Isomers Practice Problems

hello everybody my name is Iman welcome back to my YouTube channel today we&#39;re going to do practice problems that relate to isomers the first problem says which of the following does not show Optical activity now what we are looking for in the answer choices here is a race mixture race mix mixtures with equal concentrations of two enantiomers will not be optically active because the two enantiomers rotations are going to cancel each other out now looking at the answer choices here we have r two butanol S2 butanol a solution of one molar of the r and two molar of the S and then D is two molarity R to molarity s okay when we look at these answer choices answer Choice D this is a racemic mixture of two butanol because it&#39;s going to consist of equal molar amounts of r two butanol and S two butanol the R2 butanol molecule is going to make the the plane of polarized light rotate in One Direction and the S rotates it by the same angle but in the opposite direction and as a result no net rotation of polarized light is observed all right and so the correct answer here is going to be one the r alone and the S alone are going to rotate plane polarized light they are optically active and C has one mole a molar of of the r and two molar of the S so there&#39;s a higher concentration of the S so it will still predominate and move the plane polarized light it is optically active but when there is equal concentrations of the two Indian tumors this will not be optically active because the two enantiomers rotations cancel each other out perfectly all right fantastic one is d two says how many stereoisomers exist for the following aldehyde all right now how many stereoisomers first thing that we&#39;re going to have to do is we&#39;re going to figure out how many chiral centers does this molecule have and the rule is that the maximum number of stereoisomers of a compound is going to equal 2 to the power x where X is the number of chiral carbons in the compound now in this compound right here all right we have one chiral carbon this carbon has four different groups attached to it this is another chiral carbon and this is another chiral carbon there are three chiral carbons now this carbon is not chiral you notice it has two of the same groups attached to it remember a chiral Center is a carbon with four different groups attached to it there are only three chiral carbons here so then if we&#39;re trying to answer how many stereoisomers this molecule has well it has three chiral centers so it&#39;s going to be two to the power of three that&#39;s two times two times two and that&#39;s equal to eight so this molecule is going to have eight stereo isomers now here&#39;s also a thing that sometimes can be tricky all right if a molecule has an internal plane of reflection then you have to worry about miso compounds now this molecule does not have an internal plane of reflection so you don&#39;t have to worry about meso compounds in the case that there are miso compounds there might be less stereo isomers than you calculated all right so that&#39;s the only catch that you should be wary of when you are using this formula to calculate stereoisomers this works for compounds that do not have an internal plane of reflection cool two is B three says which of the following compounds is optically inactive all right so chiral molecules are optically active we know this so what we&#39;re looking for then here is we&#39;re looking for a molecule that isn&#39;t chiral or has no chiral centers or maybe has chiral centers but is secretly a miso compound because miso compounds are optically inactive now if we look at all these molecules all of them have two chiral centers and they&#39;re all highlighted in blue here all right those are all carbons with four different groups attached to them so each answer Choice has two chiral centers so what now well then the answer must lie in the fact that one of these is really a miso compound now this this answer Choice all right the the answer is C all right and this answer Choice particularly we&#39;re going to work backwards this way all right the answer choice is an example of a miso compound a compound that contains chiral centers one two but it has an internal plane of symmetry right there in the middle this side looks exactly like this side if there was a mirror placed right here all right and owing to this internal plane of symmetry the molecule is a chiral and hence optically inactive now had we started looking through the answer choices choices A and B are actually enantiomers of each other all right and you can notice you can notice that first of all they also don&#39;t have a plane of internal reflection whether you draw the line there or the line here they are there&#39;s no line or plane of symmetry I should say and the same goes for B and the same goes for D the only one that is a miso compound that has an internal plane of symmetry is C which is why the correct answer here is C since they all have chiral centers all right we&#39;re looking at option two for finding which one is optically in active all right so three is C Fantastic Four says cholesterol shown below contains how many chiral centers a chiral Center is a carbon that has four different substituents four different groups attached to it if we look at this molecule we&#39;ll just have to go through each carbon that is the workflow for this kind of problem and just count all the chiral carbons we have a chiral carbon right here it has four different groups these carbons one two three right here they all have two hydrogens so they&#39;re not chiral this carbon there&#39;s a double bond here so it&#39;s double bonded to a carbon so there&#39;s no four different groups attached to it so it&#39;s not a chiral Center all right so we only have one here this one is a chiral Center that&#39;s two so is this one three this one also has four different groups attached to it so four five six seven and eight all of these carbons have four different groups attached to them any of the other carbons here this one has two hydrogens this one has two hydrogens this one and this one have two hydrogens and so does this one so none of these carbons that I&#39;ve crossed