Even though fructose is a ketohexose (ketone-containing hexose, a six-carbon monosaccharide), it reduces Tollen's reagent and Fehling's solution. Generally, a ketone does not reduces Tollen's reagent and Fehling's solution. So, why is this possible in the case of fructose?


2 Answers 2


The presence of the hydrogen atom in the aldehyde group gives reducing properties to the aldehyde molecule. Aldehydes are easily oxidized by mild oxidizing agents such as Tollens’ reagent ($\ce{Ag+}$ in aqueous ammonia), Fehling’s reagent ($\ce{Cu^2+}$ complexed with tartrate), or Benedict’s reagent ($\ce{Cu^2+}$ complexed with citrate). Ketones do not have such reducing properties and are not oxidized under similar conditions. Therefore, Tollens’ reagent, Fehling’s reagent, or Benedict’s reagent can be used to distinguish aldehydes from ketones.

However, fructose reduces such reagents even though it contains no aldehyde group. Reduction occurs because the reagents are basic solutions and fructose is readily isomerized to a mixture of aldoses (glucose and mannose) under basic conditions.

keto–enol tautomerization of fructose

Under basic conditions, the proton alpha to the carbonyl group (aldehyde or ketone) is reversibly removed. If the resulting enolate ion reprotonates on the oxygen, an enediol intermediate results. Thus, the base-catalyzed enediol rearrangemet can move the carbonyl group up and down the chain.

Furthermore, in the enolate ion, the affected carbon atom is no longer asymmetric. Reprotonation can occur on either face of the enolate, giving either configuration. Thus, the base-catalyzed epimerization results in a mixture of stereoisomers.

Therefore, under strongly basic conditions, the combination of enediol rearrangements and epimerization leads to a complex mixture of sugars (see also: Lobry de Bruyn–van Ekenstein transformation).

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    $\begingroup$ However, everywhere, it is written only 'alpha-hydroxy ketones'. Not all alpha-hydroxy ketones are fructose, that is, not all alpha hydroxy ketones can double tautomerization to form an aldehyde. $\endgroup$ May 26, 2017 at 10:20
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    $\begingroup$ This answer nicely elaborates the mechanism stated in words in this answer. $\endgroup$ Mar 15, 2018 at 15:19

In the presence of very dilute alkalis, glucose, for example, is converted into a mixture of glucose, mannose and fructose. It is explained by the presence of an intermediate enodiol, and is also called epimerization, since a mixture of epimers is obtained. All monosaccharides are oxidized with Benedict, Fehling and Tollens reagents, also giving aldonic acids. These reactions do not serve to distinguish aldoses from ketoses, as occurred with aldehydes and ketones, since the latter also reduce the previous reagents, since, under the conditions of the tests, the ketoses equilibrate with the aldoses through an ene diol, as just seen. The carbohydrates that test positive for these tests are called reducing sugars.


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