Carbohydrates Questions in English

(A) Glucose is the simplest carbohydrate,$i.e.$,a monosaccharide.
The rest,such as cellulose and starch,are polysaccharides.

(A) The correct answer is $A$.
$Cellulose$ is a polysaccharide that cannot be digested by the human digestive system because humans lack the enzyme required to break the $\beta$-glycosidic linkages present in it.
It is a constituent of our diet as it is found in plant-based foods like fruits,vegetables,and grains,acting as dietary fiber.

(B) Lactose is the disaccharide present in milk. It is composed of one molecule of $D-(+)-glucose$ and one molecule of $D-(+)-galactose$ linked by a $\beta-glycosidic$ linkage.

(B) Carbohydrates are a rich source of energy. They are broken down into glucose,which provides energy for various metabolic processes in the body.

(A) . Sucrose is a non-reducing sugar because it lacks a free hemiacetal or hemiketal group. Therefore,it does not reduce Benedict's solution.

(A) The linkage that connects individual monosaccharide units in polysaccharides is known as a $Glycosidic$ linkage. This is a type of covalent bond formed by a dehydration reaction between the hydroxyl group of one monosaccharide and the anomeric carbon of another.

(A) Blood sugar refers to the concentration of glucose in the blood. $Glucose$ is the primary source of energy for the body's cells and is transported through the bloodstream. Therefore,blood sugar is synonymous with $Glucose$.

(B) Glucose $(C_6H_{12}O_6)$ is an aldohexose.
In its open-chain structure,it contains one aldehydic group $(-CHO)$ at the $C-1$ position.
It also contains five alcoholic groups $(-OH)$ attached to the $C-2$ through $C-6$ carbon atoms.
Therefore,glucose possesses both aldehydic and alcoholic functional groups.

(C) The hydrolysis of sucrose produces an equimolar mixture of $D-(+)$-glucose and $D-(-)$-fructose.
Since the specific rotation of $D-(-)$-fructose $(-92.4^{\circ})$ is greater in magnitude than that of $D-(+)$-glucose $(+52.7^{\circ})$,the resulting mixture has a net negative rotation.
Therefore,the mixture is laevorotatory,and this phenomenon is known as the inversion of cane sugar.

(A) . Mutarotation is defined as the change in the specific optical rotation of a freshly prepared solution of an optically active compound (like $\alpha-D$-glucose or $\beta-D$-glucose) until it reaches a constant equilibrium value.
This occurs due to the interconversion between the two anomeric forms through an open-chain intermediate.

(C) The correct statement is that starch is a polymer of $\alpha$-glucose.
Starch is a polysaccharide composed of two components: amylose and amylopectin,both of which are polymers of $\alpha$-$D$-glucose units.

(C) An aldohexose is a monosaccharide that contains an aldehyde group (aldose) and six carbon atoms (hexose).
$D$-glucose is a classic example of an aldohexose,as it has the formula $C_6H_{12}O_6$ and contains an aldehyde group at the $C_1$ position.
Cellulose is a polysaccharide,sucrose is a disaccharide,and raffinose is a trisaccharide.

(B) Starch is a polysaccharide,which is a polymer of $\alpha$-glucose units.
Upon complete hydrolysis,starch breaks down into its monomeric units,which is glucose.

(A) Raffinose $({C_{18}}{H_{32}}{O_{16}})$ is a trisaccharide.
${C_{18}}{H_{32}}{O_{16}} + 2{H_2}O \to {C_6}{H_{12}}{O_6} (\text{glucose}) + {C_6}{H_{12}}{O_6} (\text{fructose}) + {C_6}{H_{12}}{O_6} (\text{galactose})$.

(B) disaccharide is a sugar formed when two monosaccharides are joined by a glycosidic linkage.
Examples of disaccharides include sucrose,lactose,and maltose.
Galactose is a monosaccharide,not a disaccharide.
It combines with glucose to form the disaccharide lactose through a condensation reaction.

(C) $a)$. Molisch's test is a chemical test used to check for the presence of carbohydrates in a given analyte.
$b)$. The Biuret test is a chemical test used for detecting the presence of peptide bonds in proteins.
$c)$. Fehling's test is used to distinguish between reducing and non-reducing sugars. Reducing sugars react with Fehling's solution to form a red precipitate of $Cu_2O$,whereas non-reducing sugars do not react.
$d)$. Millon's test is used to detect the presence of soluble proteins (specifically tyrosine residues).

