Carbohydrates Questions in English

(C) Honey is a natural substance produced by honey bees from the nectar of flowers. The nectar primarily contains sucrose,which is broken down by the enzymes in the bee's saliva into simple sugars,mainly glucose and fructose. Fructose is the most abundant sugar found in honey,making it sweeter than sucrose.

(C) Aldoses contain an aldehyde group $(-CHO)$,while ketoses contain a ketone group $(>C=O)$.
$Br_2$ water is a mild oxidizing agent that oxidizes the aldehyde group of an aldose to a carboxylic acid (gluconic acid) but does not oxidize ketoses.
Therefore,$Br_2$ water can distinguish between aldoses and ketoses.
Tollen's reagent and Fehling's solution are oxidizing agents that react with both aldoses and ketoses (due to the tautomerization of ketoses into aldoses in alkaline medium),so they cannot distinguish between them.
Thus,only $I$ is the correct reagent.

(D) Aspartame is a common artificial sweetener used in soft drinks and cold foods because it is unstable at cooking temperatures.
It is approximately $160$ times sweeter than sucrose (table sugar).

(B) Cellulose is a polysaccharide consisting of a linear chain of several hundred to many thousands of $\beta$-glucose units.
These units are linked together by $\beta(1 \rightarrow 4)$-glycosidic bonds.
Unlike starch,which contains $\alpha$-glucose,cellulose provides structural support to plant cell walls due to its linear,rigid structure.

(B) The iodine test is a chemical test used to detect the presence of starch.
Starch is a polysaccharide that forms a helical structure,which can trap iodine molecules $(I_2)$ within its coils to form a starch-iodine complex.
This complex produces a characteristic deep blue color.
When the solution is heated,the helical structure of starch unwinds,releasing the iodine molecules,and the blue color disappears.
Upon cooling,the starch molecules re-form their helical structure,trapping the iodine again,and the blue color reappears.
Therefore,starch gives a positive iodine test.

(C) Glucose is a $6$-carbon sugar,hence it is a hexose.
It is a type of carbohydrate.
It contains an aldehyde group,so it is an aldose.
Glucose is a monosaccharide,not an oligosaccharide.
Oligosaccharides are carbohydrates that yield $2$ to $10$ monosaccharide units upon hydrolysis,whereas glucose cannot be further hydrolyzed into simpler sugars.

(A) Inulin is a naturally occurring polysaccharide found in the roots and tubers of various plants,such as dahlia tubers and Jerusalem artichokes.
It is a polymer composed of fructose units linked by $\beta-(2,1)$ glycosidic bonds.
Commercially,fructose is produced by the hydrolysis of inulin,which breaks the polymer down into its constituent fructose monomers.

(B) Sugars like glucose,fructose,and mannose form identical osazones when treated with excess phenylhydrazine. This occurs because the structural differences between these sugars exist only at the $C-1$ and $C-2$ positions. During the formation of osazone,these two carbon atoms are involved in the reaction,resulting in the same phenylosazone structure for all three sugars.

(A) Sugars that cannot reduce Fehling's solution and Tollen's reagent are known as non-reducing sugars.
In these sugars,the glycosidic bond is formed between the anomeric carbons of the two monosaccharide units,meaning no free aldehyde or ketone group is available.
Sucrose is a disaccharide formed by the linkage of glucose and fructose through their anomeric carbons,making it a non-reducing sugar.
In contrast,lactose,maltose,and cellobiose are reducing sugars because they possess a free hemiacetal group.

(B) Monosaccharides are the simplest carbohydrates that cannot be hydrolyzed into smaller carbohydrate units. The simplest monosaccharides are trioses (three-carbon sugars). Glyceraldehyde $(CH_{2}OH-CHOH-CHO)$ is an aldotriose and is considered the first member of the monosaccharide family.

(D) Sugars that do not reduce Tollen's reagent,Fehling's solution,or Benedict's solution are known as non-reducing sugars.
In sucrose,the glycosidic linkage is formed between the $C1$ of glucose and the $C2$ of fructose.
Since both anomeric carbons are involved in the glycosidic bond,there is no free aldehyde or ketone group available to act as a reducing agent.
Therefore,sucrose is a non-reducing sugar.

