To solve this problem, we need to analyze the reactivity of glucose with the given reagents.$\newline$Glucose is an aldose sugar with the following functional groups:$\newline$• An aldehyde group $(-CHO)$ at the first carbon.$\newline$• Multiple hydroxyl groups $(-OH)$ on the other carbons.$\newline$Let's evaluate each reagent:$\newline$A. Tollen's reagent:$\newline$• Tollen's reagent is used to test for aldehydes. Glucose, having an aldehyde group, reacts with Tollen's reagent to give a silver mirror.$\newline$B. Schiff's reagent:$\newline$• Schiff's reagent is used to detect aldehydes. However, glucose does not give a positive test with Schiff's reagent due to its cyclic hemiacetal form, which does not readily react.$\newline$C. HCN:$\newline$• Glucose reacts with HCN to form cyanohydrins. The aldehyde group in glucose reacts with HCN.$\newline$D. $NH_2OH$ (Hydroxylamine):$\newline$• Glucose reacts with hydroxylamine to form oximes. The aldehyde group in glucose reacts with $NH_2OH. \newline$E. $NaHSO_3$ (Sodium bisulfite):$\newline$• Sodium bisulfite reacts with aldehydes to form bisulfite addition compounds. However, glucose does not react with $NaHSO_3$ due to steric hindrance and its cyclic form.$\newline$Based on the analysis, glucose does not react with Schiff's reagent and sodium bisulfite.$\newline$Therefore, the correct option is Option 1: B and E.