Question

Glucose and fructose can he distinguished by:
(A) Barfoed's tests
(B) Fehling solution
(C) Bendict Solution
(D) Seliwanoff’s test

Answer 1

As we know there are a lot of tests, for distinction for different types of groups like alcoholic group, phenolic group, carbonyl group, carboxylic group, amine group and Nitro group and it we take about some distinction tests so that are2,4-dinitrophenylhydrazinc, iodoform test etc. so in this question we will discuss about distinction test of glucose and fructose.

Complete step by step answer:
Here we will explain all the tests so firstly it is Barfoed's test, it is used to detect only monosaccharide. That is this option is not valid for glucose and fructose. So, our next solution is fehling’s solution and it is used for aldehyde. So, Fehling solution is also most appropriate for glucose and fructose third one is benedict solution and it is used so, by the detection of reducing sugars.
So, by this also we cannot distinguish glucose and fructose. And our last option is Seliwanoff’s test. So, this test distinguishes between aldose and ketose.
Talking about aldose in which we have aldehyde group ${\text{ - CHO}}$ group.
Whereas ketose are those in which a carbonyl group is present; ${\text{CHO}}$ group.
And we know that -CHO group is present in glucose while ${\text{CHO}}$ is present in fructose. It means this test is used to distinguish between glucose and fructose. As we learnt previously, Seliwanoff’s test is a chemical test which distinguishes between aldose and ketose sugars. As glucose, fructose is a qualitative test, if we shake the content and after that we see the change in colour change.

So, the correct answer is Option D .

Note:
Don’t get confused between bromides and bromine water. $B{r_2}(aq)$ Is different and $B{r_2} = {H_2}O$ is different whenever $B{r_2} = {H_2}O$ is written it means that it is having more relatives as compare to $B{r_2} aq$.
Bromine water is a liquid which is used as chloroform but nowadays it is evolved as a poisonous gas.