Carbohydrates Questions in English

(C) Life on Earth is based on carbon,likely because each carbon atom can form bonds with up to four other atoms simultaneously.
This quality makes carbon well-suited to form the long chains of molecules that serve as the basis for life as we know it,such as proteins and $DNA$.

(C) $Starch$ is colourless,but it forms a characteristic $blue-black$ complex with $Iodine$.
$Starch + Iodine_{(violet)} \to Starch-Iodine_{(blue-black)}$
Therefore,$Starch$ is used as an indicator for the detection of $Iodine$.

(B) $Glucose$ $(C_6H_{12}O_6)$ and $fructose$ $(C_6H_{12}O_6)$ have the same molecular formula but differ in their functional groups.
$Glucose$ contains an aldehyde group $(-CHO)$,whereas $fructose$ contains a ketone group $(>C=O)$.
Therefore,they are functional isomers.

(C) Glucose in its open-chain aldehyde form has $4$ chiral carbon atoms (at $C-2, C-3, C-4,$ and $C-5$).
According to the formula for the number of stereoisomers,$N = 2^n$,where $n$ is the number of chiral centers.
Here,$n = 4$,so the number of stereoisomers is $2^4 = 16$.

(B) The normal blood glucose level in a healthy human,measured $8-12 \ hours$ after a meal (fasting state),typically ranges between $70 \ mg/dL$ and $100 \ mg/dL$.
Since $1 \ dL = 100 \ mL$,the amount of glucose in $100 \ mL$ of blood is approximately $80 \ mg$ to $100 \ mg$.
Among the given options,$80 \ mg$ is the correct value representing the normal fasting blood glucose level.

(A) The cell membrane is a semi-permeable membrane that allows the passage of small molecules while restricting larger ones.
$Fructose$ is a monosaccharide with a low molecular weight $(180.16 \ g/mol)$,allowing it to pass through the membrane via facilitated diffusion.
$Glycogen$,$Haemoglobin$,and $Catalase$ are macromolecules (polysaccharides or proteins) with very high molecular weights,which cannot diffuse through the cell membrane.

(A) The reaction taking place during photosynthesis is:
$6 CO_2 + 6 H_2O \rightarrow C_6H_{12}O_6 + 6 O_2$
This glucose molecule undergoes polymerization to form starch,which is a polysaccharide.
Therefore,the formation of starch in plants is a result of the process of photosynthesis.

(C) The correct answer is $(C)$.
Hydrophobic structures are typically found in substances that are non-polar or have long hydrocarbon chains that repel water.
Linseed oil,lanolin,and rubber are hydrophobic in nature as they contain long hydrocarbon chains or non-polar structures.
Glycogen is a polysaccharide consisting of many glucose units,which contain numerous hydroxyl $(-OH)$ groups.
These hydroxyl groups make glycogen hydrophilic (water-attracting) rather than hydrophobic.

(C) Maltose is a disaccharide composed of two glucose units.
On hydrolysis,it yields two molecules of glucose:
$C_{12}H_{22}O_{11} + H_2O \xrightarrow{\text{Maltase}} 2C_6H_{12}O_6$ (Glucose).

(D) The enzyme zymase is capable of fermenting various hexose sugars into ethanol and $CO_2$.
$C_6H_{12}O_6$ (Glucose) $\xrightarrow{\text{Zymase}} 2C_2H_5OH + 2CO_2$.
Similarly,invert sugar (a mixture of glucose and fructose) and fructose itself can also be fermented by zymase to produce alcohol and $CO_2$.
Therefore,the correct option is $D$.

(D) Tollen's reagent is an ammoniacal silver nitrate solution,$[Ag(NH_3)_2]^+OH^-$.
It acts as a mild oxidizing agent.
Aldehydes are oxidized to corresponding carboxylate ions by Tollen's reagent,while the silver ions are reduced to metallic silver,which deposits on the inner walls of the test tube as a '$Silver mirror$'.
Since $Glucose$ gives a positive '$Silver mirror$' test,it confirms the presence of an aldehyde group in its open-chain structure.

(D) Tollen's reagent oxidizes compounds containing an aldehyde group,such as glucose.
It also oxidizes $\alpha$-hydroxy ketones that possess a $-COCH_2OH$ group,such as fructose,because they undergo tautomerization to form an aldehyde in an alkaline medium.

