2K, Madden, I agree with that phone. Shade on Sharp, Art Nouveau. That's pretty cool, man. I saw it in their Blazers, Zion Meisters, Larry Bird, Peyton Watson, and Oostman Jain. It might have been you, Max, that said, like, you're just paying for a roster update. Yeah, I love Chell, Chell's a lot of fun. I can't really play online because people are, again, people are just too good.  Yeah, pretty much. There's Trey Young, my A-Shows for the Hawks. Again, that is to 149. Got Ty Ty Washington, Jr. to 199. Fresh paint for the Rockets. Chris Middleton, Art Nouveau, Game Use Jersey for the Bucks. You talking about Chell, man? Or are you talking about the show? Ty Ty Washington, Tyries Halliburton, Chet Holmgren-Rookie, level one and level two Christian Brown for the Knights. Yeah, I know what you mean, Michael. I know what you mean. Yep, absolutely. Yeah, the people who pause in the show, dude. Oh, my God. Oh, that's tough, dude. 96 punter. No. Purple is Anthony Davis to 49 for the Lakers. To 99, Musa Diabate, Fresh Paint for the Clips. Game Use for the Kings. Deer and Fox, Art Nouveau. Got a Watercolors, Anthony Edwards, Jaylen Williams-Rookie, level one, Andrew Nemhard. And level two, Keegan Murray. That's one there for the Kings. And cards, I'm not gonna need to lose them in the show for the first time. Yeah, that's the only thing that's gonna be rough about the show this year. James Harden-Pink, because I'm gonna have cards that I just love. To 99 for the Sixers. That's one here to 49, Brook Lopez, Master Strokes for the Bucks. That's pretty sweet. We've got Luca Game Use, artistic endeavors. That's going out to the Mads. That's one, Dallas, followed by Benedict Matherin, first steps. Ooh, that's the, oh, what is this called? This is a big hit right here. This is the Palo Bancaro. Rookie, there's one card in the set and it's Palo. Let me double check that really quick. That's big dude, let's go magic. Yeah, it's one of the super short prints and Palo's the only one in the set. Where is it, where is it? They did it with Cade last year, I remember that. Yeah, this one, Rookie Exclusive. Rookie Exclusive Palo, nice one, they're magic.  Again, Palo the only one in the set, the only card in that set, we'll get that magged. Jake Loravia Rookie and level three Walker Kessler for the Jazz. I remember Kenzie, I was toppling for Kenzie last year and we pulled the Cade of that and we were like, what is that? We had no idea. We looked it up and we were like, oh, that makes sense. Yeah, those are pretty solid dude. Bradley Beale, foil. Otto is gonna be Evan Mobley to 75, second year for the Cavs, Master Strokes. Very nice. We've got Oseman Zhang, Rookie Expression for the Thunder. First steps, Jalen Thurin, Scotty Barnes, Maestro. Jeremy Sohan, level one and level two, Marjan Bochamp for the Bucks. Four boxes left everybody, good luck again here. Kevin Durant to 149 for the Nets. Next one here for the Jazz to 199, Ochi Agbaji, heir apparent Rookie Otto for the Jazz. He looked really good toward the end of the season. He's getting a lot of minutes, he looks solid. Delan Terry, Rookie Expression for the Bulls. Do I hit my best with? Shade on Sharp, Jalen Green, Jalen Williams Rookie and level four Caleb Houston. My favorite card, I haven't played a lot of online because I'm still trying to grind my, my team still kind of stinks. Like I just have like, I love the, for some reason dude, I rake with the Corey Jolks tops now. I don't know why, but I have that at my DH dude and I've just hit like bombs with him. I don't, again, I don't know why, it's very random, but I like that card a lot. I probably hit my best with the Corey Jolks, surprisingly. Who else do I like? I don't see that, the throwback Ryan McMahon's really saw, my buddy used to torture me with that card in 2021. That's all, he used that card the entire year. That is a Anthony Edwards to 49, seven to 49. Any Julio or Juan Soto. To 49 is Tony Parker, Brushstrokes for the Spurs, nice one there, TP. We've got Jimmy Butler, Art Nouveau, Game Use Jersey for the Heat. Terry Eason, first steps. Anthony Edwards, Maestros. Ty Ty Washington, level one. And level two, Johnny Davis. The Yadir Drake is pretty solid too. The World Baseball Classic Yadir Drake. Really good contact, really, really good contact. Been doing fairly well with him. He's like my lead off and he's quick too. There's Donvimichael Foil, that is not numbered. Nice one here, 39 of 75, Deontay Murray. Impressionist, Ink for the Hawks. Deontay Murray. We've got Joel Envied, Art Nouveau, Game Use Jersey for the Sixers. Yeah, new Tatees. People are saying it's the best card in the game right now. Yanis Watercolors. Marjan Bochamp works in progress. Peyton Watson, level one. And level two, Nicola Jovic for the Heat. Last box of the break everyone, good luck here. See what we can do. Court Kings, let's see a one-on-one tonight with us. That is Kelden Johnson, Foil. We've got 275, Rick Fox, Master Strokes. Auto for the Lakers. Big Bang Theory Legend. Keegan Murray, Rookie Expressionin' for the Kings. Nice one there, Sacramento. Lastly, we've got LeBron Maestro, Dame Watercolour. Level one, Keegan Murray. And level two, Dyson Daniels. That'll do it for the break, guys. We'll do a quick recap here. We've got a Keegan Murray, Joellen B. Jimmy Butler, Dallin Terry, Osemane Jane. Luca, De'Aren Fox, Chris Middleton. Shadon Sharp, Jeremy Sohan, Karl-Anthony Towns. A.J. Griffin, Ty Ty Washington, Al Horford. Christian Coloco, Paolo. Autos, we had a Rick Fox, Deontay Murray, Tony Parker, Ochaibahi, Evan Mobley, Brooke Lopez, Musa Diabate, Ty Ty Washington, Rick Fox, Josh Christopher, Scotty Pippen Jr., Jalen Williams, Jeremy Sohan, Jayden Ivey, Sean Kemp.  Only card in this set, Halibank Arrow. Nice short print there. That'll do it for the break, guys. Thanks again for the fill. Next up, we've got football.


