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Lecture 22: Benzene and Aromatic Compounds

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part didn't start the we are going to make chapter 16 may start chapter 17 which is a short chapter I wanna start chapter 18 just as soon as I possibly can because it's a long chapter of the book chapter that's my point is we have a Monday final unfortunately and so I wanted it done as soon as we can I'm going to have a review on Friday nite I don't know what the time is yes yes that's face yeah I don't like I forgot to submit a request already which could be a problem so long on Audio denied any questions alright alright so we're talking about how to predict the right stereo chemistry the the piano product for basically dying and so what we said is that outside substituents on the dining on the dying father there are no outsider inside was so we're only talking about the diary will talk about outside substituents so outside substituents right here that's outside as outside to hydrogen during the inside so all labels just in the last instance data and then the electrode during the permitting but are all on the same side so you can you could have evolved dash like I have here or you could have them all wedged both of those will give you the correct answer right so this so I just I'm going back on the label here these are the inside ones that's that's due out those are inside the inside substituents so let's look at this 1 here we got enough aligned on we had a methyl on the 2 ends here so we want to be the outside substituent so level the hydrogen and the electron are are all going to be drawing going back again if you draw them for adults also correct that's going to go back that's going to go back a so here's our product the the sniffles going back the the out it's going back In the hydrogens going back so we need to throw a methyl here Serena have that 1st Union makes it easy to draw on which is what is pretty complicated sterical makes it much easier to draw the correct product are minors wanna show you that the and answer as the opposite relative orientation and so am a region of the product that I just drew so I want you to see with the neon sure looks like there's 2 ways we can draw the ne ensure so we're going to redraw this so let's go ahead and leave the hydrogen off because that's understood but certainly on Saturn you don't worry the hydrogen off so that's the product that we during their the New Hampshire is the mirror image so let's draw the in a mirror image right so in a draw this exact reflection in the mirror here the the yeah everybody agreed that submerged so that's what the this is the and nancial drawn using their image this and that is certainly 1 of perfectly good weights you drop meaning and work I not of slip over the union sure the you don't have to be able to do this within your head on this show you what the flip looks like a but if you can imagine if we if the double bond right here was the spine of a book and you have your book open and this is on the left hand side and all we're going to do is like as if turning page stable like this and I'm going to flip back over the understood that are in draw that so here's my spine now yeah the spine of a book right here when I flip that over everything that is coming up is not identity going back so for example here's the laser pen so I have with see the on the math along the bottom is coming up and when I flip that now the metal's going back from the because of OPS science so this is so this method will be coming up now this method will be going back that's also they ensure so that's the nearshore were made use of by 1 of 2 wall there's 3 ways to make a Nash were drawn Emeritus would be like we just did or here's your 1 and answerer and to draw this we flip the stereo chemistry at every single carbon OK so this method would be going back you know it's coming forward this is going back now it's coming forward the metals coming straight out this is going back of the other way under the other way this is the angelus by of this right here outside substituents on the dielectric during group on the diner going on on the same side of the ring in the product I choose them going back if you choose them going forward this is what you would drop so if you choose them all going forward and that's the nature so you can get in 3 different ways around on my tests the order of 1 of the Union's words and then you can just write policy and happen go through this whole thing but I wanted to kind of prove that to you and that that is inside substituents substituents and electron-withdrawing groups yeah going back to alright so very elaborate stereo chemistry for the deals all that's why it's such a valuable reaction because we it's very very stare selected now on there's 1 other thing that we have not talked about that we have to consider we're drawing deals all their products and that is geochemistry and we kind of avoided the issue you probably didn't
