Carbohydrates Questions in English

(B) Starch is a polymer of $\alpha-D$-glucose.
Amylose and Amylopectin are the two components of starch.
Lactose is a disaccharide composed of $\beta-D$-galactose and $\beta-D$-glucose.
Proteins are biopolymers composed of various types of amino acids.

(D) The enzyme maltase specifically catalyzes the hydrolysis of maltose into two molecules of $D$-glucose.
$C_{12}H_{22}O_{11} + H_2O \xrightarrow{\text{maltase}} 2C_6H_{12}O_6$ (glucose).

(A) Sucrose is a disaccharide composed of one molecule of glucose and one molecule of fructose linked by a glycosidic bond.
Upon hydrolysis with dilute $HCl$,the glycosidic bond breaks to yield an equimolar mixture of glucose and fructose.
The chemical reaction is: $C_{12}H_{22}O_{11} + H_2O \xrightarrow{H^+} C_6H_{12}O_6 (\text{glucose}) + C_6H_{12}O_6 (\text{fructose})$.
Therefore,the ratio of glucose to fructose formed is $1: 1$.

(B) Starch is a polysaccharide. Upon acid-catalyzed hydrolysis,it yields the monosaccharide glucose.
Reaction:
$(C_6H_{10}O_5)_n + nH_2O \xrightarrow[393 \ K, 2-3 \ atm]{H^+} nC_6H_{12}O_6$ (Glucose)
Therefore,the product of the above reaction is glucose.

(B) Glucose + Fructose $\longrightarrow$ Sucrose.
Sucrose is a disaccharide,which is made of $\alpha-D$-glucose and $\beta-D$-fructose joined together by a $1, 2$-glycosidic linkage.

(A) Lactose is a disaccharide composed of $\beta-D-galactose$ and $\beta-D-glucose$ units.
These units are linked by a $\beta-1,4-glycosidic$ linkage.
The structure of lactose is correctly represented by the image $250769-$s,which shows the $\beta-1,4$ linkage between the $C-1$ of galactose and $C-4$ of glucose.

(D) Reducing saccharides are carbohydrates that can act as reducing agents because they contain a free aldehyde or ketone group in their open-chain form.
$1$. Sucrose: Non-reducing sugar.
$2$. Ribose: Reducing sugar (contains a free aldehyde group).
$3$. Maltose: Reducing sugar (contains a free hemiacetal group).
$4$. Lactose: Reducing sugar (contains a free hemiacetal group).
$5$. Cellulose: Non-reducing polysaccharide.
Therefore,the reducing saccharides are $2, 3, 4$.

(D) $I$. Sucrose is a disaccharide formed by the condensation of $D$-glucose and $D$-fructose,joined by a glycosidic linkage. This statement is correct.
$II$. Cellulose is a structural polysaccharide found exclusively in the cell walls of plants. It is not present in animals. This statement is incorrect.
$III$. Lactose is a disaccharide composed of $D$-galactose and $D$-glucose units linked by a $\beta$-glycosidic bond. This statement is correct.
Therefore,statements $I$ and $III$ are correct.

(A) In $D-glucose$ $(X)$,the hydroxyl group at $C-5$ attacks the aldehyde group at $C-1$ to form a six-membered pyranose ring (hemiacetal).
In $D-fructose$ $(Y)$,the hydroxyl group at $C-5$ attacks the ketone group at $C-2$ to form a five-membered furanose ring (hemiketal).
Therefore,in both cases,the hydroxyl group at $C-5$ participates in the ring formation.

(D) nucleoside consists of a nitrogenous base attached to a pentose sugar.
In $RNA$,the pentose sugar is ribose,which has a hydroxyl $(-OH)$ group at both the $2'$ and $3'$ positions.
$RNA$ contains the nitrogenous bases Adenine $(A)$,Guanine $(G)$,Cytosine $(C)$,and Uracil $(U)$.
Thymine $(T)$ is found in $DNA$,not $RNA$.
Looking at the options:
Option $A$ has $T$ (Thymine),which is characteristic of $DNA$.
Option $B$ and $C$ show a deoxyribose sugar (missing the $-OH$ group at the $2'$ position),which is characteristic of $DNA$.
Option $D$ shows a ribose sugar (with $-OH$ groups at both $2'$ and $3'$ positions) attached to Uracil $(U)$,which is a characteristic nucleoside of $RNA$ (Uridine).