(A) Animal starch is the common name for glycogen.
Glycogen is a multi-branched polysaccharide of glucose that serves as a form of energy storage in animals,fungi,and bacteria.
It is primarily synthesized and stored in the liver and muscle cells of animals.

(D) Dextrins are low molecular weight carbohydrates produced by the hydrolysis of starch. They are widely used in various industries.
$1$. They serve as a base for adhesives due to their sticky nature.
$2$. They are used in the confectionery industry for glazing and thickening.
$3$. They are used in the paper industry for sizing,which improves the surface quality of the paper.
Therefore,all the given options are correct uses of dextrins.

(C) The test described is the $Molisch$ test,which is a sensitive chemical test for the presence of carbohydrates.
When an aqueous solution of a carbohydrate is treated with an alcoholic solution of $\alpha$-naphthol followed by the addition of concentrated $H_{2}SO_{4}$ along the sides of the test tube,a purple or violet-coloured ring is formed at the junction of the two layers.
This reaction occurs because the acid dehydrates the carbohydrate to form furfural or its derivatives,which then react with $\alpha$-naphthol to produce the characteristic violet-coloured complex.

(B) The chemical formula of glucose is $C_6H_{12}O_6$.
Its open-chain structure is $CHO-(CHOH)_4-CH_2OH$.
In this structure,there is one aldehyde group $(-CHO)$ at the $C-1$ position.
The carbon at the $C-6$ position is attached to a primary hydroxyl group $(-CH_2OH)$,meaning it is a primary $OH$ group.
The carbons at positions $C-2, C-3, C-4$,and $C-5$ are each attached to one hydroxyl group $(-OH)$,and these are secondary $OH$ groups.
Therefore,glucose contains one primary $OH$ group and four secondary $OH$ groups.

(D) Carbohydrates are defined as optically active polyhydroxy aldehydes or polyhydroxy ketones.
$1$. In polyhydroxy aldehydes (aldoses),the functional groups present are the hydroxyl group $(-OH)$ and the aldehyde group $(-CHO)$.
$2$. In polyhydroxy ketones (ketoses),the functional groups present are the hydroxyl group $(-OH)$ and the ketone group $(-C=O)$.
Since carbohydrates are characterized by the presence of multiple hydroxyl groups along with either an aldehyde or a ketone group,the most fundamental functional groups present in a typical carbohydrate are the hydroxyl group $(-OH)$ and the carbonyl group (either $-CHO$ or $-C=O$). Given the options,$-OH$ and $-CHO$ (or $-C=O$) represent the core functional components.

(C) Cane sugar (sucrose) is a disaccharide with the molecular formula $C_{12}H_{22}O_{11}$.
Upon acid-catalyzed hydrolysis, it breaks down into two monosaccharide units: glucose and fructose.
The chemical reaction is represented as: $C_{12}H_{22}O_{11} + H_2O \xrightarrow{H^+} C_6H_{12}O_6 (\text{Glucose}) + C_6H_{12}O_6 (\text{Fructose})$.
This process is also known as the inversion of cane sugar because the resulting mixture is levorotatory.

(B) Rayon,also known as artificial silk,is a regenerated fiber.
It is produced by processing natural polymers,specifically $Cellulose$,which is obtained from wood pulp or cotton linters.
The $Cellulose$ is chemically treated to form a viscous solution,which is then extruded through spinnerets to create fibers.

(A) Glucose and mannose are epimers because they differ in configuration only at the $C_{2}$ carbon atom. In glucose,the $-OH$ group at $C_{2}$ is on the right,whereas in mannose,the $-OH$ group at $C_{2}$ is on the left.

(A) Cellophane is a thin,transparent sheet of regenerated cellulose. It is produced by treating cellulose fibres with chemicals,making it a semipermeable membrane often used in laboratory experiments and packaging.

(C) Ribose is a pentose sugar,which means it contains $5$ carbon atoms,not $6$. Therefore,the statement that it has $6$ carbon atoms is incorrect. It is a polyhydroxy aldehyde (aldopentose) and exhibits optical activity due to the presence of chiral carbon atoms.

(A) Bromine water ($Br_2$ water) is a mild oxidizing agent.
When glucose $(CHO(CHOH)_4CH_2OH)$ reacts with bromine water,the aldehyde group $(-CHO)$ is selectively oxidized to a carboxylic acid group $(-COOH)$ while the secondary alcoholic groups remain unaffected.
This reaction results in the formation of gluconic acid $(COOH(CHOH)_4CH_2OH)$.