even notice about this is an important issue question you you so that where insights stitch you out 0 yeah I did make a mistake of this is outside substituents thank you yeah so yeah go fix that OK thank you for pointing that out alright so regioselectivity ISS on the next page the deals reactions regions like without unsymmetrical dying in a dienophile there 2 possible orientations yes we we are going for all constant you know what you yet coming forward I I had that completely backwards erase it you the better as hesitation like 200 returning the going forward OK were more that thing you my guys we forward yes I will use a b a bunch of e-mails thank you all the of the OK we got that right now alright now ready to talk about regional chemistry alright so and I have GTG for electron donor in a group and the WG for electoral during the and you can imagine and that we have found if we have the DG WG here there's actually 2 possible orientations OK so we can have the orientation and I'm not even gonna show Jerry chemistry here because of what I wanna do is I wanna focus on some re chemistry here selected using stereo chemistry we'll come back and the staircase history they so if we have free react like that I'm just looking right now at the relationship between electrode during your an electron donor if the reactant that that's what you would get now that they also you can imagine in last scroll up a little and dried up here so you can imagine that I could also oriented dienophile in the opposite way so here a b E. d Jane I could actually go like this right I could audience Vatican slip that oriented that way but if I do that then I get this product and that's really important because really actually get 1 of those 2 products alone we show you another example here where the electron-donating group is actually on the middle of 1 of the middle carbons are we can how orient that also to weights so EDGE here and EWG here or once again we can flip the dienophile if we flip the dienophile then this would be our product so 2 possible products here we only get 1 of them so how do we know which ones which 1 we act and that leads us into the Oracle Parallel now or pair is new terminology that we often talk of will that will talk about chapter 17 and so also is and then we get this nomenclature from aromatic chemistry so if we have a benzene ring there's too fast there's there's a lot of the 2 substituents that would be 1 possible arrangement this would be another possible arrangement and then and this would be I haven't right here so I might even drawing shows that it I don't I'm off today thing very off today OK so no greater confidence in chemistry look at the relationship between these guys here this is a one to relationship this is a one to 3 4 relationship and this is a 1 2 3 relationship so well when it's a 1 you relationship on a benzene ring we call this some or felt with someone for relationship we call it pair and weights of 1 3 relationship we call that a matter and so were borrowing from this nomenclature for benzene chemistry and saying that I'm calling this the author peril the substituents end up on a 1 2 or 1 for relationship to under 6 membered ring so that the electron donating group and electron-withdrawing group are either for tho apparent to each other so as the author parallel so let's see an end and were not knowing like we're going to need it were just looking at the relationship and here you see that that's a 1 2 relationship between the carbons that those electron-donating groups in life are during the search hatched so this is ortho and that's the 1 we get from this would be 1 2 3 4 we don't get any that this would be a matter so no matter this would be 1 2 3 4 and again it doesn't matter which direction number this is that's the 1 will get this is 1 2 3 this is matter we do not get another so questions on radiochemistry so that's another factor that we're going to have to consider yes getting 1 the the the now you'll get that nobody yeah talk i it's a complicated explanation that I'm not gonna have you guys know and so if it if you're curious about it you can come up with OK so this is something you're going to have to remember I can explain it but I don't have time will run out of time to cover chapter 18 and you will be happy with me by gender alright it has to do with the polarity of the 2 of the dying dienophile died has to do with the how the orientation are a highly oriented depends on the polarity of the ions that are involved in the groups involved are isosynthesis using Diels-Alder reaction predict them with a product in the following reaction and what we're going to go ahead and do to make this a little bit better is we're going to put a methyl group on the end here so go ahead and throw a methyl group on the end of this problem the the that make it a lot more a reflection of what you see on the final which is what we want to be ready for and when I do this as I do read geochemistry 1st then ideas steric I know we covered stereo chemistry 1st and then region chemistry but when a when predicting a product and you re geochemistry 1st and then I just error chemistry so read chemistry here thank so the division so they can orient this way of war let me draw the other way we can orient to these 2 groups are gonna keep this year the and
