My book shows this figure of Fischer formula of D-glucose:

enter image description here

I don't understand this figure, and I wonder why the upper carbon atoms in both the right and left formulas have 5 bonds.

And from where comes this hydrogen atom in the upper left side of the figure (attached to α-D-glucopyranose).

When turning D-glucose to Fischer formula, it seems there's an additional hydrogen atom, isn't it? Why and how does this happen?!!

Could someone explain how to get from the aldehyde form of glucose (the middle structure) to the β-D-Glucopyranose and the α-D-Glucopyranose?


1 Answer 1


Throw away that book, you have found a very terrible error. Okay the rest of the book may still be good, but at least black out that picture or stick a correct version of the picture over the top.

In the image below, you can see four depictions of α-D-glucose taken from Wikipedia (where a full list of authors is available). From left to right these are: Fischer projection, Haworth projection, chair depiction and ‘flat circle’ (a depiction which follows the guidelines for 2D representations of 3D structures).

Four projections of alpha-D-glucose

As you can see, there is no additional hydrogen on the anomeric carbon and it is not pentavalent.

The mechanism of ring-closure is also mentioned in sufficient detail on the corresponding Wikipedia page to which I hereby point you. Basically, the 5-hydroxy group attacks the $\ce{C=O}$ π system as a nucleophile. Proton transfer will lead to the cyclic product. (In actual aquaeous solution, the carbonyl oxygen will be protonated immediately from a nearby water molecule and one will also take away the hydroxy group’s excess proton, so the two are not the same.)


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