out these two are double bonded to each other so they&#39;re attached to carbon twice so they don&#39;t have four different groups attached to them this carbon has three hydrogens this carbon has two hydrogens this carbon is attached to two methyl groups so it&#39;s also not a unique carbon all right and this carbon&#39;s attached to three hydrogens so this methyl group is also not a chiral Center so this cholesterol molecule has eight chiral centers all right and that&#39;s going to be answer Choice C so 4 is C beautiful five says which isomer of the following compound is the most stable all right and we&#39;re looking at some chair conformations Here For A and B and then this is good old boat conformation so we&#39;re looking at some chair conformations all right we know that for cyclohexane the chair conformation is the most stable so in just comparing the chair in this boat both out of the question here all right C is incredible because it is in the more unstable boat conformation and if C is incorrect so is D so now we&#39;re stuck between A and B we have these two chair conformations all right we have these two chair confirmations and there are substituents at Carbon one and two for both of them now here in answer Choice a this methyl group is in an axial position and so is this methyl axial and axial in answer Choice B both of the methyl groups are in an equatorial position all right now the molecule that has both of these methyl groups in Equatorial position are going to be more stable all right we talked about this in the lecture how um axial substituents create more non-bonded stress and in cyclohexane molecules with multiple substituents the largest substituents will usually take the equatorial position to minimize strain so having both of these methyl groups in Equatorial position is going to minimize strain and it&#39;s going to lead to a more stable conformation so the answer Choice here is B this molecule okay is a chair conformation again in which the two equatorial methyl groups are trans to each other all right and because the axial methyl hydrogens do not compete for the same space as the hydrogens attached to the ring this conformation ensures the least amount of steric strain Choice a is going to be less stable because those diaxial methyl group hydrogens are closer to the hydrogens on the ring and that&#39;s going to cause greater steric strength all right so that&#39;s another way of rephrasing that to think about it the answer choice for 5 here is B beautiful so yeah you can think about it as oh what are the where are the bulky groups is the bulky groups are in the equatorial position that&#39;s better that creates a more stable configuration and you can also think about it as other interactions for example if your methyl group is right here in an axial position is going to have that 1 3 diaxial interaction within hydrogen group all right whereas a methyl group in the equatorial position won&#39;t have that one three dioxyl interaction fantastic six says the following reaction results in blank a says retention of relative configuration and a change in the absolute configuration B says a change in the relative and absolute configurations C says retention of the relative and absolute configurations and D says retention of the absolute configuration and a change in the relative configuration so what we&#39;re noticing in this reaction is um we are forming hydrochloric acid and then there&#39;s these new bonds being formed to create this molecule now the relative configuration is retained because the bonds of the stereocenter are not broken all right so the relative configuration is retained because the bond of the stereocenter the chiral carbon are not broken thus the position of groups around the chiral carbon are maintained now the absolute configuration is also retained because both the reactant and the product are going to have the same configuration it&#39;s R all right and we can double check that ourselves all right if we look at this chiral carbon and we number things appropriately with the appropriate priorities this is one this is two this is three and this is four all right your fourth group um is near a dashed line all right your fourth group is not on a dash but it&#39;s near it so what we&#39;re going to do is we&#39;re going to figure out the one two three all right that gives us counterclockwise which is s and then we do the opposite of that remember that was our rule so this is r okay when the fourth group is not on a dash and you don&#39;t know how to rotate figure out the one two three and it&#39;s the opposite of whatever that is all right this was one of the rules that we covered in the lecture all right if we do the same thing here all right if we do the same thing here we have one two three four again um the same kind of numbering scheme and the fourth is not on the dash but it&#39;s near it so we do the one two three that&#39;s s and then we do the opposite so the relative configurations retained because the bonds on the chiral carbon are not broken and then the absolute configurations also retained because the configuration of that chiral centers are in both of the react in both the reactant and the product so the answer is going to be C retention of the relative and absolute configuration there&#39;s retention in both the relative and the absolute configuration fantastic seven says the following molecules are considered to be blank all right these compounds are non-superimposable mirror images all right and you can make this analysis a bit easier we can rotate structure two 180 degrees and it&#39;ll look like structure will look like this structure sorry I don&#39;t know why I said two and one if you wanted to take this one and call it one and then this one call it two I should have written that before I started talking all right if you take this too and you rotate it 180 degrees all right you can begin to see and compare the two now if that&#39;s hard and I personally don&#39;t ever rely on my rotation ability especially when I am stressed in exam what we do what we can do is assign R and S configuration at these chiral centers all right and then start to compare all right start to compare these chiral