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Compound has one chiral carbon and no other chiral centers; One enantiomer of compound A has specific rotation, [a], of+252 Is this the (+)-enantiomer or the (-)-e...

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Compound has one chiral carbon and no other chiral centers; One enantiomer of compound A has specific rotation, [a], of+252 Is this the (+)-enantiomer or the (-)-enantiomer? What is the specific rotation of the other enantiomer? Can you say which enantiomer has the R configuration at the chiral carbon?What is the observed specilic rotation; [aJobs, ofa mixture of the two enantiomers of = has an enantiomeric excess (ee:) of 4O% of the compound A that (-)-enantiomer? What is the observed specific

Compound has one chiral carbon and no other chiral centers; One enantiomer of compound A has specific rotation, [a], of+252 Is this the (+)-enantiomer or the (-)-enantiomer? What is the specific rotation of the other enantiomer? Can you say which enantiomer has the R configuration at the chiral carbon? What is the observed specilic rotation; [aJobs, ofa mixture of the two enantiomers of = has an enantiomeric excess (ee:) of 4O% of the compound A that (-)-enantiomer? What is the observed specific rotation; [aJobs; ofa A and 45% of the ()-enantiomer? mixture of 55% of the (+)-enantiomer of compound



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Optical isomerism arises due to difference in the rotation of plane polarized light, that is, due to optical rotation by asymmetric or chiral molecules. A molecule having minimum one stereogenic centre, without any kind of symmetry and with a non-superimposable mirror image is capable to show optical isomerism. Stereoisomers that are non-superimposable mirror images are called enantiomers while non-mirror image ones are called diastereomers. Prostaglandin $\mathrm{E}_{1}$ is a compound generated by the body to regulate a number of phenomenons like fever blood, clothing etc. Which of the following is correct about it? (a) It contains 4 -chiral C-atoms. (b) The geometrical configuration at double bond in it is (E). (c) It contains $5 \mathrm{sp}^{2}$ hybridized $\mathrm{C}$ - atoms. (d) Both (a) and (b)

These four years we have So identify which one is which one is more? Okay. In the first option this molecule is that hi? Because This center have four groups exports and CS two words and CS to feel that's why you'd have a parent. Because all the four groups are different. And the second only to This center will not be Cairo Center because CBS two CS towards our scene. Okay. So it will not be carried in the second. Awesome. This will be carol because Cairo center, it will not be carol center. Any carbon. Do not have any four different groups. But in the second molecule in this center the center is um Cara Centre Hydrogen CS three. We are in this computer group. It's not a different group. That's why Carlos Santana. Okay. And the third option this morning, you have carol center. This is the this is the center. But the second one if you have a plane of symmetry with respect to the success there. So it will be optically inactive or then you know that tell center and the fourth molecules, this is the bridgehead carbon, bridge and carbon is not considered as the parent carbon. But in the second molecules, this carbon is that carol carbon. So this will be Colonel Sanders. Okay, so correct. Option will be. Thank you.

First, let's do old possible damn usual cycle prep pains. One. Then it was like nothing we want to says Damn it! To miss recycled propane Run to dance dime. It'll cycle appropriate in another run to drills. Dan, it's encyclopedia. Yeah, A. And see a constitutional ice arrests because he contains one coordinated carbon and see doesn't hell run. This difference is enough to kill that Settles are connected in different orders and you can see in beer. Also, institutional ice amiss with the same. It is, um, B and C. The are their streamers. They're both want to date me. It'll cycle per pains. So ends are connected in the same order that scare my service. But they're not mirror images so that that's German. See, indeed, they are sterile ice emus, and there are mirror images. It means CNG are an internist. Come, Bond is a cardinal. It was a lame similarly, but I know we'll meet about this bank Indyk Ordinary carbon B is a cardinal because a blend of seeming to you kind of through the myrtle seeking Here they are, Kyle, they don't have a plan, will see me here. And there are in your images and then your friend see you alone would be optical active on the kind of compounds alone are after Cally active. So si ngl What compounds here playing the symmetry? It can't be. Yeah, boarding boils and be they will every different boarding points There are institutional ice and rest B and C are the experiments. They also have different boring points. Seon year There are in London, Miss. They will have same boiling points and same properties in being in the garden environment. Yeah. Which one is in the compound? Be? Is it music? Umberg Music Compound is a a Kyle Come bold with Kyle Centers Because of the plan of symmetry those kyle centers of the irritation cancel each other Jew miss just C l d There were no particular. It's a serious Mr Irritation off seeing gear is equal but in the opposite directions. So the 11 Mr will cancel all rotation and the mission will articulate You want me? Sure, being in C be hers. Law Tradition in the sea has tradition, so that nature we'll have an optical rotation because of the presence full kind of comparison. See

