Carbohydrates Questions in English

(A) The plant cell wall is primarily composed of $Cellulose$.
$Cellulose$ is a structural polysaccharide and is considered a complex carbohydrate because it provides structural support and protection to plant cells.
It consists of a linear chain of several hundred to many thousands of $1,4-$linked $\beta-D-Glucose$ units.

(C) The presence of glucose (sugar) in urine is detected using Benedict's test.
Benedict's reagent contains $Cu^{2+}$ ions which are reduced to $Cu^+$ ions in the presence of a reducing sugar like glucose,forming a red precipitate of cuprous oxide $(Cu_2O)$.
The reaction is: $\text{Glucose} + \text{Benedict's solution} \to \text{Red colour } (Cu_2O)$.

(C) When sucrose $(C_{12}H_{22}O_{11})$ is heated with concentrated nitric acid $(HNO_3)$,it undergoes oxidation to form oxalic acid $(C_2H_2O_4)$ as the final product.
$C_{12}H_{22}O_{11} + 18[O] \xrightarrow{conc. HNO_3} 6C_2H_2O_4 + 5H_2O$

(D) Amylopectin is a branched-chain polymer of $\alpha$-$D$-glucose units. Due to its high molecular weight and branched structure,it is insoluble in water. Therefore,the correct option is $D$.

(C) Ribose is a monosaccharide with five carbon atoms,known as a pentose sugar. Its molecular formula is $C_5H_{10}O_5$.
Since it contains five carbon atoms,the statement that it has six carbon atoms is incorrect.
Ribose is a polyhydroxy aldehyde (aldose),meaning it contains multiple hydroxyl $(-OH)$ groups and an aldehydic $(-CHO)$ functional group.
Due to the presence of chiral carbon atoms,it exhibits optical activity.

(C) The molecular formula of maltose is $C_{12}H_{22}O_{11}$.
From the formula,it is clear that one molecule of maltose contains $11$ oxygen atoms.
Therefore,the correct option is $(C)$.

(A) Sucrose is a disaccharide composed of one molecule of $\alpha-D$-glucose and one molecule of $\beta-D$-fructose.
These two monosaccharide units are linked by a glycosidic linkage between $C1$ of $\alpha-D$-glucose and $C2$ of $\beta-D$-fructose.
Therefore,the systematic $IUPAC$ name of sucrose is $\alpha-D$-glucopyranosyl-$\beta-D$-fructofuranoside.

(B) Sucrose is dextrorotatory in nature with a specific rotation of $+66.5^{\circ}$.
Upon hydrolysis,it produces an equimolar mixture of $D-(+)$-glucose (dextrorotatory,$+52.5^{\circ}$) and $D-(-)$-fructose (laevorotatory,$-92^{\circ}$),resulting in an overall laevorotatory mixture known as invert sugar.
The hydrolysis reaction is:
$C_{12}H_{22}O_{11} + H_2O \xrightarrow{H^+} C_6H_{12}O_6 (\text{Glucose}) + C_6H_{12}O_6 (\text{Fructose})$

(A) Sucrose is dextrorotatory with a specific rotation of $+66.5^{\circ}$.
On hydrolysis,it produces an equimolar mixture of $D-(+)$-glucose $(+52.5^{\circ})$ and $D-(-)$-fructose $(-92.4^{\circ})$.
The net rotation of the resulting mixture is $\frac{+52.5^{\circ} + (-92.4^{\circ})}{2} = -19.95^{\circ}$.
Since the net rotation is negative,the mixture is laevorotatory and is known as invert sugar.

(B) Sucrose is a non-reducing sugar because both its anomeric carbons (the $C1$ of glucose and $C2$ of fructose) are involved in the formation of the glycosidic linkage. Therefore,it does not have a free hemiacetal or hemiketal group to reduce Tollens' reagent or Fehling's solution.

(D) Sucrose is a disaccharide formed by the condensation of $\alpha-D-glucose$ and $\beta-D-fructose$.
In the structure of sucrose,the glucose unit is in the pyranose form (containing a hemiacetal group) and the fructose unit is in the furanose form (containing a hemiketal group).
However,in the disaccharide linkage,the glycosidic bond is formed between the anomeric carbons of both monosaccharides.
Therefore,sucrose does not contain a free aldehydic $(-CHO)$ or free ketonic $(>C=O)$ group in its open-chain form,as both are involved in the glycosidic linkage.
Thus,the correct answer is $(d)$.

