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A.) Draw 2-chloro-3-fluorobutane, closely examine itsconnectivity, and then draw the mirror image. Draw two moremolecules by following these directions. Make a repl...

Question

A.) Draw 2-chloro-3-fluorobutane, closely examine itsconnectivity, and then draw the mirror image. Draw two moremolecules by following these directions. Make a replica of thefirst (molecule #1) you created. Switch the positions of any twogroups on carbon 2 to make molecule # 3. Build the mirror image ofmolecule # 3 to make the last molecule required for the series.Double check the procedure, there should be four molecules. What isthe relationship between molecules # 1 and # 2? Does the samerelat

A.) Draw 2-chloro-3-fluorobutane, closely examine its connectivity, and then draw the mirror image. Draw two more molecules by following these directions. Make a replica of the first (molecule #1) you created. Switch the positions of any two groups on carbon 2 to make molecule # 3. Build the mirror image of molecule # 3 to make the last molecule required for the series. Double check the procedure, there should be four molecules. What is the relationship between molecules # 1 and # 2? Does the same relationship exist between molecules # 3 and # 4? What is the relationship between molecules # 1 and # 3 or # 1 and # 4? Be sure to draw all molecules and identify the configurations of all chiral centers. B.) Take the models from the previous question and replace all the fluorine atoms with chlorine to prepare models of 2,3-dichlorobutane. What is the absolute configuration on carbons 2 and 3? What is the relationship between the molecules? What specific term applies to these molecules?



Answers

Draw structural formulas for the molecules listed below. Draw the mirror image of each. Circle the molecules with mirror-image isomers.
a. $\mathrm{CH}_{4}$
b. $\mathrm{CFH}_{3}$
c. $\mathrm{CClFH}_{2}$
d. CBrClFH

All right, So this question asks us to identify the Carol Carbone in this two. Beautiful. So I'm gonna go ahead and draw to you to know, just so we can see what it looks like. We have the O age group on that second carbon right there, and then we could just finish the rest of the compound. So when we go ahead and look the Onley carbon that has four different substitute groups coming off of it would be this second compound. And we have our substance humans being the first being this O H group, the second being this hydrogen, the third being this metal group in the fourth being this Ethel group right here. So this carbon that second carbon is our Cairo carbon. Now, if we were to draw a mirror image of this molecule, we can pretend that our mirror is right here and we're just flipping it over so we can go ahead and start with our ch three whenever ch two. Then we have that see with the oxygen group attached to your the hydroxy group attached to it. And then finally, our last ch three. Now these two molecules, this one right here in this one right here are actually non superimposed a bull. Because if you were to stack these two molecules on top of each other, they would not produce the same image. So these two molecules are non super imposible on to one another.

Of one. This is Ricky, and today we're working on Problems 67 from Chapter 22. We're gonna be drawing the structures of all the geometric I signatures of this complex. Ion is well, is drawing from your image with any molecules that are Cairo. Just a refresher. A geometric Jaisalmer is ice more of a compound with wagon's having different special arrangements about the airline. There are special types of geometric I sinners. Namely, says Trans, in fact, mur. All right, so for simplicity's sake, let's huh? Let's write out a generalized formula where metal like in a is too be too C two, where a is equal to H 20 b is equal to you and h three and C is equal to see now, too rather just the chloride ion, which there are two of them. And because there are two of each, um, the Lycans could be either cysts with one another or trans with one another. So for the different I sinners we have, or for the different possible combinations, we can have Oh, trans meaning each two O's up at the pool opposite of one another, seals pointed to the back. Other seals pointed out front, then along the other access, Um, ammonia is coming out at us while having the other running along the other or along the back. And in this case, the strip of the all trans I simmer is super imposible onto its mirror image. Meaning does not have an optical iceman. Next up, we can have h 20 trends or for simplicity's sake. We're going a trans be and see sis. And if you ever forget what I'm talking about when I say maybe you're seeing gotta look see for you up here. All right? So let's try. So about the metal ion water at the polls, trans with one another. And, um, now both chloride ion or chlorine. An ammonium will be cysts with one another or with their respective pears. So I've been age three over here now, and it's three going back. This is another case where we had we can superimpose our image. Hi. So, next up we have the case. Where be he's Trans A and C or cysts. And in this case, we have our metal ion water here. Then you'll have to use your a bit of, um, your imagination to visualize this in three D space. But trust me when I say that water assist in this configuration and, um, down here who ride? I'm not over here. Chloride says, says then ammonia is trans. And if you're wondering, there's no difference between my shaded and my unshaded wedges. Huh? I'm honestly more a fan of using the unshaded ones. But for school, I always had to use the shaded in, which is all right. Our next I simmer is the case where see this Trans and A and B or assists. So what this looks like is we have metal eye on the middle up at the top God, our water, the bottom pool pneumonia. And in this case, a and beer cysts, meaning water and ammonia RCIs. So we're gonna throw out power ammonia here. Our water come out back here. System built on another. Then the chlorine is trans, so it exists on opposite sides off the metal. Once again. This is super imposible in our last case is where a be see Oh, sis. And in this case you have it's too Oh, assist with H 20 three. Is this with on H three and cool ride. Assist with itself. Now, in this case, we do have an optical I simmer. So if we throw out our merry image pools stay the same. Are you going back? We're coming out, Dad. Sha water Going back. Pneumonia coming out at you. In this case, we have, um, and anti murders because these are not super imposible in themselves. So if this video is helpful when I'll see you in the next one