To answer A and B First, we're gonna drop models of the following. So for number one, our model looking like this, we have the carbon bonded to hydrogen. That carbon is double bonded to another carbon, which is double bonded, another carbon, and that carbon is bonded to to hide regions. There must be an answer. Questions A and beef A is asking whether it has an asymmetric centre. So we look at our carbon to carbon 12 and three. Here, Carmen one is bonded to hydrogen, and so was carbon three. And since those two, there's two hydrogen it doesn't have or distinct groups, and that is also because it has a double bond. So it only has wanted to three groups, and two of them are the same. So a is no, there are no asymmetric centers. Preppy were asked if the compound is Cairo and the answer for be it is no content is not Cairo, and that's because the mirror image of this molecule is not distinguishable from the originally look exactly the same. And also because there are no, um, asymmetric centers run over to we draw the molecule heavy carbon. Long did 23 hydrogen carbon is bonded to another carbon, which is double bonded to a carbon back home. That carbon is double bonded to another carbon which is bonded to a hydrogen in another carbon with three high virgins. So this is our structure for number two for a the determining if it has any symmetric centers, the carbons air Here, here, here, here, in here the carbons of the ends are bonded to three hydrogen. So those groups or not unique So the first and last carbons are not asymmetric centers, and all the other carbons and middle are involved in double bonds. So that means it cannot form for district groups. So the answer for a is no and so, for the answer will be whether or not the compound is Carol. The answer is yes, it is Cairo. Since the carbons on either side of the double bond are attached to hide regions and see you three groups and those are different groups with a mirror image will be just

