Carbohydrates Questions in English

(A) Glucose gives a silver mirror test (Tollens' test) with ammoniacal silver nitrate due to the presence of the $-CHO$ (aldehyde) group in its open-chain structure.
$CH_2OH(CHOH)_4CHO + 2[Ag(NH_3)_2]^+ + 3OH^- \rightarrow CH_2OH(CHOH)_4COO^- + 2Ag(s) + 4NH_3 + 2H_2O$
The formation of metallic silver $(Ag)$ results in the silver mirror.

(D) In the cyclic structure of glucopyranose,the carbon atom that is part of the hemiacetal group is known as the anomeric carbon. This carbon is the one that was the carbonyl carbon (aldehyde group) in the open-chain form of glucose. In glucopyranose,this is the $C-1$ carbon,which is bonded to both an oxygen atom in the ring and a hydroxyl group $(-OH)$.

(C) ribonucleoside consists of a ribose sugar molecule attached to a nitrogenous base.
In a ribose sugar $(C_5H_{10}O_5)$,there are $-OH$ groups at the $C_2$,$C_3$,and $C_5$ positions.
When the ribose sugar forms a nucleoside,the $-OH$ group at the $C_1$ position is replaced by the nitrogenous base.
Therefore,the remaining $-OH$ groups are at the $C_2$,$C_3$,and $C_5$ positions.
Thus,the total number of $-OH$ groups in a ribonucleoside is $3$.

(B) The sugar molecule present in an $RNA$ nucleotide is $\beta-D-ribose$.
This is a pentose sugar,which means it contains $5$ carbon atoms in its structure.
Therefore,the total number of carbon atoms in the sugar molecule of an $RNA$ nucleotide is $5$.

(B) Raffinose is a trisaccharide with the formula $C_{18}H_{32}O_{16}$.
On complete hydrolysis,it yields one molecule each of glucose,galactose,and fructose.
The reaction is: $C_{18}H_{32}O_{16} + 2H_2O$ $\rightarrow C_6H_{12}O_6 (\text{glucose}) + C_6H_{12}O_6 (\text{galactose}) + C_6H_{12}O_6 (\text{fructose})$.

(D) Stachyose is a tetrasaccharide with the molecular formula $C_{24}H_{42}O_{21}$.
Upon complete hydrolysis,one mole of stachyose yields two moles of galactose,one mole of glucose,and one mole of fructose.
Therefore,$n$ moles of stachyose will yield $2n$ moles of galactose,$n$ moles of glucose,and $n$ moles of fructose.

(A) Fructose $(C_6H_{12}O_6)$ is a laevorotatory ketohexose.
It exhibits laevorotation with a specific rotation of $[\alpha]_D^{20} = -92.4^{\circ}$.
Due to this property,fructose is commonly referred to as laevulose.

(A) Saccharic acid is formed by the oxidation of both the terminal $-CHO$ and $-CH_2OH$ groups of glucose to $-COOH$ groups.
The structure of saccharic acid is $HOOC-(CHOH)_4-COOH$.
It contains two carboxyl $(-COOH)$ groups at the ends and four hydroxyl $(-OH)$ groups attached to the carbon chain.
Therefore,the correct statement is that it contains two carboxyl and four hydroxyl groups.

(A) Explanation:
- Glucose is an aldohexose,meaning it is a $6$-carbon sugar with an aldehyde group.
- It is a reducing sugar,not nonreducing,because it can reduce Fehling's solution and Tollens' reagent.
- It contains $4$ chiral carbon atoms,not $3$.
- The ring structure of glucose is a hemiacetal,not a hemiketal,because it forms from the reaction of an aldehyde group with a hydroxyl group.

(D) In animals,excess glucose is stored as glycogen,which is a polysaccharide that serves as an energy reserve.
Glycogen has a highly branched structure consisting of glucose units.
Linear chains of glucose units are connected by $\alpha-1,4$-glycosidic linkages.
Branch points in the glycogen molecule are formed by $\alpha-1,6$-glycosidic linkages.

