Carbohydrates Questions in English

(B) $Br_{2}$ water is a mild oxidizing agent.
Glucose $(CHO(CHOH)_{4}CH_{2}OH)$ on oxidation with $Br_{2}$ water selectively oxidizes the aldehyde group $(-CHO)$ to a carboxylic acid group $(-COOH)$,resulting in the formation of gluconic acid $(COOH(CHOH)_{4}CH_{2}OH)$.
The reaction is:
$CHO(CHOH)_{4}CH_{2}OH + [O] \xrightarrow{Br_{2}, H_{2}O} COOH(CHOH)_{4}CH_{2}OH$

(D) Glucose $(C_5H_{11}O_5CHO)$ acts as a reducing sugar and reacts with Fehling solution ($Cu^{2+}$ in alkaline medium).
During this reaction,the aldehyde group of glucose is oxidized to a carboxylic acid group,forming gluconic acid $(C_5H_{11}O_5COOH)$.
Simultaneously,the $Cu^{2+}$ ions in the Fehling solution are reduced to $Cu^+$ ions,which precipitate as red cuprous oxide $(Cu_2O)$.
The chemical reaction is: $C_5H_{11}O_5CHO + 2Cu^{2+} + 5OH^- \rightarrow C_5H_{11}O_5COO^- + Cu_2O + 3H_2O$.
Thus,both cuprous oxide and gluconic acid are produced.

(D) The reaction of glucose with phenylhydrazine proceeds in three steps:
$1$. Glucose reacts with one molecule of phenylhydrazine to form phenylhydrazone.
$2$. The phenylhydrazone then reacts with a second molecule of phenylhydrazine,which acts as an oxidizing agent,to form an intermediate keto-imine.
$3$. Finally,this intermediate reacts with a third molecule of phenylhydrazine to form the stable osazone.
Thus,a total of $3$ molecules of phenylhydrazine are required for the formation of osazone from one molecule of glucose.

(C) The constituents of the given carbohydrates are as follows:
$1$. Sucrose: Glucose + Fructose
$2$. Maltose: Glucose + Glucose
$3$. Lactose: Glucose + Galactose
$4$. Raffinose: Glucose + Fructose + Galactose
$5$. Stachyose: Glucose + Fructose + $2$ Galactose
Therefore,both $Raffinose$ and $Stachyose$ contain $Galactose$ as one of their constituent units.

(D) Sucrose $(C_{12}H_{22}O_{11})$ on hydrolysis produces $1$ mole of glucose and $1$ mole of fructose.
$C_{12}H_{22}O_{11} + H_2O \rightarrow \text{Glucose} + \text{Fructose}$.
Maltose $(C_{12}H_{22}O_{11})$ produces $2$ moles of glucose on hydrolysis.
$C_{12}H_{22}O_{11} + H_2O \rightarrow 2 \text{Glucose}$.
Therefore,maltose produces double the quantity of glucose compared to sucrose.

(A) $1$. Hydrolysis of $1 \ mole$ of $Sucrose$ $(C_{12}H_{22}O_{11})$ yields $1 \ mole$ of $Glucose$ and $1 \ mole$ of $Fructose$.
$2$. Hydrolysis of $1 \ mole$ of $Lactose$ $(C_{12}H_{22}O_{11})$ yields $1 \ mole$ of $Glucose$ and $1 \ mole$ of $Galactose$.
$3$. Hydrolysis of $1 \ mole$ of $Maltose$ $(C_{12}H_{22}O_{11})$ yields $2 \ moles$ of $Glucose$.
$4$. Comparing the glucose yield: $Sucrose$ gives $1 \ mole$,$Lactose$ gives $1 \ mole$,and $Maltose$ gives $2 \ moles$.
$5$. Therefore,$Sucrose$ and $Lactose$ both produce $1 \ mole$ of glucose per mole of carbohydrate.

(A) Hydrolysis of sucrose $(C_{12}H_{22}O_{11})$ in the presence of an acid or the enzyme invertase yields an equimolar mixture of glucose and fructose.
This mixture is known as invert sugar because the sign of optical rotation changes from dextrorotatory $(+)$ to levorotatory $(-)$ during the process.
Therefore,sucrose is the correct answer.

(A) Sucrose is a disaccharide formed by the condensation of one molecule of $\alpha-D$-glucose and one molecule of $\beta-D$-fructose.
The glycosidic linkage is formed between the $C-1$ of $\alpha-D$-glucose and the $C-2$ of $\beta-D$-fructose.
This is a non-reducing sugar because both the anomeric carbons are involved in the linkage.

(B) Cellulose is a linear polysaccharide consisting of a long chain of $D$-glucose units joined by $\beta-1,4$-glycosidic linkages.
These linkages are formed between the $C1$ carbon of one glucose unit and the $C4$ carbon of the adjacent glucose unit in the $\beta$-configuration.

