{"id":2484,"date":"2018-06-19T20:32:56","date_gmt":"2018-06-19T20:32:56","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/suny-potsdam-organicchemistry\/chapter\/3-6-compounds-with-multiple-chiral-centers\/"},"modified":"2018-08-06T11:39:55","modified_gmt":"2018-08-06T11:39:55","slug":"4-4-molecules-with-multiple-chiral-centers","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-potsdam-organicchemistry\/chapter\/4-4-molecules-with-multiple-chiral-centers\/","title":{"raw":"4.4. Molecules with multiple chiral centers","rendered":"4.4. Molecules with multiple chiral centers"},"content":{"raw":"
So far, we have been analyzing compounds with a single chiral center.\u00a0 Next, we turn our attention to those which have multiple chiral centers. We'll start with some stereoisomeric four-carbon sugars with two chiral centers.\r\n

\"\"<\/p>\r\nTo avoid confusion, we will simply refer to the different stereoisomers by capital letters.\r\n\r\nLook first at compound A below.\u00a0 Both chiral centers in have the R<\/em> configuration (you should confirm this for yourself!).\u00a0 The mirror image of Compound A is compound B, which has the S <\/em>configuration at both chiral centers.\u00a0 If we were to pick up compound A, flip it over and put it next to compound B, we would see that they are not<\/em> superimposable (again, confirm this for yourself with your models!).\u00a0 A and B are nonsuperimposable mirror images: in other words, enantiomers.\r\n

\"\"<\/p>\r\nNow, look at compound C, in which the configuration is S<\/em> at chiral center 1 and R<\/em> at chiral center 2.\u00a0 Compounds A and C are stereoisomers: they have the same molecular formula and the same bond connectivity, but a different arrangement of atoms in space (recall that this is the definition of the term 'stereoisomer).\u00a0 However, they are not<\/em> mirror images of each other (confirm this with your models!), and so they are not<\/em> enantiomers.\u00a0 By definition, they are diastereomers <\/strong>of each other.\r\n\r\nNotice that compounds C and B also have a diastereomeric relationship, by the same definition.\r\n\r\nSo, compounds A and B are a pair of enantiomers, and compound C is a diastereomer of both of them.\u00a0 Does compound C have its own enantiomer?\u00a0 Compound D is the mirror image of compound C, and the two are not superimposable.\u00a0 Therefore, C and D are a pair of enantiomers.\u00a0 Compound D is also a diastereomer of compounds A and B.\r\n\r\nThis can also seem very confusing at first, but there some simple shortcuts to analyzing stereoisomers:\r\n

\r\n

Stereoisomer shortcuts<\/strong><\/p>\r\nIf all<\/em><\/strong> of the chiral centers are of opposite R\/S configuration between two stereoisomers, they are enantiomers.\r\n\r\nIf at least one, but not all<\/em><\/strong> of the chiral centers are opposite between two stereoisomers, they are diastereomers.\r\n\r\n(Note: these shortcuts to not take into account the possibility of additional stereoisomers due to alkene groups)<\/em>\r\n\r\n<\/div>\r\nHere's another way of looking at the four stereoisomers, where one chiral center is associated with red and the other blue.\u00a0 Pairs of enantiomers are stacked together.\r\n

\"\"<\/p>\r\nWe know, using the shortcut above, that the enantiomer of RR must be SS - both chiral centers are different.\u00a0 We also know that RS and SR are diastereomers of RR, because in each case one - but not both - chiral centers are different.\r\n

