Question
Question: Glucose reduces Fehling's solution to: (A) Copper (B) Black cupric oxide (C) Reddish-brown cu...
Glucose reduces Fehling's solution to:
(A) Copper
(B) Black cupric oxide
(C) Reddish-brown cuprous oxide
(D) Mixture of copper and cupric oxide
Solution
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 CuSO4containing 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 CuSO4(called Fehling’s solution ‘A’) until blue ppt. of Cu(OH)2first 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+2Cu2++5OH−→RCOO−+Cu2O↓+3H2O
The formula of glucose is CH2OH(CHOH)4CHO. So, when the glucose reacts with Fehling’s solution it forms red ppt and gluconic acid. The reaction is given:
CH2OH(CHOH)4CHO+2CuO→CH2OH(CHOH)4COOH+Cu2O(red 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.