So the first molecule is S one bromo oops S one Promo one Clara Butane. So let's just ignore stereo chemistry first and draw this out so we know what it looks like. Yeah, butane here, and then on the one position you have a bro mean and a chlorine. You also have a hydrogen here. So clearly this is our stereo center. Right? And so let's give this an S. Configuration. We take our hydrogen, we want it pointing back and then we want the other three groups to go counterclockwise. Because this is s right. The priority of the groups are as follows. Highest priority is brimming because that's a high smaller mass. That is chlorine, that is carbon. Then it's this hydrogen here. So the hydrogen points back. Yeah, bro. Mean here, then you have a chlorine here. Remember we're going counterclockwise and then you have this beautiful this beauty the alcan chain here. It's a carbon points back goes counter clockwise. All right, Let's look at the next one. We're looking at two R 3, R 2, 3, Declare A. Pantene. So again, let's not even consider the stereo chemistry first. So, we know what this looks like. It should look like this. So, you've got to stereo centers here and here. Let's consider the first one. Do you want the hydrogen pointing backwards? And then the priority of the other groups are as follows. You've got chlorine, highest priority this al caine chain. This carbon is going to have a higher priority than this one. Okay, for so the hydrogen points back and when the chlorine faces this way, then you see That the priorities go clockwise, which is what we want. So one 23 clockwise. Look at the other. Yeah, Statue for this. one. One 234 So, if this hydrogen is pointing back and this chlorine is pointing forward, then the chlorine. Um then to three groups go clockwise. So 1, 2, 3. So, this is what um If both of them were the our configuration should look like for the third part, we're looking at an A Cairo is summer of 12 dimethyl cyclo vaccine, 12 dime ethel cyclone vaccine. So, in order for this compound to be a chi role, it's got to have a mirror plane. All right. No, the only way you can do that is if it looks like this. Similarly, with number four, we were asked to find and a Cairo is summer Of 1 to Die Bro, Mo Cyclo Butane. We want to mirror plane again or sorry. Um Because we want we want it to be Cairo this time. We want the bro means facing different directions. So this would be Cairo all because they don't have a mirror plane. Mhm. And then for five, sorry, it got a bit ahead of myself. Earlier there you want to a Cairo is summers of two of 345 tri metal help attain 123 4567. So in order for this to be a Cairo all you want to mirror plane right? So you could for example Have them all facing one direction. This would be a Cairo. You can also have them like so. Okay and these are both a Cairo and you know they're miso compounds because they have this mirror plane here. Okay. And that's the problem.

A molecule of the waste urine. And if we draw the simple Lewis structure, we will see that it is tribunal plainer around the carbon tetra hydro or tribunal, a parameter around each nitrogen. The bond angles around the nitrogen zehr 109.5, but around the carbon 120 degrees for the whole molecule to have a all bond angles at 120 degrees. And the double bond must resonate between the carbon and nitrogen, said not just between the carbon and oxygen. So here's an example of residents structure, where at any given time, um, this nitrogen would have 120 degree bond angles on, um, also the oxygen this carbon up here. And then if the double bond were to be placed here instead, then this nitrogen would also have some tribunal plainer geometry. So the best explanation is that double bond must resonate from the carbon oxygen to the two carbon nitrogen


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