Answer

Glucose contains five and sucrose contains eight -OH groups. These -OH groups form H-bonds with water. Due to this extensive intermolecular H -bonding, glucose and sucrose are soluble in water irrespective of the fact that their molecular masses are $180 {~g} {~mol}^{-1}$ and $342 {~g} {~mol}^{-1}$ respectively.

In contrast, benzene and cyclohexane are simple molecules having low molecular masses. Even then they are insoluble in water. The reason being that these compounds do not contain -OH groups and hence do not form H -bonds with water.

Answer

Lactose is a diassacharide. On hydrolysis, it gives two molecules of monosaccharides, i.e., one molecule of each of D-(t)-glucose and D-(+)-galactose.

$ \underset{\text { Lactose }}{{C_{12}} {H_{22}} {O_{11}}}+{H_2} {O} \longrightarrow \underset{\text { D-(+)-Glucose }}{{C_6} {H_{12}} {O_6}}+\underset{\text { D-(+)-Galactose }}{{C_6} {H_{12}} {O_6}} $

Answer

The cyclic hemiacetal form of glucose contains an OH group at ${C}-1$ which gets hydrolysed in aqueous solution to produce the open chain aldehydic form which then reacts with ${NH}_2 {OH}$ to form the corresponding oxime. Thus, glucose contains an aldehydic group. In contrast, when glucose is reacted with acetic anhydride, the OH group at ${C}-1$, along with the four other OH groups at ${C}-2, {C}-3, {C}-4$ and ${C}-6$ form a pentaacetate.

Since the pentaacetate of glucose does not contain a free OH group at ${C}-1$, it cannot get hydrolysed in aqueous solution to produce the open chain aldehydic form and hence glucose pentaacetate does not react with ${NH}_2 {OH}$ to form glucose oxime. This proves that glucose pentaacetate does not contain the aldehyde group.