centers now let&#39;s look at these all right let&#39;s look at these and let&#39;s start figuring out how we want to approach this problem all right the way that we want to approach this problem is by first and foremost I&#39;m just going to do a 180 degree on this molecule right here okay and so all that means is it&#39;s going to look like it is turned 180 degrees so this is going to be ch3 here and this is going to be ch3 here so all we&#39;re going to do is rotate we&#39;re rotating it 180 degrees about this all right and we get the ch3 here now it&#39;s no longer on a wedge it&#39;s on a dash and it&#39;s no longer on a dash it&#39;s on a wedge now we can compare this and this appropriately all right this chiral Center is the same as this chiral Center and then this chiral Center is the same as this chiral center now just looking at it they look like enantiomers because on the purple sites one is on a dash and one is on a wedge and on the red sides one is a dash and also and the other ones on a wedge so they&#39;re opposites of each other in in the at the purple site and they&#39;re the Opposites of each other on the right side but to ensure that we&#39;re right what we can do is determine the RNs configuration all right we can determine the r and s configuration of each site so if we look at this purple site right here all right we have a hydrogen that&#39;s right here we have a methyl group right that&#39;s right here this will be one all right this will be two this will be 3 and the hydrogen will be four and it&#39;s on a dash so then all we have to do is figure out the one two three rotation that&#39;s clockwise so this is r if we do the same here all right it&#39;s going to be the same numbering but now one two three but now the hydrogen is not on a dash it&#39;s on a wedge so we figure out the one two two three rotation and we do the opposite of it so this would be R but since the hydrogen is not on a dash but it&#39;s near it we have to do the opposite so this is not r it&#39;s going to be the opposite of that it will be s all right and then if we figure out the next site the red site all right this is going to be in um let&#39;s do it let&#39;s do our priorities here all right so at the red at the red or now blue dot all right let&#39;s assign priority there&#39;s a hydrogen right here this is going to be one this is going to be two the methyl group is three the hydrogen is four all right we&#39;re going to figure out our one two three rotation that&#39;s an S but because the hydrogen is not on a dash it&#39;s next to a dash we have to do the opposite of that so it&#39;s going to be R and then if we do the same thing here this is going to be an S so at the purple site their opposites are and S and at the at the blue site they&#39;re also opposites they&#39;re r and s all right so that means because they&#39;re opposites at each of the chiral centers they&#39;re going to be in near tumors all right fantastic so that means 7 is a beautiful let&#39;s do eight 8 says plus glyceraldehyde and minus glyceraldehyde refer to the r and s forms of two three dihydroxy propanal respectively these molecules are considered blank all right so plus and minus glyceraldehyde or in other words R and S 2 3 dihydroxy propanol just based off of the naming gives it away our enantiomers one is has an R chiral Center the other one has an S and enantiomers are non-superimposable mirror images each has only one chiral Center and clearly within the name they&#39;re given the opposite configuration one has R and one has s so they have the opposite absolute configurations in these molecules so they&#39;re going to be enantiomers eight is a beautiful nine says consider e two butane and Z2 butane this is a pair of what type of isomers all right now E2 butene all right can also be called trans e refers to trans all right and Z refers to CIS so these can be also called trans tubutine and CIS to butene now as such they are CIS trans isomers that is correct and remember that CIS trans isomers are a subtype of diastereomers in which the position of the substituents differ about an immovable Bond and diastereomers are molecules that are non-mirage non-mirror image stereoisomers all right and so they are CIS and trans and they are diastereomers these are not enanti tumors because they are not mirror images of each other all right there are cysts and trans and their diastereomers all right and so the correct answer here is going to be C I don&#39;t know why B is highlighted the correct answer here is going to be c one and two are true all right this e and Z2 butene are CIS and trans isomers and CIS and trans isomers are a category of diastereomers are a subtype of diastereomers all right so 9 is C beautiful let&#39;s do 10 10 says three methyl pentane and hexane are related in that they are blank let&#39;s go ahead and draw these out because they&#39;re very easy three-methylpentane so pentane that&#39;s five carbon chain and at the three position there&#39;s a methyl and hexane is just a six carbon chain now the molecular formula for three-methylpentane is c6h14 and for hexane it&#39;s c6h14 so h14 they both have the same molecular formula but clearly from the drawings they are they have completely different structures and so things that have the same molecular formula but are different in their atomic connectivity are called constitutional isomers all right so the answer to 10 here is going to be beautiful 11 says are two chloro S three bromobutane and S2 chloro S3 bromobutane are blank all right we don&#39;t even have to draw these all right we don&#39;t have to draw these because it&#39;s telling us for this first name we have at the at the two chlora we have an r and at the three bromo it&#39;s an s all right and then for this other molecule at the two chloro we have an S and at the three bromo we have an S all right so if we&#39;re comparing the same chiral centers all right the things that we circled in green are at the same location in these two molecules all right and then things we highlight in blue are going to be oh sorry guys and the things we highlight in blue right here are going to be at the same chiral Center in these two two different molecules all right now