(B) The sugar present in $DNA$ is $\beta-D-2-$deoxyribose.
In contrast,the sugar present in $RNA$ is $\beta-D-$ribose.

(C) The synthetic detergent used in toothpaste is typically an anionic detergent,such as sodium lauryl sulfate.
Animal starch is known as glycogen,which is a polysaccharide that serves as the form of energy storage in animals.
Therefore,$X$ is anionic and $Y$ is glycogen.

(B) The artificial sweetener sucralose is a $trichloro$ derivative of sucrose.
It is stable at cooking temperatures and does not provide calories to the body.

(C) Saccharic acid is also known as $(2R, 3S, 4S, 5S)-2,3,4,5-$tetrahydroxyhexanedioic acid.
Its chemical structure is $HOOC-(CHOH)_4-COOH$.
As shown in the structure,saccharic acid has four hydroxyl groups attached to the four central carbon atoms.
Therefore,the total number of hydroxyl groups is $4$.

(A) The reaction sequence is as follows:
$1$. Glucose reacts with $HCN$ to form a cyanohydrin $(X)$. The aldehyde group $(-CHO)$ is converted into a cyanohydrin group $(-CH(OH)CN)$.
$2$. The cyanohydrin $(X)$ undergoes acid-catalyzed hydrolysis $(H^+/H_2O)$ to convert the cyano group $(-CN)$ into a carboxylic acid group $(-COOH)$.
$3$. The starting material is glucose,which has $6$ carbon atoms. The addition of $HCN$ adds one carbon atom,resulting in a chain of $7$ carbon atoms.
$4$. The final product $Y$ is a heptanoic acid derivative with hydroxyl groups on carbons $2, 3, 4, 5, 6,$ and $7$. Thus,the $IUPAC$ name is $2,3,4,5,6,7$-hexahydroxyheptanoic acid.

(A) Cellulose is a linear polysaccharide composed of $ \beta $-$D$-glucose units joined through $1,4$-glycosidic linkages.
These glucose units are linked by glycosidic bonds between the $C_1$ carbon of one unit and the $C_4$ carbon of the adjacent unit,where the anomeric $C_1$ carbon maintains a $ \beta $-configuration.

(D) The given structure is a six-membered ring containing an oxygen atom,which identifies it as a pyranose form.
In the Haworth projection of $D$-glucose,the anomeric carbon $(C1)$ has the $-OH$ group pointing downwards,which corresponds to the $\alpha-$configuration.
The other hydroxyl groups at $C2, C3,$ and $C4$ are arranged such that they match the structure of $D$-glucose.
Specifically,the $-OH$ at $C2$ is down,at $C3$ is up,and at $C4$ is down.
Therefore,the structure represents $\alpha-D-(+)-$ glucopyranose.

(C) - and $L$- configurations are relative configurations assigned with respect to $D(+)$-glyceraldehyde.
Both $D(+)$-glucose and $D(-)$-fructose have the same $D$-configuration because the hydroxyl group at the chiral carbon farthest from the carbonyl group is on the right side in their Fischer projection.
However,$D(+)$-glucose is dextrorotatory $(+)$,whereas $D(-)$-fructose is levorotatory $(-)$.
Therefore,Assertion $(A)$ is true,but Reason $(R)$ is false.

(D) The $D$ or $L$ configuration of a monosaccharide is determined by the position of the $-OH$ group on the chiral carbon furthest from the carbonyl group (the $C_5$ carbon in fructose).
If the $-OH$ group is on the right side,it is the $D$-isomer.
If the $-OH$ group is on the left side,it is the $L$-isomer.
Structure $II$ represents $D$-fructose because the $-OH$ group at $C_5$ is on the right.
Structure $IV$ represents $L$-fructose because the $-OH$ group at $C_5$ is on the left.
Therefore,the pair representing $D$-fructose and $L$-fructose is $II$ and $IV$.