(A) Sucrose is a disaccharide composed of $\alpha$-$D$-glucose and $\beta$-$D$-fructose units.
In the structure of sucrose,the glucose unit exists in the pyranose form (six-membered ring),while the fructose unit exists in the furanose form (five-membered ring).
Therefore,the fructose molecule in sucrose exists as furanose.

(C) Fructose is laevorotatory.
It rotates the plane of polarized light in an anticlockwise direction.
Thus,it is also called levulose and is named as $D-(-)$-fructose.

(C) Glucose contains an aldehyde group $(-CHO)$ in its open-chain structure,which acts as a reducing agent.
Ammoniacal silver nitrate,also known as $Tollen's$ reagent,is used to identify aldehydes.
Glucose reacts with $Tollen's$ reagent to form a silver mirror,confirming the presence of the aldehyde group.
The reaction is: $R-CHO + 2[Ag(NH_3)_2]^+ + 3OH^- \rightarrow R-COO^- + 2Ag + 2H_2O + 4NH_3$.

(C) Sucrose is a disaccharide composed of $1$ unit of $\alpha-D-glucose$ and $1$ unit of $\beta-D-fructose$ linked by a glycosidic bond.
Upon hydrolysis,the glycosidic linkage breaks,yielding $1$ molecule of glucose and $1$ molecule of fructose.

(A) Disaccharides are oligosaccharides formed by the condensation of two monosaccharide units.
Lactose is a disaccharide composed of one molecule of $D$-glucose and one molecule of $D$-galactose.
Starch and cellulose are polysaccharides,while glucose is a monosaccharide.
Therefore,lactose is the correct answer.

(C) Glucose contains just $1$ ring ($1$ carbohydrate unit) so it is a monosaccharide.
It is not an oligosaccharide.
Oligosaccharides contain $2-10$ carbohydrate units.
Therefore,glucose cannot be classified as an oligosaccharide.

(B) $Glucose$ reacts with $3$ moles of phenylhydrazine $(C_6H_5NHNH_2)$ to form a crystalline derivative known as glucosazone.
The reaction is:
$CH_2OH(CHOH)_3CHO + 3C_6H_5NHNH_2 \rightarrow CH_2OH(CHOH)_3C(=NNHC_6H_5)CH=NNHC_6H_5 + C_6H_5NH_2 + NH_3 + 2H_2O$

(C) The hydrolysis of canesugar (sucrose) yields an equimolar mixture of glucose and fructose.
$C_{12}H_{22}O_{11} + H_2O \xrightarrow{H^{+}} C_6H_{12}O_6 (\text{Glucose}) + C_6H_{12}O_6 (\text{Fructose})$

(A) Monosaccharides,also called simple sugar,are the simplest form of sugar and the most basic units of carbohydrates.
They cannot be further hydrolyzed to simpler chemical compounds.
Glucose is an important monosaccharide in that it provides both energy and structure to many organisms.

(B) Monosaccharides are the simplest carbohydrates and cannot be hydrolysed into smaller,simpler carbohydrate units.

(D) monosaccharide containing an aldehyde group $(-CHO)$ is called an aldose.
In an aldose,the carbonyl group is located at the end of the carbon chain.
Conversely,a monosaccharide containing a ketone group $(C=O)$ is called a ketose.

(A) Monosaccharides are classified based on the functional group present in their structure.
If a monosaccharide contains an aldehyde group $(-CHO)$,it is called an aldose.
If a monosaccharide contains a keto group $(>C=O)$,it is known as a ketose.

(D) Starch is a polysaccharide that forms a complex with iodine,resulting in a characteristic dark blue colour.
$Starch + I_2 \to \text{Blue complex}$

(D) disaccharide is a carbohydrate that yields two molecules of monosaccharides on hydrolysis.
$Glucose$ and $Fructose$ are monosaccharides.
$Raffinose$ is a trisaccharide.
$Maltose$ is a disaccharide formed by the condensation of two $Glucose$ units linked by an $\alpha-1,4-glycosidic$ linkage.