(C) disaccharide is a sugar formed when two monosaccharides are joined by a glycosidic linkage.
Glucose,Ribulose,and Arabinose are monosaccharides.
Lactose is a disaccharide found in milk,consisting of galactose joined to glucose by a $\beta-1,4-$glycosidic linkage.
Therefore,the correct option is $C$.

(B) Glucose is a reducing sugar because it contains a free aldehyde group.
When glucose is heated with Fehling's solution,the aldehyde group is oxidized to a carboxylate group,and $Cu^{2+}$ ions are reduced to $Cu^+$ ions.
The reaction produces a red precipitate of cuprous oxide $(Cu_2O)$.
Reaction: $\text{Glucose} + \text{Fehling's solution} \to \text{Gluconic acid} + Cu_2O \text{ (Red precipitate)}$.

(D) Glycolysis is a metabolic pathway that occurs in the cytosol of the cell.
In this process,one molecule of glucose $(C_6H_{12}O_6)$ is broken down into two molecules of pyruvate $(CH_3COCOOH)$ through a series of enzymatic reactions.
It is an anaerobic process that yields a net gain of $2$ $ATP$ and $2$ $NADH$ molecules.

(A) . Glucose is a monosaccharide,while the others are polysaccharides. Therefore,glucose is the simplest sugar.

(A) Glucose and mannose are epimers because they differ in configuration only at the $C-2$ carbon atom. Isomers that differ in configuration at only one stereocenter are known as epimers.

(B) The hydrolysis of $Maltose$ $(C_{12}H_{22}O_{11})$ yields two molecules of $Glucose$ $(C_6H_{12}O_6)$.
$C_{12}H_{22}O_{11} + H_2O \xrightarrow{\text{Hydrolysis}} 2C_6H_{12}O_6$ $(Glucose)$.

(B)

Sugar	Relative sweetness
$Sucrose$	$100$
$Glucose$	$74$
$Lactose$	$16$
$Fructose$	$173$

Among the given sugars,$Fructose$ has the highest relative sweetness value of $173$.
Therefore,it is the sweetest sugar.
Correct option is $B$.

(D) Glucose and fructose are both monosaccharides,meaning they cannot be hydrolyzed into simpler carbohydrates. Therefore,option $A$ is incorrect. Both glucose and fructose contain carbonyl groups that can be oxidized by Tollen's reagent to form a silver mirror. Specifically,glucose is an aldose and fructose is a ketose,but fructose isomerizes to glucose and mannose in the presence of the basic Tollen's reagent,allowing it to also give a positive silver mirror test. Since both exhibit these properties,there is no difference between them regarding these specific options. Thus,the correct answer is $D$.

(B) Fructose $(C_6H_{12}O_6)$ is a ketohexose containing $3$ chiral carbon atoms.
The number of optical isomers is given by the formula $2^n$,where $n$ is the number of chiral centres.
Here,$n = 3$,so the number of optical isomers = $2^3 = 8$.
Therefore,the correct option is $(B)$.

(A) . Glucose is a monosaccharide with the chemical formula $C_6H_{12}O_6$,not a disaccharide. Therefore,the statement that glucose is a disaccharide is false.

(C) nucleoside is composed of a nitrogenous base attached to a sugar molecule (an aldopentose).
Upon hydrolysis,a nucleoside breaks down into its constituent parts,which are an aldopentose and a heterocyclic base.
Note that a nucleotide,unlike a nucleoside,contains a phosphate group in addition to these components,which would yield orthophosphoric acid upon hydrolysis.

(B) chiral carbon is a carbon atom that is bonded to four different groups.
In the cyclic structure of $\beta-D-(+)$-glucose,there are five carbon atoms that are bonded to four different groups (marked with dots in the structure).
These are the carbons at positions $C1, C2, C3, C4,$ and $C5$.
Therefore,the total number of chiral carbons in $\beta-D-(+)$-glucose is $5$.

(B) $Amylose$ is a polysaccharide,which is a polymer of $\alpha-D-glucose$ units linked by $\alpha-1,4-glycosidic$ bonds.

