Carbohydrates Questions in English

(D) Assertion $A$ states that Amylose is insoluble in water,which is incorrect because Amylose is a water-soluble component of starch.
Reason $R$ states that Amylose is a long linear molecule with more than $200$ glucose units,which is correct as it consists of $200$ to $1000$ $\alpha-D-(+)$-glucose units held by $\alpha$-glycosidic linkages.
Therefore,$A$ is incorrect but $R$ is correct.

(C) The reaction of $D$-glucose with ammoniacal $AgNO_3$ (Tollen's reagent) is an oxidation reaction.
The aldehydic group $(-CHO)$ at the $C_1$ position of $D$-glucose is oxidized to a carboxylic acid group $(-COOH)$,forming $D$-gluconic acid.
This reaction confirms the presence of a free aldehydic group in glucose,making it a reducing sugar.
The correct structure corresponds to option $C$.

(D) When $\alpha-D-(+)$-glucose is dissolved in water,it undergoes mutarotation.
This process involves the conversion of $\alpha-D-(+)$-glucose into an open-chain structure,which then cyclizes to form $\beta-D-(+)$-glucose.
At equilibrium,the solution contains a mixture of both $\alpha-D-(+)$-glucose (approx. $36\%$) and $\beta-D-(+)$-glucose (approx. $64\%$),with only a negligible amount of the open-chain form.
Therefore,the major constituents present in the solution are $\alpha-D-(+)$-glucose and $\beta-D-(+)$-glucose.

(B) and $L$ before the name of any compound commonly indicate the relative configuration. For assigning the configuration of monosaccharides,it is the lowest asymmetric carbon atom (chiral center) that is compared to glyceraldehyde. For glucose,if the $-OH$ group on the lowest asymmetric carbon $(C-5)$ is on the right side,it is assigned $D$-configuration. If the $-OH$ group is on the left side,it is assigned $L$-configuration. Both $D$ and $L$ configurations are non-superimposable mirror images of each other. Therefore,the structure of $L-(-)$-glucose is the mirror image of $D-(+)$-glucose.

(A) -glucose is an aldose sugar. When it is treated with a mild oxidizing agent like bromine water ($Br_2$ water),the aldehyde group $(-CHO)$ is selectively oxidized to a carboxylic acid group $(-COOH)$ without affecting the secondary alcoholic groups. This reaction produces gluconic acid.

(D) By analyzing the Haworth projections:
$(P)$ is $\alpha-D-(+)-$Glucopyranose,which matches $(iii)$.
$(Q)$ is $\beta-D-(+)-$Glucopyranose,which matches $(iv)$.
$(R)$ is $\alpha-D-(-)-$Fructofuranose,which matches $(i)$.
$(S)$ is $\beta-D-(-)-$Fructofuranose,which matches $(ii)$.
Therefore,the correct matching is $P$ $\rightarrow iii, Q$ $\rightarrow iv, R$ $\rightarrow i, S$ $\rightarrow ii$.

(C) Seliwanoff's test is a chemical test used to distinguish between aldose and ketose sugars.
Ketoses are more rapidly dehydrated in the presence of concentrated acid to form $5$-hydroxymethylfurfural compared to aldoses.
This intermediate then reacts with resorcinol to produce a cherry-red colored complex.
Therefore,Assertion $(A)$ is true.
However,the mechanism involves acid-catalyzed dehydration,not $\beta$-elimination to form furfural. Thus,Reason $(R)$ is false.

(A) $(i)$ Formation of a tetraacetate with $Ac_2O$ indicates that compound $A$ contains four $-OH$ groups.
$(ii)$ Oxidation of $A$ with $Br_2-H_2O$ yields an acid $C_5H_{10}O_6$,which confirms the presence of an aldehyde group $(-CHO)$ in $A$,as $Br_2-H_2O$ specifically oxidizes aldoses to aldonic acids.
$(iii)$ Reduction of $A$ with $HI$ gives isopentane $(2-methylbutane)$,which indicates the carbon skeleton is branched (isopentane structure).
$(iv)$ Combining these facts,the structure must be a branched pentose. The structure in option $A$ represents $2,3,4-trihydroxy-2-(hydroxymethyl)butanal$,which fits the molecular formula $C_5H_{10}O_5$ and the branched skeleton.