In this lecture, we will discuss a very important class of molecules called metal allyl complexes. These metal allyl complexes are indeed very complex, because they adopt a variety of interesting structures and they undergo a variety of interesting reactions as well. So, before we plunge into metal allyl complexes, let us just get the context in which we are talking about them. We started by looking at molecules where we had a formal metal carbon double bond. These are molecules like carbenes and also metal carbonyls, where you do have a metal carbon double bond, which can be written in the valence bond formalism. You can use a double bond to indicate the type of interaction between the metal and the ligand. The important thing is that there is only one carbon attached to the metal. Following that, we also discussed metal alkyl complexes. These are complexes where there is a metal carbon sigma bond and there is no formal pi bond between the carbon and the metal. So, these are typified by the example that I have given you here. Any alkyl compound, metal alkyl compound represents a case where you have only a sigma bond between the organic ligand and the metal. We also talked about metal allyl complexes, where two carbons are interacting equally with the metal atom. Although we do not indicate these interactions specifically, you can see that these are molecules in which two carbons are bonded to the metal. So, we write them as hapticity as 2 and indicated with eta 2. This eta 2 symbol indicates that two carbon atoms of the ligand are interacting with the metal atom. Today, we are going to talk about metal allyl complexes, which are interesting because they have two forms which are predominantly found in the literature. One in which the three carbon atoms of the allyl moiety are interacting with the metal. In almost equidistant, the metal is almost equidistant from all three carbon atoms. In other words, carbon atom 1, 2 and 3 are interacting equally with the metal. So, the hapticity is 3. So, we indicated with the eta 3 notation, eta with the superscript 3 to indicate that three carbon atoms are interacting with the metal simultaneously. The same molecule can also interact with the metal through only one carbon atom. When it does so, we have a eta 1 allyl species. This is also an allyl compound, but this is a eta 1 allyl, where there is only a sigma bond between the metal and the carbon moiety. The double bond, which is present on the carbon species is not interacting with the metal. We shall see specific examples of these molecules as we go through this lecture. So, in the context of the ligand space that we are examining in organometallic chemistry, you will notice that the sigma molecules are all listed above the central line, where carbon monoxide is falling. So, if you can see this central line and all the molecules, which have only a sigma bond or a sigma and a pi bond are indicated above this line. All those which are interacting with the metal with only a pi fashion are indicated below this line. So, these are the type of ligands, where you have a sigma bond as well as a pi bond in the molecule. So, today we will be discussing the allyl complexes in organometallic chemistry. As I mentioned earlier, allyl complexes are in fact very complicated because of the way in which they interact with the metal. There are a variety of ways to make them and today we will concentrate on the ways by which we can make these allyl complexes. Later on, we will talk about the reactivity of these metal complexes. So, today we will try to cover the ways by which we can make these allyl complexes and the type of structures they adopt in solution and in the solid state. First, let us talk about a simple substitution reaction. In these reactions, which two examples are given here in this slide. If you treat a metal halide with an allyl magnesium bromide, which is a simple Grinard reagent, allyl bromide with magnesium turnings will give you allyl magnesium bromide. This allyl magnesium bromide is predominantly C 3 H 5 minus. In other words, you can think of it as if it is C H 2 C H C H 2 minus. This allyl anion is what we are talking about is a predominant species that is formed when you have allyl magnesium bromide. Although it may not be in an isolated form, that is the net species which reacts with the nickel bromide. So, it displaces a B R minus and so what you end up getting in this reaction is M G B R 2 as at the end of this reaction magnesium bromide would precipitate out and a bis allyl complex of nickel is formed. Now, the complex that is formed can be written in this fashion. The anionic allyl groups are indicated here as being coordinated to the nickel atom in a sandwich fashion. Although we draw it in a flat two dimensional representation like this, the molecule is in fact like a sandwich and we will talk about this in a few minutes. The same reaction can be done with a variety of metal halides and this allows you to prepare homoleptic metal allyl complexes. Homoleptic again let me repeat, there are molecules in which there is only one type of a ligand. So, in this particular case you have only allyl ligands and zirconium because the oxidation state is plus 4. You have 4 allyl molecules which are attached to the zirconium. Nickel is divalent and so we have 2 allyl groups on the nickel. The nickel group in the periodic table, nickel palladium platinum form very interesting allyl molecules, allyl complexes and I will show you a structure of allyl palladium complex to start with. So, that you can appreciate the type of interaction that is there between the metal and the allyl moiety. So, this is the complex in which we have a metal atom which is sandwiched, which is sandwiched between 2 allyl groups and you will notice that the allyls are staggered. In other words, one case the central atom has got a hydrogen pointing upward and in the other case the central atom has got a hydrogen pointed downwards which I am indicating with an arrow. So, here is one hydrogen pointing upward that is the central carbon atom and we can label all the atoms and you can see that the 3 carbon atoms are equally bonded to the nickel which is present here. So, this is a case where you have the 3 carbon atoms almost equally bonded to the central nickel and you can you will be able to see these distances in a moment. I will measure them this is 2.026 to the terminal carbon 2.026 is a distance between the terminal carbon and the central carbon is at a distance of 2 angstroms. So, you can see that they are almost the same distance the central carbon atom and the terminal carbon atom and you can see that the 2 allyl groups are in fact in a sandwich like fashion between the central metal atom here. So, you can appreciate this 3 dimensional structure and the way in which the allyl group is interacting with the metal. You will also notice that the allyl group is almost planar if I have aligned it in such a way that it is almost planar all the hydrogen atoms and the carbon atoms are in a approximate plane and between these 2 planes the metal atom is centrally located. The carbon atom that is in the center is slightly elevated or slightly away from the metal atom or so it appears, but you can see from the distance that it is in fact almost equally placed from the metal as the terminal carbon atoms. So, these are some very interesting molecules that are formed by the metal allyl species and you can have with zirconium you will have 4 of the allyl groups it will be arranged in a tetrahedral fashion. The only difference is that because it is a tetrahedron there are more steric crowding, but nevertheless zirconium being a large metal is able to accommodate all 4 allyl groups around it. So, let us just get back to our presentation here. Let us move on if you instead of doing a simple substitution reaction we can also do a trans metallation reaction. I distinguish the substitution reaction from the trans metallation reaction purely because in the trans metallation reaction I have a allyl tin group which is a main group allyl complex which is sigma bonded where the allyl group is sigma bonded to the tin. This allyl group is not heavily polarized as allyl