Statement I Despite having aldehyde group, glucose does not give Schiff test. Statement II Glucose exists in…
Statement I Despite having aldehyde group, glucose does not give Schiff test.
Statement II Glucose exists in $\alpha$ and $\beta$ crystalline forms.
Both statements I and II are incorrect.
Both statements I and II are correct.
Statement I is correct but statement II is incorrect.
Statement I is incorrect but statement II is correct.
Solution
Glucose does not react with Schiff's reagent because the aldehyde group in glucose is not free. It is involved in the formation of hemiacetal. Glucose exists in two different crystalline anomeric forms, alpha D-glucose and beta D-glucose.
Thus, statements I and II are correct.