(A) The specific rotation of $D-(+)$-glucose in aqueous solution is $+52.7^{\circ}$.
This value is obtained after the process of mutarotation,where the equilibrium mixture of $\alpha-D$-glucose $(+112^{\circ})$ and $\beta-D$-glucose $(+19^{\circ})$ is formed.

(D) The ring structure of glucose is formed by the intramolecular hemiacetal formation.
This involves the reaction between the aldehyde group at $C_1$ and the hydroxyl group at $C_5$ to form a six-membered pyranose ring.

(D) Glucose reacts with acetic anhydride to form glucose pentaacetate. The reaction is as follows:
$CHO-(CHOH)_4-CH_2OH + 5(CH_3CO)_2O \rightarrow CHO-(CHOCOCH_3)_4-CH_2OCOCH_3 + 5CH_3COOH$
This reaction confirms the presence of five hydroxyl $(-OH)$ groups in the glucose molecule.

(D) The ring structure of fructose is a hemiketal,not a hemiacetal.
Fructose $(C_6H_{12}O_6)$ is a laevorotatory ketohexose.
It acts as a reducing sugar due to the presence of an $\alpha$-hydroxy ketone group which can tautomerize to an aldehyde in basic solution.

(D) Glucose exists in two cyclic forms,$\alpha$ and $\beta$,which are formed by intramolecular hemiacetal formation between the aldehyde group and one of the alcoholic groups at $C-5$ within the molecule.
This cyclic structure is responsible for the mutarotation and the existence of two anomers.
Thus,the open chain structure of glucose does not explain the existence of $\alpha$ and $\beta$ forms of glucose.

(B) The specific rotation of $D$-fructose is $-92.4^{\circ}$.
Since the value is negative,it indicates that fructose is levorotatory,which is why it is commonly referred to as levulose.

(A) In an acetylation reaction,the $H$ atom of an $(-OH)$ group is replaced by an acetyl group $(-COCH_3)$.
This results in an increase in molecular mass by $[(12+16+12+3 \times 1)-1] = 42 \ u$ per $-OH$ group.
Given the total increase in molecular mass is $84 \ u$.
Therefore,the number of $-OH$ groups $= \frac{84 \ u}{42 \ u} = 2$.
Among the given options,an aldotriose (e.g.,glyceraldehyde) contains two alcoholic $-OH$ groups,which upon acetylation will increase the molecular mass by $2 \times 42 \ u = 84 \ u$.

(A) non-reducing sugar is a carbohydrate that does not have a free aldehyde or ketone group to act as a reducing agent.
$Sucrose$ is a disaccharide composed of $Glucose$ and $Fructose$ units linked by a glycosidic bond between their respective anomeric carbons ($C1$ of $Glucose$ and $C2$ of $Fructose$).
Since both anomeric carbons are involved in the glycosidic linkage,$Sucrose$ cannot open to form an aldehyde or ketone group,making it a non-reducing sugar.
$Maltose$,$Lactose$,and $Glucose$ all contain at least one free anomeric carbon,making them reducing sugars.

(A) Fructose is a ketohexose with the molecular formula $C_6H_{12}O_6$.
In its cyclic form,known as fructofuranose,the ring is formed by the reaction between the ketone group at $C-2$ and the hydroxyl group at $C-5$.
This creates a five-membered ring containing four carbon atoms and one oxygen atom.
Therefore,the $C-2$ and $C-5$ atoms are linked through an oxygen atom to form the furanose ring structure.

(D) Ribose is a monosaccharide,therefore it cannot be further hydrolysed into smaller carbohydrate units.

(D) Fructose is a ketohexose that is levorotatory in nature,which is why it is commonly known as laevulose.

