Year 12s: what will be the first way you research your future options? >>

Benedict's Test: Testing for carbohydrates question

Hi,

I'm doing some homework and it asks about how I would distinguish various carbohydrate samples. One of the questions asks how I would identify a mixture of sucrose and glucose solutions.

I'm a bit unsure. The Benedict's test for reducing sugars would of course prove positive because of the glucose. However, if I then did the Benedict's test for non-reducing sugars with another 2cm3 of the sample, surely the result would still be positive due to the glucose?

To reword the question, how do I determine that BOTH sucrose and glucose are present?

Thanks!!

Reply 1

macpatgh-Sheldon

Hi, you are correct in using Benedict's solution to test for glucose You would carry out that test on a measured, say 5 cm3 sample, and check what colour you get; it is likely to be green or yellow or even orange if there is a moderate amount of glucose in the mixture.

Then you take another sample of the mixture (the same volume as before (5 cm3, say), add dilute HCl to it, leave it for several minutes, then repeat Benedict's test. This time the colour will indicate a greater amount of reducing sugar e. g. orange if it was green or yellow in the first test, otherwise brick red.

The explanation is that the hydrochloric acid catalyses the breakdown of sucrose to glucose and fructose; this means that in the second test there would be more reducing sugar present, and this would give a stronger result with Benedict's test.

If the first test gives a brick red result (indicating that there is a substantial amount of glucose in the original mixture), then you have two options; 1. you can dilute the mixture with water, and repeat both tests OR
2. You can do the two tests on the original mixture and use a colorimeter to estimate the amount pf copper (I) oxide produced (which will indicate the amount of reducing sugar present); there will be a greater degree of colouration (higher colorimeter reading after adding dilute HCl.

Hope this helps.