(A) The given Fischer projection of sugar $[X]$ is $D-$mannose. In $D-$mannose,the $-OH$ groups at $C-2$,$C-3$,$C-4$ and $C-5$ are on the left,left,right and right respectively in the Fischer projection. When converting to the Haworth structure,groups on the left side of the Fischer projection are placed above the ring,and groups on the right side are placed below the ring. For $\alpha -D-$mannopyranose,the $-OH$ group at the anomeric carbon $(C-1)$ is below the ring. Thus,the correct Haworth structure corresponds to option $A$.

(B) $1$. $X$ is a tetrose $(C_4H_8O_4)$ with two chiral carbons and a $-CHO$ group (positive Schiff's test).
$2$. Acetylation yields a triacetate,confirming three $-OH$ groups.
$3$. The $L$-isomer of a tetrose with two chiral centers and the given reactivity corresponds to $L$-threose.
$4$. Oxidation of $L$-threose with $HNO_3$ gives $L$-tartaric acid $(A)$.
$5$. Reduction of $L$-tartaric acid with $NaBH_4$ gives a chiral compound $(B)$.
$6$. Based on the Fischer projection for $L$-threose,the correct structure is represented in option $B$.

(B) The reaction of $D-(+)-\text{Glyceraldehyde}$ with $HCN$ followed by hydrolysis and oxidation with $HNO_3$ is a Kiliani-Fischer synthesis sequence.
$1$. The addition of $HCN$ to the aldehyde group of $D-(+)-\text{Glyceraldehyde}$ creates a new chiral center,resulting in two diastereomeric cyanohydrins.
$2$. Hydrolysis of these cyanohydrins converts the $-CN$ group into a $-COOH$ group,forming two diastereomeric aldonic acids.
$3$. Oxidation with $HNO_3$ further oxidizes the terminal $-CH_2OH$ group to a $-COOH$ group,resulting in two dicarboxylic acids (aldaric acids).
$4$. One of the resulting products is $meso-\text{tartaric acid}$ (or a derivative),which is optically inactive due to an internal plane of symmetry.
$5$. The other product is an optically active isomer (e.g.,$L-(+)-\text{tartaric acid}$ derivative).
Thus,the final products are one optically inactive $(meso)$ product and one optically active product.

(D) The correct matching is as follows:
$A$. Starch is a polymer of $\alpha$-glucose $(II)$.
$B$. Cellulose is a polymer of $\beta$-glucose $(III)$.
$C$. Nucleic acid is a polymer of nucleotides $(I)$.
$D$. Protein is a polymer of $\alpha$-amino acids $(IV)$.
Therefore,the correct sequence is $A-II, B-III, C-I, D-IV$.

(A) Fehling's reagent is used to detect the presence of reducing sugars.
Reducing sugars contain a free hemiacetal or hemiketal group,which allows them to act as reducing agents.
$Sucrose$ is a non-reducing sugar because it is a disaccharide formed by the linkage of $Glucose$ and $Fructose$ through their respective anomeric carbons,leaving no free hemiacetal group.
Therefore,$Sucrose$ does not give a reddish-brown precipitate with Fehling's reagent.
$Lactose$,$Glucose$,and $Maltose$ are all reducing sugars and will give a positive test.

(B) $1$. Glucose does not react with $NaHCO_3$ because it is not acidic enough to evolve $CO_2$ gas. Thus,$A-II$.
$2$. Glucose on oxidation with concentrated $HNO_3$ gives saccharic acid (glucaric acid). Thus,$B-IV$.
$3$. Glucose on prolonged heating with $HI$ gives $n$-hexane,indicating the presence of a straight chain of six carbon atoms. Thus,$C-III$.
$4$. Glucose on oxidation with bromine water gives gluconic acid. Thus,$D-I$.
Therefore,the correct matching is $A-II, B-IV, C-III, D-I$.

(A) -Glucose has the configuration $OH$ on the right at $C-2, C-4, C-5$ and on the left at $C-3$.
$L$-Glucose is the enantiomer of $D$-Glucose,meaning all chiral centers have the opposite configuration.
Therefore,in $L$-Glucose,the $OH$ groups are on the left at $C-2, C-4, C-5$ and on the right at $C-3$.
The structure corresponding to this is shown in option $A$.