(D) The structure of glucose is $CHO-(CHOH)_4-CH_2OH$.
In this structure,the $-CH_2OH$ group is a primary alcoholic group because the carbon atom is attached to only one other carbon atom.
There is $1$ such primary alcoholic group.
The four $-CHOH$ groups are secondary alcoholic groups because each carbon atom is attached to two other carbon atoms.
Thus,there are $4$ secondary alcoholic groups.
Therefore,the number of primary and secondary alcoholic groups is $1$ and $4$ respectively.

(D) The open-chain structure of glucose is $CHO-(CHOH)_4-CH_2OH$.
In this structure,the four $-CHOH-$ groups are attached to two other carbon atoms,making the hydroxyl groups attached to these carbons secondary $(2^{\circ})$ hydroxyl groups.
The $-CH_2OH$ group at the end is a primary $(1^{\circ})$ hydroxyl group.
Therefore,there are $4$ secondary hydroxyl groups in glucose.

(B) trisaccharide is a carbohydrate that yields three monosaccharide units upon hydrolysis.
$Raffinose$ is a trisaccharide composed of galactose,glucose,and fructose units.

(D) Raffinose is a trisaccharide with the molecular formula $C_{18}H_{32}O_{16}$. Upon hydrolysis,it yields one molecule of each of the following monosaccharides: $D$-Glucose,$D$-Fructose,and $D$-Galactose.

(B) An aldohexose is a monosaccharide containing six carbon atoms and an aldehyde group $(-CHO)$.
Glucose has the molecular formula $C_6H_{12}O_6$ and contains an aldehyde group at the $C-1$ position,making it an aldohexose.
Ribose is an aldopentose $(C_5H_{10}O_5)$.
Fructose is a ketohexose $(C_6H_{12}O_6)$ containing a ketone group.
Threose is an aldotetrose $(C_4H_8O_4)$.

(C) trisaccharide is a carbohydrate that yields three molecules of monosaccharides upon hydrolysis.
$1$. Stachyose is a tetrasaccharide,yielding four monosaccharide units.
$2$. Sucrose is a disaccharide,yielding two monosaccharide units.
$3$. Raffinose is a trisaccharide,yielding three monosaccharide units $(C_{18}H_{32}O_{16} + 2H_2O \longrightarrow 3C_6H_{12}O_6)$.
$4$. Ribose is a monosaccharide itself.
Therefore,the correct answer is Raffinose.

(B) The reaction of glucose with $Br_2$ water (bromine water) is a mild oxidation reaction.
$Br_2$ water acts as a mild oxidizing agent that specifically oxidizes the aldehydic group $(-CHO)$ of glucose to a carboxylic acid group $(-COOH)$,resulting in the formation of gluconic acid.
This specific oxidation confirms the presence of an aldehydic group in the glucose molecule.

(D) Amylopectin is a branched-chain polymer of $\alpha-D-glucose$.
The linear chain of glucose units is formed by $\alpha-(1 \rightarrow 4)$ glycosidic linkages.
The branching points are formed by $\alpha-(1 \rightarrow 6)$ glycosidic linkages.
Therefore,the correct linkages for the chain and branches are $\alpha-(1 \rightarrow 4)$ and $\alpha-(1 \rightarrow 6)$ respectively.

(A) The false statement is that cellulose is a constituent of cell walls in animal cells.
Cellulose is a polysaccharide that makes up the cell walls of plant cells,but it is not found in animal cells.

(C) When glucose is heated with $HI$ for a long time,it undergoes reduction to form $n$-hexane.
The reaction is:
$CHO(CHOH)_4 CH_2 OH \xrightarrow{\Delta, HI} CH_3-(CH_2)_4-CH_3$
This reaction indicates that all six carbon atoms in glucose are linked in a straight chain.