they&#39;re giving us the chiral centers at the same points in that same order for both of these molecules are and S for One S and S for the other now here at this first position their opposites R and S but here at the second position they&#39;re the same we know that molecules that have a mix of opposite configurations at some chiral centers and the same configurations at other chiral centers is what defines diastereomers all right so the answer here is B these two molecules are stereoisomers of one another but they&#39;re not non-superimposable mirror images therefore they are diastereomers all right and note that these molecules differ by at least one but not all chiral carbons okay 11 is B fantastic 12 says a scientist takes a 0.5 molarity solution of an unknown pure dextro rotary organic molecule and it places it in a test tube with a diameter of one centimeter he observes that a plane of polarized light is is rotated 12 degrees under these conditions what is the specific rotation of this molecule all right so remember that the equation for specific rotation is this one right here all right in this example this alpha alpha observed is plus 12 degrees all right and um remember that dextra rotary or clockwise rotation is considered positive which is why we can confidently say it&#39;s plus 12 that&#39;s what they tell us here all right Dexter rotary that gets a positive value all right we also know that the concentration is 0.5 molar and the path link is one centimeter which can be converted to decimeter which is what it should be in terms of units so things properly uh calculate for this kind of equation so one centimeter is just 0.1 this meter all right and then we can calculate from this setup the um specific rotation of this molecule we have plus 12 over 0.5 multiplied by 0.1 all right so we have 12 over 0.5 times 0.1 all right that&#39;s 5 0.05 12 over 0.05 that&#39;s going to be 240 degrees and it&#39;s going to be positive all right so that&#39;s going to be answer Choice D12 is d all right so this is an important formula to remember awesome 13 says um omprazole is a proton pump inhibitor commonly used in gastro esophageal reflux disease when Omeprazole erasmic mixture went off patent pharmaceutical companies began to manufacture esome prozal the S enantiomer of omoprozal by itself given one molar solution of Omeprazole and ESO esomeprazole which solution would likely exhibit Optical activity all right now Ray Smith mixtures like ohm prosal contain equal molar amounts of two enantiomers all right that&#39;s what they&#39;re describing here for Omeprazole all right it&#39;s equal amounts of two enantiomers R and S for example all right and that means that they&#39;re not going to have any observed Optical activity each of the two and anti mirrors causes rotation in opposite directions and so essentially their effects cancel out but what they&#39;re saying here is that esomeprazole is this s enantiomer all right and so it only contains one of the two enantumers and it will cause rotation of plane polarized light all right and so the answer here is going to be B 13 is B antastic let&#39;s go to 14 14 says 2r3s 2 3 dihydroxy butane dioic acid and 2s3r23 dihydroxybutane dioic acid are blank one says miso compounds two says the same molecule three says Indian tumors all right now we can draw out these structures all right we can draw out these structures the two names describe the same molecule all right now what we notice here is if we rotate about this Bond right here if we draw this out two three dihydroxybutan dioic acid if we rotate it well this molecule will look like is a molecule that will have an internal plane of symmetry all right a molecule that&#39;s going to have a internal plane of symmetry so that&#39;s an important thing to keep in mind all right so we can draw out these structures the two names describe the same molecule which also happens to be a meso compound because it does contain a plane of symmetry these compounds are not enantiomers because they are superimposable mirror images of one another not non-superimposable mirror images all right so these are this is going to be these are miso compounds they are the same molecule they&#39;re not different all right and so the correct answer here is going to be c one and two are true statements fantastic and last but not least 15 says if the methyl group of butane are 120 degrees apart as seen in the Newman in the Newman projection this molecule is in its blank all right so if the methyl groups of butane are 120 degrees apart all right this molecule is in blank in butane the position at which the two methyl groups are 120 apart is in an eclipse conformation it&#39;s not a totally eclipsed conformation but it&#39;s just an eclipse conformation I&#39;m going to go actually back to the notes where we drew out this all right right here actually oops I missed it no I didn&#39;t go back up further sorry guys all the way over here all right when the two methyl groups are 120 degrees apart all right it&#39;s going to be this conformation right here this is when they were 180 degrees and then when we rotated that back carbon 60 degrees now there 120 degrees apart this is an eclipsed conformation now this is an eclipse confirmation where we have methyl near hydrogens all right there are Eclipse with hydrogens not in the totally eclipsed conformation where the two methyl groups are on top of each other all right so this is an eclipse conformation this has a moderate amount of energy although not as high as a totally eclipsed conformation in which the two methyl groups are zero degrees apart and so it&#39;s going to be a middle energy eclipsed form which means that 15 is going to be C all right and with that we&#39;ve completed the problems for this set let me know if you have any questions comments concerns down below other than that good luck happy studying and have a beautiful beautiful day future doctors