minus and tin plus and it can react with the palladium chloride. So, in such a way that you can transfer one allyl group to the palladium. When you do that and you reacted with 4 molecules of a triphenyl phosphine you end up with a very interesting complex where you have palladium coordinated to allyl group and a triphenyl phosphine moiety and a chloride. This molecule has been studied extensively because it reacts with nucleophiles and it allows you to carry out a nucleophilic substitution on what is apparently an allyl group which is originally allyl almost like an allyl minus and in spite of that it can undergo reactions with nucleophiles. The metal allyl usually should be the, if you want to do a trans metallation reaction the metal allyl should be little more electropositive than the metal to which you want to transfer it to. Similarly, the metal to which you want to transfer it to palladium, if it is a palladium chloride then the chloride group is transferred to the group which is transferring the allyl group. So, that should like the chlorine better. So, in this case that is a tin. So, this turns out to be an exchange of two groups and that is why we call it a trans metallation. Now, we will just take a look at the palladium allyl triphenyl phosphine complex because that is also a system where you have 3 allyl groups pointed towards the metal atom and here is the molecule which has got a triphenyl phosphine. So, the phosphorous atom is shown in bright orange here. The phosphorous atom is bright orange the metal itself is the metal is present in the metal is present just in the as in the other case in the nickel case it is present in almost equidistant from the 3 carbon atoms which are interacting with it and the allyl group is almost planar. There is a small deviation of the hydrogen from the plane in which the 4 the 2 CH 2 units are present. So, the central hydrogen is moving away from this plane away from the metal atom. This turns out to be unique for allyl complexes one of the hydrogen atoms of the central hydrogen atom moves away from this plane not very much may be by a few degrees, but nevertheless it moves away. So, that the bond distance between the carbons and the palladium and the orbitals are oriented in the right way to have maximum interaction. We will talk about this interaction little later, but you can see that this is the molecule in which the 3 carbon atoms the 3 carbon atoms are indicated here in grey. The 3 carbon atoms are interacting equally with the palladium and the chlorine atom is coordinated to the palladium as well. So, let us get back to our presentation now. If you want to prepare an allyl moiety apart from trans metallation and apart from the substitution reaction that we talked about we can also carry out what is called an oxidative addition reaction. These are reactions which we have talked about little earlier. If you have a metal halide or metal anion and an alkyl halide the metal anion can carry out a nucleophilic substitution reaction on the chloride. In principle this is MnCO5 minus. So, we can talk about it as MnCO5 minus and sodium plus and this MnCO5 minus now makes a nucleophilic substitution reaction and SN2 reaction basically and it forms an intermediate which is a sigma allyl complex. So, in the sigma allyl complex that we have indicated here we have a carbon metal sigma bond and this double bond which is present on the allyl moiety. These 2 carbons are not interacting with the metal. We will see a crystal structure of one of these molecules shortly, but once this sigma allyl complex is formed you will notice that these 2 carbon atoms this double bond can give 2 electrons to the metal and on the metal you have carbon monoxide units. So, this carbon monoxide is also a 2 electron donor. So, one can do a substitution reaction where the carbon monoxide is replaced with this double bond in a internal attack. So, that is exactly what happens you have loss of carbon monoxide and the formation of an allyl manganese complex and this allyl manganese complex now has 3 carbon atoms bonded to the metal just as we have shown in the earlier crystal structures. So, in a initial step you do form a sigma allyl complex, but the sigma allyl complex is fleeting in nature. It is not stable it will quickly rearrange to give you the eta 3 molecule or the tri hapto allyl species which we are able to isolate finally. Now, it is possible to have only the sigma allyl attached to the transition metal in some complexes and if you want to see a structure of this I will now show you a structure of this molecule where you have only a sigma allyl species which is attached to the here is a platinum complex not the manganese complex that we talked about earlier, but here is a platinum complex which has got only the carbon atom the terminal carbon atom attached to the metal. So, here we will let us take a look at the distances which are present this is a platinum which is attached to the terminal carbon and that distance that distance if you measure this distance let us measure this distance again between the platinum and the carbon is 2.086 angstroms. So, this is the carbon which is attached to the metal whereas, in the same way if you measure the distance between the terminal carbon atom. Let me show this to you in this orientation. So, that you can see the two distances very easily here is the distance between the carbon atom which is attached to the platinum and that is 2.086 angstroms. The carbon atom which is further away which I have marked in yellow now that is at the distance of 3.874 angstroms. In other words the double bond these two carbon atoms which I have indicated for you in yellow now this carbon atom and the second carbon atom are not really interacting with the platinum. So, this is a clear example of a sigma allyl complex where only the terminal carbon atom is interacting with the metal and the two carbons which have the double bond are not interacting with the metal at all. So, the carbon carbon double bond is clearly indicated by a short 1.3 angstrom distance between the two atoms and the two atoms which are bonded through a sigma bond are close to 1.5 angstroms or 1.485 angstroms. So, it is very clear that only isolated double bond is present and carbon metal distance is close in one case 2.086 in the other case it is 3.8 angstroms. So, very clear indication that it is a sigma allyl complex. So, it is interesting that you can have a variety of different coordination modes between the allyl group and the metal atom and we have seen three instances one in which it is like a sandwich one in which it is pi bonded and attached to other ligands as well and the third example where it is only sigma bonded to the metal atom. So, because all these three forms of the allyl complex are equally accessible we have very complicated chemistries which can result from the interaction of these three forms. Let us proceed further with synthesis of metal allyl complexes from alkenes this time. These are some very interesting reactions that happen when you treat an alkene compound alkene organic alkene which has got an allylic hydrogen which can be abstracted by the metal and in this case it is a palladium 2 complex palladium pdcl4 2 minus which is interacting with the allyl group. It first forms a olefin complex and this olefin complex is indicated here. Here is a olefin complex only the double bond is interacting with the metal, but because you have a hydrogen which is adjacent to the metal. You can have what is called abstraction of a hydrogen and that abstraction of a hydrogen can result in a formal oxidative addition resulting in this palladium complex which is formally in a plus 4 oxidation state. Now, the oxidation state of plus 4 is not common with palladium but nevertheless it has been observed. So, it is possible that this reaction goes through an oxidative addition and then it can rapidly eliminate a molecule of hydrogen chloride and that is exactly what is represented here. The isolated complex is a palladium chloride dimer because allyl group is negatively charged species. You can see in this molecule that palladium has retained the plus 2 oxidation state reverted back