# What reagents can be used to distinguish pentose and hexose sugars?

I have been asked how I can differentiate D-glucose and D-xylose using the following reagents / reactions:

1. Orcinol
2. Benedict's test
3. Ammoniacal silver nitrate
4. $$\ce{FeCl3}$$
5. $$\ce{C6H5SO2Cl}$$, excess $$\ce{NaOH}$$
6. $$\ce{NaOH}$$, $$\ce{H2O}$$, heat
7. t-Butyl chloride, $$\ce{AlCl3}$$
8. $$\ce{Br2}$$ in water, dark
9. $$\ce{NH2OH.HCl}$$, $$\ce{KOH}$$, $$\ce{FeCl3}$$
10. Resorcinol
11. 2,4-DNPH
12. Iodoform test

Usually, chemical tests identify certain functional groups in molecules which react in a particular way. However, here the two compounds only differ in the number of carbons and don't have any real difference in their functional groups, so what reaction can be used to differentiate them?

• react with 2,4-DNPH then a careful measurement of the melting point of the phenylhydrazone and compare against literature values – Waylander Jul 7 at 7:05
• I also noticed that the compounds appear to be open-form sugars. Maybe only one of them will form a cyclic form? – TAR86 Jul 7 at 8:54

You are correct that there are few differences between glucose and xylose simply based on functional groups. The only real difference is that one has six carbons and the other five; they are members of what we normally call hexose and pentose sugars.

This is not my area of expertise, but a quick Google search suggests that these can be differentiated using what is known as Bial's test. In the presence of concentrated HCl, pentose sugars such as xylose undergo acid-catalysed dehydration to form an aromatic molecule, furfural:

Furfural can then be reacted with orcinol to form a characteristically blue or green species, which is taken as a positive result indicating the presence of a pentose:

Because hexose sugars are also capable of existing in a cyclic hemiacetal form, they are also capable of undergoing a very similar reaction. The difference is that hexose sugars do not form furfural, but rather a derivative of furfural with an extra hydroxymethyl group ($$\ce{CH2OH}$$):

This extra hydroxymethyl group supposedly leads to a different colour being observed (Wikipedia says: "the related hydroxymethylfurfural from hexoses may give a muddy-brown or gray solution, but this is easily distinguishable from the green color of pentoses").

According to the same Wikipedia page, the reagents you need to use are therefore:

... 0.4 g orcinol, 200 ml of concentrated hydrochloric acid and 0.5 ml of a 10% solution of ferric chloride.

I'm not going to address all of the reagents, but I want to highlight a specific approach in this partial answer.

Occasionally, knowing some history of chemistry is quite useful when thinking about how to solve problems. In this specific case, I direct you to the work of Emil Fisher (chemistry Nobel 1902), who determined the structure of all hexose sugars. He did this by building up the sugars from smaller subunits while keeping track of symmetries.

This immediately suggests a solution where one performs reduction (or oxidation) of the aldehydes. In other words, make the both ends of each molecule look the same. This is a valid strategy because the underlying symmetries of glucose and xylose are different.

The penta-ol is meso and therefore achiral, while the hexa-ol is chiral and optically active since you started with one enantiomer.

Thus, any reagent that can produce the bis-acid or bis-aldehyde without epimerizing any stereocenters along the way or any reagent that can reduce the aldehyde to an alcohol will provide a suitable method of differentiation in this specific example.