(B) The two forms of $D$-glucopyranose are $\alpha-D-(+)$-glucopyranose and $\beta-D-(+)$-glucopyranose.
These are known as anomers.
Anomers are a pair of stereoisomers that differ in configuration only around the anomeric carbon (the $C_1$ carbon in glucose).

(B) . Sucrose is a disaccharide composed of an $\alpha-D$-glucopyranose unit and a $\beta-D$-fructofuranose unit.
These two monosaccharide units are linked by an $\alpha,\beta-1,2$-glycosidic linkage between the $C-1$ of the glucose unit and the $C-2$ of the fructose unit.

(D) Amylose is a linear polymer of glucose units linked by $\alpha-1,4$-glycosidic bonds.
Cellulose is a linear polymer of glucose units linked by $\beta-1,4$-glycosidic bonds.
Amylopectin and Glycogen are both branched polymers of glucose.
Amylopectin contains $\alpha-1,4$-glycosidic bonds in the chain and $\alpha-1,6$-glycosidic bonds at branch points.
Glycogen is highly branched and also contains $\alpha-1,4$ and $\alpha-1,6$-glycosidic bonds.
Since both $B$ and $D$ are branched,but Glycogen is the most characteristic branched storage polysaccharide in animals,the question typically refers to the most branched structure or expects both. However,in standard multiple-choice contexts,Glycogen is the primary example of a highly branched polymer.

(C) Cellulose is a linear polysaccharide composed of $\beta-D-glucose$ units joined by $\beta-1,4-glycosidic$ linkages.
Upon complete hydrolysis,cellulose breaks down into its constituent monomer,which is $D-glucose$.

(B) Cellulose is a linear polymer of $\beta-D-glucose$ units.
Each glucose unit in cellulose contains three hydroxyl $(-OH)$ groups (at $C_2$,$C_3$,and $C_6$ positions).
In tri-nitro cellulose,all three hydroxyl groups are esterified with nitric acid $(HNO_3)$.
Therefore,each glucose unit in tri-nitro cellulose has $0$ free $-OH$ groups,but the question asks for the number of $-OH$ groups present in the glucose unit structure before nitration or the number of sites involved in the formation of the tri-nitro derivative.
Since it is 'tri-nitro' cellulose,all $3$ $-OH$ groups are substituted.

(A) Carbohydrates are polyhydroxy aldehydes or polyhydroxy ketones.
This means they contain multiple hydroxyl $(-OH)$ groups and either an aldehyde $(-CHO)$ group or a ketone $(>C=O)$ group.
Therefore,the functional groups present are hydroxyl $(-OH)$ and carbonyl ($>C=O$ or $-CHO$).

(D) Starch is a mixture of two polysaccharides,$Amylose$ and $Amylopectin$.
$Amylose$ is a linear polymer of $\alpha-D-glucose$ units linked by $C1-C4$ glycosidic linkages.
$Amylopectin$ is a branched-chain polymer.
$Glycogen$ is also a highly branched polymer found in animals.
Therefore,$Amylose$ is the correct linear polymer.

(C) The reaction of glucose with phenylhydrazine involves the formation of glucosazone.
In this process,$1$ mole of glucose reacts with $3$ moles of phenylhydrazine.
The reaction proceeds as follows:
$C_6H_{12}O_6 + 3C_6H_5NHNH_2 \rightarrow C_6H_{10}O_4(NNHC_6H_5)_2 + C_6H_5NH_2 + NH_3 + 2H_2O$.
Thus,$3$ moles of phenylhydrazine are consumed.

(C) Glucose and mannose differ in configuration only at the $C-2$ carbon atom.
Such diastereomers that differ in configuration at only one stereocenter are known as epimers.
Therefore,glucose and mannose are $C-2$ epimers.

(B) Glucose $(C_6H_{12}O_6)$ is a monosaccharide because it cannot be further hydrolyzed into simpler carbohydrates.
It contains six carbon atoms,so it is a hexose.
It contains an aldehyde group,so it is an aldose.
It is a type of carbohydrate.
Therefore,it cannot be classified as an oligosaccharide,as oligosaccharides yield $2-10$ monosaccharide units upon hydrolysis.

(B) The two forms of $D$-glucopyranose,namely $\alpha-D$-glucopyranose and $\beta-D$-glucopyranose,differ only in the configuration of the hydroxyl group at the $C-1$ carbon atom.
These specific types of diastereomers,which differ in configuration at the hemiacetal or hemiketal carbon (anomeric carbon),are known as anomers.