hello everybody my name is Iman welcome back to my YouTube channel today we&amp;#39;re going to do practice problems that relate to isomers the first problem says which of the following does not show Optical activity now what we are looking for in the answer choices here is a race mixture race mix mixtures with equal concentrations of two enantiomers will not be optically active because the two enantiomers rotations are going to cancel each other out now looking at the answer choices here we have r two butanol S2 butanol a solution of one molar of the r and two molar of the S and then D is two molarity R to molarity s okay when we look at these answer choices answer Choice D this is a racemic mixture of two butanol because it&amp;#39;s going to consist of equal molar amounts of r two butanol and S two butanol the R2 butanol molecule is going to make the the plane of polarized light rotate in One Direction and the S rotates it by the same angle but in the opposite direction and as a result no net rotation of polarized light is observed all right and so the correct answer here is going to be one the r alone and the S alone are going to rotate plane polarized light they are optically active and C has one mole a molar of of the r and two molar of the S so there&amp;#39;s a higher concentration of the S so it will still predominate and move the plane polarized light it is optically active but when there is equal concentrations of the two Indian tumors this will not be optically active because the two enantiomers rotations cancel each other out perfectly all right fantastic one is d two says how many stereoisomers exist for the following aldehyde all right now how many stereoisomers first thing that we&amp;#39;re going to have to do is we&amp;#39;re going to figure out how many chiral centers does this molecule have and the rule is that the maximum number of stereoisomers of a compound is going to equal 2 to the power x where X is the number of chiral carbons in the compound now in this compound right here all right we have one chiral carbon this carbon has four different groups attached to it this is another chiral carbon and this is another chiral carbon there are three chiral carbons now this carbon is not chiral you notice it has two of the same groups attached to it remember a chiral Center is a carbon with four different groups attached to it there are only three chiral carbons here so then if we&amp;#39;re trying to answer how many stereoisomers this molecule has well it has three chiral centers so it&amp;#39;s going to be two to the power of three that&amp;#39;s two times two times two and that&amp;#39;s equal to eight so this molecule is going to have eight stereo isomers now here&amp;#39;s also a thing that sometimes can be tricky all right if a molecule has an internal plane of reflection then you have to worry about miso compounds now this molecule does not have an internal plane of reflection so you don&amp;#39;t have to worry about meso compounds in the case that there are miso compounds there might be less stereo isomers than you calculated all right so that&amp;#39;s the only catch that you should be wary of when you are using this formula to calculate stereoisomers this works for compounds that do not have an internal plane of reflection cool two is B three says which of the following compounds is optically inactive all right so chiral molecules are optically active we know this so what we&amp;#39;re looking for then here is we&amp;#39;re looking for a molecule that isn&amp;#39;t chiral or has no chiral centers or maybe has chiral centers but is secretly a miso compound because miso compounds are optically inactive now if we look at all these molecules all of them have two chiral centers and they&amp;#39;re all highlighted in blue here all right those are all carbons with four different groups attached to them so each answer Choice has two chiral centers so what now well then the answer must lie in the fact that one of these is really a miso compound now this this answer Choice all right the the answer is C all right and this answer Choice particularly we&amp;#39;re going to work backwards this way all right the answer choice is an example of a miso compound a compound that contains chiral centers one two but it has an internal plane of symmetry right there in the middle this side looks exactly like this side if there was a mirror placed right here all right and owing to this internal plane of symmetry the molecule is a chiral and hence optically inactive now had we started looking through the answer choices choices A and B are actually enantiomers of each other all right and you can notice you can notice that first of all they also don&amp;#39;t have a plane of internal reflection whether you draw the line there or the line here they are there&amp;#39;s no line or plane of symmetry I should say and the same goes for B and the same goes for D the only one that is a miso compound that has an internal plane of symmetry is C which is why the correct answer here is C since they all have chiral centers all right we&amp;#39;re looking at option two for finding which one is optically in active all right so three is C Fantastic Four says cholesterol shown below contains how many chiral centers a chiral Center is a carbon that has four different substituents four different groups attached to it if we look at this molecule we&amp;#39;ll just have to go through each carbon that is the workflow for this kind of problem and just count all the chiral carbons we have a chiral carbon right here it has four different groups these carbons one two three right here they all have two hydrogens so they&amp;#39;re not chiral this carbon there&amp;#39;s a double bond here so it&amp;#39;s double bonded to a carbon so there&amp;#39;s no four different groups attached to it so it&amp;#39;s not a chiral Center all right so we only have one here this one is a chiral Center that&amp;#39;s two so is this one three this one also has four different groups attached to it so four five six seven and eight all of these carbons have four different groups attached to them any of the other carbons here this one has two hydrogens this one has two hydrogens this one and this one have two hydrogens and so does this one so none of these carbons that I&amp;#39;ve crossed out these two are double bonded to each other so they&amp;#39;re attached to carbon twice so they don&amp;#39;t have four different groups attached to them this carbon has three hydrogens this carbon