(A) The correct matches are as follows:
$A$. $\alpha$-Glucose and $\alpha$-Galactose are $C-4$ epimers,so $A-IV$.
$B$. $\alpha$-Glucose and $\beta$-Glucose differ only in the configuration at the anomeric carbon $(C-1)$,so they are anomers,$B-III$.
$C$. $\alpha$-Glucose (an aldohexose) and $\alpha$-Fructose (a ketohexose) are functional isomers,so $C-I$.
$D$. $\alpha$-Glucose (hexose) and $\alpha$-Ribose (pentose) belong to the same homologous series (carbohydrates),so $D-II$.
Therefore,the correct sequence is $A-IV, B-III, C-I, D-II$.

(C) The four nitrogenous bases present in $DNA$ are adenine,guanine,cytosine,and thymine.
Option $A$ shows the structure of adenine.
Option $B$ shows the structure of thymine.
Option $D$ shows the structure of cytosine.
Option $C$ shows a structure that does not correspond to any of the standard $DNA$ bases (it appears to be a modified purine base not found in $DNA$). Therefore,the structure in option $C$ is incorrect.

(A) The structure provided is $D$-glucose.
$1$. $Br_2$ water is a mild oxidizing agent that oxidizes the $-CHO$ group to a $-COOH$ group,forming gluconic acid (a monocarboxylic acid),not a dicarboxylic acid. Thus,this statement is incorrect.
$2$. Glucose exists primarily in cyclic hemiacetal forms in solution,where the concentration of the free aldehyde group is extremely low. Therefore,it does not give Schiff's test.
$3$. $D$-glucose has $4$ chiral (asymmetric) carbon atoms (at $C_2, C_3, C_4, C_5$). This statement is correct.
$4$. In aqueous solution,$D$-glucose exists in equilibrium with its $\alpha$ and $\beta$ pyranose cyclic forms. This statement is correct.

(A) $Glucose$ is soluble in water due to the presence of multiple hydroxyl $(-OH)$ groups which facilitate extensive hydrogen bonding with water molecules. The aldehyde group does not primarily account for its high solubility.
$Glucose$ exists in equilibrium between open-chain and cyclic (pyranose) isomeric forms in its aqueous solution.
$Glucose$ contains $6$ carbon atoms,making it a hexose,and possesses an aldehyde functional group,making it an aldose. Thus,it is an aldohexose.
$Sucrose$ is a disaccharide composed of one unit of $\alpha-D-glucose$ and one unit of $\beta-D-fructose$ linked by a glycosidic bond.
Therefore,the statement that $Glucose$ is soluble in water because of the aldehyde group is incorrect.

(B) Glucose contains an aldehyde group $(-CHO)$,but it exists primarily in a cyclic hemiacetal form in equilibrium with a small amount of the open-chain form.
Due to this cyclic structure and the low concentration of the free aldehyde group,glucose does not undergo certain reactions typical of aldehydes.
Specifically,glucose does not give Schiff's test $(B)$ and it does not form the hydrogen sulphite addition product with sodium bisulphite ($NaHSO_3$,$E$).
Therefore,the correct reagents are $B$ and $E$.

(C) $Statement-1$ is True: Glucose is a reducing sugar and reacts with Fehling's solution to form a reddish-brown precipitate of cuprous oxide $(Cu_2O)$.
$Statement-2$ is False: The reaction of glucose with Fehling's solution produces gluconic acid and cuprous oxide $(Cu_2O)$,not cupric oxide $(CuO)$.
The chemical reaction is:
$C_6H_{12}O_6 + 2Cu^{2+} + 5OH^- \longrightarrow C_6H_{11}O_7^- + Cu_2O \downarrow + 3H_2O$

(A) Cellulose is a linear polymer of $\beta-D$-glucose units joined by $\beta-1,4$-glycosidic linkages.
Each glucose unit in cellulose contains three hydroxyl $(-OH)$ groups (at $C_2, C_3,$ and $C_6$ positions).
Upon acetylation with excess acetic anhydride in the presence of catalytic $H_2SO_4$,all three hydroxyl groups are converted into acetyl $(-OAc)$ groups,resulting in cellulose triacetate.
The structure must show the $\beta-1,4$-linkage (where the glycosidic oxygen is in the equatorial position relative to the ring) and the presence of $-OAc$ groups at all three positions on each glucose unit.
Option $A$ correctly depicts the $\beta-1,4$-glycosidic linkage and the acetylation of all three hydroxyl groups on each glucose unit.