to the plus 2 oxidation state. We started with palladium in plus 2 and then we formed an olefin complex and in the olefin complex we had an oxidative addition, abstraction of the palladium and that results in a species which is palladium in the plus 4 oxidation state. You have 2 chlorides, 1 hydride and 1 allyl species. So, this species rapidly loses HCl that is does a reductive elimination and forms the palladium 2 complex which we have indicated here. So, this is one way in which we can understand the formation of an allyl complex from an alkene. Let us talk about an alternative way in which we can understand this reaction. Here I have formed the double bond. You have an olefin complex which is formed between the palladium and the alkene. After it forms the palladium olefin complex, the hydrogen which is present on the allylic position, this position it can be removed from the molecule as H plus. In other words, the pair of electrons which are held between the hydrogen and the carbon are donated to the palladium and that is indicated here and the chloride leaves from the coordination sphere of the metal. You should remember that palladium first had 2 extra chlorides it was PdCl 4 2 minus and in the initial coordination of the alkene we lost 1 chloride and that is what is indicated here. So, we have a mono negative species to start with and if you lose an H plus and simultaneously lose a Cl minus, this hydrogen goes off as H plus, this hydrogen goes off as H plus and this chloride goes off as Cl minus. We will be still left with a negatively charged species, but notice we have not increased the oxidation state of the palladium during the course of this reaction. As a result, what we have ended up with is a palladium 2 species in this case also, here also palladium is in the plus 2 oxidation state. So, you have an allylic group 2 chlorides and a net negative charge and as a result this molecule can now lose a Cl minus and form the dimeric species which I have formed in the previous case also. So, in order to form this beautiful palladium allyl dimer palladium chloro dimer, you need you can do it in 2 different you can understand it in 2 different ways. One case you do not need to do an oxidative addition and a reductive elimination and in the other case you do end up with an oxidative addition and the palladium 4 complex which rapidly eliminates HCl. Because topologically you cannot distinguish the 2 processes it is not possible to figure out exactly how the reaction is happening, but nevertheless the final product formed is the same. So, let us this the alternative yet another alternative way of understanding this is to think of it as a sigma bond metathesis. Now, after forming the palladium olefin complex one can think of this H as leaving as HCl. So, you have a sigma bond metathesis palladium forms a new bond between the carbon and the metal and a new bond between hydrogen and chlorine and the old bond that was present the old bond which was present between the carbon and hydrogen palladium and chlorine are broken. So, in the sigma bond metathesis 2 sigma bonds are broken 2 new sigma bonds are formed and hydrogen hydrochloric acid is eliminated and you form the complex. So, whether you think of it as a sigma bond metathesis or you think of it as a loss of a proton and a chloride either way you would end up with no oxidation state change on the palladium and the formation of a palladium allyl chloro complex. So, these are 2 different ways of understanding the reaction without an oxidative addition. Now, oxidative addition to palladium definitely happens in situation where you reduce the palladium 2 here again we start with Na 2 PdCl 4 palladium is in the plus 4 oxidation plus 2 oxidation state Na 2 PdCl 4 and you reduce it with carbon monoxide. Carbon monoxide or SNCl 2 tin chloride tin in plus 2 oxidation state can reduce palladium to palladium 0 and tin will get oxidized to palladium 4 sorry tin will get tin will get oxidized to tin 4 and will become SNCl 4 and palladium will get reduced to palladium 0. Now, palladium 0 can oxidatively add allyl chloride. So, this is a simple reaction which will also lead to the chloro palladium allyl complex which we can represent like this. So, this is the dimeric species that we are talking about the same complex which we have been showing in the last 3 slides. So, you will notice that here palladium 0 has changed to palladium plus 2. So, this is an oxidative addition to palladium 0 and it is promoted either by an initial reduction of the palladium 2 to palladium 0 with carbon monoxide or with tin. Both of them are known to cause this reaction to proceed forward. Now, it is always possible to use palladium metal or any metal in a 0 oxidation state when you do a co condensation reaction with a ligand. Then you can make the allyl complexes and in the case of co condensation reactions the advantages the metal is present in a very highly reactive state. In this case we are using palladium as an example, but when you have a benzyl chloride. So, here you have benzyl chloride and you have a very reactive very reactive metal atom. You can undergo an oxidative addition and this is in fact a molecule which should have formed only a sigma complex because you have an aromatic molecule. You have an aromatic molecule and you have the possibility for oxidatively adding it to the halogen. So, this is what you would expect, but surprisingly the double bond which is present on the benzene ring just this double bond also interacts with the metal. So, you have an allyl complex with the benzene ring participating in the reaction and partly it will lose the aromaticity. Surprisingly there are many bonds which can do this. So, benzyl halides can oxidatively add to the metal atom. On oxidative addition one of the double bonds interacts with the metal atom to form a allyl compound in which it is a eta 3 molecule. The hapticity is 3 and it is looking just like other allyl complexes. Let us take a look at this molecule because this is in fact a very strange compound where you can have the benzyl interacting with the metal atom. So, we now take a look at the benzyl case. So, here is an example which has been taken from molybdenum compound. We will not worry about the remaining ligands on the molybdenum, but you can see that there is a very nice aromatic ring which is present. This aromatic ring is interacting in such a fashion with the molybdenum such that 3 of the carbon atoms this C 1, C 4 and C 14 all of them are interacting with the molybdenum in an equal fashion. So, if you want to measure the distance you can see that this carbon is at a distance of 2.2 angstroms. The first carbon the benzyl carbon is at a distance of 2.2 angstroms and I have indicated it here. The central carbon which is the C 4 is at a distance only at a distance of 2.3 angstroms. So, 2 carbon atoms are interacting almost in a equidistant fashion from the molybdenum atom. The third atom is at a distance of 2.48 angstroms. So, although it is slightly away you can see that all three carbon atoms are interacting with the molybdenum and it looks like a allyl complex. In fact, from the electron count you can see that this double bond is in fact interacting with the molybdenum atom. To show that there is bond alternation you can look at the carbon carbon bond distance of the bond which is trans or which is opposite this and that is clearly 1.42 angstroms and it is much longer compared to the other two double bonds. So, there is some bond fixation or the delocalization is completely destroyed. Two bonds are short 1.3 to 8 angstroms one bond is very long. So, that this really looks like a complex in which the allyl moiety is interacting one of the double bonds has decided to interact with the metal in such a way that the aromaticity of this benzene ring is destroyed. So, the gain that is there for this molecule in going from this aromatic system to this non aromatic system is because it can satisfy the molybdenum requirement for two extra electrons and those two extra electrons are coming from the pi bond which is present here. So, it is interesting that you can have this alternative way of interacting with a metal even if