L.6 FEHLING'S TEST: REDUCING SUGARS

INTRODUCTION:

We start this experiment un 27th of october, but is not finished yet because we did it approximately the last 15 minutes of class. But this experiment is related with saccharides.
First of all, the fehling's solution is a chemical test to differentiate between reducing and non-reducing sugars. Is based on the reaction of a functional group of sugar molecules with Fehling's reagent. 

Fehling's A: is a blue aqueous solution of cooper(II) sulphate.

Fehling's B: clear and colourless solution of potassium sodium tartrate and sodium hydroxide. 

When both are mixed a deep colour solution can be seen. 

When a sugar has reducing ability. the mixture turns from deep blue to green colour suspension with a red precipitate. Some sugars are capable of reducing cooper II ions to cooper I ions. This reducing ability is useful in classifying sugars. When the sugar to be tasted is added to the Fehling's solution and the mixture is heated, some sugars can be oxidized and the Fehling's mixture can reduced. 

The goal is identify reducing sugars, comprehend redox reactions and understand the relation between structure ability of some sugars. 

MATERIALS:

- Test tube rack

- 10mL Pipet

- Distilled water

- 5 test tubes

- 5 Spatula

- Lactose

- Maltose

- Glucose

- Sucrose

- Starch

- Fehling's A and B solutions

- HCl

PROCEDURE:

First you will determine with sugars give a positive test with Fehling's reagent and then, by testing the reaction of some organic molecules containing only a single functional group, you should be able to deduce which functional group og sugars is reacting with Fehling's reagent.

1.  Take 5 test tubes and label: G ,M, S, L, ST.
2.  Put 2mL of water in each test tube.
3.  With different spatulas put a small amount of each sugar. Dissolve the suggar.
4.  Add 2mL of Fehling's A solution and then Fehling's B.
5.  Place each test-tube in a boiling water bath (250mL beaker on a hotplate stirrer).
6.  Observe what is happening.

Starch Hydrolysis: 

Hydrolysis is the reaction of a compound with water. As you know, starch is a polymer, consisting of many units of α-D-glucose covalently linked together.

1. Place 2mL of 1% starch in test tube and add o.5mL of 3M HCl. Mix and place this mixture in a boiling water bath for 10 minutes.
2. After 10 minutes, remove the tube from the waterbath and let it cool. Neutralize this solution with 1M NaOH and mix well. 
3. transfer 8-10 drops of this solutionto a small test tube.
4. Add 1mL of Fehling's A solution and 1mL of Fehling's B.
5. Heat for a few minutes in a boiling water bath.
6. Record your observations. compare the results of this test with your results for unhydrolyzed tarch in the step 1 of this experiment. 
7. You can test the absence of starch with iodine solution too. 

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 it is free and the sacharide have the reducing power.

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

When the starch is hydrolized it turns into glucose. But in the Fehling's experiment the starch doesn't have reducing power, the OH is not free. 

however, the glucose is a monosacharide and has the OH free, like the other solutions. 

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

 Glucose, have a free OH that's because they are monosacharide, this happens because we hydrolyzed fructose and we obtain glucose. 

4.-What does "reducing sugars" mean? 

A reducing sugar is any sugar that either has an aldehyde group or is capable of forming one in solution through  isomerism. The aldehyde functional group allows the sugar to act as a reducing agent, for example in the Tollen's test or Benedict's reagent, or the Maillard reaction important in the browning of many foods. The cyclic hemiacetal forms of aldoses can open to reveal an aldehyde and certain ketoses can undergo tautomerization to become aldoses. However, acetals, including those found in polysaccharide linkages, cannot easily become free aldehydes.