Carbohydrates Questions in English

(D) reducing sugar is a carbohydrate that possesses a free aldehyde or ketone group,allowing it to act as a reducing agent.
In $Sucrose$,the glycosidic linkage is formed between $C_1$ of $\alpha$-$D$-glucose and $C_2$ of $\beta$-$D$-fructose.
Since both the anomeric carbons are involved in the linkage,there is no free aldehyde or ketone group available.
Therefore,$Sucrose$ is a non-reducing sugar.

(C) $\alpha -D(+)$ Glucose and $\beta -D(+)$ Glucose are stereoisomers that differ in configuration only at the $C-1$ carbon atom,which is known as the anomeric carbon.
Such isomers are specifically referred to as anomers.

(B) When glucose $(C_6H_{12}O_6)$ is treated with bromine water ($Br_2$ water), which is a mild oxidizing agent, the aldehyde group $(-CHO)$ is oxidized to a carboxylic acid group $(-COOH)$.
This reaction results in the formation of gluconic acid $(C_6H_{12}O_7)$.
The reaction is: $CHO(CHOH)_4CH_2OH + [O] \xrightarrow{Br_2 \, \text{water}} COOH(CHOH)_4CH_2OH$.

(NONE) Sucrose is a disaccharide composed of $\alpha-D$-glucose and $\beta-D$-fructose. This is a correct match.
Bakelite is a polymer formed by the condensation of formaldehyde and phenol. This is a correct match.
Lactose is a disaccharide formed by $\beta-D$-galactose and $\beta-D$-glucose. The option $C$ states $\beta-D$-glucose and $\beta-D$-galactose,which is technically correct in terms of components,but the standard representation for lactose is $\beta-D$-galactose linked to $\beta-D$-glucose. However,looking at the options,all are chemically correct. Re-evaluating: Sucrose is $\alpha-D$-glucose + $\beta-D$-fructose (Correct). Bakelite is phenol + formaldehyde (Correct). Lactose is $\beta-D$-galactose + $\beta-D$-glucose (Correct). $PHBV$ is $3$-hydroxybutanoic acid + $3$-hydroxypentanoic acid (Correct). Given the context of typical textbook questions,if this is a multiple choice question where one must be incorrect,there might be a typo in the question or options. Assuming the question asks for the incorrect match,and all are correct,the question is flawed. However,based on standard curriculum,all these statements are factually correct.

(A) Bromine water $(Br_2/H_2O)$ is a mild oxidizing agent.
It selectively oxidizes the aldehyde group $(-CHO)$ of glucose to a carboxylic acid group $(-COOH)$.
This reaction results in the formation of gluconic acid,which is represented by the formula $COOH-(CHOH)_4-CH_2OH$.
It does not oxidize the primary or secondary alcohol groups present in the molecule.

(C) $Br_2/H_2O$ is a mild oxidizing agent that can oxidize an aldose (like glucose) to a carboxylic acid (gluconic acid) but cannot oxidize a ketose (like fructose).
Since glucose contains an aldehyde group,it undergoes oxidation,whereas fructose,which contains a ketone group,does not react under these conditions.
Therefore,$Br_2/H_2O$ is used to differentiate between glucose and fructose.

(D) Lactose is a disaccharide sugar derived from galactose and glucose.
The glycosidic linkage between galactose and glucose is $\beta-1 \rightarrow 4$ linkage.
Lactose makes up around $2-8 \%$ of milk by weight.
The infants of mammals feed on milk rich in lactose,which is further digested by the enzyme lactase in the intestine.
So,the correct answer is 'Lactose'.

(B) $Glucose$ $(C_6H_{12}O_6)$ contains an aldehyde group $(-CHO)$ as its functional group.
$Fructose$ $(C_6H_{12}O_6)$ contains a ketone group $(>C=O)$ as its functional group.
Since both have the same molecular formula but different functional groups,they are classified as functional group isomers.

(D) Glucose is an optically active molecule that exists in both $(+)$-form and $(-)$-form (as isomers).
However,a meso-form is not possible in glucose because it lacks a plane of symmetry due to its chiral centers and specific molecular structure.

