Monosaccharides - D and L Isomerism

All monosaccharides come in dextrorotary (D) and levorotary (L) forms. You may have heard these terms before when learning about optical activity. Another way to descrbise these D and L forms are enantiomers of each other.


Video Transcript

Now that we know some of the basics about monosaccharides, I want to focus on a new property which is the chirality, also known as the absolute configuration. Now, it turns out that some parts of monosaccharide chirality are actually much easier than chirality of other molecules, but then other elements of the chirality are really confusing and honestly a little bit messed up, so what IÕm going to do in this section is IÕm going to clear both of those sides of chirality for you, okay? So, letÕs go ahead and get started. So, monosaccharides can come in one or two forms, either in the dextroratory form, also known as the D form or the levoratory form, L form. And if you guys remember, these terms are borrowed from your chirality section of your text book, where we learnt about optical activity, and we learnt that dextrorotary molecules rotate like clockwise and levorotary molecules rotate like counter clockwise in a polarimeter. Remember that another property about optical active molecules is that opposite rotations were always enantiomers of each other, so the positive and the negative rotations were always going to be perfect enantiomers of each other and thatÕs a parallel that holds true for this section, the D enantiomer or the D sugar is always going to be an enantiomer of the old sugar, so this is true. Unfortunately, this is where the parallels from optical activity end for sugars, because it turns out that when sugars where discovered it was based on a lot of gas work by Emil Fischer back in the 1800Õs, he didnÕt have a polimeter, he didnÕt have all the monosaccharides, a lot of this was simply gas work. So, it turns out that he made this rules of D and L before he actually knew that there were a lot of exceptions to that rule, so it turns out that D doesnÕt specifically mean positive rotation anymore, L doesnÕt mean negative rotation, so youÕll be saying: ÒWell, Johnny then what does it mean?Ó Well, IÕm about to explain that too but just so you know there is no chiralation anymore between D being a positive rotation and L being a negative rotation, you should just think of them as two categories of sugars, okay? Now, the way you determine D or the way you determine L on a monosaccharide is decided by the penultimate carbon, such a fancy word, what does that mean? Penultimate is just a word that means second to last, okay? So, the penultimate carbon is simply going to be the last chiral carbon of your molecule and thatÕs going to determine if itÕs a D or itÕs an L. Now, notice of this penultimate carbon is really important, itÕs the C-5 carbon usually but it depends on the length of the monosaccharide, in a six carbon sugar itÕs going to be the C-5 but itÕs very important, itÕs going to be used as what we call stereo descriptor, hopefully I spell this right, stereo descriptor later in this chapter. So, these two words are going to be synonyms of each other; penultimate carbon and stereo descriptor carbon are synonyms of each other, so just keep that in mind, okay? I am going to use both of those words interchangeable. So, how do you know if it is a D or an L? Well, thankfully this is one of the really easy parts of monosaccharide chirality, all that is is this: the D configuration is going to be any sugar that has the penultimate or stereo descriptor OH pointed to the right, okay? So, letÕs look at the glucose here. Glucose which-, OH is the stereo descriptor, itÕs going to be the last chiral carbon, so itÕs going to be this one right here, that is the last chiral carbon, itÕs the one thatÕs furthest down, like in it happens to be C-5 here, we are going to learn how to number a carbon soon or number a sugar soon, but in this case this would be 1, 2, 3, 4, 5, okay? So, thatÕs my penultimate carbon or my stereo descriptor and we notice that if it faces to the right, thatÕs going to be a D-glucose, so IÕm going to-, just put that right there, this is D glucose, and then if it faces to the left, thatÕs going to be an L-glucose. And that makes it so easy guys, because you can just remember that L is left, okay? D is right, L is left, you have no excuse to mess that up, because the LÕs itÕs a double L, okay? Now guys, it turns out that usually that means that itÕs going to be, the D is going to be an R configuration on that last sugar or on that last chiral center, and usually L is going to be an S at that position, okay? But there are exceptions to that as well, so just to you know if this works for all sugars but for other molecules, biomolecules like amino acids, itÕs not going to work anymore. So, for right now you can use that rule but the rule that I really want you to remember is just right and left, that always holds true, okay? So, let me just kind of prove to you that R and S works here, but then just donÕt worry about it for other types of molecules. So for example, hereÕs this chiral center and what I would see is that-, let me just erase these numbers, so that we can pick our priorities, so our priorities would be 1, the carbon-, I mean the O is number 1, then this must be number 2, this must be number 3 and this must be priority number 4, so that means that my rotation goes from 1 to 2, from 2 to 3, I ignore 4, it looks like itÕs an S but since my 4 is on the horizontal, that means thatÕs going to be an R, okay? So, remember guys, what IÕm doing is IÕm using the rules of determining chirality of Fischer projections to determine thatÕs an R. So, what IÕm doing here is IÕm showing you guys that yes, the D happens to be an R but is not always work for other types of biomolecules, okay? The one thing that you should always remember is that to the right is D and to the left is L, and thatÕs always going to hold true, cool. Alright guys, so weÕre done with this concept, letÕs move on to a practice problem.

Problem: Provide the generic name, including stereochemistry, for the following monosaccharide:


Monosaccharides - D and L Isomerism Additional Practice Problems

The bronchodilator ephedrine is erythro -2-(methylamino)-1-phenylpropan-1-ol. The decongestant

pseudoephedrine is threo -2-(methylamino)-1-phenylpropan-1-ol.

(a) Draw the four stereoisomers of 2-(methylamino)-1-phenylpropan-1-ol, either as Fischer projections or as three-dimensional representations (dotted lines and wedges).

(b) Label ephedrine and pseudoephedrine. What is the relationship between them?

(c) Label the D and L isomers of ephedrine and pseudoephedrine using the Fischer–Rosanoff convention.

(d) Both ephedrine and pseudoephedrine are commonly used as racemic mixtures. Ephedrine is also available as the pure levorotatory 1-2 isomer (Biophedrine®), and pseudoephedrine is also available as the more active 1+2 isomer (Sudafed®). Can you label the 1-2 isomer of ephedrine and the 1+2 isomer of pseudoephedrine?

Watch Solution

Which configuration (R or S) does the bottom asymmetric carbon have for the D series of sugars? Which configuration for the L series?

Watch Solution

Draw and name the enantiomers of the sugars shown in Figure 23-2. Give the relative configuration (D or L) and the sign of the rotation in each case.

Watch Solution

What are the systematic names of the following compounds? Indicate the configuration (R or S)of each asymmetric center.

a. D-glucose

b. D-mannose

c. D-galactose

d. L-glucose

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a. Are D-erythrose and L-erythrose enantiomers or diastereomers?

b. Are L-erythrose and L-threose enantiomers or diastereomers?

Watch Solution

Indicate whether each of the following structures is d -glyceraldehyde or l -glyceraldehyde, assuming that the horizontal bonds point toward you and the vertical bonds point away from you ( Section 4.6 ) :

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The structure of D-mannose is shown below. Draw the structure of L-mannose. 

Watch Solution

Pick the ‘D’ sugar.

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Given that compound A is D-glucose, which of the following structures corresponds to L-glucose?

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Which stereochemical description best fits the following carbohydrate?

1) (+)–D–4–(R)

2) (–)-L–4–(S)

3) (–)–D–4–(R)

4) (+)–L–4–(S)

5) D–4–(R) You can’t determine +/– from a chemical structure.

Watch Solution