(C) $\alpha-D$-Fructofuranose is a monosaccharide. It has a five-membered furanose ring structure. In the $\alpha$-anomer,the hydroxyl group $(-OH)$ at the anomeric carbon $(C-2)$ is on the same side as the $-CH_2OH$ group at $C-5$ in the Haworth projection,or specifically,the $-OH$ group at $C-2$ is pointing downwards. The molecular formula is $C_6H_{12}O_6$. The correct structure is shown in the solution image.

(D) Reducing sugars are those carbohydrates that contain a free hemiacetal or hemiketal group,which can act as a reducing agent.
$(A)$ Sucrose is a non-reducing sugar because both anomeric carbons are involved in the glycosidic linkage.
$(B)$ Maltose is a reducing sugar as it contains a free hemiacetal group.
$(C)$ Lactose is a reducing sugar as it contains a free hemiacetal group.
$(D)$ Fructose is a reducing sugar because it can tautomerize to glucose in alkaline solution due to the presence of an $\alpha$-hydroxy ketone group.
Therefore,$(B)$,$(C)$,and $(D)$ are reducing sugars.

(C) glycosidic linkage is a type of covalent bond that joins a carbohydrate (sugar) molecule to another group,which may or may not be another carbohydrate.
Monosaccharides like Fructofuranose,Glucopyranose,and $\beta-D$-fructose do not contain glycosidic linkages because they are single sugar units.
Maltose is a disaccharide formed by the condensation of two $D$-glucose units.
In maltose,the two glucose residues are joined by an $\alpha-1,4$-glycosidic linkage between the $C-1$ of one glucose unit and the $C-4$ of the other.

(A) Amylose is a polysaccharide composed of $\alpha-D$-glucose units linked together by $\alpha(1-4)$ glycosidic linkages.
Looking at the provided structures,the structure that correctly depicts the $\alpha(1-4)$ glycosidic linkage between two $\alpha-D$-glucose units in a repeating polymer chain is represented by option $A$.

(A) Glucose on prolonged heating with $HI$ gives $n$-hexane $(CH_3CH_2CH_2CH_2CH_2CH_3)$,which is compound '$C$'.
In the Wurtz reaction,two alkyl halide molecules react with sodium metal to form an alkane.
To obtain $n$-hexane $(C_6H_{14})$ via Wurtz reaction,the alkyl halide '$D$' must be $n$-propyl chloride $(CH_3CH_2CH_2Cl)$.
The reaction is: $2CH_3CH_2CH_2Cl + 2Na \rightarrow CH_3CH_2CH_2CH_2CH_2CH_3 + 2NaCl$.

(B) Glucose $(CHO(CHOH)_4CH_2OH)$ on oxidation with mild oxidizing agents like $Br_2$ water gets converted to gluconic acid $(COOH(CHOH)_4CH_2OH)$. Here,the aldehyde group $(-CHO)$ is oxidized to a carboxylic acid group $(-COOH)$.
Gluconic acid $(COOH(CHOH)_4CH_2OH)$ on oxidation with strong oxidizing agents like $HNO_3$ gets converted to saccharic acid $(COOH(CHOH)_4COOH)$. Here,the primary alcoholic group $(-CH_2OH)$ is oxidized to a carboxylic acid group $(-COOH)$.
Thus,the functional groups involved in the respective conversions are $-CHO$ and $-CH_2OH$.