Meso forms<\/h3>\r\nIf the two chiral centers in a molecule contain the same four groups, then it is possible that one center is just an \"internal reflection\" of the other.\u00a0 (Since the groups are the same, we will color both of them red when we indicate R\/S.)\u00a0 This produces an internal mirror plane, which means that the overall molecule is achiral (despite the two chiral centers!).\u00a0 Such forms are referred to as meso<\/em><\/strong>.\u00a0 For example, with butane-2,3-diol we may have a stereoisomer where the left hand side of the molecule matches the right hand side, shown as SR in the diagram.\u00a0 We also have a mirror image form which is RS, but if you look carefully you will notice that this is simply the SR isomer flipped upside down.\u00a0 Thus both forms are identical and it (they) is meso.\r\n\r\nIn addition, there are two chiral stereoisomers - the SS and the RR isomers.\u00a0 These are chiral, and not meso.\r\n\r\n\"\"\r\n

More than two chiral centers<\/h3>\r\nNow, let's extend our analysis to a sugar molecule with three chiral centers.\u00a0 Going through all the possible combinations, we come up with eight total stereoisomers - four pairs of enantiomers.\r\n

\"\"<\/p>\r\nLet's draw the R<\/i><\/span>R<\/span>R stereoisomer. Being careful to draw the wedge bonds correctly so that they match the RRR<\/i> configurations, we get:\r\n

\"\"<\/p>\r\nNow, using the above drawing as our model, drawing any other stereoisomer is easy.\u00a0 If we want to draw the enantiomer of R<\/em>RR, we don't need to try to visualize the mirror image, we just start with the RRR<\/em> structure and invert the configuration at every<\/em> chiral center to get SSS<\/em>.\r\n

\"\"<\/p>\r\nTry making models of R<\/em>RR and SSS<\/em> and confirm that they are in fact nonsuperimposable mirror images of each other.\r\n\r\nThere are six diastereomers of R<\/em>RR.\u00a0 To draw one of them, we just invert the configuration of at least one, but not all three, of the chiral centers.\u00a0 Let's invert the configuration at chiral center 1 and 2, but leave chiral center 3 unchanged.\u00a0 This gives us the SSR<\/em> configuration.\r\n

\"\"<\/p>\r\nOne more definition at this point: diastereomers which differ at only a single chiral center are called epimers<\/strong>.\u00a0 For example, R<\/em>RR and SRR <\/em>are epimers:\r\n

\"\"<\/p>\r\nThe RRR and SSR<\/em> stereoisomers shown earlier are diastereomers but not<\/em> epimers because they differ at two<\/em> of the three chiral centers.\r\n

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\r\n

Exercise<\/h3>\r\n1. a) Draw the structure of the enantiomer<\/em> of the S<\/em>RS stereoisomer of the sugar used in the previous example.\r\n\r\nb) List (using the X<\/em>XX format, not drawing the structures) all of the epimers of SRS.<\/em>\r\n\r\nc) List all of the stereoisomers that are diastereomers, but not epimers, of SRS.\r\n\r\n[reveal-answer q=\"591622\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"591622\"]\r\n\r\na) Starting with the RRR stereoisomer (which is given in the example), we flip the first and third chiral center to get SRS.\u00a0 The enantiomer of the SRS stereoisomer is that in which all three chiral centers are flipped: the RSR stereoismer.<\/span>\r\n\r\n \r\n

\"\"<\/span><\/p>\r\nb) Epimers of the SRS stereoismer are RRS, SSS, and SRR (in each case, one of the three chiral centers has been flipped)<\/span>\r\n\r\nc) How to find the compounds that are diastereomers of the SRS stereoisomer, but not epimers? Start with the list of the eight possible stereoisomers given in the example.\u00a0 Cross out SRS itself, and its enantiomer RSR (determined in part (a) above).\u00a0 Then cross out the three epimers we found in part (b).\u00a0 We are left with three isomers: RRR, SSR, and RSS. Each of these have one chiral center in common with SRS, and two that are flipped.<\/span>\r\n

\"\"<\/p>\r\n

[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\nThe epimer term is useful because in biochemical pathways, compounds with multiple chiral centers are isomerized at one specific center by enzymes known as epimerases<\/strong>.\u00a0 Two examples of epimerase-catalyzed reactions are below.\r\n