keeping the dying in the same of orientation and I'm just going to flip the the dienophile 1 of these 2 is correct 1 of them is not alright so then of and in order to figure out which 1 I get I'm going to connect the 2 ends and on the number so this is where our new sigma bond is going to be so I don't even need to draw the product here to do this read your chemistry because that's where new stable bonds in to the right where that started this and what I wanna do is I wanna do the electron-donating groups on the dying electro on the diner fall on the the the the dying has to electron-donating groups which 1 is stronger this is strong because it has long hair OK this 1 is stronger because it has all here so that's the 1 that's going to determine the outcome this is the stronger electron-donating yeah you will see that again in chapter 18 so by the time you get 2 of the Chapter 18 you'll definitely be able to know that that's the stronger 1 right away so that's our stronger electron-donating group and then over here is our electron withdrawing groups what's the relationship 1 2 3 4 that's will look like they're that's good that's clear right over here this would be 1 to it 3 that's matter we're not going to get that to In the veil and is thank so that's the Radiochemistry now we need to do stereo chemistry alright Sarah chemistry says who are rule is the the outside substituents and the electoral during are on the same side in the rain so here's our outside so we don't really have been set well we have a hydrogen outside there's too much of this here so we don't even need to worry about that carbon is it's got 2 hydrogens interelectrode during your promise inside is methyl electron during the the but was not an electron-withdrawing group so the it's electron-withdrawing group that has the p orbitals that talks self underneath to get the secondary interactions was needed to be on the same side right so our product is let's try methyl going back leto during group going back what about the middle on the dienophile and set up a down it's up why is it up because it's trends its opposite on opposite sides of that so we know change that's scary contrition and have this and then we take a look at that is that car it's Cairo so we will write plus the so the story control questions anybody the yes to all the really which ch through this 1 this 1 right here if in the eye this was if this knuckle was faced here the this not was sincere that it would be going back also here yes and are right alright so expect to see 1 like that on the final 1 maybe 2 you will see here and alright so this 1 here is a little bit different because it's a it's asking for what dying dienophile needed to make the following product that's a little harder isn't it in some ways so step struck 3 hours around the ring starting with the pi bond you can go clockwise secure clockwise I don't know why I always declined counterclockwise with I don't know what that means for sure that means something about me the the the right handed or something and then I always seem to go get counterclockwise there alright so that I'm going to tell us where the electrons are going and we get a user can be made from a retrosynthetic arrow so as you can see and really breaking this high water and is this sigma bond right here because those electrons are going to make it a double bond here were and these electrons on this time are moving over so that is going to be a double bond there were breaking this 1 right here and then I this in the double bond there because it's already a single why we're adding another pair of electrons so that's going to be a sigma bond so let's draw a skeleton and then we'll look at where we can get the correct stereo chemistry so here's are dying here's our file because a notice that I wanna dienophile this group in this group are on opposite sides that means that they will have to be trained to each other on all the dying of the so here's a dienophile the trans has to be because of their on opposite sides here alright also notice the orientation of between the deuterium In the electron-withdrawing group on that Jerry no files they're on opposite sides aren't they what is that mean about the deuterium the it's inside substituent so the Turin has to be this way and the manifold and doesn't matter doesn't is it down and that's just going to be right here so let's make sure we put that on the right part of place 1 2 3 because messes 7 discussion on the slope here because so those would be the 2 things that we would combine to make that questions on that 1 anyway through various chapter 16 women start on chapter 17 I mean by this the I think the the the