you have a aromatic system part of the aromatic system can interact with the metal and form a pi allyl complex. So, these are called pi allyl complexes or tri hapto species or eta 3 complexes all three nomenclatures are used in the book. So, let us move on let us take a look at another way of making allyl complexes and in this case what I am shown you is a system in which there is a formal oxidative addition oxidation state change and these are molecules in which there is a manganese in a plus 1 manganese is present in a plus 1 oxidation state these two are carbon monoxide units. So, what happens is there is a olefin complex and in this olefin complex there is an allylic OH group. This OH group can be protonated with a strong acid like HBF4 and if it protonates it and loses a molecule of water after protonation if it loses a molecule of water then formally this allylic moiety will become allylic cation. Now, all three carbon atoms of this allyl group all three carbon atoms which I am going to indicate with green atoms all the three carbon atoms can now interact with the metal atom with the manganese in a eta 3 fashion. So, because this is an eta 3 interaction and formally we talk about allyl as allyl minus the carbon is always considered as more electronegative than a metal you end up with a oxidative addition reaction. Because manganese is in the plus 1 oxidation state in this molecule after protonation and loss of water molecule you will end up with a cationic species and this cationic species now has got two negatively charged molecules or ligands one is C p minus and the other is the allyl minus. So, as a result what you end up with is a manganese in plus 3 oxidation state. So, here manganese is in the plus 3 oxidation state and you started with manganese in plus 1. So, you formally carried out a oxidative addition reaction a similar situation happens when you protonate a diene complex here is an iridium complex which is butadiene unit is coordinated to the iridium here also iridium is in the plus 1 oxidation state and after the reaction after protonation iridium goes to plus 3 oxidation state. Now, in order to show this reaction exactly how it is reacting I have indicated this double bond this is a C H 2 unit here this C H 2 unit if it is protonated it will become a methyl group and that is that methyl group is now indicated initially it was indicated in green C H 2 and now it is become a C H 3 group and once it becomes a C H 3 group you will notice that this carbon can equally interact with the iridium and all three carbon atoms the two carbon atoms that were part of the diene and one carbon atom that was part of the in both all three of them can interact with the iridium and so now you have the possibility of an allyl complex. So, here also it is a eta 3 allyl complex and iridium because you have protonated it you need to add a plus charge to this molecule and if you count the number of anionic groups present on the iridium there are already two anionic groups there is a plus 1 charge. So, iridium is in the plus 3 oxidation state. So, these are cases where you can do an oxidative addition with a proton oxidative addition with a proton resulting in a change a formal change in the oxidation state of the metal and the formation of an allyl molecule. So, in both cases you get a iridium 3 sorry iridium 3 and manganese 3 species and both cases you have a eta 3 coordination. Now, there is another anomalous reaction which we should talk about and that is the formal addition of a HCOCO4 that is a formal addition of a metal hydride a classic example is HCOCO4. We have discussed this earlier, but hydrides are anomalous because many times they will react as if they are protons the hydrogen reacts as if it is a proton. So, HCOCO4 is a strong acid in fact and if this metal hydride formal metal hydride is added to a diene it ends up informing a eta 3 allyl complex it ends up informing a eta 3 allyl complex and here I have a cobalt example you will notice that when you add the hydrogen the hydrogen can be added to the diene to the diene. So, that it forms this allyl complex, but the metal group can be either oriented away from the central hydrogen. So, if you talk about the central hydrogen as a unique hydrogen now the metal group can be added in such a way that it is pointed in the same direction. So, this metal group is now sin with respect to the central hydrogen or it can be anti to the central hydrogen. So, this results in the formation of two different molecules either a sin or an anti molecule. One last example where you have a hydrogen transfer which results in the formation of allyl molecule. Here you have a metal group and one of these hydrogens can be transferred to the nickel such that you form an allyl molecule. This molecule is in fact a dynamic system at temperatures above 40 degrees you have the formation of a eta 3 allyl complex. If you cool the molecule if you cool the reaction system and then you end up with a olefin complex. So, this tells you how the energy required for transferring the hydrogen from the alkene moiety from the allylic position of the alkene to the metal is very very small. So, you are able to transfer the hydrogen very easily and form a eta 3 allyl molecule. So, here is yet another example which is formally an oxidative addition. In this particular case there is no negative charge, but still C P C O C O 2 is a electron rich species. This is a electron rich species and this electron rich species does a nucleophilic attack on this allyl and as a result the iodide goes out and it forms a species forms a intermediate in which you have eta 3 allyl coordinated to the cobalt and iodide is a counter ion. But during the course of this reaction iodide attacks the cobalt again this iodide moiety attacks the cobalt again and displaces the carbon monoxide. So, that you end up with end up with a cobalt neutral compound. This is a positively charged species and now you have a completely neutral molecule in which cobalt is in the plus 3 oxidation state. Here also it is an oxidative addition, but nevertheless earlier we looked at a manganese compound which is negatively charged and it underwent or it attacked allyl halide displays the chloride. Here we have a neutral molecule which is carrying out to nucleophilic attack on an allyl halide which is a allyl iodide species. So, a variety of reactions have been shown. These are examples which we will talk about later where allyl molecule is formed from a larger ring system. So, we will skip this here is a species where you again have a diene and instead of adding a hydride instead of adding h plus or h minus we are adding an R minus. This R minus can be R M G B R and this R M G B R essentially does a nucleophilic attack on the double bond here and it results in the formation of an allyl molecule. So, there are a variety of ways probably more than 10 ways we have considered in which we can form an allyl molecule. Now, what we have to look at is the structure and bonding in these molecules. I have drawn out the allyl system which is a 3 carbon system in which there are 3 pi bonds or 3 p orbitals which are not utilized in bonding. So, if you look at this particular system you will notice that the 3 p orbitals can combine together to form 3 molecular pi molecular orbitals out of which this is the lowest energy orbital in which all 3 of these lobes have the same sign. Now, the best way to illustrate this is to is to show is to overlap these molecules overlap these orbitals here on my on the right side of the screen you have the metal orbitals on the right side of the screen you have the metal orbitals on the left side you have the allyl group and the allyl orbitals and the best way to understand which orbitals would overlap with the metal orbitals will overlap with the pi orbitals of the allyl group. One has to just move these groups of these groups to the right position and see if they will form an overlap a positive overlap and you will notice that the 4 s orbital can form a nice overlap with new if the allyl group is present let us say in the x y plane. So, if the allyl group is present in the x y plane and if the metal is present in the z axis