(A) Optical activity in organic molecules is exhibited by compounds that possess at least one chiral carbon atom and lack a plane of symmetry (i.e.,they are chiral).
All three structures provided ($I$: Ribose,$II$: Arabinose,$III$: Xylose) are aldopentoses with the general formula $CHO-(CHOH)_3-CH_2OH$.
In these molecules,the carbons at positions $2$,$3$,and $4$ are chiral centers because they are bonded to four different groups.
Since these molecules contain chiral centers and do not have an internal plane of symmetry,they are optically active.
Therefore,all three pentoses exhibit optical activity.

(B) The open-chain structure of glucose is $CHO-(CHOH)_4-CH_2OH$.
In this structure,there is one aldehyde group $(-CHO)$ at the $C-1$ position.
There are four secondary alcoholic groups $(-CHOH)$ at positions $C-2, C-3, C-4,$ and $C-5$.
There is one primary alcoholic group $(-CH_2OH)$ at the $C-6$ position.
Therefore,glucose contains one primary $-OH$ group and four secondary $-OH$ groups.

(B) The two forms of $D$-glucopyranose,namely $\alpha$-$D$-glucopyranose and $\beta$-$D$-glucopyranose,differ only in the configuration of the hydroxyl group at the $C-1$ carbon atom.
Such isomers,which differ in configuration only at the anomeric carbon ($C-1$ in aldoses),are known as anomers.

(D) Benedict's reagent is a chemical reagent used to detect the presence of reducing sugars,such as glucose,in urine.
It contains copper$(II)$ sulfate $(CuSO_4)$,sodium citrate,and sodium carbonate.
When heated with a reducing sugar,the blue $Cu^{2+}$ ions are reduced to red copper$(I)$ oxide $(Cu_2O)$ precipitate,indicating a positive test for glucose.

(D) Sucrose is a disaccharide composed of $D-(+)$-glucose and $D-(-)$-fructose linked by a glycosidic bond.
It is a non-reducing sugar because both anomeric carbons are involved in the glycosidic linkage.
On hydrolysis,it yields an equimolar mixture of $D-(+)$-glucose and $D-(-)$-fructose.
Aspartame is an artificial sweetener made from aspartic acid and phenylalanine,not from sucrose.
Therefore,the statement that sucrose gives aspartame on heating is false.

(B) When glucose reacts with an excess of phenylhydrazine $(C_6H_5NHNH_2)$,it forms a crystalline derivative known as glucosazone.
The reaction involves the oxidation of the $C-2$ carbon and the formation of a hydrazone linkage.
The overall reaction is: $C_6H_{12}O_6 + 3C_6H_5NHNH_2 \rightarrow C_{18}H_{22}N_4O_4 + C_6H_5NH_2 + NH_3 + 2H_2O$.

(D) Anomers are a specific type of diastereomers (epimers) found in cyclic sugars.
They differ in configuration only at the anomeric carbon atom,which is the $C_1$ carbon in aldoses like glucose.
Specifically,$\alpha$-$D$-glucose and $\beta$-$D$-glucose are anomers because they differ in the orientation of the hydroxyl group $(-OH)$ attached to the $C_1$ carbon.

(C) Glucose $(C_6H_{12}O_6)$ contains an aldehydic group $(-CHO)$ at the $C_1$ position.
Tollens' reagent is an ammoniacal silver nitrate solution,$[Ag(NH_3)_2]^+OH^-$.
When glucose reacts with Tollens' reagent,the aldehydic group is oxidized to a carboxylic acid group (gluconic acid),and the silver ions are reduced to metallic silver,which forms a silver mirror on the inner walls of the test tube.
This reaction is a characteristic test for the presence of an aldehydic group.

(A) Disaccharides are carbohydrates that yield two molecules of monosaccharides on hydrolysis.
Their general formula is derived from the condensation of two monosaccharide units $(C_6H_{12}O_6)$ with the loss of one water molecule $(H_2O)$.
Calculation: $2 \times (C_6H_{12}O_6) - H_2O = C_{12}H_{24}O_{12} - H_2O = C_{12}H_{22}O_{11}$.
Thus,the general formula for a disaccharide is $C_{12}H_{22}O_{11}$.