height so I'm in my you mind noticed maybe you didn't notice that as you've been working product problems all year you might notice that benzene rings don't seem to react in any of the reactions and talk about right they don't do things that we did in chapter and they don't seem to react with anything we have things then Zoellick reacting but we never really seen anything that's actually attacked the aromatic ring and there's a good reason for that so we're going to talk about that in this chapter so I when know about benzene the straw the structure of benzene we know that it's a resident stabilized so we can force these zeros around in a circle here this residents stabilized we have 2 equivelent resonance structures 2 equivalent energetically the same the same energy to equivalent a resonance structures so if we're going to draw a hybrid Ch it would look like this dotted sedated and starts in a circle that's what the hybrid will look like so you'll see in some textbooks where and when they draw benzene ring a draw a circle in the middle we're not going to draw that way but that's where that comes from from that hybrid hybrid alright so in the resonance hybrid 6 electrons from the 3 pi bonds are d localized over the ring the dashed lines on mean that the pilot residue shared equally by all 6 carbons the dashed lines also indicate that all carbon carbon bonds have the same length so they absolutely do have the same length and this would be a typical carbon-carbon bond would be 1 point 5 3 angstroms and typical carbon carbon double bond would be 1 . 3 4 angstroms and this is going to be something in between and for benzene all bond lengths equal all gone so equal but 1 . 3 9 angstroms so is somewhere between a single and double bond in life the alright so so far so good hybridization in along the orbitals this brings us back to chapter 1 all carbons are S P 2 hybridized is all carbons are part of the carbon carbon double bond all carbons are S P 2 hybridized it is therefore a trigonal planar and the wanted bond angles 120 alright so it turns out that benzene we have continuous overlap
and over all of the peer roles so if we drew a free drop benzene on its side so this would be floating above your page perpendicular to your page and we know we have an unhybridized p orbital on every single carbon here and what we have is continuous overlap find the top and the bottom so when we're in that when we were doing the Oenomaus chapter we talked about the fact that we call we have like a donut of of overlapping electron density on the top and on the bottom we saw the consequences for that and in in what we saw that hydrogens that bonded to that to that on benzene ring are actually from New pretty far down killed because of that so continuous overlap the yeah but the p orbitals the so from like a donut on the top the and a donor on the bonds alright and you may have heard the word error Letizia aromatic rings had thrown around but maybe you don't really know exactly what that means because we think it we we think of aromatic in 2 different ways something that smells nice notes aromatic and we also think of it as now you think of it differently in this chapter it's it's something that's very very stable think questions of our the alright so like I said you know we just kind maybe in the back your mind kind and honest that benzene rings me into any the reactions that we talked about for alkenes so we don't we do bromine no reaction constant ethoxide no reaction hydration no reaction ozonolysis no reaction hydrogenation all reactions so this looks like we're not that very much to talk about but we actually are because benzene rings have their own different chemistry that's really separate from what we've already done so far it turns out you can hydrogenate this room this ring forcing conditions hydrogenation of Thundera forcing conditions so a hundred atmospheres of H 2 this is done also in a bomb when you have this high pressures like that activated Nicholas stronger than regular article and so we can capitalism you he doesn't want to be hydrogenated which means it's really stable right if it's so hard to react with know in reactions that's been really stable and of course no deals alter within the yeah alright so and we know where we can all but this is everything we know about benzene so far and is in the so the question we wanna answer is on is resonance stabilization is that enough to explain the special to stability of bands and that's what we're going to try to answer right now the alright so to answer this question was predict expected resonance stabilization of abandoning compare this with the experimental resonance stabilization obtain when benzene is hydrogenated will see for right or not OK so we have done in this 1st example here we have 1 isolated alkenes 1 isolated alkene and if we hydrogenate out unsaturated molecules are they all go to the same product we can predict the stability just based on measuring the heat that's given off when we do that reaction are a somewhat isolated alkene we hydrogenate that to get this year the yeah the heat that we give off we would measure this in a calorie matter is minus 28 . 6 K. counts for more if we have 2 isolated alkenes the and we hydrogen to isolated alkanes we sit at the same product very convenient so now we can compare stability delta h naught is equal to what we would expect to
