this meeting is being recorded in section 7.6 uh we're going to review uh this concept of alen stability and uh this will lead us on our way to be able to predict products of uh these types of reactions specifically elimination reactions to produce alkenes if we have a um a general understanding of the stability of our different um product s that could be produced so in this first slide here we're going to look at steric strains specifically um with the two possible isomers of buttin so we know that we have four carbons and a double bond um somewhere in this compound um so we see this trans two buttine over here where the two uh methyl groups uh surrounding uh these two positions here the two vinyl positions are uh far apart from one another whereas with the uh CIS orientation of uh Tu buttine uh we have what's indicated here as steric strain so these two methyl groups are closer to one another and um this generally is uh less stable um because it would be higher energy to produce this confirmation of the compound so this isomer uh again is generally going to be less stable um because of that steric strain and we can see that this um in evidence here through um our process of heat of combustion comparison uh so what we have down here is the trans isomer up top the CIS isomer below and um we see that there's actually more energy release so a um a higher magnitude here this 2686 over 2682 um in this uh enthalpy change overall so we know that the Cs isomer must have been at a higher energy um so this reveals the relative stability of these two isomers so um you may be wondering how will this help you predict products um so generally uh the more stable the lower energy um representation uh of an alken would be our major product so um that's what we see here um this representation of the heat of combustion shown um we have the same product so we know that we are at uh the same energy at the end point here um but our CIS isomer would have to start 4 K per mole higher than our trans isomer to produce these changes in enthalpy so there is um another way to uh rank alken stability besides um Cy versus trans comparisons um we can also look at whether or not our um alken is monos substituted so here's our double bond here uh we have right three hydrogens so only one position has a substituent so mono substituted um we see D substituted here if we fill in the hydrogens tri substituted and then our most stable compound would be a Tetra substituted alken which would have four alkal groups which are highlighted here um and uh the reasoning behind this is uh something that we call hyper conjugation and you'll just need to know um very a very brief definition that the increase in alkal groups um help to stabilize the CC Pi Bond um via this term that we use called hyper conjugation um so in the end the simple way to look at this the more alkal groups the more stable the alken so when we talk about uh cyclic alken so cyc alen stability um whenever we have a cyclic alken with less than seven carbons um only the Cy alkenes are stable um and it would require uh too much energy to keep a trans confirmation possible um some of these Rings um definitely could not even exist um like the cyclopropene here with a transbond um and there's a specific rule that we say uh Brett's rule here when we talk about bridged uh byy alen so this specific rule uh again applies to bicyclic compounds by um cyclic alkenes and uh if we have a bridge head here um with a double bond um this compound um is not very stable um if we have less than uh eight carbons we cannot have a bridge head definitely not with a trans uh Pi Bond here right unless one of the Rings has at least eight carbons um so um briefly the geometry um at the bridge head will prevent this uh overlap of the p orbital so um it would be strained and it wouldn't be able to um keep that uh that positioning so um again the larger ring um can be stable so we'll continue with um a discussion of about elimination reactions next time