Draw every stereoisomer for 1,2-difluoro-1,2-dimethylcyclopent...

Question

Draw every stereoisomer for 1,2-difluoro-1,2-dimethylcyclopentane

Accepted Answer

Welcome back everyone in this example. We are given 13 dye bromo, cyclo butane which has two stereo members were told each of which has a different chemical and physical properties and how many stereo customers are possible for one to die. Bmo cycle butane. And are any of them non polar. So we're given an entire different molecule here because we have 12 as our assignments for the location of our substitue ints. And so what we will draw is we're going to recognize that according to the name given in the prompt, we have cyclo butane meaning we have a ring. And because we have the suffix being butte, we would say that we would recall that this means we have a ring of four carbon atoms. And so drawing our ring, we have four carbons in a rink where we should recognize that that carbon one which we can make any of these carbons. But let's just go with this corner here as carbon one. We have our first substitue in as our first bromine atom. So we have one bromine atom here and then at carbon to which we can make carbon to here. We have a second substitution which is our second bromine atom. So we need to draw the stereo its members of this structure here. And drawing out its stereo. I guess we can recognize that we can use stereo chemistry which we recall is using our dashes and wedges to represent our substitue ints here. And so we can draw our booming at carbon one coming out of the plane of our paper towards us where we would recall, we would use a wedge to represent something coming out of the plane of the paper. So we have our wedge here Representing our bromine atoms constituent on carbon one. And what that means for carbon too is that we would have to put it on a dash meaning it goes towards the plane of the paper away from us. So we would represent that on a dash here and so that's browning our second constituent at carbon two on a dash. And this is going to represent our first stereo. Now, what we should recognize about this stereo I summer is that we have two different types of stereo chemistry here. We have a wedge and we have a dash representing our bro mean substitue ints. And we would recall that this because they are different. They have different stereo chemistry. This would be what we recall as trans. So this is a trans of our one to die bro, mo. Cyclo butane. And we would label this as trans 12. We don't really need a dash there. So we would just say trans 12. Di bromo cyclo butane? S the name here. Now we should recall that. This will also have an anti tumor and we want to recall that in an anti tumor is a mirror image. So this will still only count as one transit. However, we can have also drawn our above summer where we have instead our bromine at carbon one on a dash and then the carbon too now on a wedge. So we could have also drawn it this way. But this is just an anti mir. So we're going to and let's just make that need er Sorry about that. So This is still going to count as just one stereo. I summer of our 1 to Die. Burma, Cyclo, Butane. And that is because we're considering this second drawing as in an anti tumor because it's a mirror image of what we drew above. So going on to our second type of stereo I summer, we would recall that because we did the trans version of our molecule above. We can draw aces version where we have our system er where each blooming substitue in has the same stereo chemistry. So we can draw both of our blooming substitue ints with the same stereo chemistry where we could both have each of them on wedges coming out of the plane of our paper towards ourselves. So we have both of these roaming substitue ints on wedges here. And sorry about that, I'm pretty bad at drawing wedges. So we have the second wedge with our second roaming substitution. And as we can see this is a system are because these two browning substitutions have the same cereal chemistry. They're both on wedges. So this is our cis di bromo cyclo butane and as far as polarity going back to our first stereo, we should recognize and recall that based on our electro negativity trend on our periodic tables, which we recall increases from the left to the right on our periodic table. We would recognize that we have a dipole moment going towards our bromine atoms. Because roaming is we recall more electro negative than our carbon atoms here. And so we have no symmetry. Because to cancel out these dipole moments, we would need a bromine atom on a wedge either on this side or over here on this side. So because we have carbon atoms over here and because these sides are brimming atoms, these dipole moments are not going to cancel out. So this is going to be labeled as a polar molecule. And going to our second stereo I summer we still would have those dipole moments going towards our bromine atom. And so this is also going to be considered polar because we would need bromine atoms on this side of our molecule with the same stereo chemistry to cancel out these dipole moments. And so for our final answers, we can confirm that we have our first stereo I Summer which is our trans I summer here and then we have our second stereo I Summer which is our system. Er So we have a total of two I summers total to stereo Somers Total for 1 2 di bromo cyclo butane and all our polar. And so this statement here is going to be our final answer to complete this example. So I hope that everything I reviewed was clear. So each of the highlighted stereo I summers are what we would consider our two I simmers for one to die bro mo cyclo butane. And we confirm that they are all polar because they have dipole moments which do not cancel because there's no symmetry in our molecules going towards our bromine atoms. So I hope everything I reviewed was clear. If you have any questions, leave them down below and I'll see everyone in the next practice video.

Draw every stereoisomer for 1,2-difluoro-1,2-dimethylcyclopentane

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