at this position if the metal is present in the minus z axis along the z axis, but in the minus part of the z axis then you can see very clearly that the s 4 s 4 p z and the 3 d z 3 d z squared orbital all of them are capable of interacting with the lowest energy pi molecular orbital. So, the lowest energy pi molecular orbital will have 2 electrons which will have 2 electrons and these 2 electrons can interact or can be donated to the 4 s which will be m t the 4 p z and the 3 d z squared if it is m t as well. So, you can realize that if you are talking about an allyl anion as we formally talk about there are 2 electrons in this orbital also. So, these 2 electrons can be in fact interacted with or these 2 orbitals on the metal the 4 p x and the 3 d x z both of them can interact with the with the second pi molecular orbital which will also have 2 electrons if it is an allyl anion. So, the 2 donor orbitals on the organic system the 2 pi molecular orbitals on the organic system can donate electron density to the metal orbitals which are indicated here. Now, comes the highest energy molecular orbital which is the third pi m o and this is invariably m t. So, whether you talk about an allyl anion or an allyl cation the last highest occupied molecular orbital is always m t and that orbital can interact with the 4 p y orbital and the 4 p y orbital can interact and the 3 d x y z orbital can also interact. The 3 d y z orbital is the one in which the 2 the 2 lobes are pointed are going in and out of the plane of the screen. So, you can see some the second lobe is almost hidden by the front lobe. So, these 2 orbitals can interact nicely with the m t orbital now because you can have electron density on the 3 d y z. The 3 d y z can in fact donate electron density back into the m t orbitals of the pi and this will lead to stabilization of the pi system through back donation of electron density. Now, it is also possible that we talk about the allyl group as a allyl cation in which case the 3 d x z can also donate electron density back into the allyl group. So, there can be a lot of give and take the electron count on the metal electron count on the metal is important electron count on the allyl group is also important. You will remember that we generated the allyl group starting with an allyl alcohol by protonation and elimination of water. In that case it is very obvious that we are talking about an allyl cation which is coordinated to the metal. In that case we can say confidently say that the 2 orbitals both orbitals will be m t. But formally we always talk about it as if it is an allyl anion and so the back donation from the metal to the allyl pi star orbitals must be happening only from the 3 d y z. Now, the 2 d x squared minus y squared and the d x y orbitals do not have suitable combinations with the allyl orbitals and they are not interacting with the metal they are not interacting with the allyl group. But all other groups all other the s p and the 3 d orbitals are capable of interacting with the allyl moiety. You will also realize that these 3 orbitals interact with the allyl group in a sigma fashion. These 3 groups interact in the sigma fashion and so the allyl moiety can freely rotate about the z axis. So, if you think about the allyl moiety as sitting in the x y plane then the metal can rotate along the z axis can spin about its axis freely. Because the sigma there is a strong sigma bond which is formed between the allyl group and 3 orbitals on the metal. However there are 2 orbitals which are interacting in a pi fashion and that is the orbitals which are marked as 3 d x z and the 4 p x. These 2 orbitals if you rotate the allyl group with respect to the metal atom along the around the z axis. So, around this axis if we rotate it if you rotate the metal atom then what will happen is that we will lose the second p the pi orbital interacting with the 4 p x and the 3 d x z. So, these 2 orbitals will stop interacting or rather the overlap will go to 0 and the interaction energy will become 0. So, the extent to which these 2 interaction these pi interactions are important determines how much you can rotate the allyl group with respect to the metal atom. Just like we discussed the olefin case. Now having discussed the structure of the allyl molecules we must talk about the fluxionality and the reactivity. Let us take a look at the fluxionality of allyl complexes. We will discuss this in 2 dimensions when you show this figure and then we will try to illustrate it with a small video clip when we look at how this allyl group can react. So, here is a metal which is shown interacting in pi fashion and the plane in which the allyl group is present. This is the x y plane that we are talking about and this is the z axis. So, this is the z axis that we are talking about. So, when you rotate the metal atom with respect to this plane and you rotate the metal with respect to this along this axis along the z axis we can keep the sigma interactions intact the pi interactions will fail. Now it is possible for the metal pi a slight movement towards one direction. So, it can move towards a high carbon atom 1 and the carbon atom 1 is a 1 which is bearing h a 1 and h s 1. The a indicates the fact that it is anti to the central hydrogen. Central hydrogen is marked as h c and h a indicates that it is anti to the central hydrogen. S indicates that it is sin to the central hydrogen. So, this is anti and this is sin to the central hydrogen. So, these two carb, these two hydrogens will come closer to the metal if the carbon one is interacting with the metal in a sigma fashion. So, this is a pi allyl eta 3 system and this is a eta 1 system. So, we can go from a eta 3 to a eta 1 system fairly rapidly by a simple movement of the metal along this axis. So, that let us just mark this out. So, if you if the metal moves in this direction then it can go from the eta 3 to the eta 1 fashion. This can go back and forth very rapidly this eta 3 to eta 1 migration can happen very rapidly and it can go either to C 1 or it can go to C 3. So, we let us just mark this as C 3. So, either you can go to C 3 or it can go to C 1 both of them are equivalent and as a result you can see how the molecule can go from sigma complex to a pi complex and can go to this alternatively. In addition what happens is if the metal rotates around this bond if the metal rotates around this bond then from the minus z axis it can move to the plus z plane and if this happens then the allyl group is intact in the same position allyl group is present in the x y plane. But then the metal is moved from the minus z position to the plus z position. So, the metal can come back to the central position and so it will become a eta 3 system, but now the metal is in the plus z position. So, this type of a metal flip can result in a new complex and you will notice that because we have maintained the sigma bond what was originally H A 1 has now become has now come to a different position it has come it has formally become sin to the central hydrogen. So, the position of the sin and the anti-hyrogens have got interchanged as a result of this flipping of this of the metal from one side of the allyl group to the other side. So, this is called a metal flip and in the following lecture we will talk about how the allyl group can flip. So, there are two things that we are talking about one is a fact that the metal can go from an eta 3 geometry to a eta 1 geometry that also leads to a fluxional behavior and it will rapidly inter convert the two carbon C 1 and C 3. There is another fluxional behavior and that is rotation of the metal or rotation about this carbon bond carbon C 1 C 2 bond about the C 1 C 2 bond. If you have a flip then the metal goes from the plus z to the minus z or the minus z to the plus z and as a result you have interchanging of two hydrogens. The two hydrogens in which to which the metal was attached will now get interchanged and what was anti will become sin and what was sin will become anti with respect to the central hydrogen. So, you have a fairly interesting fluxional behavior and because these are rotations they can happen in the time scale of 10 power minus 13 to 15 seconds.