has two hydrogens this carbon is attached to two methyl groups so it&amp;#39;s also not a unique carbon all right and this carbon&amp;#39;s attached to three hydrogens so this methyl group is also not a chiral Center so this cholesterol molecule has eight chiral centers all right and that&amp;#39;s going to be answer Choice C so 4 is C beautiful five says which isomer of the following compound is the most stable all right and we&amp;#39;re looking at some chair conformations Here For A and B and then this is good old boat conformation so we&amp;#39;re looking at some chair conformations all right we know that for cyclohexane the chair conformation is the most stable so in just comparing the chair in this boat both out of the question here all right C is incredible because it is in the more unstable boat conformation and if C is incorrect so is D so now we&amp;#39;re stuck between A and B we have these two chair conformations all right we have these two chair confirmations and there are substituents at Carbon one and two for both of them now here in answer Choice a this methyl group is in an axial position and so is this methyl axial and axial in answer Choice B both of the methyl groups are in an equatorial position all right now the molecule that has both of these methyl groups in Equatorial position are going to be more stable all right we talked about this in the lecture how um axial substituents create more non-bonded stress and in cyclohexane molecules with multiple substituents the largest substituents will usually take the equatorial position to minimize strain so having both of these methyl groups in Equatorial position is going to minimize strain and it&amp;#39;s going to lead to a more stable conformation so the answer Choice here is B this molecule okay is a chair conformation again in which the two equatorial methyl groups are trans to each other all right and because the axial methyl hydrogens do not compete for the same space as the hydrogens attached to the ring this conformation ensures the least amount of steric strain Choice a is going to be less stable because those diaxial methyl group hydrogens are closer to the hydrogens on the ring and that&amp;#39;s going to cause greater steric strength all right so that&amp;#39;s another way of rephrasing that to think about it the answer choice for 5 here is B beautiful so yeah you can think about it as oh what are the where are the bulky groups is the bulky groups are in the equatorial position that&amp;#39;s better that creates a more stable configuration and you can also think about it as other interactions for example if your methyl group is right here in an axial position is going to have that 1 3 diaxial interaction within hydrogen group all right whereas a methyl group in the equatorial position won&amp;#39;t have that one three dioxyl interaction fantastic six says the following reaction results in blank a says retention of relative configuration and a change in the absolute configuration B says a change in the relative and absolute configurations C says retention of the relative and absolute configurations and D says retention of the absolute configuration and a change in the relative configuration so what we&amp;#39;re noticing in this reaction is um we are forming hydrochloric acid and then there&amp;#39;s these new bonds being formed to create this molecule now the relative configuration is retained because the bonds of the stereocenter are not broken all right so the relative configuration is retained because the bond of the stereocenter the chiral carbon are not broken thus the position of groups around the chiral carbon are maintained now the absolute configuration is also retained because both the reactant and the product are going to have the same configuration it&amp;#39;s R all right and we can double check that ourselves all right if we look at this chiral carbon and we number things appropriately with the appropriate priorities this is one this is two this is three and this is four all right your fourth group um is near a dashed line all right your fourth group is not on a dash but it&amp;#39;s near it so what we&amp;#39;re going to do is we&amp;#39;re going to figure out the one two three all right that gives us counterclockwise which is s and then we do the opposite of that remember that was our rule so this is r okay when the fourth group is not on a dash and you don&amp;#39;t know how to rotate figure out the one two three and it&amp;#39;s the opposite of whatever that is all right this was one of the rules that we covered in the lecture all right if we do the same thing here all right if we do the same thing here we have one two three four again um the same kind of numbering scheme and the fourth is not on the dash but it&amp;#39;s near it so we do the one two three that&amp;#39;s s and then we do the opposite so the relative configurations retained because the bonds on the chiral carbon are not broken and then the absolute configurations also retained because the configuration of that chiral centers are in both of the react in both the reactant and the product so the answer is going to be C retention of the relative and absolute configuration there&amp;#39;s retention in both the relative and the absolute configuration fantastic seven says the following molecules are considered to be blank all right these compounds are non-superimposable mirror images all right and you can make this analysis a bit easier we can rotate structure two 180 degrees and it&amp;#39;ll look like structure will look like this structure sorry I don&amp;#39;t know why I said two and one if you wanted to take this one and call it one and then this one call it two I should have written that before I started talking all right if you take this too and you rotate it 180 degrees all right you can begin to see and compare the two now if that&amp;#39;s hard and I personally don&amp;#39;t ever rely on my rotation ability especially when I am stressed in exam what we do what we can do is assign R and S configuration at these chiral centers all right and then start to compare all right start to compare these chiral centers now let&amp;#39;s look at these all right let&amp;#39;s look at these and