(C) The acid hydrolysis of sucrose is represented by the following chemical equation:
$C_{12}H_{22}O_{11} + H_2O \rightarrow C_6H_{12}O_6 \text{ (Glucose)} + C_6H_{12}O_6 \text{ (Fructose)}$
From the stoichiometry of the reaction,$1 \ mole$ of sucrose produces $1 \ mole$ of glucose and $1 \ mole$ of fructose.
The molar mass of sucrose $(C_{12}H_{22}O_{11})$ is calculated as: $(12 \times 12) + (22 \times 1) + (11 \times 16) = 144 + 22 + 176 = 342 \ g/mol$.
Therefore,to produce $1 \ mole$ of glucose,$1 \ mole$ of sucrose is required,which corresponds to $342 \ g$.

(C) According to the chemical properties of glucose,prolonged heating of glucose with $HI$ and red phosphorus at $100^{\circ} C$ results in the reduction of all hydroxyl groups and the aldehyde group to a straight-chain alkane,specifically $n$-hexane.
$CH_2OH(CHOH)_4CHO \xrightarrow{HI / \text{Red } P, 100^{\circ} C} CH_3(CH_2)_4CH_3$
The formation of $n$-hexane confirms that glucose has a straight-chain open structure.

(A) Starch is a polysaccharide with the formula $(C_6H_{10}O_5)_n$.
When starch is boiled with dilute $H_2SO_4$ under high pressure,it undergoes hydrolysis to produce glucose $(A)$.
The reaction is: $(C_6H_{10}O_5)_n + nH_2O \xrightarrow{\text{dil. } H_2SO_4, \text{high pressure}} nC_6H_{12}O_6$.
Glucose $(C_6H_{12}O_6)$,on prolonged heating with hydroiodic acid $(HI)$,undergoes reduction to form $n$-hexane $(B)$,which is $CH_3-(CH_2)_4-CH_3$.
Therefore,the correct option is $A$.

(A) The cyclic structure of glucose is supported by the following facts:
$(i)$ Glucose does not give Schiff's test because the $-CHO$ group is involved in hemiacetal formation and is not free.
$(ii)$ Glucose exists in two different crystalline forms ($\alpha$-$D$-glucose and $\beta$-$D$-glucose) due to the formation of an anomeric carbon.
$(iv)$ Pentaacetate of glucose does not react with hydroxylamine because the aldehyde group is not free (it is in the cyclic hemiacetal form).
Statement $(iii)$ describes the oxidation of glucose to saccharic acid,which is a property of the open-chain structure,not the cyclic structure.
Therefore,statements $(i)$,$(ii)$,and $(iv)$ support the cyclic structure of glucose.

(A) When glucose is oxidised by a strong oxidising agent like $HNO_3$,both the terminal aldehyde group $(-CHO)$ and the primary alcoholic group $(-CH_2OH)$ are oxidised to carboxylic acid groups $(-COOH)$.
This results in the formation of a dicarboxylic acid known as saccharic acid or glucaric acid.
The reaction is represented as:
$CHO(CHOH)_4 CH_2OH + [O] \xrightarrow{Conc. HNO_3} COOH(CHOH)_4 COOH + H_2O$.

(C) Glucose $(CHO(CHOH)_4CH_2OH)$ on oxidation with concentrated nitric acid $(HNO_3)$ gets oxidized to a dicarboxylic acid,which is known as saccharic acid (glucaric acid).
The reaction is as follows:
$CHO(CHOH)_4CH_2OH \xrightarrow{HNO_3} COOH(CHOH)_4COOH$
Thus,$HNO_3$ is the reagent that oxidizes glucose into saccharic acid.
Hence,option $(C)$ is the correct answer.

(C) The carbohydrate $(A)$ is $Sucrose$ $(C_{12}H_{22}O_{11})$.
Upon hydrolysis with dilute $HCl$,$Sucrose$ yields $D-(+)-Glucose$ and $D-(-)-Fructose$.
$Glucose$ $(B)$ is an aldose,which on reaction with bromine water $(Br_2/H_2O)$ is oxidized to gluconic acid $(Z)$,a monocarboxylic acid.
$Fructose$ $(C)$ is a ketohexose.
Thus,$(A)$ is $Sucrose$.

