While numbering the furanose ring from the left are we supposed to place the atoms on right of the Fischer projection to the top? Isn't it just the opposite to what we do while numbering from the right?

enter image description here

In the image the atoms/groups lying on the right of the 2nd Fischer structure is present to the bottom of every furanose carbon of the 3rd image (chronologically). So aren't they supposed to be present at the bottom even if we start from the left (referring to structure A)? But I find Wikipedia says they are supposed to be at the top.

  • $\begingroup$ Tl;dr: It doesn’t matter how you do it, as long as it is consistent and correct. $\endgroup$
    – Jan
    Commented Jun 14, 2016 at 19:23
  • $\begingroup$ @Jan I don't get you. $\endgroup$
    – Tyto alba
    Commented Jun 14, 2016 at 19:28
  • $\begingroup$ chemistry.stackexchange.com/questions/44332/… $\endgroup$
    – Mithoron
    Commented Jun 14, 2016 at 19:37
  • $\begingroup$ Also, why are you linking to sucrose rather than to fructose? I thought, your question was about fructose? (I know it is present in sucrose, but still.) $\endgroup$
    – Jan
    Commented Jun 14, 2016 at 20:43
  • $\begingroup$ It was the only wikipedia page where the structure of Beta-D-fuctose is provided where it is numbered from the left. $\endgroup$
    – Tyto alba
    Commented Jun 14, 2016 at 21:12

1 Answer 1


A Fischer structure is one way of depicting structures such as sugars or amino acids to quickly draw them and have a definite way of seeing stereochemistry on every carbon. The rule behind Fischer structures is that the highest oxidised atom should be at the top or as close as possible to the top. The following convention is used to turn tetrahedral carbons into a square environment:

Convention used in Fischer projections

This allows you to assign absolute configurations to every carbon atom.

When instead drawing the same structure as a ring like in your lower images, you just need to take care that every carbon has the same configuration. (Note that your drawing of β-D-fructose is wrong; it should be drawn with a very wavy line like in my image below, because of the constraints of Fischer projections.)

α-D-fructose β-D-fructose

Once you have gotten there, you are free to rotate the ring around as much as you wish while retaining the absolute configuration at every stereocentre. Note especially, that a rotation of $180^\circ$ of my structure of β-D-fructose will lead to you one on the left. If you don’t believe me, print it out, turn the paper around and then nudge the forwards-pointing part of the ring downwards below the tetrahydrofuryl-oxygen.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.