let&amp;#39;s start figuring out how we want to approach this problem all right the way that we want to approach this problem is by first and foremost I&amp;#39;m just going to do a 180 degree on this molecule right here okay and so all that means is it&amp;#39;s going to look like it is turned 180 degrees so this is going to be ch3 here and this is going to be ch3 here so all we&amp;#39;re going to do is rotate we&amp;#39;re rotating it 180 degrees about this all right and we get the ch3 here now it&amp;#39;s no longer on a wedge it&amp;#39;s on a dash and it&amp;#39;s no longer on a dash it&amp;#39;s on a wedge now we can compare this and this appropriately all right this chiral Center is the same as this chiral Center and then this chiral Center is the same as this chiral center now just looking at it they look like enantiomers because on the purple sites one is on a dash and one is on a wedge and on the red sides one is a dash and also and the other ones on a wedge so they&amp;#39;re opposites of each other in in the at the purple site and they&amp;#39;re the Opposites of each other on the right side but to ensure that we&amp;#39;re right what we can do is determine the RNs configuration all right we can determine the r and s configuration of each site so if we look at this purple site right here all right we have a hydrogen that&amp;#39;s right here we have a methyl group right that&amp;#39;s right here this will be one all right this will be two this will be 3 and the hydrogen will be four and it&amp;#39;s on a dash so then all we have to do is figure out the one two three rotation that&amp;#39;s clockwise so this is r if we do the same here all right it&amp;#39;s going to be the same numbering but now one two three but now the hydrogen is not on a dash it&amp;#39;s on a wedge so we figure out the one two two three rotation and we do the opposite of it so this would be R but since the hydrogen is not on a dash but it&amp;#39;s near it we have to do the opposite so this is not r it&amp;#39;s going to be the opposite of that it will be s all right and then if we figure out the next site the red site all right this is going to be in um let&amp;#39;s do it let&amp;#39;s do our priorities here all right so at the red at the red or now blue dot all right let&amp;#39;s assign priority there&amp;#39;s a hydrogen right here this is going to be one this is going to be two the methyl group is three the hydrogen is four all right we&amp;#39;re going to figure out our one two three rotation that&amp;#39;s an S but because the hydrogen is not on a dash it&amp;#39;s next to a dash we have to do the opposite of that so it&amp;#39;s going to be R and then if we do the same thing here this is going to be an S so at the purple site their opposites are and S and at the at the blue site they&amp;#39;re also opposites they&amp;#39;re r and s all right so that means because they&amp;#39;re opposites at each of the chiral centers they&amp;#39;re going to be in near tumors all right fantastic so that means 7 is a beautiful let&amp;#39;s do eight 8 says plus glyceraldehyde and minus glyceraldehyde refer to the r and s forms of two three dihydroxy propanal respectively these molecules are considered blank all right so plus and minus glyceraldehyde or in other words R and S 2 3 dihydroxy propanol just based off of the naming gives it away our enantiomers one is has an R chiral Center the other one has an S and enantiomers are non-superimposable mirror images each has only one chiral Center and clearly within the name they&amp;#39;re given the opposite configuration one has R and one has s so they have the opposite absolute configurations in these molecules so they&amp;#39;re going to be enantiomers eight is a beautiful nine says consider e two butane and Z2 butane this is a pair of what type of isomers all right now E2 butene all right can also be called trans e refers to trans all right and Z refers to CIS so these can be also called trans tubutine and CIS to butene now as such they are CIS trans isomers that is correct and remember that CIS trans isomers are a subtype of diastereomers in which the position of the substituents differ about an immovable Bond and diastereomers are molecules that are non-mirage non-mirror image stereoisomers all right and so they are CIS and trans and they are diastereomers these are not enanti tumors because they are not mirror images of each other all right there are cysts and trans and their diastereomers all right and so the correct answer here is going to be C I don&amp;#39;t know why B is highlighted the correct answer here is going to be c one and two are true all right this e and Z2 butene are CIS and trans isomers and CIS and trans isomers are a category of diastereomers are a subtype of diastereomers all right so 9 is C beautiful let&amp;#39;s do 10 10 says three methyl pentane and hexane are related in that they are blank let&amp;#39;s go ahead and draw these out because they&amp;#39;re very easy three-methylpentane so pentane that&amp;#39;s five carbon chain and at the three position there&amp;#39;s a methyl and hexane is just a six carbon chain now the molecular formula for three-methylpentane is c6h14 and for hexane it&amp;#39;s c6h14 so h14 they both have the same molecular formula but clearly from the drawings they are they have completely different structures and so things that have the same molecular formula but are different in their atomic connectivity are called constitutional isomers all right so the answer to 10 here is going to be beautiful 11 says are two chloro S three bromobutane and S2 chloro S3 bromobutane are blank all right we don&amp;#39;t even have to draw these all right we don&amp;#39;t have to draw these because it&amp;#39;s telling us for this first name we have at the at the two chlora we have an r and at the three bromo it&amp;#39;s an s all right and then for this other molecule at the two chloro we have an S and at the three bromo we have an S all right so if we&amp;#39;re comparing the same chiral centers all right the things that we circled in green are at the same location in these two molecules all right and then things we highlight in blue are going to be oh sorry guys and the things we highlight in blue