Question

Ketones do not reduce Fehling's solution and Tollen's reagent while fructose containing ketonic group does. Why?

Hint:

Remember: "Ketose becomes aldose in base." The transient enediol tautomer lets fructose behave like a reducing sugar.

Answer 1

Ordinary ketones are not readily oxidised by Fehling's or Tollen's reagents, hence they are non-reducing.
Fructose (a ketohexose), however, in alkaline medium undergoes enediol formation and the Lobry de Bruyn–van Ekenstein rearrangement to give the corresponding aldoses (glucose/mannose), which are reducing and therefore reduce \([\mathrm{Ag(NH_3)_2}]^+\) (Tollen's) and \(\mathrm{Cu^{2+}}\) (Fehling's).
\[ \mathrm{Fructose\ (keto)} \xrightleftharpoons[\mathrm{OH^-}]{\text{enediol}} \mathrm{Glucose/Mannose\ (aldehydo)} $\Rightarrow$ \text{reduces Fehling's & Tollen's} \]