(B) When glucose $(C_6H_{12}O_6)$ is oxidized with dilute nitric acid $(HNO_3)$,both the terminal aldehyde group $(-CHO)$ and the primary alcoholic group $(-CH_2OH)$ are oxidized to carboxylic acid groups $(-COOH)$.
This results in the formation of saccharic acid (also known as glucaric acid),which is a dicarboxylic acid with the structure $COOH-(CHOH)_4-COOH$.

(A) Glucose contains an aldehyde group $(-CHO)$ at the $C-1$ position and a primary alcohol group $(-CH_2OH)$ at the $C-6$ position.
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 $(-COOH)$,forming gluconic acid.
Therefore,the carbon atom at the $C-1$ position is converted to a $-COOH$ group.

(B) Glucose contains an aldehyde group $(-CHO)$ at $C-1$ and a primary alcoholic group $(-CH_2OH)$ at $C-6$.
When glucose is treated with dilute nitric acid $(dil. HNO_3)$,it undergoes oxidation to form a dicarboxylic acid known as saccharic acid (or glucaric acid).
In this reaction,both the terminal aldehyde group and the primary alcoholic group are oxidized to carboxylic acid $(-COOH)$ groups.
The reaction is represented as:
$CHO-(CHOH)_4-CH_2OH + [O] \xrightarrow{dil. HNO_3} COOH-(CHOH)_4-COOH$
Thus,the correct reagent is dilute nitric acid.

(C) Glucose $(CHO-(CHOH)_4-CH_2OH)$ and gluconic acid $(COOH-(CHOH)_4-CH_2OH)$ both undergo oxidation with dilute $HNO_3$ to form saccharic acid $(COOH-(CHOH)_4-COOH)$.
In this reaction,the terminal primary alcoholic group $(-CH_2OH)$ is oxidized to a carboxylic acid group $(-COOH)$.
Since only one $-CH_2OH$ group is present in the glucose molecule and it gets converted to $-COOH$,this reaction confirms the presence of one primary alcoholic group in glucose.

(A) Starch is a polysaccharide that acts as the main storage carbohydrate in plants.
Upon complete hydrolysis,starch breaks down into its monomeric units.
The monomeric unit of starch is $\alpha-D-glucose$.

(A) Glucose $(CHO(CHOH)_4CH_2OH)$ on oxidation with strong oxidizing agents like dilute nitric acid $(HNO_3)$ gets oxidized to a dicarboxylic acid known as saccharic acid (or glucaric acid).
The reaction is:
$CHO(CHOH)_4CH_2OH + [O] \xrightarrow{\text{Dil. } HNO_3} COOH(CHOH)_4COOH$
Thus,dilute $HNO_3$ is the correct reagent.

(A) When glucose is heated with concentrated $HI$ (hydroiodic acid) for a long time,it undergoes reduction to form $n$-hexane.
This reaction confirms the presence of a straight chain of six carbon atoms in the glucose molecule.
The reaction is as follows:
$CHO-(CHOH)_4-CH_2OH \xrightarrow{HI, \Delta} CH_3-(CH_2)_4-CH_3$ (n-hexane).

(A) When glucose is heated with $HI$ (hydroiodic acid) and red phosphorus,it undergoes reduction to form $n$-hexane $(CH_3(CH_2)_4CH_3)$.
This reaction confirms that the six carbon atoms in glucose are linked in a straight chain.

(C) When glucose is treated with dilute nitric acid $(HNO_3)$,both the aldehyde group $(-CHO)$ and the primary alcoholic group $(-CH_2OH)$ are oxidized to carboxylic acid groups $(-COOH)$.
This reaction results in the formation of a dicarboxylic acid known as saccharic acid (also called glucaric acid).
The reaction is represented as:
$CHO-(CHOH)_4-CH_2OH + [O] \xrightarrow{Dil. HNO_3} COOH-(CHOH)_4-COOH$
Thus,the correct product is saccharic acid.