This is the Asus M3 Wired ROG Chakram Core, Wired ROG Carries, and presently the Wireless ROG Gladius III. You can check out our review of each mouse at the links in the video description. We use all our review mice for a minimum of one month and a half so there is plenty of time for us to get rid of the unboxing hype so that it doesn't interfere with our genuine feelings about the mouse. So that our review to you guys is useful and holistic and not just a buy this and not that. I'm Rafalef Mark for Sugar, the only PC shop in the Philippines with no BS warranty and this is our Ultimate Mouse Highlights Guide for 2021 and our review of the Wireless ROG Gladius III. The investment in a good mouse might be the cheapest and also the most productive investment you can make for your computer setup. Regardless of if you are a gamer or not, it doesn't cost as much as a new graphics card or sometimes even a decent keyboard. In fact, if you only had the budget to invest in either a good mouse or a keyboard, hands down. Invest in a good mouse. Don't make my mistake when I bought a mouse which was too small for my hands. Let's do a quick product review of the Wireless Gladius III first. It and the Carries look extremely similar and most of everything I said about the Carries are true for the Gladius as well. They both have a very similar sculpted hand design which is very comfortable to use for medium to large hand users, me being the latter. And they made this excellent design choice of making the entire scroll wheel RGB rather than just the sides. This design choice is unique I believe to ROG and you won't see this on other premium mouse brands such as Razer, SteelSeries and Corsair. The RGB designs of which still just color the sides of the scroll wheel rather than the wheel itself. Their difference in design is the etching on the left side which reminds me of computer code from The Matrix. If you look closer, it's the words ROG, dare to play, Gladius and Republic of Gamers in small font which are lit up through the RGB within the mouse. Another difference is that the Gladius has a button next to the scroll wheel which you can use to change your sensitivity in case you are sniping or switching from sniper to assault. I'll talk more about this later. Worthy to point out however is that the Carus is ever so slightly smaller and shorter than that of the Gladius 3. The Gladius 3 is wireless thus it comes with a 3GHz sensor. In terms of actual real world usage, I couldn't tell the difference between the Carus and the Gladius 3. Both of them are very accurate and glide across a mousepad wonderfully for extreme FPS gaming precision. The wired Carus weighs in at 62g while the wireless Gladius 3 weighs in at 79g. These mice are extremely light on their feet. To put things in context, the ROG Chakram Core clocks in at 111g or close to double the weight of the Carus. The battery life of the Gladius 3 is also great, lasting close to 4 days of moderate to heavy usage and charges up via USB-C in practically no time at all. I didn't religiously count how long it takes to charge each mouse but if a 30 minute charge can keep it going for 2 days straight then that's more than acceptable for mostly everyone's needs. It also comes with the Paracore or what I like to call the Shule style wire which I think every hardware developer must begin using as their standard. It's just plain awesome how freely it bends and feels in the hand. We would give the Gladius 3 a 8 out of 10 primarily for its minimalist look, subtle yet novel RGB design and for it being wireless and USB-C chargeable. The only drawback is the cost of it being $6500 or for being the most expensive mouse we have ever reviewed. Then again, if you are buying ROG, you already know that quality doesn't come cheap. I only wish it had one feature which exists on the very cheapest of ASUS mice which I'll talk about in a bit. Alright, now let's get into the specifics of all the other mice. In terms of productivity, I highly recommend the ROG Chakram Core over all the mice here because it has 9 programmable switches, even if it only has 6 buttons. The price here is the small thumbstick which is well ergonomically placed which allows you to move it up, down, left and right, each direction of which can be programmed to any key you want. Having 9 programmable switches on a gaming mouse is really helpful for getting work done and I personally liked having it so I can minimize all my windows during times of emergency in case I was looking at something I shouldn't be looking at. It works great as a gaming mouse as well and you can bring down the overall weight of the mouse from 111g to 97g, something I didn't attempt to do because I don't have the courage to take apart a mouse I don't own just yet. Note that the weight of the carry is at 62g and the Gladius at 79g still makes these mice a lot lighter by far. I reviewed the wired version of the Chakram however it does have a wireless version which is double the price. The runner up mice for productivity is the Corsair Wireless Iron Claw simply because it is literally studded with buttons of all shapes, sizes and placements. The button placements also feel quite natural when you begin using the mouse. Weighing in at 105g and it's almost batmobile tank like design, this guy feels extremely sluggish compared to the carries and the Gladius 3. However, while it seems sluggish as a right hand user, I find the grip where I rest my thumb to be extremely comfortable because of how it seems to massage my thumb. In terms of sheer FPS gaming, the carries, Gladius 3 and the Asus M3 must be on your shortlist simply because these mice were bred to be light responsive and meant to help dominate the FPS field. While the Asus M3 might be the cheapest mouse in the entire list costing only 980 pesos, it has one feature which the other two don't and that is the ability to change mouse sensitivity both ways. Two buttons allow you to lower and increase the mouse sensitivity at your choosing rather than the Gladius 3 which only has one button on top allowing you to change mouse sensitivity in one direction. The worst is the carries in which there is no button on top at all and you can only change sensitivity for it by flipping it on its back and pressing the button there and even then it only changes sensitivity also in one direction. I promise you that you will be dead well before you get the desired sensitivity you need in combat. So yes, while the carries and Gladius 3 are flashier, pricier and lighter than the M3, they lack a key tactical element which is indispensable for any professional gaming mouse. The fact that this is offered on such a budget friendly mouse makes this a no brainer option for young gamers on a tight budget. The M3 is a great gaming mouse for its price range and I extremely recommend it for gamers on a tight budget. The M3 should serve as a reminder to Asus that one of their cheapest mice offers something not even some of their most expensive products do. For those who don't know, Asus owns ROG so the fact that the design teams didn't talk about adding this basic feature into the ROG mice line leaves me a little bit baffled. However, as an FPS gamer, I'm not really a sniper and so I don't change my sensitivity as often. However, it's nevertheless an option I would much prefer to have as a safety blanket. If you're looking for the mouse to end all mice, you may want to look into the Keras and the Gladius because they are built to be deconstructed. They come with replaceable mouse switches which means that when the time comes that your mouse begins to suffer double clicking, you can open it up and place in new switches. It is important to note that ROG does not provide clearance structures on how this is done and I was too afraid to try it out myself. Nevertheless, some YouTubers have been able to swap out their switches and even comment that the Japanese Omron switches given for free with the Keras feel better than the switches that come pre-installed with it. In short, buying these mice will last you a long time because