Biomolecules

Monosaccharide oxidises on reacting with HNO₃ to form carboxylic acids. Primary alcohols also oxidise to carboxylic acid alongwith aldehyde.

Complete reaction is as follows :

The enantiomer of product (P) has rotation −52.7^o . The two enantiomers (c and d) with their reactions are as follows :

The explanation of given statements are as follows :

(a) Two six membered cyclic hemiacetal form of D-(+)- glucose are called anomers.



Both are anomers.

(b) Oxidation of glucose in presence of Br₂ water gives gluconic acid.



(c) Monosaccharides can not be hydrolysed into polyhydroxy aldehydes and ketones.

(d) Hydrolysis of sucrose gives D-glucose and L-fructose.



Hence, options (a, c, d) are correct.

Invert sugar is prepared by acid catalyzed hydrolysis of sucrose.

${C_{12}}{H_{22}}{O_{11}} + {H_2}O\buildrel {HCl} \over \longrightarrow \underbrace {\mathop {{C_6}{H_{12}}{O_6}}\limits_{D - ( + ) - Glu\cos e} + \mathop {{C_6}{H_{12}}{O_6}}\limits_{D - ( - ) - Fructose} }_{Invert\,sugar}$

Specific rotation of invert sugar is

[$\alpha$]_mix = 0.5 $\times$ (+52) + 0.5 $\times$ ($-$92) = + 26 $-$ 46 = $-$20$^\circ$

On reaction with Br₂ water, invert sugar forms gluconic acid as one of the products. Br₂ water oxidises glucose into gluconic acid and fructose is not oxidised by it.

Reducing sugars contain cyclic hemiacetal or hemiketal groups in equilibrium with the open chain from having a free $-$CHO or $-$C = O group. Non-reducing sugars contain stable acetal or ketal structures their cyclic structures cannot be opened into an open chain form having a free carbonyl group. In X the glyco-sidic linkage is in between two anomeric carbon atoms while in Y it is only with one anomeric carbon, the other one is free. So, X is non-reducing while Y is reducing.