(D) reducing sugar is a carbohydrate that possesses a free aldehyde or ketone group,which allows it to act as a reducing agent.
$Glucose$,$fructose$,and $galactose$ are all monosaccharides that contain a free carbonyl group (aldehyde in $glucose$ and $galactose$,and a ketone group in $fructose$ that can tautomerize to an aldehyde in alkaline solution).
Therefore,all of these are reducing sugars.

(A) The formation of osazone involves the reaction of $D-glucose$ with $3$ molecules of phenylhydrazine.
First,$D-glucose$ reacts with one molecule of phenylhydrazine to form $D-glucose$ phenylhydrazone.
Second,the second molecule of phenylhydrazine acts as an oxidizing agent,oxidizing the $C-2$ hydroxyl group to a keto group.
Third,the third molecule of phenylhydrazine reacts with the newly formed keto group to form $D-glucosazone$.
The involvement of only the first two carbon atoms is due to the formation of a stable chelated structure (hydrogen bonding) between the phenylhydrazone group at $C-1$ and the imine group at $C-2$,which prevents further reaction with other hydroxyl groups.

(C) Mutarotation is the change in the optical rotation of a chiral center in solution due to the equilibrium between the $\alpha$ and $\beta$ anomers of glucose. This process is catalyzed by both acids and bases,but it also occurs spontaneously in a neutral solvent like water. Therefore,glucose shows mutarotation in a neutral solvent.

(D) The structure of $Glucose$ is $CHO-(CHOH)_4-CH_2OH$.
$1$. It contains one aldehyde $(-CHO)$ group at $C_1$.
$2$. It contains five hydroxyl $(-OH)$ groups attached to different carbon atoms.
$3$. It contains one primary alcoholic group $(-CH_2OH)$ at $C_6$.
Therefore,all the given statements are correct.

(B) Glycogen is a multi-branched polysaccharide of glucose that serves as a form of energy storage in animals and fungi.
This polysaccharide structure represents the main storage form of glucose in the human body.

(B) Maltose is a disaccharide composed of two units of glucose.
The hydrolysis of maltose is represented as:
$Maltose \xrightarrow{Hydrolysis} \text{Glucose} + \text{Glucose}$

(A) Fructose is a naturally occurring monosaccharide found in many fruits,which is why it is commonly known as fruit sugar.

(B) Carbohydrates are defined as optically active polyhydroxy aldehydes or polyhydroxy ketones,or compounds that produce such units on hydrolysis.
They are organic compounds organized in the form of aldehydes or ketones with multiple hydroxyl groups attached to the carbon chain.
The general empirical structure for carbohydrates is $(CH_2O)_n$.

(C) Glucose $(C_6H_{12}O_6)$ and fructose $(C_6H_{12}O_6)$ have the same molecular formula but different structural arrangements.
Therefore,they are classified as isomers.

(C) The hydrolysis of sucrose yields an equimolar mixture of $D-(+)$-glucose and $D-(-)$-fructose.
This process is called inversion because the optical rotation of the solution changes from dextrorotatory $(+)$ to levorotatory $(-)$ as the reaction proceeds.
The resulting mixture is known as invert sugar.

(A) Starch is a polysaccharide,which is a polymer of $\alpha$-glucose units.
It is formed by the condensation of glucose units linked by glycosidic bonds.
The general formula of starch is $(C_6H_{10}O_5)_n$.
It exists in two forms: amylose and amylopectin.
Amylose is a linear polymer of $\alpha$-$D$-glucose units linked by $\alpha-1,4$-glycosidic bonds.
Amylopectin is a branched polymer of $\alpha$-$D$-glucose units,where the main chain is linked by $\alpha-1,4$-glycosidic bonds and branching occurs via $\alpha-1,6$-glycosidic bonds.

(B) The correct answer is $(B)$.
Carbohydrates are polyhydroxy aldehydes or ketones or substances that yield these on hydrolysis.
The simplest carbohydrates are trioses,which contain $3$ carbon atoms.
For example,Glyceraldehyde $(HOCH_2-CH(OH)-CHO)$ is the simplest aldose sugar containing $3$ carbons.