times minus 28 . 6 0 if we have 20 . 6 for the 1st of the if we have to evolve in their isolated we which is expected to be 2 times that and indeed it is it's a minus 57 . 2 K counts formal alright so now if we do the same experiment except now we have conjugated double bonds the so this 1 here as too far well what happened 2 conjugated double bonds we know that when double bonds are conjugated the more stable so we might expect this to give off less than 57 . 2 indeed it does so here's the product delta h knot equals 0 minus 55 . 4 K cows from all the so and alteration offer to isolated double bonds is 2 times are minus 20 points x equals minus 57 . 2 K. cows from all actual delta H for 2 conjugated double bonds is minus 55 . 4 the difference is the resonance energy alright so we have a residence resonance energy for 2 conjugated double bonds is minus 55 . 4 why is a minus 57 . 2 so we we want is the absolute value of the difference here the as equal to 1 . 8 K. accounts for more so that is the resonance stabilization for 2 conjugated double bonds question so far 1 . 8 it for the solid graphically graphically in just a minute it will be easier for you to see what's going on OK so on the next page so I wanna expect for balancing if 2 times you don't want someone flaky comes from all residents energy than if we had another double bond with winning the twice that would you expect twice that OK the I should have known when it was 1 . 8 that's are working theory here of wave 1 thing is 3 . 6 k and so that all all we did to get that was we added another conjugated double bond in in this right so let's see what we actually get to see for prediction is correct again we get we did we got is forcing conditions so we're likely not right prediction here notice extreme conditions if it was just 1 . 8 K accounts for more per mole more stable we wouldn't have to use activated nickel and 100 atmospheres of of hydrogen OK so let's see what we actually do get delta h naught equals minus 49 . 8 K cows per mole expect adult they shop for 3 isolated double bonds is some 3 times minus 20 . 6 that's equal to minus 85 . 8 we are very far away from 85 . 8 actual it is minus 49 . so the actual resonance energy here is the absolute difference of the absolute value the difference between these 2 guys so whatever you wanna put away put up 1st it doesn't really matter here we I suppose I should do it the other way 49 . 8 was still absolutely absolute value here as Akershus which which side around can have a negative number for a present stabilization 36 K accounts for role by adding to single simply adding another double bond there we have 36 10 times what we would expect OK so something something's going on here that we can't explain with simple resonance stabilization so here's what we actually have your
graphically of our work that we're doing is we're hydrogenating each of these so they're all going to the same part cyclohexanone that's the energy per cycle hexane we start here with 1 double bond releases mice 128 . 6 to isolated double bonds twice 28 . 6 I'm now we have them conjugating we expect our minds to sample 2 but it's actually a little more stable it only gives off minus 50 of 5 . 4 so this is actually this is at a lower energy and now we would expect here all way up here this is actually lower than both of the 2 movies OK so there's a really big big thing going on here but so the difference between the actual and the expected the it's no wonder this thing is not is not very reaction is really really stable now and it just to give you an example of why have how this is very different here but we have cis compare this guy here this thing here we have 3 conjugated double bonds what is a resonance stabilization here versus fencing and the resonance staged stabilization here is 5 3 . 