(A) Sucrose is a disaccharide,not a polymer. Upon acid or enzymatic hydrolysis,it yields only two molecules of monosaccharides.
Sucrose $\xrightarrow{H^+ \text{ or invertase}} D(+)\text{-glucose} + D(-)\text{-fructose}$

(C) Cellulose is a linear polysaccharide consisting of a large number of $D$-glucose units joined by $\beta$-glycosidic linkages.
Upon complete hydrolysis,cellulose yields only $D$-glucose molecules.

(C) The process of mercerisation involves treating cotton fibers with a concentrated solution of $NaOH$ (sodium hydroxide).
This treatment causes the cellulose fibers to swell,which increases their affinity for dyes and imparts a silky lustre to the fabric.
Therefore,the swelling of cellulose is caused by $Aq. \ NaOH$.

(C) Sterile gauze or medical cotton is prepared by the oxidation of cellulose using $NO_2$ (nitrogen dioxide) gas. This process converts the primary hydroxyl groups of cellulose into carboxylic acid groups,resulting in oxidized cellulose,which is used as a hemostatic agent in medicine.

(A) Cellulose is a natural polymer of $\beta-D-glucose$ units. The general formula for cellulose is represented as $({C_6}{H_{10}}{O_5})_n$,where $n$ is the degree of polymerization.

(A) Inulin is a naturally occurring polysaccharide consisting of fructose units.
Its general formula is $(C_6H_{10}O_5)_n$ (often represented as a polymer of fructose).
However,for the specific context of the question provided,the value of $n$ for inulin found in the roots of Dahlia is approximately $30$.
Therefore,the correct option is $A$.

(A) $Starch$ is a polymeric carbohydrate consisting of numerous glucose units joined by glycosidic bonds.
It consists of two types of molecules: the linear and helical $\alpha-amylose$ and the branched $amylopectin$.
Depending on the plant,$starch$ generally contains $20$ to $25 \%$ $amylose$ and $75$ to $80 \%$ $amylopectin$ by weight.

(C) Explanation: Cellulose is a structural component of the cell walls in plants.
It is a polysaccharide that consists of several $D-glucose$ units joined by glycosidic bonds.
Specifically,it is formed by the $\beta-1,4-glycosidic$ linkage of $\beta-D-glucose$ units.
Therefore,cellulose is a polymer of $\beta-D-glucose$.

(D) . The change in optical rotation of a freshly prepared solution of sugar with time is known as mutarotation.
$(\alpha-D$-Glucose$) ([\alpha] = +112^o, 36\%) \rightleftharpoons$ Equilibrium mixture $([\alpha]_D = +52^o, 0.02\%) \rightleftharpoons$ $(\beta-D-$Glucose$)$ $([\alpha]_D = +19^o, 64\%)$.
Glucose exists in two anomeric forms,$\alpha$ and $\beta$. When either of these pure forms is dissolved in water,it undergoes interconversion until an equilibrium mixture is reached,resulting in a change in the observed optical rotation.

(C) is the correct answer.
Inulin is a polysaccharide carbohydrate which is stored in the roots of Dahliya.

(C) The term carbohydrate was derived from the fact that many of them have the general formula $C_x(H_2O)_y$,which represents them as hydrates of carbon.
Therefore,the correct option is $C$.

(C) Benedict's solution is a chemical reagent used to test for the presence of reducing sugars,which contain a free aldehyde or ketone group.
It contains copper$(II)$ sulfate $(CuSO_4 \cdot 5H_2O)$,sodium carbonate,and sodium citrate.
The active species in Benedict's solution is the $Cu^{+2}$ ion,which is complexed with citrate ions to keep it in solution at an alkaline $pH$.
During the reaction with a reducing sugar,the aldehyde group is oxidized to a carboxylic acid,and the $Cu^{+2}$ ion is reduced to $Cu^{+1}$ in the form of copper$(I)$ oxide $(Cu_2O)$,which appears as a brick-red precipitate.

