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I am really finding it difficult to identify the correct configuration(s) of a given monosaccharide, in fact my friends and I have been trying to solve this, much to our annoyance.

In one of our classes, our teacher said, in cyclic form:

If $\ce{CH2OH}$ is below the ring its configuration is L, and if its above the ring its D.
If $\ce{C1}$'s $\ce{OH}$ is below the ring its alpha configuration and if its above its beta configuration .

My friends seem to agree with the above explanation, however my understanding is:

If all groups of $\ce{OH}$ and $\ce{-CH2OH}$ is reversed from D form, its L.
If $\ce{C1}$'s $\ce{OH}$ is on the same side of the ring as $\ce{-CH2OH}$ its beta else its alpha.

So I want to know if these assumptions are right:

  1. D and L in linear form:

    • If the last chiral carbon's OH (one before the last carbon) is on the right its D and if its on the left its L. so if you want to make for example L glucose from D, you should reverse ALL groups.

  2. D and L in cyclic form:

    • It's hard mainly but can be deduced by making linear form into cyclic and cyclic into linear and comparing them or reversing all carbon branches.

  3. Alpha and Beta in cyclic form:

    • If the carboxyl carbon's $\ce{OH}$ is on the same side of last ring carbons branch ($\ce{CH2OH}$ in glucose, $\ce{COO-}$ in glucoronate) it's neta, else it's alpha. So in D form, if $\ce{OH}$ is above it's beta else it is alpha. In L form, if $\ce{OH}$ is above its alpha else its beta.

Also when i see 3D linear form of Glucose it doesn't match with 2D at all: 2D view of this site isn't like what I find in books I can't find the match between 2D and 3D view neither.

I'm still confused because of following examples:

1-Alpha-L-Glucose
enter image description here

2-Alpha-D-Glucose
enter image description here

3-Beta-D-Glucuronate
enter image description here

4-Beta-L-Iduronic Acid
enter image description here

these just don't match.

also Harper biochemistry image is this:
enter image description here

but lehningers is this:
enter image description here

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It’s easy to understand your frustrations, however you need to understand the basic stereochemistry and conformations of sugars.

A monosaccharide is designated D if the hydroxyl group on the highest numbered asymmetric carbon is drawn to the right in a Fischer projection, as in D-glyceraldehyde:

enter image description here

Note that the designation D or L merely relates the configuration of a given molecule to that of glyceraldehyde and does not specify the sign of rotation of plane-polarized light.

As I covered this explanation already on one of my posts:Are glucose and galactose cis-trans isomers of each other?

D sugars have the same absolute configuration at the asymmetric center farthest removed from their carbonyl group as does D-glyceraldehyde. The L sugars, in accordance with this convention, are mirror images of their D counterparts

Cyclic forms

The designation α indicates that the hydroxyl group at the anomeric center (hemiacetal or hemiketal carbon) is, in a Fischer projection, on the same side as the hydroxyl attached at the farthest chiral center, whereas β indicates that these hydroxyl groups are on opposite sides:

enter image description here

In the α anomer, the $\ce{OH}$ substituent to the anomeric carbon is on the opposite side of the sugar ring from the $\ce{CH2OH}$ group at the chiral center that designates the D or L configuration ($\ce{C5}$ in hexoses). The other anomer is known as the β form.

The Haworth projections represent the stereochemistry of sugars more realistically than do the Fischer projections, which is why they are preferred:

For a D sugar, any group that is written to the right of the carbon in a Fischer projection has a downward direction in a Haworth projection; any group that is written to the left in a Fischer projection has an upward direction in a Haworth projection.

The terminal $\ce{-CH2OH}$ group, which contains the carbon atom with the highest number in the numbering scheme, is shown in an upward direction.

Similarly in the α-anomer, the hydroxyl on the anomeric carbon is on the opposite side of the ring from the terminal $\ce{-CH2OH}$ group (i.e., pointing down). In the β-anomer, it is on the same side of the ring (pointing up).

enter image description here

References

  1. Lehninger Principles of Biochemistry
  2. Voet and Voet biochemistry
  3. Biochemistry Grisham
  4. Biochemistry Campbell and Farrel
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  • $\begingroup$ From the given explanation above, you can easily deduce if your previous assumptions were correct:) $\endgroup$
    – xavier_fakerat
    Nov 5, 2017 at 6:13
  • $\begingroup$ Why is d-glucuronate alpha when c1 oh id down while L-iduronate is alpha when c1 oh is down? $\endgroup$
    – Nemexia
    Nov 5, 2017 at 16:43
  • $\begingroup$ Sorry for late response. What determines whether a given sugar is d or l is the furthest assymetric (chiral carbon) and whether alpha or beta is anomeric carbon (i.e C2) so since the hydroxyl groups on d-glucuronate or l-iduronate on C2 lie on right side (same side with anomeric carbon) they both alpha hence OH groups is dowward in haworth projection, however the furthest chiral carbon on d-glucuronate is to the right in fischer projection like d-glyceraldehyde but on the left side with l-iduronate (like l-glyceraldyde) $\endgroup$
    – xavier_fakerat
    Nov 7, 2017 at 18:05
  • $\begingroup$ im still confused. can you compare alpha and beta in these?: Beta-D-Glucuronate,Beta-D-Glucose,Beta-L-Glucose they are different :( $\endgroup$
    – Nemexia
    Nov 18, 2017 at 19:28
  • $\begingroup$ "As I covered this explanation already on one of my posts" The blockquote following this line is unavailable in that answer you linked to. Are you sure you are directly quoting your answer or are there some changes? $\endgroup$
    – Gaurang Tandon
    Mar 11, 2018 at 13:54

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