(B) Glucosazone is formed by the reaction of glucose with phenylhydrazine $(C_6H_5NH-NH_2)$.
This reaction involves the formation of an osazone,where three molecules of phenylhydrazine are used for one molecule of glucose.
The reaction proceeds through the formation of a phenylhydrazone intermediate followed by oxidation and further reaction with phenylhydrazine.

(A) The open chain structure of glucose is $CH_2OH-(CHOH)_4-CHO$.
It contains $4$ chiral carbon atoms.
The number of optical isomers is calculated using the formula $2^n$,where $n$ is the number of chiral carbon atoms.
Number of optical isomers $= 2^4 = 16$.

(A) In polysaccharides,the linkage connecting monosaccharides is called a glycosidic linkage.
This is a covalent bond formed between the hemiacetal or hemiketal group of one monosaccharide and the hydroxyl group of another monosaccharide.
Therefore,the correct answer is $A$.

(D) non-reducing sugar is a carbohydrate that does not contain a free aldehyde or ketone group,meaning it cannot reduce Tollens' reagent or Fehling's solution.
Sucrose is a disaccharide where the glycosidic linkage is between the anomeric carbons of glucose and fructose,leaving no free hemiacetal or hemiketal group.
Starch and cellulose are polysaccharides where the anomeric carbons are involved in glycosidic linkages,making them non-reducing as well.
Therefore,all the given options are non-reducing sugars.

(C) Glucose is an aldohexose with the structure $CH_2OH(CHOH)_4CHO$.
It contains $4$ chiral carbon atoms.
The number of stereoisomers is given by the formula $2^n$,where $n$ is the number of chiral centers.
Here,$n = 4$.
Therefore,the number of stereoisomers $= 2^4 = 16$.

(B) Mutarotation is the change in the optical rotation due to the change in the equilibrium between two anomers when the corresponding stereocenters interconvert.
$A$ sugar must have a free hemiacetal or hemiketal group (anomeric carbon) to undergo mutarotation.
$Maltose$,$Mannose$,and $Fructose$ are reducing sugars that possess a free anomeric carbon,allowing them to open and close their rings,thus exhibiting mutarotation.
$Sucrose$ is a non-reducing sugar because the glycosidic bond is formed between the anomeric carbons of both $Glucose$ and $Fructose$ ($C_1$ of $Glucose$ and $C_2$ of $Fructose$).
Since there is no free anomeric carbon available,the ring cannot open,and therefore,$Sucrose$ does not show mutarotation.

(D) Cellulose is the most abundant organic compound found in nature. It is a major structural component of the cell walls of plants.

(C) Sucrose is a disaccharide with the molecular formula $C_{12}H_{22}O_{11}$.
Upon hydrolysis in the presence of an acid or the enzyme invertase,it breaks down into its constituent monosaccharides.
The reaction is: $C_{12}H_{22}O_{11} + H_2O \rightarrow C_6H_{12}O_6 \text{ (glucose)} + C_6H_{12}O_6 \text{ (fructose)}$.
Thus,one molecule of sucrose yields one molecule of $D-(+)$-glucose and one molecule of $D-(-)$-fructose.

(C) Glucose reacts with methanol $(CH_3OH)$ in the presence of dry $HCl$ to form methyl glucoside.
This reaction involves the formation of an acetal from the hemiacetal structure of glucose.
Since glucose forms a glycoside,it confirms that glucose exists in a cyclic structure (hemiacetal form) rather than an open-chain aldehyde form.

(D) The phenomenon of change in optical rotation of a freshly prepared solution of an optically active sugar (like glucose) with time until it reaches a constant value is known as $Mutarotation$.

(C) The enzyme amylase acts as a catalyst for the hydrolysis of starch.
Starch is a polysaccharide made up of glucose units.
In the presence of amylase,starch undergoes hydrolysis to produce maltose,which is a disaccharide consisting of two glucose units.