right here are going to be at the same chiral Center in these two two different molecules all right now they&amp;#39;re giving us the chiral centers at the same points in that same order for both of these molecules are and S for One S and S for the other now here at this first position their opposites R and S but here at the second position they&amp;#39;re the same we know that molecules that have a mix of opposite configurations at some chiral centers and the same configurations at other chiral centers is what defines diastereomers all right so the answer here is B these two molecules are stereoisomers of one another but they&amp;#39;re not non-superimposable mirror images therefore they are diastereomers all right and note that these molecules differ by at least one but not all chiral carbons okay 11 is B fantastic 12 says a scientist takes a 0.5 molarity solution of an unknown pure dextro rotary organic molecule and it places it in a test tube with a diameter of one centimeter he observes that a plane of polarized light is is rotated 12 degrees under these conditions what is the specific rotation of this molecule all right so remember that the equation for specific rotation is this one right here all right in this example this alpha alpha observed is plus 12 degrees all right and um remember that dextra rotary or clockwise rotation is considered positive which is why we can confidently say it&amp;#39;s plus 12 that&amp;#39;s what they tell us here all right Dexter rotary that gets a positive value all right we also know that the concentration is 0.5 molar and the path link is one centimeter which can be converted to decimeter which is what it should be in terms of units so things properly uh calculate for this kind of equation so one centimeter is just 0.1 this meter all right and then we can calculate from this setup the um specific rotation of this molecule we have plus 12 over 0.5 multiplied by 0.1 all right so we have 12 over 0.5 times 0.1 all right that&amp;#39;s 5 0.05 12 over 0.05 that&amp;#39;s going to be 240 degrees and it&amp;#39;s going to be positive all right so that&amp;#39;s going to be answer Choice D12 is d all right so this is an important formula to remember awesome 13 says um omprazole is a proton pump inhibitor commonly used in gastro esophageal reflux disease when Omeprazole erasmic mixture went off patent pharmaceutical companies began to manufacture esome prozal the S enantiomer of omoprozal by itself given one molar solution of Omeprazole and ESO esomeprazole which solution would likely exhibit Optical activity all right now Ray Smith mixtures like ohm prosal contain equal molar amounts of two enantiomers all right that&amp;#39;s what they&amp;#39;re describing here for Omeprazole all right it&amp;#39;s equal amounts of two enantiomers R and S for example all right and that means that they&amp;#39;re not going to have any observed Optical activity each of the two and anti mirrors causes rotation in opposite directions and so essentially their effects cancel out but what they&amp;#39;re saying here is that esomeprazole is this s enantiomer all right and so it only contains one of the two enantumers and it will cause rotation of plane polarized light all right and so the answer here is going to be B 13 is B antastic let&amp;#39;s go to 14 14 says 2r3s 2 3 dihydroxy butane dioic acid and 2s3r23 dihydroxybutane dioic acid are blank one says miso compounds two says the same molecule three says Indian tumors all right now we can draw out these structures all right we can draw out these structures the two names describe the same molecule all right now what we notice here is if we rotate about this Bond right here if we draw this out two three dihydroxybutan dioic acid if we rotate it well this molecule will look like is a molecule that will have an internal plane of symmetry all right a molecule that&amp;#39;s going to have a internal plane of symmetry so that&amp;#39;s an important thing to keep in mind all right so we can draw out these structures the two names describe the same molecule which also happens to be a meso compound because it does contain a plane of symmetry these compounds are not enantiomers because they are superimposable mirror images of one another not non-superimposable mirror images all right so these are this is going to be these are miso compounds they are the same molecule they&amp;#39;re not different all right and so the correct answer here is going to be c one and two are true statements fantastic and last but not least 15 says if the methyl group of butane are 120 degrees apart as seen in the Newman in the Newman projection this molecule is in its blank all right so if the methyl groups of butane are 120 degrees apart all right this molecule is in blank in butane the position at which the two methyl groups are 120 apart is in an eclipse conformation it&amp;#39;s not a totally eclipsed conformation but it&amp;#39;s just an eclipse conformation I&amp;#39;m going to go actually back to the notes where we drew out this all right right here actually oops I missed it no I didn&amp;#39;t go back up further sorry guys all the way over here all right when the two methyl groups are 120 degrees apart all right it&amp;#39;s going to be this conformation right here this is when they were 180 degrees and then when we rotated that back carbon 60 degrees now there 120 degrees apart this is an eclipsed conformation now this is an eclipse confirmation where we have methyl near hydrogens all right there are Eclipse with hydrogens not in the totally eclipsed conformation where the two methyl groups are on top of each other all right so this is an eclipse conformation this has a moderate amount of energy although not as high as a totally eclipsed conformation in which the two methyl groups are zero degrees apart and so it&amp;#39;s going to be a middle energy eclipsed form which means that 15 is going to be C all right and with that we&amp;#39;ve completed the problems for this set let me know if you have any questions comments concerns down below other than that good luck happy studying and have a beautiful beautiful day future doctors

Transcript for:Isomers Practice Problems

Transcript for:
Isomers Practice Problems