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 $D-series$.

(B) Lactose is a disaccharide composed of one molecule of $D-(+)$-galactose and one molecule of $D-(+)$-glucose.
The two monosaccharide units are linked together through a $\beta - 1, 4$-glycosidic linkage.
This means the $C-1$ of the galactose unit is connected to the $C-4$ of the glucose unit via an oxygen atom in the $\beta$-configuration.

(B) Maltose is a disaccharide formed by the condensation of two molecules of $\alpha - D - \text{glucose}$.
The glycosidic linkage is formed between the $C1$ of one glucose unit and the $C4$ of the other glucose unit.
Since both glucose units are in the $\alpha$ configuration, the bond is an $\alpha - 1, 4 - \text{glycosidic linkage}$.

(D) $1$. Glucose reacts with acetic anhydride to form glucose pentaacetate.
$2$. This reaction indicates that there are $5$ hydroxyl $(-OH)$ groups present in the glucose molecule.
$3$. The reaction is: $C_6H_{12}O_6 + 5(CH_3CO)_2O \rightarrow C_6H_7O(OCOCH_3)_5 + 5CH_3COOH$.

(B) Glucose reacts with hydroxylamine $(NH_2OH)$ to form an oxime. The formation of an oxime indicates the presence of a carbonyl group (aldehyde or ketone) in the glucose molecule.
Reaction: $CHO(CHOH)_4CH_2OH + NH_2OH \rightarrow CH=NOH(CHOH)_4CH_2OH + H_2O$.

(C) Step $1$: Bromine water ($Br_2$ water) is a mild oxidizing agent.
Step $2$: It specifically oxidizes the aldehyde group $(-CHO)$ of $glucose$ to a carboxylic acid group $(-COOH)$.
Step $3$: The reaction is: $CHO(CHOH)_4CH_2OH + [O] \xrightarrow{Br_2/H_2O} COOH(CHOH)_4CH_2OH$.
Step $4$: The resulting product is $Gluconic \ acid$.

(B) $Maltose$ is a carbohydrate formed by the condensation of two molecules of $D-glucose$. Since it yields two molecules of monosaccharides upon hydrolysis, it is classified as a disaccharide.

(D) $1$. In the open-chain structure of glucose, the aldehyde group is at $C-1$.
$2$. The hydroxyl $(-OH)$ group at $C-5$ attacks the carbonyl carbon $(C-1)$ to form a stable six-membered pyranose ring.
$3$. This intramolecular reaction between the aldehyde group at $C-1$ and the hydroxyl group at $C-5$ results in the formation of a hemiacetal structure.
$4$. Therefore, the $C-5$ carbon atom is involved in the ring closure.

(A) Step $1$: Starch is a polysaccharide composed of glucose units.
Step $2$: Starch consists of two components, amylose and amylopectin, both of which are polymers of $\alpha-D-glucose$.
Step $3$: These units are linked together by $\alpha-glycosidic$ linkages.
Step $4$: Therefore, the monomer unit of starch is $\alpha-D-glucose$.

(B) In the nomenclature of carbohydrates, the prefix '$D$' or '$L$' is used to specify the configuration of the chiral carbon atom furthest from the carbonyl group. It is based on the configuration of glyceraldehyde. Therefore, '$D$' signifies the configuration of the molecule.

(C) $1$. Ribose is a pentose sugar with the formula $C_5H_{10}O_5$, where each carbon atom is attached to a hydroxyl $(-OH)$ group.
$2$. Deoxyribose is a derivative of ribose with the formula $C_5H_{10}O_4$.
$3$. The prefix 'deoxy' indicates the removal of an oxygen atom.
$4$. In deoxyribose, the hydroxyl group at the $2'$ position of the ribose sugar is replaced by a hydrogen atom, meaning the oxygen atom is missing at the $2'$ carbon.
$5$. Therefore, the $2'$ carbon atom lacks the oxygen atom.

(D) The hydrolysis reaction of sucrose is: $C_{12}H_{22}O_{11} + H_2O \rightarrow C_6H_{12}O_6 (\text{glucose}) + C_6H_{12}O_6 (\text{fructose})$.
One mole of sucrose $(342 \text{ g})$ reacts with one mole of water $(18 \text{ g})$ to produce one mole of glucose $(180 \text{ g})$ and one mole of fructose $(180 \text{ g})$.
Total mass of products = Mass of glucose + Mass of fructose = $180 \text{ g} + 180 \text{ g} = 360 \text{ g}$.

(D) $1$. An aldohexose is a monosaccharide containing six carbon atoms and an aldehyde group $(-CHO)$.
$2$. $Ribose$ is an aldopentose ($5$ carbons).
$3$. $Fructose$ is a ketohexose ($6$ carbons with a ketone group).
$4$. $Lactose$ is a disaccharide.
$5$. $Glucose$ $(C_6H_{12}O_6)$ contains six carbon atoms and an aldehyde group, making it an aldohexose.

(B) $1$. Carbohydrates are classified based on the number of sugar units produced upon hydrolysis.
$2$. Monosaccharides like $Ribose$, $Fructose$, and $Glucose$ cannot be further hydrolyzed into simpler carbohydrates.
$3$. Oligosaccharides are carbohydrates that yield $2$ to $10$ monosaccharide units upon hydrolysis.
$4$. $Sucrose$ is a disaccharide (a type of oligosaccharide) which on hydrolysis yields one molecule of $Glucose$ and one molecule of $Fructose$ $(C_{12}H_{22}O_{11} + H_2O \rightarrow C_6H_{12}O_6 + C_6H_{12}O_6)$.

(C) $1$. $Sucrose$ is a disaccharide composed of $glucose$ and $fructose$.
$2$. $Maltose$ is a disaccharide composed of two units of $glucose$.
$3$. $Lactose$ is a disaccharide composed of $glucose$ and $galactose$.
$4$. $Raffinose$ is a trisaccharide composed of $galactose$, $glucose$, and $fructose$.
$5$. Since both $lactose$ and $raffinose$ contain $galactose$ as a constituent, the correct pair is $Lactose$ and $Raffinose$.

(C) Step $1$: Monosaccharides contain a free aldehyde or ketone group, which makes them reducing sugars. Thus, Statement $I$ is correct.
Step $2$: Sucrose is a disaccharide composed of glucose and fructose linked by a glycosidic bond between their anomeric carbons. It lacks a free aldehyde or ketone group, making it a non-reducing sugar. Therefore, it cannot reduce ammoniacal silver nitrate solution (Tollens' reagent). Thus, Statement $II$ is incorrect.