CBSE Class 12 Chemistry Biomolecules NCERT Solutions

The solubility of a compound in water is largely determined by its ability to form hydrogen bonds with water molecules. Glucose possesses five hydroxyl (-OH) groups, and sucrose has eight hydroxyl (-OH) groups. These numerous -OH groups allow both glucose and sucrose to form extensive hydrogen bonds with water molecules. This strong interaction overcomes the forces holding the solute molecules together and the forces between water molecules, leading to their dissolution in water.

In contrast, cyclohexane and benzene are nonpolar organic compounds that do not contain hydroxyl groups or other polar functionalities capable of forming hydrogen bonds. Consequently, they cannot interact favorably with polar water molecules and are therefore insoluble in water.

Lactose is a disaccharide. Upon hydrolysis, it breaks down into its constituent monosaccharides. Lactose is specifically composed of one molecule of β-D-galactose and one molecule of β-D-glucose, linked by a β-1,4 glycosidic bond.

The hydrolysis reaction can be represented as:

$$C\_\{12\}H\_\{22\}O\_\{11\}\; \backslash text\{\; (Lactose)\}\; +\; H\_2O\; \backslash longrightarrow\; C\_6H\_\{12\}O\_6\; \backslash text\{\; (D-Glucose)\}\; +\; C\_6H\_\{12\}O\_6\; \backslash text\{\; (D-Galactose)\}$$

Therefore, the expected products of lactose hydrolysis are D-glucose and D-galactose.

The presence of an aldehyde group in D-glucose is indicated by its reaction with hydroxylamine (NH₂OH) to form an oxime. This reaction occurs because, in an aqueous medium, glucose exists in equilibrium between its cyclic hemiacetal form and its open-chain aldehydic form. The open-chain form readily reacts with hydroxylamine.

However, when D-glucose is treated with acetic anhydride ((CH₃CO)₂O), it forms D-glucose pentaacetate. In this derivative, all five hydroxyl groups of glucose, including the one at the anomeric carbon (C1), are acetylated. The acetylation of the hydroxyl group at the anomeric carbon converts the cyclic hemiacetal structure into a stable cyclic acetal structure. This process prevents the formation of the open-chain aldehydic form.

Since the open-chain aldehydic form is not available in D-glucose pentaacetate, it does not react with hydroxylamine to form an oxime. This absence of reaction with hydroxylamine is evidence that the aldehyde group is masked or absent in the pentaacetate derivative.

The reaction can be summarized as:

Glucose (cyclic form) + (CH₃CO)₂O → Glucose Pentaacetate (stable cyclic acetal)

Glucose Pentaacetate + NH₂OH → No reaction (No oxime formation)