just like a car with a flat tire, you can always change out the tires. Note however that if your problem lays elsewhere such as the scroll wheel or the RGB then there is no quick fix for that and if you lose the replaceable switches, which is easy to do, then you wouldn't benefit from this longevity. In terms of gaming on a budget, I would highly recommend the Asus M3 mouse because it retails for less than a thousand pesos and it gets so many practical things done correctly. Such as quick mouse sensitivity change. Changing sensitivity isn't just important for shooters but also for productivity, especially if you are working with two or more monitors. The extra screen real estate definitely requires a boost and while you can eventually reach your desired sensitivity, the other mice require you to go through the entire cycle before you do and if you miss it then you have to keep pressing all over again. For a wireless budget option, the Corsair Harpoon priced at $2,795 might seem appealing at first but for only a couple of pesos more at $3,950, you could already get the Corsair Wireless Ironclaw which is so much more ergonomic and comfortable to use than the Harpoon. If you have big hands, stay clear from the Corsair Wireless Harpoon. Your hand will cramp up eventually. While it may seem heavy at first, the Corsair Ironclaw is precise enough that I've never felt that I missed any shots when doing intense gaming and in terms of productivity, the many button options and the ability to change my sensitivity in both directions makes me recommend it as the better wireless gaming mouse option on the market. The biggest limitation of our study is that I haven't gamed the Steel Series or Razer mouse in a long time and so it's entirely possible you might find the perfect fit for your hand and budget in a different brand or ASUS mouse model. What I can't say however is that regardless of brand, aim for a mouse which is naturally comfortable to the touch and practical for what you need to use it for. I might have come down hard on the carers and gladiators for the lack of an additional sensitivity button but they are nevertheless serious contenders because of how light they are. It is because of these mice that I actually began researching on the different weight of a mouse in addition to their other features. If you like this video, please consider liking and subscribing to Hardware Sugar where we review anything gamer related and sometimes chairs. We want to give an extremely special thanks to our top fans who help make all of our work possible. ITX addict, Rafael James, Ian Meru, Liam Magnae, Richard Ongkinko, John Rubin Ottia, and Christian Espinosa. It's good seeing all of you so regularly during our streams and again, thank you so much for the support.


Good morning guys, Jax's back, feels a little weird to talk after so long of being absent. As I said before, had a little surgery, just kind of minor and just kind of relaxing and resting and just kind of, you know, taking naps. Anyways, this morning we're going to meet up with Chris Hotel Prius for some donuts. He's in Portland, I'm in Portland, and we'll be good to see him. Good morning sir. What's up man? What's up fam? Long time no see. Yeah, no kidding. Yeah, what are you doing in Portland? Just working on a house remodel, buddy of mine just bought a house and so he's got a basement, he wants to turn it into an Airbnb so he can call. There you go. Hashtag basement camper. I'm gonna come up with a name, yeah.  There is a trusty, trusty mobile, the Hotel Prius, which by the way, I don't know if you guys saw this but there was a video out there that got a couple hundred thousand views or a couple million views on your whole project, right? What was that on? Was it BuzzFeed? Oh BuzzFeed, yeah, it almost took two million, that's what. That's crazy. Yeah. So if you guys want any info on how to travel in a very small space like a car and make it work, you should watch this guy's video. What's it called? Uh, I live in my car, I think. I didn't name it, BuzzFeed named it, I didn't like it but you know, they're BuzzFeed so we can't argue with them, right? Yeah, yeah. But yeah, I live in my car and then you see this Google Hotel Prius inside it. There you go. Alright guys, me and Chris, Hotel Prius, we're just hanging out for a bit, discussing strategies, our strategy, well my strategy, his strategy for life and content ideas, things like that and we're going to be parting ways here shortly, unfortunately. Try to get back to where it is right. That's right. If you guys haven't seen what he's doing before because I know there's a lot of new subscribers. He's doing some glamping stuff in the backyard of his place in Austin and people can find you with Austin Glamping on Instagram. Austin Glamping, still a work in progress on the on the gram but there's a link to my Airbnb so if you ever want to stay in Austin, if you know Jax, promo code Jax10off. Yeah, that'll work, right? Let's do it. Yeah, you can come stay. That's sweet. Let's do it. I guess this is farewell and parked down that way. Yeah, good to see you. Yeah, likewise. I will see you when I pass through Texas at some point. Alright, see you. Alright, see ya. Powell Books, I met up with Kendra. Kendra is a, I would say, she's doing some stuff on yoga and dancing and health and wellness and all kinds of fun stuff so we're just taking a little stroll through some Sunday market here and it looks pretty awesome.  I'm going to head back to the bus. I also need to continue working on the bus. There is a video that I did before this which wasn't a great video and I haven't posted it yet. I basically installed the propane into the bus and so next up I got to install the water tank and the heater. And then we will be in action. So I'm going to try and work as much as possible on that today. And yeah, so see you over at the house. Alright ladies and gentlemen, it is now the evening time. Just kind of did a few errands. I had a good lunch, ate some dinner, relaxed a little bit, watched Anthony Bourdain, Parks Unknown on CNN. And yeah, I just want to thank Chris and Kendra for meeting up with me today. That was a lot of fun. Chris and I had some really good conversations about different types of videos and it's funny because he nailed it right on the head because I was actually thinking about doing more specific topics like single topic videos way shorter than these vlogs but more informational because I've learned so much about life on the road after, jeez, after quite a while. So yeah, so I'm glad that we were on the same page with that. And let me know in the comments if you think that is a good idea as well. I know some people do videos where they talk about things and I just feel like some of the information just gets buried in these vlogs because it's hard to have like five different titles for the same thing and it's hard to choose words that describe everything in the vlog. That might be a good rant, that might be useful and I feel like I should elaborate on those rants but it's helpful to people. So yeah, so that would be cool.  It should be a pretty busy day. I need to start heading south towards Vidcon. Basically got to rip up some 2x4s, raise it up just a tiny bit I think with some 4x4s and then I need to put this gigantic water tank back there. Then I got to drill out some holes because those are sealed. And then I have that pump I got installed. It should be relatively easy. It's just I got to move the bed. It's kind of heavy. Then get over to Lowe's and get some custom cut wood and resane it and stuff like that. And also I need to get some wood glue and fix that up. So I got a busy day tomorrow for sure. And so we will see what goes on. I promise tomorrow will be a full vlog. Even if I have errands you guys are coming with me. I'm feeling better after my surgery, all is well. I'm ready to hit the road again. And I apologize for the little blackout here but I'm sure you guys can totally understand that obviously my health is kind of important. We'll see you guys in Manana. Thanks for watching.