(C) The chemical formula for glucose is $CH_2OH-(CHOH)_4-CHO$.
In the open-chain structure of glucose,there are $6$ carbon atoms in total.
The carbon atom at position $1$ is part of the aldehyde group $(-CHO)$,and the carbon atom at position $6$ is part of the primary alcohol group $(-CH_2OH)$.
The carbon atoms at positions $2, 3, 4,$ and $5$ are each bonded to four different groups ($-H, -OH, -CHO/CH_2OH,$ and the rest of the carbon chain).
Therefore,these $4$ carbon atoms are asymmetric (chiral) centers.
Thus,the total number of asymmetric carbon atoms in a glucose molecule is $4$.

(C) Glucose is a reducing sugar because it contains a free aldehyde group.
Ammoniacal $AgNO_{3}$ is known as Tollens' reagent.
When glucose reacts with Tollens' reagent,the aldehyde group is oxidized to a carboxylate group,and silver ions are reduced to metallic silver,forming a characteristic 'silver mirror' on the inner walls of the test tube.
Therefore,ammoniacal $AgNO_{3}$ is used to identify glucose.

(C) Glucose and fructose have the same molecular formula $C_{6}H_{12}O_{6}$.
Glucose contains an aldehyde group $(-CHO)$,while fructose contains a ketone group $(>CO)$.
Since they have the same molecular formula but different functional groups,they are classified as functional isomers.

(D) Anomers of glucose are cyclic diastereomers (epimers) that differ in configuration only at the anomeric carbon,which is $C-1$ in glucose.
These exist in two forms,$\alpha$-$D$-glucose and $\beta$-$D$-glucose,depending on the orientation of the hydroxyl $(-OH)$ group at the $C-1$ position.

(D) The chemical formula of fructose is $CH_2OH \cdot CO \cdot (CHOH)_3 \cdot CH_2OH$.
$1$. It contains $5$ hydroxyl $(-OH)$ groups.
$2$. It contains $3$ secondary alcoholic groups $(-CHOH-)$ and $2$ primary alcoholic groups $(-CH_2OH)$.
$3$. It contains one keto group $(>C=O)$ at the $C-2$ position.
Therefore,all the given statements are correct.

(D) In an aqueous solution, glucose exists in an equilibrium mixture of its cyclic forms and a small amount of the open-chain form.
Specifically, it exists as an equilibrium between $\alpha-D$-glucose $(approx 36 \%)$, $\beta-D$-glucose $(approx 64 \%)$, and a very small fraction of the open-chain form $(approx 0.02 \%)$.
These forms are interconvertible through a process called mutarotation.

(B) When a reducing sugar is heated with Fehling's solution,the $Cu^{2+}$ ions in the solution are reduced to $Cu^{+}$ ions.
This reaction results in the formation of cuprous oxide $(Cu_{2}O)$,which appears as a reddish-brown or brick-red precipitate.
Therefore,the correct colour of the precipitate is red.

(A) Glucose gives a silver mirror test (Tollens' test) with ammoniacal silver nitrate because of the presence of the $-CHO$ group (aldehyde group) in its open-chain structure.
The reaction is as follows:
$CH_2OH(CHOH)_4CHO + Ag_2O \rightarrow CH_2OH(CHOH)_4COOH + 2Ag \downarrow$ (silver mirror)
Here,the aldehyde group is oxidized to a carboxylic acid group,and silver ions are reduced to metallic silver,forming the silver mirror.

(D) The formation of osazone from glucose involves the reaction with three molecules of phenylhydrazine $(C_6H_5NHNH_2)$.
$1$. The first molecule of phenylhydrazine reacts with the aldehyde group of glucose to form glucose phenylhydrazone,releasing water $(H_2O)$.
$2$. The second molecule of phenylhydrazine acts as an oxidizing agent,oxidizing the $C-2$ hydroxyl group to a carbonyl group,while itself being reduced to aniline $(C_6H_5NH_2)$ and ammonia $(NH_3)$.
$3$. The third molecule of phenylhydrazine reacts with the newly formed carbonyl group at $C-2$ to form the final osazone product,releasing another molecule of water $(H_2O)$.
Thus,a total of three molecules of phenylhydrazine are required.