(B) Maltose is a disaccharide composed of two $\alpha-D$-glucose units linked by an $\alpha(1 \rightarrow 4)$-glycosidic bond.
Upon hydrolysis,maltose yields two molecules of $\alpha-D$-glucose.
Option $A$ is correct as both units are $\alpha-D$-glucose.
Option $B$ is incorrect because maltose does not contain fructose; it is composed of two glucose units.
Option $C$ is correct because both glucose units have a free anomeric carbon,making them reducing sugars.
Option $D$ is correct as the linkage is indeed $1,4$-glycosidic.
Therefore,the incorrect statement is $B$.

(D) Lactose,i.e.,milk sugar,is composed of a $\beta-D$-galactose unit and an $\alpha-D$-glucose unit joined by a $\beta$-glycosidic linkage between $C-1$ of galactose and $C-4$ of the glucose unit. Thus,Statement $I$ is incorrect because it mentions $\alpha-D$-galactose instead of $\beta-D$-galactose.
Sucrose,i.e.,cane sugar,is composed of $\alpha-D$-glucose and $\beta-D$-fructose units. These units are joined by an $\alpha,\beta$-glycosidic linkage between $C-1$ of glucose and $C-2$ of fructose. Thus,Statement $II$ is correct.

(C) Maltose is a disaccharide composed of two $\alpha-D$-glucose units linked by an $\alpha-1,4$-glycosidic bond.
Since it has a free hemiacetal group,it acts as a reducing sugar.
Option $C$ is incorrect because it describes the composition of lactose,not maltose.

(B) reducing sugar is a carbohydrate that can act as a reducing agent due to the presence of a free aldehyde or ketone group.
All monosaccharides,such as $Ribose$ and $Fructose$,are reducing sugars.
Among disaccharides,$Lactose$ is a reducing sugar because it contains a free hemiacetal group,whereas $Sucrose$ is a non-reducing sugar because its glycosidic linkage involves the anomeric carbons of both glucose and fructose units.
Therefore,$II$ (Ribose),$III$ (Lactose),and $IV$ (Fructose) are reducing sugars.
Hence,the correct option is $B$.

(D) Sucrose and lactose both are disaccharides.
In sucrose,the anomeric $C_1$ of $\alpha-D$-glucose and $C_2$ of $\beta-D$-fructose are involved in the $1,2$-glycosidic linkage,leaving no free anomeric carbon.
So,it is a non-reducing sugar.
In lactose,the anomeric $C_1$ of $\beta-D$-galactose is involved in the $1,4$-glycosidic linkage,but the anomeric $C_1$ of $\beta-D$-glucose remains free with an $-OH$ group,which exhibits reducing properties.

(D) Cellulose is a polysaccharide consisting of many glucose units linked by $(\beta)-1,4$-glycosidic bonds.
Human beings lack the specific enzyme (cellulase) required to break these $(\beta)-1,4$-glycosidic linkages.
Therefore,cellulose cannot be digested in the human body.
Option $A$ is incorrect because starch hydrolysis yields glucose.
Option $B$ is incorrect because lactose hydrolysis yields glucose and galactose.
Option $C$ is incorrect because the oxidation of glucose is an exothermic process that releases energy.

(B) Lactose is a disaccharide composed of $\beta-D(+)$-galactose and $\beta-D(+)$-glucose units.
These two monosaccharides are linked together by a $\beta-1,4$-glycosidic linkage.
It is a reducing sugar because the hemiacetal group at $C-1$ of the glucose unit is free.

(A) Sucrose undergoes hydrolysis in the presence of dilute aqueous sulphuric acid to produce an equimolar mixture of $D-(+)$-glucose and $D-(-)$-fructose.
The chemical reaction is:
$C_{12}H_{22}O_{11} + H_2O \xrightarrow{H_2SO_4} C_6H_{12}O_6 + C_6H_{12}O_6$
($D-(+)$-glucose) ($D-(-)$-fructose)
The molar ratio of the products is $1:1$.
Sucrose is dextrorotatory,but the resulting mixture is laevorotatory because the specific rotation of $D-(-)$-fructose $(-92.4^{\circ})$ is greater in magnitude than that of $D-(+)$-glucose $(+52.7^{\circ})$.