\"\"<\/p>\r\nWe know that enantiomers have identical physical properties and equal but opposite magnitude specific rotation.\u00a0 Diastereomers, in theory at least, have different<\/em> physical properties \u2013 we stipulate \u2018in theory\u2019 because sometimes the physical properties of two or more diastereomers are so similar that it is very difficult to distinguish between them.\u00a0 In addition, the specific rotation values of diastereomers are unrelated \u2013 they could be the same sign or opposite signs, similar in magnitude or very dissimilar.\r\n

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Exercises<\/h3>\r\n1. The sugar below is one of the stereoisomers that we have been discussing.\r\n

\"\"<\/p>\r\nThe only problem is, it is drawn with the carbon backbone in a different orientation from what we have seen.\u00a0 Determine the configuration at each chiral center to determine which stereoisomer it is.\r\n\r\n[reveal-answer q=\"953111\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"953111\"]\r\n\r\nThis is the SRR stereoisomer:\r\n

\"\"<\/span><\/p>\r\n[\/hidden-answer]\r\n\r\n2. Draw the enantiomer of the xylulose-5-phosphate structure in the previous figure.\r\n\r\n[reveal-answer q=\"519396\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"519396\"]\"\"<\/span>[\/hidden-answer]\r\n\r\n3. The structure of the amino acid D<\/span>-threonine, drawn without stereochemistry,\u00a0 is shown below. D<\/span>-threonine has the (S) configuration at both of its chiral centers.\u00a0 Draw D<\/span>-threonine, it's enantiomer, and its two diastereomers.\r\n

\"\"<\/p>\r\n[reveal-answer q=\"944385\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"944385\"]\"\"<\/span>[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\nHere is some more practice in identifying isomeric relationships. D<\/span>-glucose is the monosaccharide that serves as the entrance point for the glycolysis pathway and as a building block for the carbohydrate biopolymers starch and cellulose.\u00a0 The open-chain structure of the sugar is shown below.\r\n

\"\"<\/p>\r\nBecause D<\/span>-glucose has four chiral centers, it can exist in a total of 24<\/sup> = 16 different stereoisomeric forms: it has one enantiomer and 14 diastereomers.\r\n\r\nNow, let's compare the structures of the two sugars D<\/span>-glucose and D<\/span>-gulose, and try to determine their relationship.\r\n

\"\"<\/p>\r\nThe two structures have the same molecular formula and the same connectivity, therefore they must be stereoisomers.\u00a0 They each have four chiral centers, and the configuration is different at two of these centers (at carbons #3 and #4).\u00a0 They are diastereomers.\r\n\r\nNow, look at the structures of D<\/span>-glucose and D<\/span>-mannose.\r\n

\"\"<\/p>\r\nHere, everything is the same except for the configuration of the chiral center at carbon #2.\u00a0 The two sugars differ at only one of the four chiral centers, so again they are diastereomers, and more specifically they are epimers.\r\n\r\nD<\/span>-glucose and L<\/span>-glucose are enantiomers, because they differ at all four chiral centers.\r\n

\"\"<\/p>\r\nD<\/span>-glucose is the enantiomer commonly found in nature.\u00a0 D<\/span>-glucose and D<\/span>-fructose are not stereoisomers, because they have different bonding connectivity: glucose has an aldehyde group, while fructose has a ketone.\u00a0 The two sugars do, however, have the same molecular formula, so by definition they are constitutional isomers.\r\n

\"\"<\/p>\r\nD<\/span>-glucose and D<\/span>-ribose are not isomers of any kind, because they have different molecular formulas.\r\n

\"\"<\/p>\r\n\r\n

<\/div>\r\n
\r\n
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\r\n

Exercises<\/h3>\r\n 1. Identify the relationship between each pair of structures.\u00a0 Your choices are: not isomers, constitutional isomers, diastereomers but not epimers, epimers, enantiomers, or same molecule\"\"<\/span>\r\n[reveal-answer q=\"29238\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"29238\"]Going left to right, top to bottom:<\/span><\/span>\r\n

