Question

Which sugar does not give Benedict’s test?
A) Glucose
B) Maltose
C) Fructose
D) Sucrose

Answer 1

Hint: Benedict’s test is the test used for simple carbohydrates. It is helpful in identifying reducing sugars (monosaccharide and some disaccharides) having free ketone or aldehyde as a functional group. It is used in the diagnosis of various clinical parameters present in urine.

Complete answer:
Benedict’s test works on the principle that when the benedict’s solution and simple carbohydrate are heated together, the solution changes to orange red/brick red. This reaction takes place due to the reducing property of the simple carbohydrates.
The copper (II) ions in the benedict’s solution are reduced to copper (I) ions which are responsible for the transformation of color to orange red/brick red. The red copper (I) oxide formed is insoluble in water and thus precipitated out in the solution.
The reaction is given below:
$RCHO + 2C{u^{2 + }} + 5O{H^ - } \to RCO{O^ - } + C{u_2}O + 3{H_2}O$
The reducing sugars that show positive results with benedict’s solution are glucose, fructose, maltose etc.

The correct option is D i.e. sucrose.

Additional Information: Sucrose is a disaccharide composed of two monosaccharide molecules that are glucose and fructose. This is called as non-reducing sugar because of the absence of free ketone and aldehyde groups. Since they do not have free aldehyde and ketone groups, they are not able to reduce the copper ions present in the benedict’s solution. Thus, they do not give positive results for Benedict’s test and remain blue in color.

Note: Benedict’s solution is a deep blue colour alkaline solution which comprises anhydrous sodium carbonate, sodium citrate and copper (II) sulfate pentahydrate.
Sodium carbonate provides an alkaline condition which is necessary for redox reaction.
Sodium citrate forms a complex with copper (II) ions so that it does not deteriorate into copper (I) ions.
It has a clinical significance in determining the glucose level in urine.