(C) 'Invert sugar' is an equimolar mixture of $D-(+)$-glucose and $D-(-)$-fructose.
$(C)$ The specific rotation of 'Invert sugar' is calculated as the average of the specific rotations of its components: $\alpha_{\text{invert sugar}} = \frac{(+52.7^{\circ}) + (-92.4^{\circ})}{2} \approx -20^{\circ}$.
$(A)$ 'Invert sugar' is prepared by the hydrolysis of sucrose,not maltose.
$(D)$ $Br_2$ water oxidizes the aldehyde group of glucose to gluconic acid,not saccharic acid (which is formed by oxidation with $HNO_3$).

(A) disaccharide that cannot be oxidized by bromine water is a non-reducing sugar,meaning it lacks a free hemiacetal or hemiketal group. Sucrose is a non-reducing sugar because its glycosidic linkage involves both anomeric carbons ($C1$ of glucose and $C2$ of fructose).
Upon acid hydrolysis,sucrose yields $D-(+)$-glucose and $D-(-)$-fructose. The resulting mixture is laevorotatory because the specific rotation of $D-(-)$-fructose $(-92^{\circ})$ is greater in magnitude than that of $D-(+)$-glucose $(+52.5^{\circ})$.
Structure $A$ represents sucrose,which consists of $\alpha-D$-glucopyranose and $\beta-D$-fructofuranose units linked by an $\alpha, \beta-1,2$-glycosidic bond.

(A) In sugar $X$ (Sucrose),the glycosidic linkage is between $C1$ of $\alpha$-$D$-glucose and $C2$ of $\beta$-$D$-fructose. Both anomeric carbons are involved in the linkage,making it a non-reducing sugar.
In sugar $Y$ (Maltose),the glycosidic linkage is between $C1$ of $\alpha$-$D$-glucose and $C4$ of another glucose unit. One anomeric carbon remains free (hemiacetal),making it a reducing sugar.
Regarding the linkages: $X$ has an $\alpha$-linkage (from glucose) and $Y$ has an $\alpha$-linkage (maltose is $\alpha(1\to4)$). However,based on the provided options and standard structural analysis,$X$ is non-reducing and $Y$ is reducing (Statement $B$). The linkage in $X$ is $\alpha, \beta$ and in $Y$ is $\alpha$. Thus,$(B)$ and $(C)$ are the correct statements.

(A) The given structure represents a disaccharide where ring $(a)$ is a six-membered pyranose ring (glucose unit) and ring $(b)$ is a five-membered furanose ring (fructose unit).
The glycosidic linkage connects the anomeric carbon of the glucose unit to the fructose unit.
In the provided structure,the oxygen atom of the glycosidic linkage is directed downwards relative to the plane of the ring $(a)$,which characterizes an $\alpha$-glycosidic linkage.
Therefore,ring $(a)$ is a pyranose ring with an $\alpha$-glycosidic linkage.

(B) The given structure is a six-membered pyranose ring.
At the anomeric carbon $(C1)$,the $-OH$ group is in the equatorial (up) position,which identifies it as the $\beta$-anomer.
Since the substituent at the anomeric carbon is a hydrogen atom $(-H)$ rather than a hydroxymethyl group $(-CH_2OH)$,it is an aldohexose (specifically $\beta$-$D$-glucopyranose).
Therefore,the correct option is $B$.

(C) The given structure is an aldohexose with three chiral centers at $C-3, C-4,$ and $C-5$.
When it forms a pyranose ring,the $C-1$ carbon (the carbonyl carbon) becomes a new chiral center (anomeric carbon).
Therefore,the total number of chiral centers in the pyranose form is $3 + 1 = 4$.
The total number of stereoisomers is $2^n$,where $n$ is the number of chiral centers.
Thus,the total number of stereoisomers $= 2^4 = 16$.
However,the question specifies the $D$-configuration,which fixes the configuration at the highest numbered chiral center $(C-5)$.
With the $D$-configuration fixed,the number of stereoisomers is $2^{n-1} = 2^{4-1} = 2^3 = 8$.