Diastereomers (two chiral centers are flipped).<\/span><\/p>\r\n

Enantiomers (all five chiral centers are flipped).<\/span><\/p>\r\n

Identical (drawing is flipped vertically but they are the same structure).<\/span><\/p>\r\n

Identical (the carbon which appears to be flipped in the drawing is not<\/em> a chiral center).<\/span><\/p>\r\n

Constitutional isomers (same molecular formula, but notice that inositol does not have a ring oxygen.\u00a0 Is not a monosaccharide, it is a cyclohexane with six hydroxyl substituents.)<\/span><\/p>\r\n[\/hidden-answer]<\/span>\r\n\r\n2. <\/span>Identify the relationship between each pair of structures. \u00a0Hint<\/em> - figure out the configuration of each chiral center\"\"[reveal-answer q=\"698380\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"698380\"]\r\n\r\na) Identical.\u00a0 They have the same molecular formula and connectivity; there is only one chiral center, which is R <\/em>in both structures.<\/span>\r\n\r\nb) Enantiomers. The compound on the left is R<\/em>, compound on the right is S<\/em>. <\/span><\/span>\r\n\r\nc) Enantiomers.\u00a0 The compound on the left is R<\/em>, compound on the right is S<\/em>. <\/span><\/span><\/span>\r\n\r\nd) Enantiomers.\u00a0 The compound on the left is SS<\/em>, compound on the right is RR<\/em>.\u00a0\u00a0 <\/span><\/span><\/span><\/span>\r\n\r\ne) Identical.\u00a0 The structures are both glycerol, which is not<\/em> chiral (the left and right 'arms' are the same, so the middle carbon is not a chiral center)\u00a0 <\/span><\/span><\/span><\/span><\/span>\r\n\r\nf) Identical.\u00a0 Both structures are SS<\/em>. Also, notice that if you rotate the right-side structure 120 degrees clockwise, it becomes exactly the same as the left-side structure.[\/hidden-answer]<\/span><\/span><\/span><\/span><\/span><\/span>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\nKhan Academy video tutorial on stereoisomeric relationships\"\"<\/span>\r\n\r\nhttps:\/\/youtu.be\/aYp39AaZqvI\r\n\r\n<\/section>","rendered":"

So far, we have been analyzing compounds with a single chiral center.\u00a0 Next, we turn our attention to those which have multiple chiral centers. We’ll start with some stereoisomeric four-carbon sugars with two chiral centers.<\/p>\n

\"\"<\/p>\n

To avoid confusion, we will simply refer to the different stereoisomers by capital letters.<\/p>\n

Look first at compound A below.\u00a0 Both chiral centers in have the R<\/em> configuration (you should confirm this for yourself!).\u00a0 The mirror image of Compound A is compound B, which has the S <\/em>configuration at both chiral centers.\u00a0 If we were to pick up compound A, flip it over and put it next to compound B, we would see that they are not<\/em> superimposable (again, confirm this for yourself with your models!).\u00a0 A and B are nonsuperimposable mirror images: in other words, enantiomers.<\/p>\n

\"\"<\/p>\n

Now, look at compound C, in which the configuration is S<\/em> at chiral center 1 and R<\/em> at chiral center 2.\u00a0 Compounds A and C are stereoisomers: they have the same molecular formula and the same bond connectivity, but a different arrangement of atoms in space (recall that this is the definition of the term ‘stereoisomer).\u00a0 However, they are not<\/em> mirror images of each other (confirm this with your models!), and so they are not<\/em> enantiomers.\u00a0 By definition, they are diastereomers <\/strong>of each other.<\/p>\n

Notice that compounds C and B also have a diastereomeric relationship, by the same definition.<\/p>\n

