Glucose does not react with

Options

A. `C_(6) H_(5) NHNH_(2)`
B. `NaHSO_(3)`
C. `HCN`
D. `NH_(2)OH`

Answer

B

Explanation

The open chain form of glucose is in equilibrium with the cyclic hemiacetal form. Since reactions with `C_(6) H_(5) NHNH_(2)`, `HCN` and `H_(2)NOH` are irreversible, the equilibrium shifts to restore the low concentration (0.02%) of the aldehyde as it is depleted through reaction, and eventually all the aldose reacts.
On the other hand, bisulphite addition is reversible and hence enough open-chain aldehyde remains in equilibrium with the bisulphite adduct to satisfy the equilibrium with the hemiacetal form. Consequenlty, there is no shifts noticeable reaction as there is not shift of equilibrium which can increase the cone.of open-chain aldehyde form.