(D) $1$. $D$-glucose and $L$-glucose are enantiomers. To obtain the Fischer projection of $L$-glucose,we invert the configuration at all chiral centers of $D$-glucose.
$2$. In $D$-glucose,the $-OH$ groups are at $C2$ (right),$C3$ (left),$C4$ (right),and $C5$ (right).
$3$. In $L$-glucose,the $-OH$ groups are at $C2$ (left),$C3$ (right),$C4$ (left),and $C5$ (left).
$4$. The cyclization of $L$-glucose involves the nucleophilic attack of the $-OH$ group at $C5$ on the carbonyl carbon $(C1)$.
$5$. For the $\beta$-anomer,the $-OH$ group at the anomeric carbon $(C1)$ is on the same side as the $-CH_2OH$ group (in the Haworth projection,both are pointing up).
$6$. Based on the provided solution image,the structure of $\beta-L$-glucopyranose is correctly represented by option $D$.

(B) The reaction of an aldose with $HNO_3$ (nitric acid) oxidizes both the terminal aldehyde group and the terminal primary alcohol group to carboxylic acid groups,forming a dicarboxylic acid (saccharic acid).
$D$-Glucose gives a product with $[\alpha]_{D} = +52.7^{\circ}$.
The enantiomer of this product will have an equal and opposite optical rotation,i.e.,$[\alpha]_{D} = -52.7^{\circ}$.
To obtain the enantiomer of the product formed from $D$-Glucose,we must start with the enantiomer of $D$-Glucose,which is $L$-Glucose,or any sugar that oxidizes to the enantiomer of the product formed from $D$-Glucose.
Looking at the structures provided:
Compound $(C)$ is $L$-Gulose.
Compound $(D)$ is $L$-Glucose.
Both $L$-Gulose and $L$-Glucose,upon oxidation with $HNO_3$,yield dicarboxylic acids that are enantiomers of the product formed from $D$-Glucose,thus exhibiting a rotation of $-52.7^{\circ}$.
Therefore,the correct compounds are $(C)$ and $(D)$.

(B) and $L$ configurations are determined by the position of the $-OH$ group on the chiral carbon furthest from the carbonyl group (the $C-5$ carbon in glucose).
$D$-glucose has the $-OH$ group on the right side at $C-5$.
$L$-glucose is the enantiomer of $D$-glucose,meaning it is its non-superimposable mirror image.
Therefore,in $L-(-)$-glucose,all $-OH$ groups are inverted compared to their positions in $D-(+)$-glucose.

(A) When $D$-glucose is treated with aqueous $NaOH$,it undergoes base-catalyzed tautomerization.
This process involves the formation of a common enediol intermediate.
The enediol intermediate can then tautomerize back to form $D$-glucose,$D$-fructose,or $D$-mannose.
Therefore,the resulting mixture contains $D$-glucose,$D$-fructose,and $D$-mannose.

(A) For a monosaccharide to be correlated to $D$-glyceraldehyde,the $-OH$ group on the lowest chiral carbon (the chiral carbon farthest from the carbonyl group) must be on the right-hand side in the Fischer projection.
Looking at the structures provided in the solution image:
In structure $(A)$,the $-OH$ group on the lowest chiral carbon is on the right.
In structure $(B)$,the $-OH$ group on the lowest chiral carbon is on the right.
In structure $(C)$,the $-OH$ group on the lowest chiral carbon is on the left ($L$-configuration).
In structure $(D)$,the $-OH$ group on the lowest chiral carbon is on the right.
Thus,structures $(A)$,$(B)$,and $(D)$ have the $D$-configuration.
The total number of such structures is $3$.

(B) Statement $I$ is true: Fructose is a ketohexose and does not contain an aldehydic group. However,in the alkaline medium of Tollen's reagent,it undergoes tautomerization to form an enediol intermediate,which then rearranges to form aldoses like glucose and mannose. These aldoses contain an aldehydic group and thus reduce Tollen's reagent.
Statement $II$ is true: As shown in the reaction,in the presence of a base,fructose undergoes tautomerization to form an enediol intermediate,which further rearranges to give a mixture of glucose and mannose.