So, compounds A and B are a pair of enantiomers, and compound C is a diastereomer of both of them.\u00a0 Does compound C have its own enantiomer?\u00a0 Compound D is the mirror image of compound C, and the two are not superimposable.\u00a0 Therefore, C and D are a pair of enantiomers.\u00a0 Compound D is also a diastereomer of compounds A and B.<\/p>\n

This can also seem very confusing at first, but there some simple shortcuts to analyzing stereoisomers:<\/p>\n

\n

Stereoisomer shortcuts<\/strong><\/p>\n

If all<\/em><\/strong> of the chiral centers are of opposite R\/S configuration between two stereoisomers, they are enantiomers.<\/p>\n

If at least one, but not all<\/em><\/strong> of the chiral centers are opposite between two stereoisomers, they are diastereomers.<\/p>\n

(Note: these shortcuts to not take into account the possibility of additional stereoisomers due to alkene groups)<\/em><\/p>\n<\/div>\n

Here’s another way of looking at the four stereoisomers, where one chiral center is associated with red and the other blue.\u00a0 Pairs of enantiomers are stacked together.<\/p>\n

\"\"<\/p>\n

We know, using the shortcut above, that the enantiomer of RR must be SS – both chiral centers are different.\u00a0 We also know that RS and SR are diastereomers of RR, because in each case one – but not both – chiral centers are different.<\/p>\n

Meso forms<\/h3>\n

If the two chiral centers in a molecule contain the same four groups, then it is possible that one center is just an “internal reflection” of the other.\u00a0 (Since the groups are the same, we will color both of them red when we indicate R\/S.)\u00a0 This produces an internal mirror plane, which means that the overall molecule is achiral (despite the two chiral centers!).\u00a0 Such forms are referred to as meso<\/em><\/strong>.\u00a0 For example, with butane-2,3-diol we may have a stereoisomer where the left hand side of the molecule matches the right hand side, shown as SR in the diagram.\u00a0 We also have a mirror image form which is RS, but if you look carefully you will notice that this is simply the SR isomer flipped upside down.\u00a0 Thus both forms are identical and it (they) is meso.<\/p>\n

In addition, there are two chiral stereoisomers – the SS and the RR isomers.\u00a0 These are chiral, and not meso.<\/p>\n

\"\"<\/p>\n

More than two chiral centers<\/h3>\n

Now, let’s extend our analysis to a sugar molecule with three chiral centers.\u00a0 Going through all the possible combinations, we come up with eight total stereoisomers – four pairs of enantiomers.<\/p>\n

\"\"<\/p>\n

Let’s draw the R<\/i><\/span>R<\/span>R stereoisomer. Being careful to draw the wedge bonds correctly so that they match the RRR<\/i> configurations, we get:<\/p>\n

\"\"<\/p>\n

Now, using the above drawing as our model, drawing any other stereoisomer is easy.\u00a0 If we want to draw the enantiomer of R<\/em>RR, we don’t need to try to visualize the mirror image, we just start with the RRR<\/em> structure and invert the configuration at every<\/em> chiral center to get SSS<\/em>.<\/p>\n

\"\"<\/p>\n

Try making models of R<\/em>RR and SSS<\/em> and confirm that they are in fact nonsuperimposable mirror images of each other.<\/p>\n

There are six diastereomers of R<\/em>RR.\u00a0 To draw one of them, we just invert the configuration of at least one, but not all three, of the chiral centers.\u00a0 Let’s invert the configuration at chiral center 1 and 2, but leave chiral center 3 unchanged.\u00a0 This gives us the SSR<\/em> configuration.<\/p>\n

\"\"<\/p>\n

One more definition at this point: diastereomers which differ at only a single chiral center are called epimers<\/strong>.\u00a0 For example, R<\/em>RR and SRR <\/em>are epimers:<\/p>\n

\"\"<\/p>\n

The RRR and SSR<\/em> stereoisomers shown earlier are diastereomers but not<\/em> epimers because they differ at two<\/em> of the three chiral centers.<\/p>\n

