this meeting is being recorded in this video uh we're going to cover uh the first half of section 7.7 and that will specifically uh review uh E2 elimination reactions and the radiio selectivity so um how do we um determine what product will be produced um what region of our uh compound will have this elimination occur depends on whether or not our strong base that's used is either a um bulky or a uh small type of Base so depending on um what type of Base we use we're going to generate um one product or another so here we have our uh tertiary alkal halide our substrate and um if we have more than one beta carbon so that would be a carbon that has a beta proton um that could be deated by a strong base um we can have more than one product so we see here this is our Alpha carbon so we have this beta carbon here or this beta carbon here Each of which has um either two protons or uh three equivalent protons that could be removed so um what we see with this um non bogy base here the sodium ethoxide is that our product that is more highly substituted called the Zev product is going to be our major product this is more stable um as we've seen in uh previous discussions because it's more highly substituted um so overall this would be our more um produced product and then we have our minor product so this is minor over here this is our Hoffman product and um again that would be due to um elimination of one of these protons over here to generate the least uh substituted product possible so again depending on what proton is removed um what area or what region in your molecule um will have this elimination and uh Pi Bond formation uh occur uh that is dependent on what type of base is used so um we can uh predict or even control a reaction by carefully choosing which strong base we use so let's take a look at some uh common bases um that can be used and some data that has been produced so we see here with a uh nonbulky uh unhindered base like ethoxide here um this would favor the zit or my highly substituted product um then when we have either tur butoxide or um even this more highly um hindered type of uh sterically hindered base um then we have a majority of our product that is favored to be hofman um and again that has to do with sterix and the fact that um this uh bulky and hindered base so this is the bulky house down here um it would be unable to um deprotonate are alkal halide in an area that the more highly substituted alken would result so we will um continue with um oh this discussion here um on radio selectivity and uh sterix with some um more common um bases that are bulky down here so we have uh these non-nucleophilic bases they're not really able to function as a nucleophile because of these um esteric hindering um groups around um the um lone pairs that function as a base so we have our potassium turut oxide sometimes that's written this way um then we have diisopropylamine in the middle you can see that bulky uh and then we have our triethyl Amun which has that nitrogen surrounded by three ethyl groups so let's see if we can U predict products for these three reactions shown below um so we're told that they undergo um E2 mechanisms so uh step one we should find our Alpha carbon and then look for any beta carbons close by um and then our third down here so uh this beta carbon does not have any protons uh attached that could be uh removed in this elimination um so we'll discount any products that could be formed um at that site through elimination um but we do have this uh non- bulky nonsterically hindered base so we expect that our zit product would be our major product so um that would be if um one of these protons here were removed so there's two equivalent protons so let's draw our product so we have those two U methyl groups there the other methyl group up top here the um chlorine was removed that's our leaving group and then um one proton on this side over here so this would be our major product the Z set product and then our minor product would be if one of these beta protons was removed um so we would have our less highly substitute a product result and it would look uh something like this so let's see what happens um we have the same substrate given in each of these examples here same starting material a tertiary um alkal iodide here and um first up we have a non bulky nonsterically hindered base and then down here we do uh indeed have tur butoxide so we have hysterically hindered base so um both of these will generate uh the same two products but um we're going to see that um what is the major in the first case um is going to be the minor in the second again so um again we'll identify our Alpha carbon and then we have uh these two equivalent beta carbons over here and then another one here so again we have our two beta carbons and then the beta hydrogens that we can remove so um with a uh nonsterically hindered base we should see that the more highly substituted zit of product would be our major product and then the least sterically hindered Hoffman product would be our minor product and then like I said the opposite would happen with this bulky base um so down here uh this would be the minor product due to tur butoxide being used and then over here this would be our major product so um we'll review more uh practice problems in the um later on in other videos so that we can Master how to predict products um and determine radio selectivity so we'll continue with uh part two of this section and determine uh stereo selectivity of these elimination reactions