JEE Main 2020 · 8 January · morning · Single correct · +4 / −1
Which of the following statement is not true for glucose?
- (a) Glucose gives Schiff's test for aldehyde
- (b) Glucosse reacts with hydroxylamine to form oxime
- (c) The pentaacetate of glucose does not react with hydroxylamine to give oxime
- (d) Glucose exists in two crystalline forms α and β
Answer and solution
Answer (a) Glucose gives Schiff's test for aldehyde
Solution
Chemistry ma’am · 2 steps
Glucose looks like an aldehyde on paper, but it mostly lives as a ring, so Schiff's test stays silent.
- Mostly a ring. In solution, almost all glucose is a six-membered cyclic hemiacetal. Only a tiny fraction is in the open chain form with a free -CHO group.
Statement True or false Reason (a) Gives Schiff's test False Too little free -CHO is present to restore the pink colour of Schiff's reagent (b) Forms an oxime with NH₂OH True The ring opens, and the free -CHO is used up as it reacts, so more ring keeps opening (c) Pentaacetate gives no oxime True Acetylating the ring -OH at C1 locks the ring, so no free -CHO can form (d) Two crystalline forms, α and β True These are the two anomers of the ring, differing only at C1 - So. Statement (a) is the one that is not true. Glucose does not give Schiff's test, and this is one of the facts that the open chain structure cannot explain.
The idea to keep. Schiff's test, the bisulphite reaction and the pentaacetate giving no oxime all point to one thing: glucose is mainly a cyclic hemiacetal, not a free aldehyde.
Answer from the NTA final answer key. Solution by PrepWiser.