Idea behind Benedict’s solution

Benedict’s solution gives an orange-red precipitate with reducing sugars.

A sugar is reducing if it has a free aldehyde group or can form one through open-chain/enediol form in solution.

Check each compound

(i) Glucose

Glucose is an aldose and a reducing sugar.

$\Rightarrow$ Gives orange-red precipitate.

(ii) Maltose

Maltose has one free anomeric carbon, so it is a reducing disaccharide.

$\Rightarrow$ Gives orange-red precipitate.

(iii) Sucrose

In sucrose, both anomeric carbons are involved in glycosidic bond formation. Hence it is non-reducing.

$\Rightarrow$ Does not give orange-red precipitate.

(iv) Ribose

Ribose is an aldopentose, hence a reducing sugar.

$\Rightarrow$ Gives orange-red precipitate.

(v) 2-Deoxyribose

2-Deoxyribose is also an aldose (deoxy aldopentose), so it remains reducing.

$\Rightarrow$ Gives orange-red precipitate.

(vi) Amylose

Amylose is a polysaccharide component of starch. At JEE level, starch/amylose is treated as non-reducing for Benedict’s test.

$\Rightarrow$ Does not give orange-red precipitate.

(vii) Lactose

Lactose has one free anomeric carbon, so it is a reducing disaccharide.

$\Rightarrow$ Gives orange-red precipitate.

Count compounds that do not give the test

Non-reducing compounds are:

• Sucrose
• Amylose

Therefore, the number of compounds that will not produce orange-red precipitate with Benedict’s solution is

$2$

Comparison with stored answer

Stored correct answer = $2$

Our derived answer = $2$

Hence, they agree.