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\n

Exercise<\/h3>\n

1. a) Draw the structure of the enantiomer<\/em> of the S<\/em>RS stereoisomer of the sugar used in the previous example.<\/p>\n

b) List (using the X<\/em>XX format, not drawing the structures) all of the epimers of SRS.<\/em><\/p>\n

c) List all of the stereoisomers that are diastereomers, but not epimers, of SRS.<\/p>\n

Show Solution<\/span><\/p>\n
\n

a) Starting with the RRR stereoisomer (which is given in the example), we flip the first and third chiral center to get SRS.\u00a0 The enantiomer of the SRS stereoisomer is that in which all three chiral centers are flipped: the RSR stereoismer.<\/span><\/p>\n

 <\/p>\n

\"\"<\/span><\/p>\n

b) Epimers of the SRS stereoismer are RRS, SSS, and SRR (in each case, one of the three chiral centers has been flipped)<\/span><\/p>\n

c) How to find the compounds that are diastereomers of the SRS stereoisomer, but not epimers? Start with the list of the eight possible stereoisomers given in the example.\u00a0 Cross out SRS itself, and its enantiomer RSR (determined in part (a) above).\u00a0 Then cross out the three epimers we found in part (b).\u00a0 We are left with three isomers: RRR, SSR, and RSS. Each of these have one chiral center in common with SRS, and two that are flipped.<\/span><\/p>\n

\"\"<\/p>\n

<\/div>\n<\/div>\n<\/div>\n<\/div>\n

The epimer term is useful because in biochemical pathways, compounds with multiple chiral centers are isomerized at one specific center by enzymes known as epimerases<\/strong>.\u00a0 Two examples of epimerase-catalyzed reactions are below.<\/p>\n

\"\"<\/p>\n

We know that enantiomers have identical physical properties and equal but opposite magnitude specific rotation.\u00a0 Diastereomers, in theory at least, have different<\/em> physical properties \u2013 we stipulate \u2018in theory\u2019 because sometimes the physical properties of two or more diastereomers are so similar that it is very difficult to distinguish between them.\u00a0 In addition, the specific rotation values of diastereomers are unrelated \u2013 they could be the same sign or opposite signs, similar in magnitude or very dissimilar.<\/p>\n

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\n

Exercises<\/h3>\n

1. The sugar below is one of the stereoisomers that we have been discussing.<\/p>\n

\"\"<\/p>\n

The only problem is, it is drawn with the carbon backbone in a different orientation from what we have seen.\u00a0 Determine the configuration at each chiral center to determine which stereoisomer it is.<\/p>\n

Show Solution<\/span><\/p>\n
\n

This is the SRR stereoisomer:<\/p>\n

\"\"<\/span><\/p>\n<\/div>\n<\/div>\n

2. Draw the enantiomer of the xylulose-5-phosphate structure in the previous figure.<\/p>\n

Show Solution<\/span><\/p>\n
\"\"<\/span><\/div>\n<\/div>\n

3. The structure of the amino acid D<\/span>-threonine, drawn without stereochemistry,\u00a0 is shown below. D<\/span>-threonine has the (S) configuration at both of its chiral centers.\u00a0 Draw D<\/span>-threonine, it’s enantiomer, and its two diastereomers.<\/p>\n

\"\"<\/p>\n

Show Solution<\/span><\/p>\n
\"\"<\/span><\/div>\n<\/div>\n<\/div>\n<\/div>\n

Here is some more practice in identifying isomeric relationships. D<\/span>-glucose is the monosaccharide that serves as the entrance point for the glycolysis pathway and as a building block for the carbohydrate biopolymers starch and cellulose.\u00a0 The open-chain structure of the sugar is shown below.<\/p>\n

\"\"<\/p>\n

Because D<\/span>-glucose has four chiral centers, it can exist in a total of 24<\/sup> = 16 different stereoisomeric forms: it has one enantiomer and 14 diastereomers.<\/p>\n

