Step 1: Recall the anomalies of glucose used to deduce its cyclic structure: it does not give the Schiff test, does not react with \(NaHSO_3\), and its pentaacetate does not react with \(NH_2OH\).
Step 2: These facts prove the aldehyde group is not truly free but tied up as a hemiacetal (ring form).
Step 3: Reagents that still work (bromine water, phenylhydrazine, acetic anhydride) do so via the small open-chain fraction or the hydroxyl groups.
Step 4: The sodium bisulphite addition reaction, however, fails, so glucose does not react with \(NaHSO_3\).