JEE Main 2023 · 11 April · evening · Numerical answer · +4 / −0
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
Answer and solution
Answer 2
Solution
Chemistry ma’am · 3 steps
Look for a free hemiacetal or aldehyde carbon: no free one means no orange-red precipitate.
- What Benedict tests. Benedict solution contains Cu²⁺, which a reducing sugar turns into the orange-red precipitate Cu₂O. A sugar is reducing when its anomeric carbon is free, so the ring can open to an aldehyde or α-hydroxy ketone.
Compound Free anomeric carbon? Orange-red precipitate? Glucose Yes, C1 hemiacetal Yes Maltose Yes, C1 of the second glucose is free Yes Sucrose No, C1 of glucose bonded to C2 of fructose No Ribose Yes, C1 hemiacetal Yes 2-Deoxyribose Yes, only C2 lost its OH, C1 is untouched Yes Amylose No, the glucose units are locked in glycosidic links No Lactose Yes, C1 of glucose is free Yes - Why amylose fails. Amylose is a long chain of glucose units joined C1 to C4. Only one end of the whole chain is free, which is far too little to give a visible precipitate, so starch counts as non-reducing.
- So. Only sucrose and amylose give no orange-red precipitate, so the answer is 2.
The common slip: 3. Counting 2-deoxyribose as non-reducing gives 3. The missing oxygen is at C2, and its C1 aldehyde is still free.
The idea to keep. A sugar reduces Benedict or Fehling solution only if its anomeric carbon is free. Sucrose and the polysaccharides like starch are the usual non-reducing ones.
Answer from the NTA final answer key. Solution by PrepWiser.