(B) The structure of fructose is $CH_2OH-CO-(CHOH)_3-CH_2OH$.
An asymmetric carbon atom (chiral center) is a carbon atom bonded to four different groups.
In the fructose molecule,the carbons at positions $3$,$4$,and $5$ are asymmetric carbons.
Therefore,the total number of asymmetric carbon atoms in fructose is $3$.

(B) The two forms of $D$-glucopyranose are $\alpha-D-(+)$-glucopyranose and $\beta-D-(+)$-glucopyranose.
These two forms differ in configuration only at the $C-1$ carbon atom,which is the anomeric carbon.
Such stereoisomers that differ in configuration specifically at the anomeric carbon are known as anomers.

(A) Raffinose is a trisaccharide. It yields three molecules of monosaccharides upon hydrolysis.
$C_{18}H_{32}O_{16} + 2H_{2}O \xrightarrow{H^{+}} \text{glucose} + \text{fructose} + \text{galactose}$

(C) Glucose is an aldohexose with the chemical formula $C_6H_{12}O_6$.
It contains $4$ chiral (asymmetric) carbon atoms.
The number of stereoisomers for a molecule with $n$ chiral centers is given by the formula $2^n$.
Here,$n = 4$.
Therefore,the total number of stereoisomers = $2^4 = 16$.

(D) Glucose $(C_6H_{12}O_6)$ is an aldohexose,meaning it contains an aldehyde group $(-CHO)$ in its open-chain structure.
In an aqueous solution,glucose exists in an equilibrium between its cyclic hemiacetal form and its open-chain aldehyde form.
Although the concentration of the open-chain form is very low,it is sufficient to react with reagents like Tollens' reagent or Fehling's solution,which are characteristic of the aldehyde group.
Therefore,glucose exhibits many reactions typical of aldehydes due to this dynamic equilibrium.

(A) Glucose is a reducing sugar because it contains a free aldehyde group,which allows it to reduce Fehling's solution to a red precipitate of cuprous oxide $(Cu_2O)$.
Cane sugar (sucrose) is a non-reducing sugar because its glycosidic linkage involves the anomeric carbons of both glucose and fructose,leaving no free aldehyde or ketone group.
Therefore,Fehling's solution can distinguish between glucose and sucrose.

(C) The general formula for carbohydrates is $C_n(H_2O)_n$.
Acetic acid has the chemical formula $CH_3COOH$,which can be written as $C_2(H_2O)_2$.
Although it fits the general formula,it is a carboxylic acid and not a carbohydrate because it does not contain polyhydroxy aldehyde or ketone groups.

(C) Despite having an aldehyde group,glucose does not give the $2,4-DNP$ test or Schiff's test.
Furthermore,it does not form a hydrogen sulphite addition product with $NaHSO_3$.
These observations indicate that the aldehyde group is not free but is involved in the formation of a hemiacetal structure,confirming that glucose exists as a cyclic compound.

(A) Maltose is a disaccharide composed of two $\alpha$-$D$-glucose units.
These two glucose units are linked together by an $\alpha(1 \rightarrow 4)$ glycosidic bond.
This means the bond is formed between the anomeric carbon (Carbon $1$) of the first glucose molecule and the Carbon $4$ of the second glucose molecule.
Therefore,the correct option is $A$.

(A) The provided image shows a five-membered ring structure (furanose ring) with five carbon atoms in total (a pentose sugar).
Ribose is a classic example of a pentose sugar $(C_5H_{10}O_5)$ commonly found in $RNA$.
Glucose and Fructose are hexose sugars $(C_6H_{12}O_6)$,which typically form six-membered rings (pyranose) or five-membered rings with different substituents.
Raffinose is a trisaccharide,not a monosaccharide.
Therefore,the structure represents Ribose.

(B) Chitin is a complex structural polysaccharide found in the exoskeletons of arthropods and the cell walls of fungi.
It is a polymer of $N$-acetylglucosamine,which is a nitrogen-containing derivative of glucose.
Since it consists of long chains of these repeating units linked by $\beta-1,4$-glycosidic bonds,it is classified as a polysaccharide.

