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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?

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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. $\endgroup$ – Reeshabh Ranjan May 26 '17 at 10:20
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    $\begingroup$ This answer nicely elaborates the mechanism stated in words in this answer. $\endgroup$ – Gaurang Tandon Mar 15 '18 at 15:19

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