JEE Chemistry 2023 Question with Solution

Given: The compounds are glucose, maltose, sucrose, ribose, $2$-deoxyribose, amylose, and lactose.

Find: The number of compounds which will not produce orange red precipitate with Benedict solution.

Benedict’s solution is used to test for the presence of reducing sugars. Reducing sugars have a free aldehyde or ketone group that can reduce $\text{Cu}^{2+}$ to $\text{Cu}^{+}$, forming orange-red precipitate of $\text{Cu}_2\text{O}$.

From the given analysis:

• Glucose: reducing sugar
• Maltose: reducing sugar
• Sucrose: non-reducing sugar
• Ribose: reducing sugar
• $2$-deoxyribose: reducing sugar
• Amylose: non-reducing sugar
• Lactose: reducing sugar

So, the compounds that do not react with Benedict’s solution are sucrose and amylose.

Therefore, the number of such compounds is $2$.

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

Find: Which given compounds are non-reducing.

A carbohydrate gives Benedict’s test when it has a free reducing end or a free aldehyde/ketone group.

• Glucose has a free aldehyde group in its open-chain form, so it is reducing.
• Maltose has one free anomeric carbon, so it is reducing.
• Sucrose has both anomeric carbons involved in glycosidic linkage, so it is non-reducing.
• Ribose is an aldose sugar, so it is reducing.
• $2$-Deoxyribose is also a reducing sugar.
• Amylose is taken here as non-reducing in Benedict’s test.
• Lactose has a free reducing end, so it is reducing.

Hence, exactly $2$ compounds do not produce the orange-red precipitate.

Note: The solution shows a listed correct answer as $3$, but the worked explanation concludes that only two compounds—sucrose and amylose—do not react. Therefore the defensible final answer is $2$.