(D) The Tollen's reagent test is a characteristic test for the presence of an aldehyde group.
Glucose $(C_6H_{12}O_6)$ contains an aldehydic group $(-CHO)$ at the $C_1$ position.
When glucose reacts with Tollen's reagent,the aldehyde group is oxidized to a carboxylic acid group (gluconic acid),and the silver ions $(Ag^+)$ are reduced to metallic silver $(Ag)$,which forms a silver mirror on the inner walls of the test tube.
Therefore,the formation of a silver mirror confirms the presence of an aldehydic group.

(B) Ninhydrin ($2, 2-$dihydroxyindane$-1, 3-$dione) reacts with amino groups in proteins and amino acids to produce a blue-violet color. Since the compound gives a negative test with ninhydrin,it does not contain free amino groups.
Benedict's solution ($alkaline \ CuSO_4$ and citrate ions) reacts with reducing sugars (monosaccharides and some disaccharides) to produce a reddish-brown precipitate of $Cu_2O$.
Therefore,a compound that is negative for ninhydrin and positive for Benedict's solution is a monosaccharide.

(D) $1$. Molisch's test is a general test for all carbohydrates.
$2$. Benedict's test is used to identify reducing sugars,which possess a free aldehyde or ketone group.
$3$. Seliwanoff's test is specific for ketohexoses (like fructose) and distinguishes them from aldoses.
$4$. Sucrose is a non-reducing sugar (negative Benedict's test).
$5$. Fructose is a ketohexose (positive Seliwanoff's test).
$6$. Maltose is a reducing disaccharide (positive Benedict's test) and an aldose (negative Seliwanoff's test).
$7$. Therefore,the compound is Maltose.

(C) Glucose exists in a cyclic hemiacetal form in equilibrium with its open-chain structure.
When glucose is treated with methanol $(CH_3OH)$ in the presence of dry $HCl$ gas,the hemiacetal hydroxyl group (at $C_1$) reacts with methanol to form an acetal,known as a methyl glucoside.
The formation of two isomeric products,$\alpha$-methyl glucoside and $\beta$-methyl glucoside,is possible only because the $C_1$ carbon is a chiral center in the cyclic (ring) structure of glucose.
Therefore,the presence of a ring structure is responsible for the formation of these two glucosides.

(D) disaccharide,also known as a double sugar,is a carbohydrate composed of two monosaccharide units linked by a glycosidic bond.
Glucose,Fructose,and Xylose are monosaccharides.
Sucrose is a disaccharide formed by the condensation of one molecule of $Glucose$ and one molecule of $Fructose$.

(C) Both glucose and fructose have the same molecular formula,$C_6H_{12}O_6$.
Glucose is an aldohexose,while fructose is a ketohexose.
Since they share the same molecular formula but have different structural arrangements,they are functional isomers of each other.

(C) When sucrose is hydrolyzed,the direction of rotation of plane-polarized light changes from dextrorotatory to levorotatory. This change in the sign of optical rotation is why the process is known as the inversion of sucrose. The reaction is catalyzed in an acidic medium.

(C) Glycolipids are lipids with a carbohydrate attached by a glycosidic bond.
They are a group of sphingolipids that contain one or more sugar molecules.
The two sugars commonly found in glycolipids are $Glucose$ and $Galactose$.

(A) Epimerization is the process by which two compounds formed are epimers,i.e.,two compounds differ in configuration only at one chiral center,while the rest of the configuration of the two compounds is the same.
For example,glucose and galactose are epimers of each other,as they differ only in the position of the hydroxyl group at $C_4$ (a chiral carbon atom).

(A) An asymmetric carbon atom (chiral carbon) is a carbon atom that is attached to four different types of atoms or groups of atoms.
Glycolaldehyde $(HOCH_2CHO)$ has the structure $HO-CH_2-CHO$. In this molecule,the carbon atom of the $CH_2$ group is attached to two hydrogen atoms,and the carbonyl carbon is attached to a hydrogen and an oxygen atom via a double bond,meaning neither carbon is bonded to four different groups.
Therefore,Glycolaldehyde does not have a chiral carbon atom.

(B) Glyceraldehyde has been chosen as the arbitrary standard for the $D$ and $L$ notation in sugar chemistry.
This is because it contains an asymmetric carbon atom and can exist as a pair of enantiomers,which serves as the reference point for assigning relative configurations to other carbohydrates.

