L6. Fehling's Test: Reducing sugars

On Monday 27th of October after doing our last experiment we had a little bit of time left so we decided to do another experiment realted with the sacharides. Fehling's solution is a chemical test used to differentiate between reducing and non-reducing sugars. This test is based on the reaction of a functional group of sugar molecules with Fehling's regent. The objectives were to identify reducing sugars, comprehend redox reactions and understand the relation between structure and reducing ability of some sugars.

MATERIALS:
-Test tube rack
-10 mL pipet
-Distilled water
-5 test tubes
-Spatula

-Lactose
-Maltose
-Sucrose
-Glucose
-Starch
-Fehling's A and B solutions

PROCEDURE:
1-We took 5 test tubes and labeled them: G, M, S, L, SU
2-We put 2 mL of distilled water inside each tube.
3- Then, with differents spatulas we put a small amount of each sugar. 
4-After that we added 2 mL of Fehling's A solution and then Fehling's B.
5-We placed each test tube in a boiling water bath (250 mL beaker on a hotplate stirrer)
6- Then we observed what was happening.

Results, observations and conclusions:  Glucose, maltose and lactose have reducing ability because they turned from deep blue to a red colour.  However, sucrose, in which the anomeric carbons of the two units are linked together, is a non-reducing disaccharide since neither of the rings is capable of opening. Starch is not a reducing sugar either, because the first ring cannot open up, there's no hydrogen on the circled oxygen to allow for ring opening. Similarly the next ring, and the next ring, et cetera, cannot open up.

QUESTIONS

1-From your observations and the structures of the sugars given above, indicate which functional group in the sugar molecules reacts with Fehling's reagent.

The OH group is the one that reacts with Fehling's reagent because when it is free the sacharide will have the reducing power.

If the bond is monocarbonilic the sacharide will have reducing power, that's why all monosacharides have it (the OH from the C1 is always free).

 

2-Compare the results you obtained for the Fehling's test of starch and Fehling's test of hydrolyzed starch. Explain your results.

I haven't done the Fehling's test of hydrolized starch but i can deduce it:

In the Fehling's experiment the starch doesn't have a reducing power because the OH is not free but when the starch is hydrolyzed it turns into glucose and glucose has a free OH because it's a monosacharide. Also the last glucose of the starch chain will have a reducing power (the ones that are in the ends).

The starch components are alfa D glucoses: amylose alfa (1 -> 4) linear chain, and amylopectine alfa (1->6) ramifications.

 

3-Would have you obtained a Fehling's positive test if you had hydrolyzed the sucrose? Why? 

Yes i would. Hydro meaning "water", and lysis, meaning "separation" usually means the cleavage of chemical bonds by the addition of water. So this means if we hydrolyze sucrose we will obtain glucose and fructose and they both will have a free OH because they are monosacharides. 

4-What does "reducing sugars" mean? 

A reducing sugar is the one that reacts positive to the Fehling's test. This means that they are capable of reducing coper II ions to copeer I ions. When the sugar to be tested is added to the Fehling's solution and the mixture is heated, some sugars can be oxidized (to lose electrons) and the Fehling's mixture can obtain the electrons (reduced).