Ambiguous nature of aldehydic group in glucose

Question

In aqueous solution, α-anomer and β-anomer of glucose remain in equilibrium with each other along with a small amount of glucose in open chain form ($0.02~\%$). The open chain form in the solution contains a free aldehydic group and hence gives several reactions involving free aldehyde.

Though little, the open chain form of glucose reacts with many reagents which are used to treat free aldehydic groups. Therefore, glucose in aqueous solution also reacts with those reagents.

$$\ce{glucose_{(aq)} + HCN -> gluco cyanohydrin} $$

$$\ce{glucose_{(aq)} + H2NOH -> gluco oxime} $$

$$\ce{glucose_{(aq)} + H2NNHPh -> gluco phenylhydrozone} $$

$$\ce{glucose_{(aq)} + 3H2NNHPh-> gluco osazone} $$

$$\ce{glucose_{(aq)} + Br2_{(aq)}-> gluconic acid} $$

Glucose also answers Tollen's reagent test, Fehling's solution test and Benedict's solution test.

However, as interesting it may seem, glucose is a two-face molecule. Here are some reactions which glucose does not like:

$$\ce{glucose_{(aq)} + NaHSO3 -> no reaction} $$

$$\ce{glucose_{(aq)} + {2,4-DNP} -> no reaction} $$

$$\ce{glucose_{(aq)} + {Schiff's reagent} -> no reaction} $$

Why is that glucose does not react with $\ce{NaHSO3}$ or $\ce{{2,4-DNP}}$ or Schiff's reagent while it answers many other reactions where the free aldehydic group is involved?

I am looking for a convincing answer. Most answers in web tell that some are strong and some aren't strong enough. If you give enough time, all the carbonyl reactions should work with glucose because there will always be $0.02~\%$ in open chain form. If some of it gets used up, then the equilibrium will shift and hence more open chain form of glucose will be formed.

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Answer 1

The distal hydroxy groups are in a perfect distance to form pyranose (six-membered oxane) or furanose (five-membered oxolane) rings. Any equilibrium reaction that leaves the carbonyl group liberated for extended periods is prone to lose free aldehydes to the (unreactive) hemiacetal forms. Any reaction whose equilibrium is strongly product-sided or which proceeds somewhat irreversibly due to precipitation etc. will not be affected by that effect.

This explains why Fehling’s and Tollens’ solutions react with glucose: in both cases one of the products precipitates which drives the reaction to completion. None of the aldehyde that is used up in the test reaction will be regenerated. The same is true for the osazone formation which involves a redox step or the reaction with bromine which is also a redox reaction.

The reaction with sulphite to give sulphonic acids is an equilibrium: sulphite can attach but is equally quickly liberated again. However, sulphite is a rather soft nucleophile and an acceptable leaving group. The reaction is reversible and due to the formation of hemiacetals not much of the sulphonic acid will be observed. In general sulphite testing, the precipitation of the strongly hydrophilic sulphonate drives the reaction to completion but sugars themselves are already strongly hydrophilic so this driving force cannot exist.

A cyanohydrin, however, is a much more stable intermediate because cyanide is a much worse leaving group. Thus, here again we are driving a reaction to completion due to a practically irreversible step — and contrary to sulphite which is based on the same mechanism but is much more reversible.

The same difficulty occurs when attempting to generate imines: these are also prone to attack by the distal hydroxy functions giving O,N-acetals. While these may be stable products, they do not have the same features as the imines usually generated by the reaction of carbonyl compounds with amines and thus are not captured by the tests. This also prevents productive reaction of glucose with Schiff’s reagent — and additionally, Schiff’s test requires the addition of a sulphite which we learnt already is unfavourable.

The big question remains why some hydrazines react with glucose according to your list but some don’t. I have no satisfactory explanation why that would be the case. The only difference between 2,4-DNPH and phenyl hydrazine is the former’s electron-withdrawing groups that could lead to an overall slower reaction or less tendency to undergo the required redox step to form an oxazone. It is not a great explanation but maybe a starting point.