Carbohydrates Questions in English

(B) The reactions of glucose are as follows:
$1$. Glucose reacts with hydroxylamine $(NH_2OH)$ to form an oxime,known as Glucoxime $(A-IV)$.
$2$. Glucose reacts with bromine water ($Br_2$ water) to undergo mild oxidation of the aldehyde group to a carboxylic acid,forming Gluconic acid $(B-I)$.
$3$. Glucose reacts with excess acetic anhydride to form Glucose pentaacetate,indicating the presence of five hydroxyl groups $(C-II)$.
$4$. Glucose reacts with concentrated nitric acid $(HNO_3)$ to undergo strong oxidation of both the aldehyde and the primary alcohol group,forming Saccharic acid $(D-III)$.
Thus,the correct matching is $A-IV, B-I, C-II, D-III$.

(D) Sucrose is dextrorotatory $(+66.5^\circ)$.
Upon hydrolysis,it yields an equimolar mixture of $D-(+)$-glucose $(+52.5^\circ)$ and $D-(-)$-fructose $(-92.4^\circ)$.
Since the magnitude of the laevorotation of fructose $(-92.4^\circ)$ is greater than the dextrorotation of glucose $(+52.5^\circ)$,the resulting mixture is laevorotatory.
This process is known as the inversion of sugar.
Statement $I$ is correct.
Statement $II$ is incorrect because it incorrectly states that glucose is laevorotatory and fructose is dextrorotatory,whereas the opposite is true.

(B) Glucose contains an aldehyde group $(-CHO)$.
Aldehydes react with hydroxylamine $(NH_2OH)$ to form oximes.
The general reaction is: $RCHO + NH_2OH \rightarrow RCH=N-OH + H_2O$.
Therefore,glucose reacts with $NH_2OH$ to form glucose oxime.
Thus,option $(B)$ is the correct reagent.

(A) polysaccharide is a carbohydrate that consists of a number of sugar molecules bonded together.
$Ribose$ is a simple sugar or a monosaccharide (specifically a pentose sugar).
$Starch$,$Gum$,and $Glycogen$ are all examples of polysaccharides.
Therefore,$Ribose$ is not a polysaccharide.

(C) -aldotetrose has the general structure $CHO-(CHOH)_2-CH_2OH$.
Upon oxidation with concentrated $HNO_3$,the terminal aldehyde $(-CHO)$ and primary alcohol $(-CH_2OH)$ groups are oxidized to carboxylic acid $(-COOH)$ groups,resulting in a tartaric acid derivative $(HOOC-(CHOH)_2-COOH)$.
For this dicarboxylic acid to be optically inactive,it must contain a plane of symmetry,making it a meso compound.
In the Fischer projection of the $D$-aldotetrose,the $D$-configuration implies that the hydroxyl group at the chiral carbon furthest from the aldehyde group (i.e.,$C-3$) is on the right side.
For the resulting dicarboxylic acid to be meso,the hydroxyl groups must be on the same side (erythro configuration) so that a plane of symmetry exists.
Therefore,the $D$-aldotetrose must have both hydroxyl groups on the right side,which corresponds to $D$-erythrose.

(D) . Glucose exists in $\alpha$ and $\beta$ anomeric forms. (Correct)
$B$. Anomers differ in configuration only at the hemiacetal carbon $(C_1)$. (Correct)
$C$. Melting points: $\alpha$-$D$-glucose is $419 \text{ K}$ and $\beta$-$D$-glucose is $423 \text{ K}$. Thus,the melting point of $\beta$-anomer is greater than $\alpha$-anomer. (Incorrect)
$D$. Specific rotation: $\alpha$-anomer is $+112^\circ$ and $\beta$-anomer is $+19^\circ$. The values are swapped in the statement. (Incorrect)
$E$. Crystallization of glucose from hot saturated aqueous solution at $371 \text{ K}$ yields $\alpha$-$D$-glucose,and below $303 \text{ K}$ yields $\beta$-$D$-glucose. (Correct)
Therefore,statements $A, B$,and $E$ are correct.

(B) Statement $B$ is incorrect because oligosaccharides like sucrose,upon hydrolysis,yield different monosaccharide units (glucose and fructose). Monosaccharides are simple sugars that cannot be hydrolyzed further,and most (including all aldoses) are reducing sugars. Polysaccharides consist of a large number of monosaccharide units linked together,and $D-(+)$-glucose exists in an equilibrium between its open-chain and cyclic forms.

(A) Statement $I$ is true because $\alpha$ and $\beta$ isomers of $D-(+)-glucose$ are indeed anomers,differing only in the configuration at the anomeric carbon $(C1)$.
Statement $II$ is also true; if we look at the Fischer projections of $D-glucose$ and $D-fructose$,the configurations of the chiral carbons at $C3$,$C4$,and $C5$ are identical in both,as both belong to the $D-series$.