(D) The nitrogenous bases found in nucleic acids are Adenine $(A)$,Guanine $(G)$,Thymine $(T)$,Cytosine $(C)$,and Uracil $(U)$.
$DNA$ contains $A, G, T,$ and $C$.
$RNA$ contains $A, G, C,$ and $U$.
Uracil is present in $RNA$ instead of Thymine,which is found in $DNA$. Therefore,Uracil is present in $RNA$ only.

(A) Statement $(I)$ is correct: Starch is a polymer of $\alpha-D(+)$-glucose.
Statement $(II)$ is incorrect: Amylose is the water-soluble component of starch.
Statement $(III)$ is incorrect: Amylose is a long unbranched chain polymer of $\alpha-D(+)$-glucose.
Statement $(IV)$ is correct: Cellulose is a straight chain polymer of $\beta-D(+)$-glucose units.
Therefore,the correct statements are $(I)$ and $(IV)$.

(D) Ribose is an aldopentose that reacts with phenylhydrazine $(Ph-NH-NH_2)$ to form a characteristic osazone derivative.
In $2$-deoxyribose,the $-OH$ group at the $C-2$ position is replaced by a hydrogen atom.
The formation of an osazone requires the presence of an $\alpha$-hydroxy carbonyl group (or a group that can be oxidized to it).
Since $2$-deoxyribose lacks the $-OH$ group at the $C-2$ position,it cannot form the stable osazone structure under standard conditions,allowing it to be differentiated from ribose.

(D) In an aqueous solution,glucose exists in a dynamic equilibrium between its open-chain form and two cyclic (pyranose) forms,known as $\alpha-D-(+)-glucopyranose$ and $\beta-D-(+)-glucopyranose$.
The equilibrium mixture consists of approximately $36\%$ of $\alpha$-anomer,$64\%$ of $\beta$-anomer,and a very small amount (about $0.02\%$) of the open-chain form.
Therefore,glucose exists in all three forms in equilibrium.

(B, C, D) Both the given molecules contain $6$ carbon atoms with an aldehyde group,so these are called aldohexoses.
In $X$,the $-OH$ group attached to the second-last carbon ($C$-$5$) is on the right-hand side,so it is a $D$-sugar.
In $Y$,the $-OH$ group attached to the second-last carbon ($C$-$5$) is on the left-hand side,so it is an $L$-sugar.
Since $X$ and $Y$ are non-superimposable mirror images of each other,they are enantiomers.
Therefore,statements $B$,$C$,and $D$ are correct.

(D) Tollen's reagent reacts with reducing sugars that possess a free anomeric $-OH$ group,which allows for mutarotation and the formation of an open-chain aldehyde form.
Structure $(A)$ is $\alpha$-$D$-glucopyranose,which has a free anomeric $-OH$ group at the $C1$ position,making it a reducing sugar that reacts with Tollen's reagent.
Structure $(B)$ is methyl $\alpha$-$D$-glucopyranoside,which is an acetal (the anomeric $-OH$ is replaced by an $-OCH_3$ group).
Since acetals cannot undergo mutarotation to form an open-chain aldehyde,they do not react with Tollen's reagent.
Therefore,only structure $(B)$ will not react with Tollen's reagent.

(A) sugar is non-reducing if it does not have a free hemiacetal or hemiketal group,meaning both anomeric carbons are involved in the glycosidic linkage.
In the structure shown in image $278743-a$ (sucrose),the glycosidic linkage is between the $C1$ of $\alpha-D-glucose$ and the $C2$ of $\beta-D-fructose$.
Since both anomeric carbons are involved in the linkage,there is no free hemiacetal group,making it a non-reducing sugar.
The other structures $(278743-b, 278743-c, 278743-d)$ contain at least one free anomeric carbon (hemiacetal group),making them reducing sugars.