(A) The sugar present in $DNA$ is $\beta-2$-deoxy-$D$-ribose.
$1$. It is a pentose sugar because it contains $5$ carbon atoms ($A$ is correct).
$2$. It exists in furanose form (a $5$-membered ring),not pyranose form ($B$ is incorrect).
$3$. It has the $D$-configuration ($C$ is correct).
$4$. It is a reducing sugar because it has a hemiacetal group at the $C1$ position,which can open to form an aldehyde ($D$ is correct).
$5$. In $DNA$,it exists in the $\beta$-anomeric form,not the $\alpha$-anomeric form ($E$ is incorrect).
Therefore,statements $A, C,$ and $D$ are correct.

(B) Statement $I$ is false because $D$-glucose pentaacetate does not contain a free aldehyde group,which is required for the reaction with $2,4$-dinitrophenylhydrazine ($2,4$-$DNP$).
Statement $II$ is true because starch undergoes hydrolysis in the presence of dilute acid at high temperature and pressure to yield glucose.

$(A)$ Sucrose $\rightarrow \alpha 1-\beta 2$ Glycosidic linkage.
$(B)$ Maltose $\rightarrow \alpha 1-4$ Glycosidic linkage.
$(C)$ Lactose $\rightarrow \beta 1-4$ Glycosidic linkage.
$(D)$ Amylopectin $\rightarrow \alpha 1-4$ and $\alpha 1-6$ Glycosidic linkage.
Therefore, the correct matching is $A-III, B-I, C-IV, D-II$.

(C) Starch $\xrightarrow{H^+ / H_2O}$ Glucose (Incorrect,starch yields glucose).
$(B)$ Cane sugar (Sucrose) $\xrightarrow{H^+ / H_2O}$ Glucose + Fructose (Correct,$50\%$ each).
$(C)$ Milk sugar (Lactose) $\xrightarrow{H^+ / H_2O}$ Glucose + Galactose (Correct).
$(D)$ Amylopectin $\xrightarrow{H^+ / H_2O}$ Glucose (Incorrect,amylopectin is a polymer of glucose).
$(E)$ Amylose $\xrightarrow{H^+ / H_2O}$ Glucose (Correct,amylose is a linear polymer of glucose).
Therefore,the correct options are $(B), (C)$ and $(E)$ only.

(B) The correct matches are as follows:
$A$. Amylose: It is a linear polymer of $\alpha-D$-glucose units linked by $\alpha-C_1-C_4$ glycosidic linkage,found in plants $(IV)$.
$B$. Cellulose: It is a linear polymer of $\beta-D$-glucose units linked by $\beta-C_1-C_4$ glycosidic linkage,found in plants $(I)$.
$C$. Glycogen: It is a branched polymer of $\alpha-D$-glucose units with $\alpha-C_1-C_4$ and $\alpha-C_1-C_6$ linkages,found in animals $(II)$.
$D$. Amylopectin: It is a branched polymer of $\alpha-D$-glucose units with $\alpha-C_1-C_4$ and $\alpha-C_1-C_6$ linkages,found in plants $(III)$.
Therefore,the correct matching is $A-IV, B-I, C-II, D-III$.

(A) The $D$ and $L$ configuration of monosaccharides is determined by the position of the $-OH$ group on the chiral carbon furthest from the carbonyl group (the $C-5$ carbon in fructose).
In $D$-fructose,the $-OH$ group at $C-5$ is on the right side.
In $L$-fructose,the $-OH$ group at $C-5$ is on the left side.
Comparing the given options,the structure representing $L$-fructose is the one where the $-OH$ group at the $C-5$ position is oriented to the left.

(B) Sucrose on hydrolysis yields $D-(+)-\text{glucose}$ and $D-(-)-\text{fructose}$.
Statement $I$ is incorrect because it incorrectly lists $D-(+)-\text{fructose}$ instead of $D-(-)-\text{fructose}$.
Statement $II$ is correct because the equimolar mixture of $D-(+)-\text{glucose}$ and $D-(-)-\text{fructose}$ obtained from sucrose hydrolysis is known as invert sugar.

