# How do I arrive at correct furanose structures when starting from a Fischer projection?

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?

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.

• Tl;dr: It doesn’t matter how you do it, as long as it is consistent and correct. – Jan Jun 14 '16 at 19:23
• @Jan I don't get you. – Tyto alba Jun 14 '16 at 19:28
• chemistry.stackexchange.com/questions/44332/… – Mithoron Jun 14 '16 at 19:37
• 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.) – Jan Jun 14 '16 at 20:43
• It was the only wikipedia page where the structure of Beta-D-fuctose is provided where it is numbered from the left. – Tyto alba Jun 14 '16 at 21:12

## 1 Answer

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:

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.)

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.