Now, let’s compare the structures of the two sugars D<\/span>-glucose and D<\/span>-gulose, and try to determine their relationship.<\/p>\n

\"\"<\/p>\n

The two structures have the same molecular formula and the same connectivity, therefore they must be stereoisomers.\u00a0 They each have four chiral centers, and the configuration is different at two of these centers (at carbons #3 and #4).\u00a0 They are diastereomers.<\/p>\n

Now, look at the structures of D<\/span>-glucose and D<\/span>-mannose.<\/p>\n

\"\"<\/p>\n

Here, everything is the same except for the configuration of the chiral center at carbon #2.\u00a0 The two sugars differ at only one of the four chiral centers, so again they are diastereomers, and more specifically they are epimers.<\/p>\n

D<\/span>-glucose and L<\/span>-glucose are enantiomers, because they differ at all four chiral centers.<\/p>\n

\"\"<\/p>\n

D<\/span>-glucose is the enantiomer commonly found in nature.\u00a0 D<\/span>-glucose and D<\/span>-fructose are not stereoisomers, because they have different bonding connectivity: glucose has an aldehyde group, while fructose has a ketone.\u00a0 The two sugars do, however, have the same molecular formula, so by definition they are constitutional isomers.<\/p>\n

\"\"<\/p>\n

D<\/span>-glucose and D<\/span>-ribose are not isomers of any kind, because they have different molecular formulas.<\/p>\n

\"\"<\/p>\n

<\/div>\n
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Exercises<\/h3>\n

1. Identify the relationship between each pair of structures.\u00a0 Your choices are: not isomers, constitutional isomers, diastereomers but not epimers, epimers, enantiomers, or same molecule\"\"<\/span><\/p>\n

Show Solution<\/span><\/p>\n
Going left to right, top to bottom:<\/span><\/span><\/p>\n

Diastereomers (two chiral centers are flipped).<\/span><\/p>\n

Enantiomers (all five chiral centers are flipped).<\/span><\/p>\n

Identical (drawing is flipped vertically but they are the same structure).<\/span><\/p>\n

Identical (the carbon which appears to be flipped in the drawing is not<\/em> a chiral center).<\/span><\/p>\n

Constitutional isomers (same molecular formula, but notice that inositol does not have a ring oxygen.\u00a0 Is not a monosaccharide, it is a cyclohexane with six hydroxyl substituents.)<\/span><\/p>\n

<\/div>\n<\/div>\n

<\/span><\/p>\n

2. <\/span>Identify the relationship between each pair of structures. \u00a0Hint<\/em> – figure out the configuration of each chiral center\"\"<\/p>\n

Show Solution<\/span><\/p>\n
\n

a) Identical.\u00a0 They have the same molecular formula and connectivity; there is only one chiral center, which is R <\/em>in both structures.<\/span><\/p>\n

b) Enantiomers. The compound on the left is R<\/em>, compound on the right is S<\/em>. <\/span><\/span><\/p>\n

c) Enantiomers.\u00a0 The compound on the left is R<\/em>, compound on the right is S<\/em>. <\/span><\/span><\/span><\/p>\n

d) Enantiomers.\u00a0 The compound on the left is SS<\/em>, compound on the right is RR<\/em>.\u00a0\u00a0 <\/span><\/span><\/span><\/span><\/p>\n

e) Identical.\u00a0 The structures are both glycerol, which is not<\/em> chiral (the left and right ‘arms’ are the same, so the middle carbon is not a chiral center)\u00a0 <\/span><\/span><\/span><\/span><\/span><\/p>\n

f) Identical.\u00a0 Both structures are SS<\/em>. Also, notice that if you rotate the right-side structure 120 degrees clockwise, it becomes exactly the same as the left-side structure.<\/div>\n<\/div>\n

<\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n<\/div>\n<\/div>\n<\/div>\n

Khan Academy video tutorial on stereoisomeric relationships\"\"<\/span><\/p>\n