(A) Maltose is a disaccharide composed of two $ \alpha-D-glucose $ units.
These two glucose units are linked together by an $ \alpha-1,4-glycosidic $ linkage.
In this linkage,the $ C1 $ of one glucose molecule is connected to the $ C4 $ of the other glucose molecule.

(A) Amylose is a linear polymer of $\alpha-D$-glucose units.
These glucose units are linked together by $\alpha-1,4$-glycosidic linkages.
Amylopectin,on the other hand,contains both $\alpha-1,4$ and $\alpha-1,6$-glycosidic linkages due to its branched structure.

(C) Cellulose is a polysaccharide consisting of a linear chain of several hundred to many thousands of $\beta(1 \rightarrow 4)$ linked $D$-glucose units.
Humans lack the necessary enzymes (cellulases) to hydrolyze the $\beta$-glycosidic linkages in cellulose.
Therefore,it cannot be digested and acts as dietary fiber (roughage),which adds bulk to the food and facilitates bowel movement in the human digestive system.

(B) Maltose is a disaccharide composed of two $\alpha-D-glucose$ units.
The glycosidic linkage is formed between the $C-1$ of one $\alpha-D-glucose$ molecule and the $C-4$ of another $\alpha-D-glucose$ molecule.
Therefore,it is an $\alpha-1,4$ glycosidic linkage.

(A) Sucrose is a disaccharide formed by the condensation of one molecule of $\alpha$-$D$-glucose and one molecule of $\beta$-$D$-fructose.
The glycosidic linkage is formed between the $C-1$ of $\alpha$-$D$-glucose and the $C-2$ of $\beta$-$D$-fructose.
Since the reducing groups (aldehyde of glucose and ketone of fructose) are involved in this bond,sucrose is a non-reducing sugar.

(C) $Maltose$ and $Lactose$ are disaccharides.
$Raffinose$ is a trisaccharide.
$Stachyose$ is a tetrasaccharide.

(B) Sucrose is a disaccharide formed by the condensation of one molecule of $\alpha$-$D$-glucose and one molecule of $\beta$-$D$-fructose.
The glycosidic linkage is formed between the $C-1$ of $\alpha$-glucose and the $C-2$ of $\beta$-fructose.
This bond involves the anomeric carbon atoms of both monosaccharides,which is why sucrose is a non-reducing sugar.

(C) Lactose is a disaccharide composed of $\beta-D-galactose$ and $\beta-D-glucose$.
The glycosidic linkage is formed between the $C-1$ of $\beta-D-galactose$ and the $C-4$ of $\beta-D-glucose$.
Therefore,the linkage is a $\beta-1,4-glycosidic$ linkage.

(D) $Stachyose$ is a tetrasaccharide.
Upon hydrolysis,it yields four monosaccharide units: two units of $D-galactose$,one unit of $D-glucose$,and one unit of $D-fructose$.
$Glycogen$ and $Cellulose$ are polysaccharides,while $Ribose$ is a monosaccharide.

(A) Lactose is a disaccharide composed of $\beta$-$D$-galactose and $\beta$-$D$-glucose units.
In lactose,the glycosidic linkage is formed between the $C-1$ carbon of $\beta$-$D$-galactose and the $C-4$ carbon of $\beta$-$D$-glucose.
Therefore,the linkage is a $\beta-1,4$-glycosidic linkage.

(D) Sucrose,maltose,and lactose are disaccharides,which consist of two monosaccharide units.
Raffinose is a trisaccharide,which consists of three monosaccharide units (galactose,glucose,and fructose).

(D) Sucrose is a disaccharide with the formula $C_{12}H_{22}O_{11}$.
When treated with concentrated $H_2SO_4$,it undergoes dehydration.
The concentrated $H_2SO_4$ acts as a strong dehydrating agent and removes water molecules from the sucrose.
The reaction is: $C_{12}H_{22}O_{11} \xrightarrow{\text{conc. } H_2SO_4} 12C + 11H_2O$.
Thus,the products obtained are sugar charcoal $(12C)$ and water $(11H_2O)$.

(B) The chemical equation for the acid hydrolysis of sucrose is:
$C_{12}H_{22}O_{11} + H_2O \xrightarrow{H^{+}} C_6H_{12}O_6 (\text{glucose}) + C_6H_{12}O_6 (\text{fructose})$
From the stoichiometry of the reaction,$1 \ mol$ of sucrose reacts with $1 \ mol$ of $H_2O$ to produce $1 \ mol$ of glucose and $1 \ mol$ of fructose.
Therefore,to obtain $1 \ mol$ of glucose,we require $1 \ mol$ of $H_2O$.
The molar mass of water $(H_2O)$ is $18 \ g/mol$.
Mass of water = $\text{moles} \times \text{molar mass} = 1 \ mol \times 18 \ g/mol = 18 \ g$.