(C) When glucose is treated with bromine water ($Br_2$ water),it undergoes mild oxidation.
In this reaction,the aldehydic group $(-CHO)$ of glucose is selectively oxidized to a carboxylic acid group $(-COOH)$,resulting in the formation of gluconic acid.
The reaction is:
$CHO(CHOH)_4CH_2OH + [O] \xrightarrow{Br_2/H_2O} COOH(CHOH)_4CH_2OH$

(C) Glucose reacts with hydroxylamine $(NH_2OH)$ to form an oxime,which is known as glucoxime.
The reaction is:
$CHO(CHOH)_4CH_2OH + NH_2OH \rightarrow CH=NOH(CHOH)_4CH_2OH + H_2O$
Therefore,the reagent used is hydroxylamine.

(A) The chemical formula of saccharic acid (also known as $D$-glucaric acid) is $C_6H_{10}O_8$.
Based on the structure provided:
- There are two $COOH$ groups,each containing $2$ oxygen atoms ($2 \times 2 = 4$ oxygen atoms).
- There are four $OH$ groups attached to the carbon chain,each containing $1$ oxygen atom ($4 \times 1 = 4$ oxygen atoms).
- Total number of oxygen atoms $= 4 + 4 = 8$.

(A) Glucose contains $5$ hydroxyl $(-OH)$ groups ($4$ secondary and $1$ primary).
When glucose reacts with excess acetic anhydride,all $5$ hydroxyl groups are acetylated to form glucose pentaacetate.
The reaction is as follows:
$CHO(CHOH)_4CH_2OH + 5(CH_3CO)_2O \xrightarrow{Pyridine} CHO(CHOCOCH_3)_4CH_2OCOCH_3 + 5CH_3COOH$
As shown in the reaction,$5$ moles of acetic acid $(CH_3COOH)$ are produced for every mole of glucose reacted.

(B) The reaction of glucose with hydrogen cyanide $(HCN)$ is a nucleophilic addition reaction. The carbonyl group $(CHO)$ of glucose reacts with $HCN$ to form a cyanohydrin derivative known as glucose cyanohydrin. The reaction is represented as follows:
$CHO(CHOH)_4CH_2OH + HCN \rightarrow CN-CH(OH)(CHOH)_4CH_2OH$

(A) When glucose is treated with bromine water ($Br_{2}$ water),it undergoes mild oxidation.
The aldehydic group $(-CHO)$ of glucose is oxidized to a carboxylic acid group $(-COOH)$,while the secondary alcoholic groups remain unaffected.
This reaction produces gluconic acid.
The reaction is:
$CHO(CHOH)_{4}CH_{2}OH + [O] \xrightarrow{Br_{2}/H_{2}O} COOH(CHOH)_{4}CH_{2}OH$
(Glucose) $\rightarrow$ (Gluconic acid)

(C)
When glucose is treated with bromine water ($Br_2$ water),it undergoes mild oxidation to form gluconic acid.
In this reaction,the aldehydic group $(-CHO)$ is selectively oxidized to a carboxylic acid group $(-COOH)$,while the secondary alcoholic groups remain unaffected.
This reaction confirms that the carbonyl group in glucose is an aldehydic group.
The reaction is:
$CHO(CHOH)_4CH_2OH + [O] \xrightarrow{Br_2/H_2O} COOH(CHOH)_4CH_2OH$
(Glucose $\rightarrow$ Gluconic acid)

(D) The chemical formula of Stachyose is $C_{24}H_{42}O_{21}$.
Therefore,a single molecule of Stachyose contains $24$ carbon atoms.

(C) Glucose $(C_6H_{12}O_6)$ on oxidation with mild oxidizing agent like bromine water ($Br_2$ water) yields gluconic acid $(COOH(CHOH)_4CH_2OH)$.
This reaction confirms the presence of an aldehyde group $(-CHO)$ in glucose,as bromine water is a mild oxidant that specifically oxidizes the aldehyde group to a carboxylic acid group without affecting the alcoholic groups.