I am really struggling with converting pyrano/furanosides into their Fisher projection counterparts. Especially when they are drawn like in the shown projection. When do you know to connect the anomeric carbon? Do you always need to show the connection like in the attached photo?

With this in mind, please could someone walk me through how to draw isopropyl beta-D-mannopyranoside or a similar structure as a fisher projection ? Or point me towards a resource that could help?

  • $\begingroup$ Who said these two representations are equivelent? $\endgroup$ Sep 26, 2022 at 3:24
  • $\begingroup$ My apologies I should have been more clear. I am not saying these are equivalent. I am giving an example of how a pyranoside is drawn occasionally and asking how do you know when to draw it in that form vs its acyclic? Then asking for help applying that reasoning to converting isopropyl-d-mannopyranoside into a fischer $\endgroup$
    – Abigail
    Sep 26, 2022 at 11:05
  • $\begingroup$ First of all, mannose is an aldohexose. Your compound is a pentose, which has two anomeric carbons. You may need to get the correct structure for isopropyl $\beta$-D-mannopyranoside (is this the correct name?). $\endgroup$ Sep 26, 2022 at 17:44
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    $\begingroup$ Perhaps the first thing is to know the person's name. It is Emil Fischer. In the left hand structure, all OH's ar equatorial except for the one at C2, which is axial. You need to work on stereochemistry in 6-membered rings and carbohydrates. I'll post some useful links in a separate Comment. Oh! Welcome to ChemSE! $\endgroup$
    – user55119
    Sep 26, 2022 at 17:44
  • $\begingroup$ Abigail: Try these links. DrawingCyclohexanes: ursula.chem.yale.edu/~chem220/chem220js/… ursula.chem.yale.edu/~chem220/chem220js/STUDYAIDS/… Carbohydrates: ursula.chem.yale.edu/~chem220/chem220js/… $\endgroup$
    – user55119
    Sep 26, 2022 at 17:50

1 Answer 1


Abigail: I think that you should have enough information from the links (vide infra) I provided to get you through most of what bothers you. It takes time to get these issues straight. Structures 1 represent the isopropoxy pyranosides of D-glucose ($\ce{C5}$-(R)). Assuming a chair conformation in 1a, all of the groups alternate, i.e.,up,down,up, etc., around the ring. All of the substituents are equatorial. Structure 1b is the chair conformation of 1a. The isopropoxy group is in the β-configuration, meaning that it is above the plane of the ring as drawn in 1a and 1b. The Fischer projection of this D-glucopyranoside is shown in structure 1c. By convention, the isopropoxy group of the β-configuration is drawn on the left side of the Fischer projection.

As @Mathew has noted, your left hand structure is a pentose. But if the hemiacetal OH were replaced by -$\ce{CH2OH}$ of the same configuration, then structures 2 would be the same as structures 1 except for the configuration at $\ce{C2}$. The $\ce{C2}$ epimer of D-glucose is D-mannose wherein the -OH group at $\ce{C2}$ is axial.

Now your right hand structure has some issues. It is a heptapyranoside as shown in structure 3a given the configurations in your structure with the addition of the configuration at the anomeric carbon based upon what your structure displays. Structure 3b is the chair conformation of 3a. Again, by convention, when the methoxy group is drawn on the right of the Fischer projection, it means that the methoxy group has the α-configuration. In keeping with the Fischer convention, structure 3c is the correct representation with the correct configurations.

Drawing Cyclohexanes:

http://ursula.chem.yale.edu/~chem220/chem220js/StudyAids.html#Conformation http://ursula.chem.yale.edu/~chem220/chem220js/STUDYAIDS/cyclohexane.pdf



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    $\begingroup$ @Abigail if this answer helped you, please accept the answer by clicking on the checkmark. $\endgroup$ Sep 27, 2022 at 10:15

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