(B) The hydrolysis products of the given carbohydrates are as follows:
$1$. Sucrose: $1$ mole of glucose + $1$ mole of fructose.
$2$. Stachyose: $2$ moles of galactose + $1$ mole of glucose + $1$ mole of fructose.
$3$. Raffinose: $1$ mole of galactose + $1$ mole of glucose + $1$ mole of fructose.
$4$. Lactose: $1$ mole of galactose + $1$ mole of glucose.
Therefore,$Stachyose$ yields the highest number ($2$ moles) of galactose molecules upon hydrolysis.

(C) Lactose is a disaccharide sugar composed of galactose and glucose units. It is commonly known as milk sugar because it is found in the milk of mammals.

(A) Sucrose is a disaccharide composed of $\alpha-D-$glucose and $\beta-D-$fructose units.
These two monosaccharide units are joined by a glycosidic linkage between $C-1$ of $\alpha-D-$glucose and $C-2$ of $\beta-D-$fructose.
This specific linkage is known as the $\alpha, \beta-1,2-$ glycosidic linkage.
Therefore,the correct answer is $A$ (Sucrose).

(B) $1$. The hydrolysis of maltose yields two molecules of glucose $(X = \text{Glucose})$.
$2$. When glucose is treated with excess dilute nitric acid $(\text{dil. HNO}_3)$,it acts as a strong oxidizing agent and oxidizes both the terminal primary alcohol group $(-CH_2OH)$ and the aldehydic group $(-CHO)$ to carboxylic acid groups $(-COOH)$.
$3$. This oxidation results in the formation of saccharic acid (also known as glucaric acid).

(A) Lactose is a disaccharide with the formula $C_{12}H_{22}O_{11}$.
On hydrolysis in the presence of an acid or the enzyme lactase,it undergoes cleavage to produce one molecule of $D-(+)$-glucose and one molecule of $D-(+)$-galactose.
The chemical reaction is:
$C_{12}H_{22}O_{11} + H_2O$ $\xrightarrow{H_3O^+ \text{ or lactase}} C_6H_{12}O_6 (\text{Glucose}) + C_6H_{12}O_6 (\text{Galactose})$
Therefore,one mole of lactose yields $1.0 \ mol$ of glucose and $1.0 \ mol$ of galactose.

(A) Raffinose is a trisaccharide with the molecular formula $C_{18}H_{32}O_{16}$.
Upon complete acid-catalyzed hydrolysis,one mole of raffinose reacts with two moles of water to yield one mole each of glucose,fructose,and galactose.
The chemical equation is: $C_{18}H_{32}O_{16} + 2H_2O$ $\xrightarrow{H^+} C_6H_{12}O_6 (\text{Glucose}) + C_6H_{12}O_6 (\text{Fructose}) + C_6H_{12}O_6 (\text{Galactose})$.

(B) Cellulose is a linear polysaccharide consisting of $D$-glucose units joined by $\beta$-glycosidic linkages between $C-1$ of one glucose unit and $C-4$ of the adjacent glucose unit.
Therefore,cellulose contains $1 \rightarrow 4$ $\beta$-glycosidic linkages.
The correct option is $(B)$.

(D) Sucrose is a disaccharide formed by the condensation of one molecule of $\alpha-D$-glucopyranose and one molecule of $\beta-D$-fructofuranose.
In this structure,the glycosidic linkage is formed between the $C-1$ carbon of $\alpha-D$-glucopyranose and the $C-2$ carbon of $\beta-D$-fructofuranose.
Therefore,the correct option is $D$.

(D) Sugars that contain a free aldehydic or ketonic group are called reducing sugars,as they can reduce Tollen's reagent or Fehling's solution.
$Ribose$,$Lactose$,and $Maltose$ are reducing sugars.
$Sucrose$ is a non-reducing sugar because its glycosidic linkage involves the anomeric carbons of both glucose and fructose,meaning no free aldehydic or ketonic group can be produced in solution.

(A) Glycogen is a natural biopolymer of monosaccharides,specifically a branched polymer consisting of linear chains of $D$-glucose residues.

