[a-fortiori]

It makes perfect sense to me that numbering of cyclic-glucose/glucopyranose is straightforward. A terminal carbon is directly attached to an acetal group, so as far as I am aware the numbering starts there.

For cyclic-fructose/fructofuranose however I cannot currently find a way to differentiate between the priority of one of side of the structure over the other. It makes sense in the linear form, where the ketone group is closer to one terminal carbon than the other, but in cyclic form there is no differentiation to be made:



Terminal carbons 1 and 6 are identical. Carbons 2 and 5 attach to the same ketal group, so what gives carbon 1 priority over carbon 6? The addition of the hydroxyl group attached to carbon 2?

I am sorry for what must appear to you as being rudimentary queries. However it is the only thing I cannot explain about the relatively simple idea of bonding glucopyranose to fructofuranose to form α-D-glucopyranosyl(1-2)β-D-fructofuranoside; I hope to confront my lecturer soon (who got carbon numbering completely wrong and is still under a false understanding) from a position of strength.

[orgopete]

Perhaps a little history is needed here. The D and L sugars have been related to D- and L-glyceraldehyde (HOCH2CHOHCHO). Glyceraldehyde has a single chirality center. If the degradation of glucose is performed by lopping off carbon after carbon, one could remove additional chirality centers until you get to just one, glyceraldehyde. Therefore, if the second to last carbon in the chain has the same configuration as D-glyceraldehyde, the sugar is a D-sugar.

At this point, it may be useful to draw Fischer projections on a sheet of paper. If you start with D-glyceraldehyde, you can draw two sugars by adding an additional CHOH to give erythrose and threose. You can continue to add carbons with altering chirality centers until you get all of the hexoses. What will be common to all sugars is the chirality center of the second to the last carbon. If you drew them with the aldehyde up, a simple pattern can be seen with the sugars in which the chirality centers are preserved in the daughter sugars upon the addition of a CHOH group.

A hexose will have four chirality centers, 16 possible compounds, 8 D-sugars and 8 L-sugars. If you maintain the numbering order, you can maintain the relationship of the chirality centers. Therefore, even though sorbitol could be numbered for either CH2OH, it would be more convenient to maintain the numbering and chirality of sorbose.

As indicated, it is easier to number the sugars from the linear form and to liken them to their Fischer projections to identify the sugar. The cyclic form adds a chirality center to the mix, and are referred to as alpha and beta forms. These hemiacetals are in equilibrium with the linear or open form. That extra oxygen should indicate the carbon that is a carbonyl group in the open form. If the fructofuranose were flipped over, you would find three of the same chirality centers as glucose. The numbering should give the highest priority to the carbonyl group and thus maintain the chirality order of the homologous sugars.

[a-fortiori]

Thank you very much for your detailed reply. I am a biology student, with only an A level in Chemistry, so I believe I have understood enough to give a simplified explanation as to the carbon numbering. I was basing my query on the notion that as both terminal carbons are equidistant from the highest priority functional group, as a whole different structure knowing which carbon to start with is impossible. Now I know such a method cannot be used, instead understanding that the open and closed forms are essentially one and the same, and to maintain correct chirality the numbering should be consistent. Thanks again.