(C) Sugar may be defined as optically active polyhydroxy aldehydes or ketones.
As it contains at least one asymmetrical (chiral) carbon and is,therefore,optically active.

(A) Glucose $(C_6H_{12}O_6)$ contains an aldehyde group $(-CHO)$ and five hydroxyl groups $(-OH)$.
When glucose is oxidized with bromine water ($Br_2$ water),the aldehyde group is selectively oxidized to a carboxylic acid group $(-COOH)$ while the hydroxyl groups remain unaffected.
This reaction produces gluconic acid,which has the molecular formula $C_6H_{12}O_7$.
Since it contains five hydroxyl groups and one carboxylic acid group,it is a pentahydroxy acid.

(B) Sucrose is a disaccharide.
Upon hydrolysis,one molecule of sucrose yields one molecule of $D-(+)$-glucose and one molecule of $D-(-)$-fructose.

(D) Disaccharides are carbohydrates that,upon hydrolysis,yield two same or different monosaccharide units.
Sucrose is the most common disaccharide,commonly known as table sugar.
It is naturally occurring and found in many plants.
The molecular formula of sucrose is $C_{12}H_{22}O_{11}$.

(A) Starch is a polysaccharide,which is a polymer of $D$-glucose units.
Upon complete hydrolysis,the glycosidic linkages in starch are broken down,resulting in the formation of $D$-glucose molecules only.

(D) Monosaccharides are the simplest form of sugar and the most basic units of carbohydrates.
They cannot be further hydrolyzed into simpler chemical compounds.
Glucose,fructose,and galactose are all examples of monosaccharides.
Therefore,the correct option is $D$.

(D) The correct answer is option $D$ i.e.,"All of these".
Polysaccharides are carbohydrate polymers,such as starch,cellulose,and glycogen,which are composed of more than $10$ monosaccharide units joined together by glycosidic linkages.
Starch,glycogen,and cellulose are all polysaccharides derived from glucose.
Maltose,sucrose,and lactose are disaccharides,whereas fructose and galactose are monosaccharides.

(C) Gun-cotton is a highly flammable compound formed by the nitration of cellulose.
When cellulose is treated with concentrated nitric acid $(HNO_3)$ in the presence of concentrated sulphuric acid $(H_2SO_4)$,it undergoes nitration to form cellulose nitrate,which is commonly known as gun-cotton.

(B) Carbohydrates (also called saccharides) are molecular compounds made from just three elements: Carbon $(C)$,Hydrogen $(H)$,and Oxygen $(O)$.

(D) The furanose structure of sugars (like fructose) is confirmed by the formation of isopropylidene derivatives.
When fructose reacts with acetone in the presence of an acid catalyst,it forms di-isopropylidene fructose,which confirms the presence of a five-membered furanose ring structure.

(A) Glucose and fructose form the same osazone.
Both glucose and fructose differ only in their configuration at carbon atoms $C-1$ and $C-2$.
During the reaction with excess phenylhydrazine,the structural differences at these two carbons are eliminated,resulting in the formation of an identical osazone derivative for both sugars.

(D) Sucrose $(C_{12}H_{22}O_{11})$ is a carbohydrate that undergoes dehydration in the presence of concentrated $H_2SO_4$.
Concentrated $H_2SO_4$ acts as a strong dehydrating agent and removes water molecules from the sugar.
The reaction is: $C_{12}H_{22}O_{11} \xrightarrow{\text{conc. } H_2SO_4} 12C + 11H_2O$.
Since the products are carbon and water,none of the given options $(A, B, C)$ are correct.

(D) The letter '$D$' indicates the configuration of a carbohydrate.
It represents the position of the $-OH$ group attached to the second-last carbon atom in the Fischer projection.
If the $-OH$ group is present on the right side,it is called '$D$'-configuration,and if it is present on the left side,it is called '$L$'-configuration.

(B) Sugars contain a carbonyl group which can undergo enolization in the presence of the $OH^-$ ions found in an alkaline medium.
This enolization process leads to the rearrangement of the sugar structure (e.g.,Lobry de Bruyn-van Ekenstein transformation).
Therefore,rearrangement is possible in an alkaline medium.

(D) Cellulose is the most abundant organic polymer found in nature,primarily as the structural component of plant cell walls.