(C) Sucrose on hydrolysis in the presence of dilute acid or the enzyme invertase gives one molecule each of glucose and fructose.
$C_{12}H_{22}O_{11} (\text{Sucrose}) + H_2O$ $\xrightarrow[\text{or invertase}]{H_3O^+} C_6H_{12}O_6 (D(+) \text{Glucose}) + C_6H_{12}O_6 (D(-) \text{Fructose})$

(D) An asymmetric carbon atom (chiral center) is a carbon atom bonded to four different groups.
$Glyceraldehyde$ $(CHO-CH(OH)-CH_2OH)$ has the structure where the central carbon is bonded to $-CHO$,$-H$,$-OH$,and $-CH_2OH$ groups.
Since all four groups are different,the central carbon is the only asymmetric carbon atom in the molecule.
Therefore,the correct option is $D$.

(D) Stachyose is a tetrasaccharide consisting of two $\alpha-D$-galactose units,one $\alpha-D$-glucose unit,and one $\beta-D$-fructose unit.
These units are sequentially linked as $gal(\alpha 1$ $\rightarrow 6)gal(\alpha 1$ $\rightarrow 6)glc(\alpha 1$ $\rightarrow 2\beta)fru$.
It is naturally found in numerous vegetables and plants.

(C) Sucrose on hydrolysis with dilute acid or the enzyme invertase yields an equimolar mixture of glucose and fructose.
$C_{12}H_{22}O_{11} + H_2O$ $\xrightarrow{\text{H}^+ \text{ or Invertase}} C_6H_{12}O_6 (\text{glucose}) + C_6H_{12}O_6 (\text{fructose})$

(C) The open-chain structure of ribose is given by the formula $CHO-(CHOH)_3-CH_2OH$.
In this structure,there are $3$ hydroxyl groups attached to the middle carbons in the $(CHOH)_3$ unit and $1$ hydroxyl group attached to the terminal $CH_2OH$ group.
Therefore,the total number of $-OH$ groups in one molecule of ribose is $3 + 1 = 4$.

(B) Nucleotides are composed of a nitrogenous base, a phosphate group, and a pentose sugar.
In $RNA$ (Ribonucleic Acid), the pentose sugar present is $\beta-D$-ribose.
In $DNA$ (Deoxyribonucleic Acid), the pentose sugar present is $\beta-D-2$-deoxyribose.
Therefore, the sugar present in the nucleotide of $RNA$ is $D$-Ribose.

(D) Polynucleotides,such as $DNA$ and $RNA$,are formed by the polymerization of nucleotides.
Each nucleotide consists of a nitrogenous base,a phosphate group,and a sugar molecule.
The sugar present in these nucleotides is a $5$-carbon sugar,known as a pentose sugar.
Specifically,$RNA$ contains $\beta-D-ribose$ and $DNA$ contains $\beta-D-2-deoxyribose$,both of which are pentose sugars.

(A) $L$-Lactic acid is primarily found in curd,produced by the fermentation of lactose by lactic acid bacteria.

(B) In the green technology developed by $Drath$ and $Frost$,adipic acid is enzymatically synthesized from $Glucose$ rather than $Benzene$,which is carcinogenic.

(A) Glyceraldehyde is an aldotriose with the structural formula $HOCH_{2}-CH(OH)-CHO$.
Counting the atoms in the structure: there are $3$ carbon atoms,$6$ hydrogen atoms,and $3$ oxygen atoms.
Therefore,the molecular formula is $C_{3}H_{6}O_{3}$.

(B) The structure of threose is $CHO-CH(OH)-CH(OH)-CH_2OH$.
$A$ chiral carbon atom is a carbon atom bonded to four different groups.
In the structure of threose,the two central carbon atoms (at positions $2$ and $3$) are each bonded to four different groups: $-H$,$-OH$,$-CHO$ (or $-CH_2OH$),and the other chiral carbon.
Therefore,there are $2$ chiral carbon atoms in the threose molecule,as indicated by the asterisks $(*)$ in the structure.

(C) Lactose is a disaccharide composed of two monosaccharide units.
These units are $β-D$-galactose and $β-D$-glucose.
The glycosidic linkage present in lactose is $β(1 \rightarrow 4)$ linkage.
Therefore,the correct option is $C$.

(D) Glucose is an aldohexose $(C_6H_{12}O_6)$.
When heated with $HI$,it undergoes reduction to form $n-hexane$,confirming the presence of a straight chain of $6$ carbon atoms.
It reacts with hydroxylamine $(NH_2OH)$ to form an oxime,confirming the presence of a carbonyl group.
However,glucose exists in a pyranose structure (a $6$-membered ring) in its cyclic form,not a furanose structure (a $5$-membered ring). Fructose exists in a furanose structure.
Therefore,the statement that glucose contains a furanose structure is incorrect.

(B) Amylopectin is a branched-chain polymer of $\alpha$-$D$-glucose units.
It consists of a linear chain formed by $C_1-C_4$ glycosidic linkages.
Additionally,it contains branching points formed by $C_1-C_6$ glycosidic linkages.
Therefore,the correct answer is $B$.