6 just what we expected for benzene K. cows per mole and this is 36 K accountable so of what we would call this ad is an unusually stable the it's unusually stable it we can't explain why it's so stable just using residents it's not adequate enough explanation for the 36 K from so are you impressed with the stability benzene if you were before maybe hopefully you are now far right so maybe our you know you can't think in this little idea in your head may ring has something to do with right namely has to do with the name something magical about that way so that maybe men were all thinking now let's see if we're right on the next page many things got something to do with our eyes so if present simulation on reverberant instability might expect similar when I just got this thing are similar system that is less stable and the answer that question cycle dying 2 conjugated double bonds in a ring and won't do here is we're gonna count pi electrons has been seen as 3 double bonds How many pi electrons 2 for each 6 electrons were an account pi electrons here so I want you to kind of try to look for a little pattern here we have a hard time finding it but this would be 4 pi electrons 2 for each double 1 this would be benzene is 6 pi electrons cycle EmOntoTag training pi electrons cycle Decopaint intend pi electrons and there was you've all way down nothing happens in between 10 and 18 anyway in which has an 18 and pilot electrons alright so here's our here's our idea here there's the ring has something to do with this let's see these other guys are especially stable the Edward quickly be frustrated here because of cycle due dying has never then isolated and purified the of the only way we can actually do not know that we can make cycle you dying is prepared at low temperature extremely low temperature spirit at prepared at low temperature and trapped in frozen argon and that's what you do when you wanna make a million reactive molecules that you can't see every temperature he tracked them in frozen argon sounds horrible doesn't I was there what's the the what temperature would frozen argon C in this anybody know that I know it's really low to somebody else no I think it's like I would say like minus 170 with something like really really cold so what I'm I'm answer that you would all agree that this is extremely unstable OK so there that goes that that idea we know 6 pi electrons has unusual stability and and really the number matters here they pi electrons and no special stability so bad things really starting to look pretty magical here no special stability every acts like a normal Polya so that means that will do still do reactions like we did in chapter 10 In addition reactions you could add bromine native running water things like that to cite the Decker pentene Turner by electrons also no special stability the reacts cycle following yeah a in and what happens is when we finally get up to eating anyway we also have unusual stability the the all right so but see what's going on we already know that conjugation is not enough to explain the special stability of benzene are ring idea seems to be out but but not completely because they're eating anyway also has an usual stability so
our conclusion is the rusty something more than simple instance stabilization explained and means for extraordinary reactivity it turns out
that there is an explanation and has to do with molecular orbitals I know you guys in love like those right I wanna make is as simple as possible bridge on a molecular orbital picture for benzene or enacted B of a CY benzene is so stable pattern we draw our do we draw the molecular orbital picture for benzene and it turns out that it's not that difficult to do and you do something called frost circle when you have a reading is where all of the but the bonds are our as in all of the answer as the to hybridize you can use sparse circle to draw the molecular orbitals so you job seeker compound with 1 of its vertices pointing down I'm a job benzene really big so the vertices where the 2 debt bonds on Christ the relative energies of the pilot corpus correspond to the relative levels of the vertices and Michael ables below the midpoint a bonding and molecular levels above the midpoint art in tight bonding so everyone and these vertices is a molecular orbital we after there and here is the midway point so just dry dash line through the midpoint here the above this is anti-bonding the below the midpoint is finding so I above the midpoint these are all the anti-bonding molecular orbitals and below the mid-point is all the bonding molecular orbitals how many an antibonding like car was to react there's 3 right to each 1 of these lines is a molecular orbital 1 2 3 what we call at least 2 like oracles have the same energy when we call that degenerate right so it is a degenerative their support regular roles these are degenerate and we have 1 2 3 bonding of 1 2 3 and type 1 alright so what we're going to do is we're gonna combines 6 p atomic orbitals so remember when we make collective oracles we combine atomic orbitals and however many orbitals we combine we get that many molecular orbitals so we combined for period the orbital in atomic orbitals we get for molecular levels at the compliance 6 we get 6 molecular orbitals so combined 60 atomic orbitals that each with 1 electron the what is the energy of those p atomic orbitals before they combine this is for the combine it would be this energy that's right in the middle here this little line here that's 1 2 3 4 5 we need 1 more so we can buy 6 the atomic orbitals we get 6 molecular orbitals and that's exactly what we have alright so far so good the now we have so he says 1 electron so let's fill up these orbitals we have 6 electrons altogether right we felt the lower energy 1st we care this stands this goes all the way back to chapter 1 degenerate orbitals we put 1 electron in each 1st and then we go back of we have more we care dispensed so that's 1 2 3 for 5 6 and it turns out that this configuration with all bonding molecular orbitals felt which we call a closed bonding shell is energetically very fair what accounts for the special stability associated with evidence there are the numbers of electrons a corresponded filled shells for smaller and larger rings and that leads us to Hubble's role and on were at a time so will save that those rule till Friday the
