This is the structure of fructose in free form:

The right moiety is the fructose as part of sucrose. The left one being glucose:

Both of the fructose are beta-D fructofuranose. But as it can be seen they both are clearly completely different structures. The anomeric hydroxyl group seems to be in an opposite position while the rest of the structure hasn't even changed shape.

Where am I getting this wrong. Thanks.

  • 1
    $\begingroup$ In sucrose, the two saccharides are linked by their anomeric carbons. $\endgroup$
    – canadianer
    May 22, 2019 at 11:05

1 Answer 1


I'll try to explain this to you, but you should really look it up here: https://pubchem.ncbi.nlm.nih.gov/compound/beta-Levulose#section=3D-Conformer&fullscreen=true

This is the structure of beta-D-fructofuranose, as a 3D conformer. On paper, or as a 2D figure, we cannot visualize the -OH groups at positions 3 and 4.

enter image description here

They appear to be inside/outside the ring, but in reality, they're not really positioned that way. Both the -OH groups are oriented in an opposite manner, but in a plane perpendicular to that of the furan ring.

So when we flip it over to represent it in sucrose, the orientation of the -OH groups in the molecule flips too.

Think of it this way: if you have a piece of cardboard (representing the ring) with a pin pierced into it (representing the -OH), and you flip the board over, the pointy side of the pin will face down too.

Moreover, if we represent it this way, the absolute configuration of the side chains at each of the carbons of fructose remain the same. You could check that using their RS configurations.

For more on RS configurations, check out : https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Chirality/Absolute_Configuration%2C_R-S_Sequence_Rules


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