Glucose reduces Fehling's solution to:
(A) Copper
(B) Black cupric oxide
(C) Reddish-brown cuprous oxide
(D) Mixture of copper and cupric oxide

Hint: Fehling’s solution is used to differentiate between the reducing and non-reducing sugar. If the sugar on reduction with Fehling’s forms a precipitate then the sugar is a reducing sugar if does not form the ppt then it is a non-reducing sugar.

Complete step by step answer:
Carbohydrates are of two types: reducing and non-reducing sugar. The reducing sugar, when treated with Fehling’s solution, forms a precipitate. This is because reducing sugar has a free aldehyde or ketone group.
Let us see the reduction of Fehling’s solution by glucose: Fehling’s solution is an alkaline solution of $CuS{{O}_{4}}$containing some Rochelle salt, i.e., sodium potassium tartrate and is prepared by adding an alkaline solution of Rochelle salt (called Fehling's solution 'B') to an aqueous solution of $CuS{{O}_{4}}$(called Fehling’s solution ‘A’) until blue ppt. of $Cu{{(OH)}_{2}}$first formed just redissolves to form a clear deep blue solution. When an aliphatic aldehyde or ketone is heated with Fehling's solution, the latter is reduced to give a reddish-brown ppt of cuprous oxide. During the reduction, the reaction which occurs is:
$RCHO+2C{{u}^{2+}}+5O{{H}^{-}}\to RCO{{O}^{-}}+C{{u}_{2}}O\downarrow +3{{H}_{2}}O$
The formula of glucose is $C{{H}_{2}}OH{{(CHOH)}_{4}}CHO$. So, when the glucose reacts with Fehling’s solution it forms red ppt and gluconic acid. The reaction is given:
$C{{H}_{2}}OH{{(CHOH)}_{4}}CHO+2CuO\to C{{H}_{2}}OH{{(CHOH)}_{4}}COOH+C{{u}_{2}}O(red\text{ }ppt.)$.
Non-reducing sugars do not have free aldehydes or ketone groups, hence they do not give Fehling's solution test.

Hence, the correct answer is an option (c)- Reddish-brown cuprous oxide.

Note: All the monosaccharides gives the Fehling’s solution test. All polysaccharides are non-reducing sugars. The glucose and monosaccharides give another test called Tollen’s reagent test in which they form a silver mirror.