Going to be going over how to draw essentially uh compounds given a certain amount of information. So we're starting with compound a. We were we were told that this has the formula of c. five H. 8. We are also told is that it is not optically active and it cannot be separated into inanity murders. So the thing that stands out in terms of that description is that we are essentially limited into as to uh perhaps how we could draw this, but we're not so limited because there are multiple ways in which we can in fact um draw this particular molecule. There's actually a couple of ways. There's more than two ways. But for this video, I'm going to be showing one possibility. You may come up with your own way in which you can draw this and then further on in the video in terms of uh compound reacting with several other substances and giving us our different products like compound B compound sing compound D. Um so again if we're uh the formula of itself C five H. A suggest in and of itself that there are two sides of unsaturated in. And if you want to think about on saturation, you think you can think about on saturation as something that has less hydrogen. Right? If we know our hydrocarbon formula or hydrocarbon formula typically would be CNN and being the number of carbons, right? H. Two N. Plus two. So let's say if we um for example are given um six carbons, if we want to draw a hydrocarbon with six carbons and then the number of hydrogen is that are needed to just for that single thing to be just a single chain of Accardo carbon. Then we would put C. Six and an H two or 2. And sorry that she went to in to end plus two. So we have six here and then two times six is 12 plus 2 2014. So their formula would be C six H 14. That is completely saturated with all the hydrogen, is that it should have. But in this case, if we have, if we're going by this formula, we have C five H eight, then going by the hydrocarbon formula, we should ideally hydrocarbon with five carbons. We should have 12 hydrogen. Um So that's why I mentioned that this has 22 sides oven saturation. Right? So one possibility in which we can draw this um is that we could draw this as a pentagon or a cycle plantain. Let's go ahead and get our marker here. Let's get this ready. We can draw it like, so this again a ring. This is a ring in and of itself. Remember the rules for a saturation or drawing saturation. A double bond counts as 1° Oven saturation A ring cancers, 1° on saturation. A triple bond counts as two degrees oven saturation. So one possibility is that we have here, um a five carbon I five carbon uh ring, for example, um and we can go ahead and start putting in our double bond as well. So that's putting a cool two degrees because we make sure that there are two sites oven saturation in this compound, given the formula. And if we want to go ahead and double check our formula, make sure that our our drawing matches the formula. Let's go ahead and put compound A. We have 123455 carbons. Now, how many hydrogen are present? There's gonna be one here, three because there's two years. So 13 5, 7 and then one extra, there's going to be eight. So this drawing does in fact match our formula. So we can go ahead and start working with this one. But doesn't match the fact that it's not optically active. Well it's not optically active. This is not a Cairo. This is not a Cairo compound. So it's something that is not Cairo is not going to be optically active. And essentially it's also not going to be separate to nancy Meyers, meaning that there is there is no asymmetric carpet present. Um So for this uh so it does meet those criteria. So we can start working with this one. And then we are also told that it reacts with rooming in order to form compound be. So compound B. Let's go ahead and type this up here. We can put me here, Compound B. Is going to have the chemical formula. Or that is going to have the formula of Let's go inside here c. five H. Eight. We are. So that's a chemical, that's the formula for compound B. And if your reaction grooming Which can be br two in carbon tetrachloride to record, remember this when you have a double bond present and you have uh blooming like this and it's a die atomic bombing. Um You are going to perform a diet halogen nation or and this is where the booming is going to be uh placed across the double bond. Um We are going to be going through the mechanism in this video. But the product of this desalination or discrimination. You go ahead and draw that cycle painting a little better. It's going to look like this. Remember there are brahmins are going to be placed right across the double bond here and let me go ahead and edit this formula just a bit because I forgot here that are brought me there should be two, bro, means not just one. So C five H eight. We are too. And again, making sure that our or drawing matches the formula, we have 12345 We have five carbons. Uh we have we're supposed to have eight hydrogen, so one 2345678 eight hydrogen. Is that that satisfies it and then to bro means that's perfect. All right, so we have compound B. And then now we're also told that compound A reacts with a with hydrogen H two um with a some other type of metal catalyst in order to give us C five H 10. So that's going to be uh compound C. So let's go ahead and put H2 H two here and we can just put a palladium as our metal catalyst so that what would give that member when you have this is going to be um uh hydration reaction essentially, you're you're you're basically going to be reversing um the saturation process that was that happened on this particular molecule. You're going to be putting just like our bombing here were replaced the BR two across the double bond. We're also going to be placing this H two across the double bond, so each hydrogen is going to be placed right across the double bond. So we have our cyclops maintain once more and we can put you don't have to put this, but this is just for emphasis that are, hydrogen is replaced right across in the formula For this compound here is going to be C5H 10. It's going to count the number of hydrogen. So there's going to be to hear so 2468 10. Perfect. That's 10 hundreds. And there's five carbon. So that does match. And let's go ahead and indicate that this is compound C. Alright, Rama's done here. And lastly, we do have a formation of compound D when it reacts with compound A. Um One thing that the thing that reacts with compound A. Is hydrochloric acid. So we can put another arrow here, we can put H. C. O. So again, just like basically the other to the reactions that we've done, our HDL is going to be added across the double bond. And in this particular case it doesn't it does not matter where we place our chlorine. Um Either here at the top position or here on this other position. It does not matter in this particular case because we have a symmetric cyclo, cyclo painting. Um So we can go ahead and draw out our cycle painting, We can just go ahead and place our chlorine here. Um and we can place our hydrogen here again. It does not matter where you put your chlorine either here or here. No other place though, it's going to be added across the double bond. Um And then the formula for this, mhm Should turn out to be when it reacts, it should be C5 each. Nine xia. Yeah. And let's go ahead and count five carbons. There's one chlorine and there's 1234567 And 8. 9. Perfect. So it looks like everything does in fact match in terms of what we needed and that's going to be your product. This is going to be compound D let's go ahead and labeled as compound D. Right here And again. So this is pretty much this is one possibility. Keep in mind that I chose to start off with this particular type of compound compound in this, uh in the way that I drew it like this there, you could have drawn it as um in a linear form. You could have drawn it with two double bonds. That would have still satisfied the fact that it it has a 2° oven saturation. Um You could have drawn a a single chain with five carbons and you could have added a triple bomb, for example. Um Those are other possibilities, but I just chose to go with this one. I was one example and I just went from there.


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