(A) The reaction of glucose with phenylhydrazine involves the formation of an osazone derivative.
One molecule of glucose reacts with $3$ molecules of phenylhydrazine.
The first molecule of phenylhydrazine reacts with the aldehyde group to form a phenylhydrazone.
The second and third molecules of phenylhydrazine are involved in the oxidation of the $C-2$ carbon and the subsequent formation of the osazone structure.
Thus,the balanced equation is: $C_6H_{12}O_6 + 3C_6H_5NHNH_2 \rightarrow C_{18}H_{22}N_4O_4 + C_6H_5NH_2 + NH_3 + 2H_2O$.
Therefore,$x = 3$.

(C) The relative sweetness of various sugars is compared to sucrose (cane sugar),which is assigned a value of $100$ in standard scales.
If the sweetness of cane sugar is taken as $10$,then the relative sweetness of glucose,which is approximately $0.74$ times that of sucrose,is calculated as $10 \times 0.74 = 7.4$.
Among the given options,$7.3$ is the closest value to the calculated sweetness of glucose.

(D) The Molisch test is a general chemical test for the presence of carbohydrates.
It involves the reaction of carbohydrates with $\alpha$-naphthol in the presence of concentrated $H_2SO_4$ to form a purple or violet ring at the interface.
Since all carbohydrates (monosaccharides,disaccharides,and polysaccharides) undergo this reaction,all of them give a positive Molisch test.

(A) In polysaccharides,individual monosaccharide units are joined together by an oxide linkage formed by the loss of a water molecule. This specific linkage is known as a $Glycosidic$ linkage.

(D) The phenomenon where $\alpha-D-$glucose and $\beta-D-$glucose undergo spontaneous interconversion in an aqueous solution to reach an equilibrium mixture is known as mutarotation.
This process involves the opening of the cyclic hemiacetal ring to form the open-chain aldehyde form,followed by re-cyclization to either the $\alpha$ or $\beta$ anomer.

(C) The sweetness of sugars is relative. Fructose is known to be the sweetest naturally occurring carbohydrate. Its relative sweetness is approximately $173$ compared to sucrose $(100)$,glucose $(74)$,and maltose $(32)$.

(C) Milk contains a disaccharide known as lactose.
Lactose is composed of one molecule of $D-(+)-glucose$ and one molecule of $D-(+)-galactose$.
It is commonly referred to as milk sugar.

(C) Glucose reacts with excess of phenylhydrazine $(C_6H_5NHNH_2)$ to form glucosazone,which is a crystalline osazone derivative.
The reaction involves the oxidation of the $C-2$ carbon and the formation of a hydrazone linkage at both $C-1$ and $C-2$ positions.

(C) $\alpha-D-$glucose and $\beta-D-$glucose are anomers of each other.
They differ in the configuration of the hydroxyl group $(-OH)$ at the $C-1$ carbon atom.
In $\alpha-D-$glucose,the $-OH$ group at $C-1$ is below the plane of the ring.
In $\beta-D-$glucose,the $-OH$ group at $C-1$ is above the plane of the ring.
Therefore,they differ due to their configuration at the anomeric carbon.

(C) The iodine test is a chemical test used to detect the presence of starch.
Starch forms a blue-black complex with iodine due to the formation of a helical structure that traps $I_2$ molecules.
Among the given options,$Starch$ is the primary substance that gives a positive iodine test.
While some other polysaccharides like glycogen give a reddish-brown color,starch is the standard answer for the iodine test in chemistry.

(D) $\alpha - D -$ glucose and $\beta - D -$ glucose 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 (the hemiacetal or hemiketal carbon),are known as anomers.
Therefore,$\alpha - D -$ glucose and $\beta - D -$ glucose are anomers.

(D) In an alkaline medium,sugars undergo a process known as Lobry de Bruyn-van Ekenstein transformation.
This process involves the rearrangement of the sugar molecule,leading to the formation of an equilibrium mixture of isomers (such as glucose,fructose,and mannose).
Therefore,sugars are unstable in alkaline media due to this rearrangement.

(B) Lactose is a disaccharide composed of $D-(+)$-galactose and $D-(+)$-glucose units linked by a $\beta-1,4$-glycosidic bond.
In lactose,the anomeric carbon of the galactose unit is involved in the glycosidic linkage,making it non-reducing.
However,the anomeric carbon of the glucose unit remains free,which allows it to exist in equilibrium with its open-chain aldehyde form.
Therefore,the glucose part of the lactose molecule is responsible for its reducing property.