(B) -Fructose is a ketohexose sugar that exists in both $\alpha$ and $\beta$-anomeric forms.
It is laevorotatory (rotates plane-polarized light to the left) and is commonly found in honey and fruits.
Therefore,the sugar $X$ is $D$-Fructose.

(C) The open-chain structure of glucose contains a free aldehyde group $(-CHO)$,which reacts with hydroxylamine $(NH_2OH)$ to form an oxime.
However,the pentaacetate of glucose does not contain a free aldehyde group because the hemiacetal hydroxyl group at $C_1$ is involved in the formation of the cyclic structure and is subsequently acetylated.
Since the cyclic structure is stable and the aldehyde group is masked in the hemiacetal form,the pentaacetate cannot react with hydroxylamine.
This specific observation confirms the existence of the cyclic structure of glucose.

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

(D) reducing sugar is a carbohydrate that possesses a free aldehyde or ketone group,or a hemiacetal/hemiketal group that can open to form such a group in solution.
$1$. Maltose (shown in option $A$) is a reducing sugar because it has a free hemiacetal group at the $C1$ position of one glucose unit.
$2$. Glucose (shown in option $B$) is an aldose and acts as a reducing sugar.
$3$. Fructose (shown in option $C$) is a ketose,but it undergoes tautomerization in alkaline solution to form glucose and mannose,thus acting as a reducing sugar.
Since all the given structures represent sugars that exhibit reducing properties,the correct answer is $D$.

(B) Epimers are diastereomers that differ in configuration at only one stereogenic center.
In the given structures,the first molecule is $D$-glucose and the second molecule is $D$-galactose.
Numbering the carbon atoms starting from the aldehyde group $(CHO)$ as $C-1$:
$C-1$: $CHO$
$C-2$: $-OH$ on right
$C-3$: $-OH$ on right
$C-4$: $-OH$ on right (in glucose) vs $-OH$ on left (in galactose)
$C-5$: $-OH$ on right
$C-6$: $CH_2OH$
Since the configuration differs only at the $C-4$ position,they are $C-4$ epimers.
Therefore,$n = 4$.

(A) In $\beta-D$-Glucopyranose,the $-OH$ group at the anomeric carbon $(C-1)$ is on the same side as the $-CH_2OH$ group at $C-5$. This means the $-OH$ group at $C-1$ is in the upward position (equatorial). Option $A$ represents this configuration correctly.

(D) hemiacetal group is formed when an alcohol reacts with an aldehyde. In disaccharides,if the anomeric carbon of one monosaccharide unit is involved in a glycosidic linkage,it loses its hemiacetal nature.
$1$. Glucose is a monosaccharide and exists in equilibrium with its cyclic hemiacetal form.
$2$. Lactose and Maltose are reducing sugars because they contain one free anomeric carbon that retains the hemiacetal group.
$3$. Sucrose is a non-reducing sugar because the glycosidic linkage is formed between the anomeric carbons of both glucose and fructose units,thereby eliminating the hemiacetal group in both.

(D) Tollen's reagent is used to detect the presence of an aldehyde group or reducing sugars.
$1$. $CH_3-CH_2-CHO$ is an aldehyde (propanal),which gives a positive Tollen's test.
$2$. $HCOOH$ (formic acid) is a unique carboxylic acid that contains an aldehyde group $(H-C=O)$ and thus acts as a reducing agent,giving a positive Tollen's test.
$3$. Glucose is a reducing sugar because it contains a free aldehyde group in its open-chain structure,which gives a positive Tollen's test.
Since all the given options (except sucrose,which is a non-reducing sugar) can produce a silver mirror,the question implies identifying which of the listed options (excluding sucrose) react. However,given the options,$HCOOH$ and Glucose are definitely positive. Propanal is also positive. Therefore,all options except Sucrose are positive. Given the structure in option $A$ is propanal,all options $A, B, C$ are positive. Thus,the correct answer is $D$.

(A) $Agar$ is a polysaccharide obtained from marine algae like $Gelidium$ and $Gracilaria$. It is widely used in laboratories as a culture medium to grow microbes and in the food industry for the preparation of ice-creams and jellies.