(C) Adenosine Triphosphate $(ATP)$ consists of an adenosine molecule bonded to three phosphate groups. The high energy released during cellular processes is primarily due to the hydrolysis of the phosphoanhydride bonds (a type of phosphate bond) between the phosphate groups. When the terminal phosphate group is removed, $ATP$ is converted into Adenosine Diphosphate $(ADP)$ and inorganic phosphate $(Pi)$, releasing a significant amount of energy: $ATP + H_2O \rightarrow ADP + Pi + \text{Energy}$.

(C) The complexity of carbohydrates increases based on the number of monosaccharide units they contain:
$1$. Triose: $A$ simple sugar containing $3$ carbon atoms (monosaccharide).
$2$. Glucose: $A$ simple sugar containing $6$ carbon atoms (monosaccharide).
$3$. Maltose: $A$ disaccharide composed of $2$ glucose units.
$4$. Oligosaccharide: Carbohydrates containing $2-10$ monosaccharide units.
$5$. Starch: $A$ polysaccharide consisting of a long chain of glucose units.
Therefore,the correct sequence of increasing complexity is: Triose,Glucose,Maltose,Oligosaccharide,Starch.

(B) dehydration reaction (also known as a condensation reaction) involves the removal of a water molecule $(H_{2}O)$ when two monosaccharides join to form a disaccharide.
When two glucose molecules $(C_{6}H_{12}O_{6})$ combine,the reaction is:
$C_{6}H_{12}O_{6} + C_{6}H_{12}O_{6} \rightarrow C_{12}H_{24}O_{12} - H_{2}O$
Subtracting the atoms of water ($2$ Hydrogen and $1$ Oxygen) from the sum of the two glucose molecules $(C_{12}H_{24}O_{12})$:
$C_{12}H_{24-2}O_{12-1} = C_{12}H_{22}O_{11}$
Therefore,the formula for maltose is $C_{12}H_{22}O_{11}$.

(A) $1$. In the given structure,observe the position of the $-OH$ group at the $C-1$ (anomeric) carbon and the $C-4$ carbon.
$2$. At the $C-1$ position,the $-OH$ group is pointing downwards,which indicates an $\alpha$-anomer.
$3$. At the $C-4$ position,the $-OH$ group is pointing downwards,which is characteristic of glucose. (In galactose,the $-OH$ group at $C-4$ points upwards).
$4$. Therefore,the structure represents $\alpha$-glucose.

(C) The iodine test is used to detect the presence of polysaccharides.
$1$. Starch forms a helical structure that traps $I_2$ molecules,resulting in a deep blue color $(P-I)$.
$2$. Glycogen has a highly branched structure that reacts with $I_2$ to produce a reddish-brown or red color $(Q-II)$.
$3$. Cellulose does not form the necessary helical structure to trap $I_2$ molecules,so it does not give a color with iodine $(R-III)$.
Therefore,the correct matching is $(P-I), (Q-II), (R-III)$.

(B) In animals,carbohydrates are stored in the form of $Glycogen$,which is a branched polysaccharide often referred to as 'animal starch'.
In plants,carbohydrates are primarily stored in the form of $Starch$,which consists of two components: $Amylose$ and $Amylopectin$.
Therefore,the correct sequence is $Glycogen$ for animals and $Starch$ for plants.

(D) The correct matches are as follows:
$P$. Chitin is a homopolymer of $N$-acetyl glucosamine $(III)$.
$Q$. Starch and glycogen are storage polysaccharides made up of $\alpha$-glucose units $(I)$.
$R$. Cellulose is a structural polysaccharide made up of $\beta$-glucose units $(II)$.
$S$. Inulin is a polymer of fructose $(IV)$.
Therefore,the correct matching is $(P-III), (Q-I), (R-II), (S-IV)$.

(B) The image shows a highly branched polysaccharide structure.
Glycogen is a storage polysaccharide in animals and is characterized by a highly branched structure,similar to amylopectin but with more frequent branching points.
Starch consists of amylose (unbranched) and amylopectin (branched,but less than glycogen).
Cellulose is a linear,unbranched polymer of glucose.
Chitin is a structural polysaccharide found in the exoskeleton of arthropods and fungal cell walls,which is also linear.
Therefore,the highly branched structure depicted is glycogen.