(D) Sucrose is a non-reducing sugar because its glycosidic linkage involves the anomeric carbons of both glucose and fructose,making it correct.
Glucose is an aldose and can be oxidized to gluconic acid by bromine water,making it correct.
Glucose is dextrorotatory,meaning it rotates plane-polarized light to the right,making it correct.
Fructose is a ketose and does not undergo oxidation with bromine water,as bromine water is a mild oxidizing agent that specifically oxidizes aldoses to carboxylic acids. Therefore,the statement that fructose is oxidized by bromine water is incorrect.

(C) Sucrose $(C_{12}H_{22}O_{11})$ undergoes dehydration when treated with concentrated sulfuric acid $(H_2SO_4)$.
The reaction is as follows:
$C_{12}H_{22}O_{11} \xrightarrow{conc. H_2SO_4} 12C + 11H_2O$
Concentrated $H_2SO_4$ acts as a strong dehydrating agent,removing water molecules from the sugar to leave behind carbon (charcoal).

(C) $1$. Glucose $(CHO(CHOH)_4CH_2OH)$ reacts with $HCN$ to form a cyanohydrin $(A)$: $CHO(CHOH)_4CH_2OH + HCN \rightarrow CH(OH)(CN)(CHOH)_4CH_2OH$.
$2$. Hydrolysis of the cyanohydrin $(A)$ yields a carboxylic acid $(B)$: $CH(OH)(COOH)(CHOH)_4CH_2OH$.
$3$. Reduction of the carboxylic acid $(B)$ with $HI$ and red phosphorus $(\Delta)$ reduces all hydroxyl groups and the carboxylic acid group to a straight-chain alkane. The resulting product is $n$-heptane $(C)$.

(B) Glucose $(CHO(CHOH)_4CH_2OH)$ reacts with mild oxidizing agents like bromine water $(Br_2/H_2O)$.
Bromine water oxidizes the aldehyde group $(-CHO)$ of glucose to a carboxylic acid group $(-COOH)$ while leaving the secondary alcoholic groups $(-CHOH-)$ unaffected.
The resulting product is gluconic acid $(COOH(CHOH)_4CH_2OH)$.

(A) Maltose is a disaccharide composed of two units of $\alpha - D -$ glucose linked together by an $\alpha - 1,4 -$ glycosidic bond.

(B) Stereoisomers that differ in configuration at only one chiral center are known as $Epimers$.
For example,$D-Glucose$ and $D-Galactose$ are epimers at the $C-4$ position.

(A) Starch is a polysaccharide that serves as a storage carbohydrate in plants. It is a polymer composed of repeating units of $\alpha-D-glucose$. Therefore,the monomer of starch is glucose.

(B) Cellulose is a linear polysaccharide of $D$-glucose units joined by $\beta$-glycosidic linkages. It is the most abundant organic substance in the plant kingdom and is the primary structural component of the plant cell wall.

(C) Fructose is a ketohexose with the molecular formula $C_6H_{12}O_6$.
Its open-chain structure contains $3$ chiral carbon atoms.
The formula to calculate the number of optical isomers is $2^n$,where $n$ is the number of chiral carbon atoms.
Here,$n = 3$.
Therefore,the number of optical isomers = $2^3 = 8$.

(B) The two forms of $D$-glucopyranose,$\alpha$-$D$-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 anomeric carbon $(C-1)$,are known as anomers.

(B) $\beta - D - \text{glucose}$ and $\beta - D - \text{galactose}$ differ in the configuration of the hydroxyl group $(-OH)$ at the $C-4$ carbon atom.
Such stereoisomers that differ in configuration at only one chiral center (other than the anomeric carbon) are known as epimers.
Specifically,they are $C-4$ epimers.

(D) Glycoproteins are conjugated proteins in which the prosthetic group is a carbohydrate.
These are proteins that contain oligosaccharide chains (glycans) covalently attached to polypeptide side-chains.

(A) Hyperglycemia is a condition characterized by an abnormally high concentration of glucose (sugar) in the blood. It is commonly associated with diabetes mellitus.