Chemische Forschung
Cycloalkane
Homöopathisches Arzneibuch
Metallatom
Kohlenstofffaser
Isotopenmarkierung
Computeranimation
Doppelbindung
Mannose
Methylgruppe
Linker
Funktionelle Gruppe
Substituent
Braunes Fettgewebe
Wasserfall
Supralitoral
Hydrierung
Elektron <Legierung>
Diene
Reaktionsführung
Quellgebiet
Reflexionsspektrum
Geochemiker
Elektronische Zigarette
Körpergewicht
Bohrium
Cycloalkane
Elektronendonator
Ethylen-Vinylacetat-Copolymere
Peroxyacetylnitrat
Computeranimation
Doppelbindung
Mannose
Stickstofffixierung
Chemische Bindung
Arginin
Methylgruppe
Lactitol
Deuterium
Reglersubstanz
Krankengeschichte
Sonnenschutzmittel
Insulin
d-Orbital
Pleuramesotheliom
Elektron <Legierung>
Reaktionsführung
Magnetometer
Geochemiker
Körpergewicht
Bukett <Wein>
Benzolring
Bohrium
Aromatizität
Chemische Forschung
Kohlenstofffaser
Malz
Hyperpolarisierung
Benzolring
Ionenpumpe
Wasserfall
Bockbier
Funktionelle Gruppe
Substituent
Braunes Fettgewebe
Differentielle elektrochemische Massenspektrometrie
Hydrierung
Diene
Gezeiten
Gangart <Erzlagerstätte>
Reflexionsspektrum
Gelöster organischer Stoff
Elektronische Zigarette
Heiße Chemie
Hydroxybuttersäure <gamma->
Cycloalkane
Homöopathisches Arzneibuch
Kohlenstofffaser
Ethylen-Vinylacetat-Copolymere
Orbital
Computeranimation
Doppelbindung
Benzolring
Chemische Struktur
Mannose
Imine
Chemische Bindung
Mesomerie
Hybridisierung <Chemie>
Molekülstruktur
Elektron <Legierung>
Atomabstand
Fluoralkene
Auxine
Fließinjektionsanalyse
Technikumsanlage
Vancomycin
Rückstand
Benzolring
Bohrium
Aromatizität
Hybridisierung <Chemie>
Hydroxybuttersäure <gamma->
Chemische Forschung
Cycloalkane
Elektronendonator
Bodeninformationssystem
Kohlenstofffaser
Zellklon
Zusatzstoff
Druckausgleich
Computeranimation
Doppelbindung
Benzolring
Mannose
Sand
Chemische Bindung
Mesomerie
Natriumdiethyldithiocarbamat
Molekül
Delta
Wasserwelle
Lösung
Alkane
Destillateur
Differentielle elektrochemische Massenspektrometrie
d-Orbital
Isomer
Hydrierung
Diene
Reaktionsführung
Bildungsenthalpie
Magnetometer
Fluoralkene
Ozonolyse
Knoten <Chemie>
Formaldehyd
Nickel
Wassertropfen
Benzolring
Krankheit
Bohrium
Hybridisierung <Chemie>
Aromatizität
Hydroxybuttersäure <gamma->
Biologisches Material
Brom
Polyadenylierung
Wasser
Zusatzstoff
Raki
Computeranimation
Doppelbindung
Additionsreaktion
Mannose
Körpertemperatur
Mesomerie
Antigen
Hexane
Natriumhydrid
Molekül
Systemische Therapie <Pharmakologie>
Cyclohexanon
Konjugate
Destillateur
Elektron <Legierung>
Fest-Flüssig-Extraktion
Reaktionsführung
Quellgebiet
Magnetometer
Tieftemperaturtechnik
Maische
Technikumsanlage
Benzolring
Darmstadtium
Bohrium
Hydroxybuttersäure <gamma->
Tillit
Lambic
Zellklon
Zusatzstoff
Orbital
Chemische Verbindungen
Therapietreue
Computeranimation
Lot <Werkstoff>
Internationaler Freiname
Atomorbital
Mannose
Chemische Bindung
Trihalomethane
Toll-like-Rezeptoren
Differentielle elektrochemische Massenspektrometrie
Wasserstand
Elektron <Legierung>
Reaktivität
Setzen <Verfahrenstechnik>
Magnetometer
Trocknung
Coffein
Ionenbindung
Technikumsanlage
Bukett <Wein>
Monomolekulare Reaktion
Fremdkörper
Benzolring
Optische Analyse
Chemischer Prozess
Periodate

Metadaten

Formale Metadaten

Titel Lecture 22: Benzene and Aromatic Compounds
Serientitel Chemistry 51B: Organic Chemistry (Winter 2015)
Teil 22
Anzahl der Teile 26
Autor King, Susan
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DOI 10.5446/21617
Herausgeber University of California Irvine (UCI)
Erscheinungsjahr 2015
Sprache Englisch

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Fachgebiet Chemie
Abstract UCI Chem 51B: Organic Chemistry (Winter 2015) Instructor: Susan King, Ph.D. This is the second quarter of the organic chemistry series. Topics covered include: Fundamental concepts relating to carbon compounds with emphasis on structural theory and the nature of chemical bonding, stereochemistry, reaction mechanisms, and spectroscopic, physical, and chemical properties of the principal classes of carbon compounds. Index of Topics: 00:59 - The Endo Rule 09:49 - Regioselectivity 15:01 - Synthesis Using the Diels-Alder Reaction 24:27 - The Structure of Benzene 30:11 - Aromaticity 46:00 - Molecular Orbital Picture for Benzene

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