The correct option is B Sucrose
Fehling's test is based on the principle that the presence of aldehydes can be detected easily due to their reducing property. The free aldehydes undergo oxidation, by reducing the deep blue solution of Fehling's reagent containing copper(II) to an easily detected red precipitate of insoluble copper oxide. Therefore this test is used only for reducing sugars, that contain the free aldehyde group. Sucrose does not contain a free aldehyde group and is a non-reducing sugar and therefore not reactive to Fehling's reagent.