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QJEEM23Ched12: JEE Main 2023 Chemistry - Class 12 JEEM23Ched12 Solution: Number of compounds from the following which will not produce…
Number of compounds from the following which will not produce orange red precipitate with Benedict solution is ___________.
Glucose, maltose, sucrose, ribose, 2-deoxyribose, amylose, lactose
Step-by-Step Verified Solution
Benedict's solution is a reagent that tests for reducing sugars. A reducing sugar is one that has a free aldehyde or ketone group, which can reduce the copper(II) ions in Benedict's reagent to copper(I) oxide, which forms a brick-red or orange-red precipitate.
Given the compounds :
1. Glucose - is a reducing sugar, as it has a free aldehyde group.
2. Maltose - is a reducing sugar. It has a free anomeric carbon that can form an aldehyde group.
3. Sucrose - is not a reducing sugar. Its glycosidic bond locks the anomeric carbons, preventing them from forming an aldehyde or ketone.
4. Ribose - is a reducing sugar, as it has an aldehyde group.
5. 2-deoxyribose - is a reducing sugar. It has a free anomeric carbon that can form an aldehyde group.
6. Amylose - is a polysaccharide and does not typically behave as a reducing sugar in Benedict's test. Even though one end of the amylose chain can technically open to reveal an aldehyde group (the so-called reducing end), the reaction would likely be much less dramatic than with true reducing sugars, and might not occur at all under typical test conditions.
7. Lactose - is a reducing sugar, as it has a free anomeric carbon that can form an aldehyde group.
Hence, out of the seven compounds listed, sucrose and amylose will not likely produce an orange-red precipitate with Benedict's solution. Thus, the number of compounds that will not produce an orange-red precipitate with Benedict's solution is 2.
Given the compounds :
1. Glucose - is a reducing sugar, as it has a free aldehyde group.
2. Maltose - is a reducing sugar. It has a free anomeric carbon that can form an aldehyde group.
3. Sucrose - is not a reducing sugar. Its glycosidic bond locks the anomeric carbons, preventing them from forming an aldehyde or ketone.
4. Ribose - is a reducing sugar, as it has an aldehyde group.
5. 2-deoxyribose - is a reducing sugar. It has a free anomeric carbon that can form an aldehyde group.
6. Amylose - is a polysaccharide and does not typically behave as a reducing sugar in Benedict's test. Even though one end of the amylose chain can technically open to reveal an aldehyde group (the so-called reducing end), the reaction would likely be much less dramatic than with true reducing sugars, and might not occur at all under typical test conditions.
7. Lactose - is a reducing sugar, as it has a free anomeric carbon that can form an aldehyde group.
Hence, out of the seven compounds listed, sucrose and amylose will not likely produce an orange-red precipitate with Benedict's solution. Thus, the number of compounds that will not produce an orange-red precipitate with Benedict's solution is 2.