(D) Lactose is a disaccharide composed of $\beta-D$-galactose and $\beta-D$-glucose units linked by a $\beta-1,4$-glycosidic bond.
On hydrolysis,it breaks down into one molecule of $\beta-D$-galactose and one molecule of $\beta-D$-glucose.

(A) non-reducing sugar is one that does not have a free aldehydic or ketonic group capable of acting as a reducing agent.
In cane sugar (sucrose),both the anomeric carbons of glucose and fructose are involved in the glycosidic linkage.
Therefore,there is no free aldehydic or ketonic group present.
In contrast,$Fructose$,$Lactose$,and $Cellobiose$ possess free functional groups or are capable of mutarotation to form them,making them reducing sugars.
Thus,cane sugar is a non-reducing sugar.

(C) Cellulose is a linear polysaccharide consisting of several hundred to many thousands of $D$-glucose units. These units are joined by $\beta$-glycosidic linkages between $C1$ of one glucose unit and $C4$ of the next.

(C) $Starch$ $\xrightarrow{\text{Diastase}}$ $Maltose$ $\xrightarrow{\text{Maltase}}$ $Glucose$.
$Starch$ is a polysaccharide that is first hydrolyzed into the disaccharide $Maltose$ by the enzyme $Diastase$,and subsequently,$Maltose$ is hydrolyzed into $Glucose$ by the enzyme $Maltase$.

(A) Aldopentoses have the general formula $CHO-(CHOH)_3-CH_2OH$.
Since the two ends of the molecule ($CHO$ and $CH_2OH$ groups) are different,there is no possibility of a plane of symmetry in the open-chain structure.
Thus,all stereoisomers of aldopentoses are chiral and optically active.
Therefore,$I$,$II$,and $III$ are all optically active.

(C) $\alpha-D$-glucose and $\beta-D$-glucose are anomers of each other.
Anomers are a type of stereoisomer that differ in configuration only at the anomeric carbon (the $C-1$ carbon in glucose).
Since they differ in the spatial arrangement of the $-OH$ group at the $C-1$ position,they are said to differ in their configuration.

(B) Blood sugar,or $D-glucose$,is the main carbohydrate found in human blood.
It serves as the primary source of energy for the body's cells.
The human body contains approximately $5 \ L$ of blood,which typically contains about $4 \ g$ of dissolved glucose.

(C) $Glucose$ contains an aldehyde group $(-CHO)$ which is capable of reducing $Tollens'$ reagent $([Ag(NH_3)_2]^+)$ to metallic silver $(Ag)$.
Since $Glucose$ reduces the $Tollens'$ reagent while itself getting oxidized to gluconic acid,it acts as a reducing agent.

(A) -fructose exists in a cyclic form known as fructofuranose.
In the cyclic structure of $D$-fructose,the ring consists of $4$ carbon atoms and $1$ oxygen atom.
Therefore,the total number of atoms in the cyclic ring is $4 + 1 = 5$.

(C) Glucose contains five hydroxyl $(-OH)$ groups. When it reacts with acetic anhydride,all five $-OH$ groups are acetylated to form glucose penta-acetate. The reaction is as follows:
$CHO-(CHOH)_4-CH_2OH + 5(CH_3CO)_2O \rightarrow CHO-(CHOCOCH_3)_4-CH_2OCOCH_3 + 5CH_3COOH$
This confirms the presence of five hydroxyl groups in glucose.

(A) In sucrose,the two monosaccharide units are joined by an $\alpha-1, 2$-glycosidic bond.
Since sucrose does not have a free hemiacetal or hemiketal carbon,it cannot open to form an aldehyde or ketone group.
Therefore,it is a non-reducing sugar.

(C) The enzyme amylase acts as a catalyst for the hydrolysis of starch.
Starch is a polysaccharide composed of glucose units.
Amylase specifically breaks down starch into the disaccharide known as maltose.
Therefore,the final product obtained is $Maltose$.

(C) In the liver,galactose is converted to glucose-$6$-phosphate via the Leloir pathway to enter the glycolytic pathway.