Carbohydrates Questions in English

(C) When glucose is heated with prolonged $HI$,it forms $n$-hexane $(CH_3-(CH_2)_4-CH_3)$.
This reaction confirms that all six carbon atoms in glucose are linked in a straight chain.

(C) The cyclic structure of glucose is called glucopyranose because it resembles the structure of the heterocyclic compound pyran,which contains a six-membered ring consisting of $5$ carbon atoms and $1$ oxygen atom. Therefore,the correct option is $C$.

(D) In $Maltose$,two $\alpha-D-glucose$ units are joined by a $C_1-C_4$ glycosidic linkage.
In $Lactose$,$\beta-D-galactose$ and $\beta-D-glucose$ are joined by a $C_1-C_4$ glycosidic linkage.
In $Cellulose$,$\beta-D-glucose$ units are joined by a $C_1-C_4$ glycosidic linkage.
In $Amylopectin$,it is a branched-chain polymer of $\alpha-D-glucose$ units. The linear chain is formed by $C_1-C_4$ glycosidic linkages,but the branching occurs due to $C_1-C_6$ glycosidic linkages. Therefore,not all units are joined by $C_1-C_4$ linkages.

(D) The reaction of glucose with $HI$ upon prolonged heating leads to the formation of $n-\text{hexane}$.
This reaction confirms that all $6$ carbon atoms in glucose are linked in a straight chain.
Therefore,the reaction $C_{6}H_{12}O_{6} \xrightarrow{HI, \Delta} n-\text{hexane}$ indicates the linear structure of glucose.
Thus,the correct option is $D$.

(C) The open-chain structure of glucose contains an aldehyde group $(-CHO)$.
Typically,aldehydes react with sodium hydrogen sulphite $(NaHSO_3)$ to form addition products.
However,glucose does not react with $NaHSO_3$,which indicates that the aldehyde group is not free.
This suggests that the aldehyde group is involved in the formation of a cyclic hemiacetal structure,confirming that glucose exists as a cyclic compound in its stable form.

(A) Glucose $(C_6H_{12}O_6)$ is an aldohexose,meaning it contains an aldehyde group $(-CHO)$.
Fructose $(C_6H_{12}O_6)$ is a ketohexose,meaning it contains a ketone group $(>C=O)$.
Since both have the same molecular formula but differ in the functional group present,they exhibit functional group isomerism.
Therefore,the correct option is $A$.

(A) The carbohydrate stored in the liver of animals is $Glycogen$.
It is often referred to as animal starch because its structure is similar to $Amylopectin$ and it serves as the primary storage form of glucose in animals.

(C) $Lactose$ is a disaccharide composed of two monosaccharide units.
Upon hydrolysis,$Lactose$ breaks down into one molecule of $D-(+)-Glucose$ and one molecule of $D-(+)-Galactose$.
The reaction is: $C_{12}H_{22}O_{11} + H_2O \rightarrow C_6H_{12}O_6 (Glucose) + C_6H_{12}O_6 (Galactose)$.

(B) Glycogen is a storage polysaccharide in animals and is highly branched,similar in structure to amylopectin but with more frequent branching points (every $8-12$ glucose units).
Amylose is a linear polymer of $\alpha$-$D$-glucose.
Cellulose is a linear polymer of $\beta$-$D$-glucose.
Amylopectin is a branched polymer of $\alpha$-$D$-glucose,but it is less branched than glycogen.

(C) sugar is considered non-reducing if it does not have a free hemiacetal or hemiketal group,meaning it cannot reduce Tollens' reagent or Fehling's solution.
Sucrose is a disaccharide composed of $D-(+)$-glucose and $D-(-)$-fructose joined by a glycosidic linkage between $C1$ of glucose and $C2$ of fructose.
Since both the anomeric carbons are involved in the glycosidic bond,sucrose does not have a free functional group to act as a reducing agent,making it a non-reducing sugar.

(D) Sucrose is a disaccharide that is dextrorotatory $(+66.5^{\circ})$.
Upon hydrolysis,it yields an equimolar mixture of $D-(+)$-glucose and $D-(-)$-fructose.
The specific rotation of $D-(+)$-glucose is $+52.5^{\circ}$,while the specific rotation of $D-(-)$-fructose is $-92.4^{\circ}$.
Since the magnitude of the laevorotation of fructose $(-92.4^{\circ})$ is greater than the magnitude of the dextrorotation of glucose $(+52.5^{\circ})$,the resulting mixture is laevorotatory.
Therefore,the correct reason is that the laevorotation of fructose is more than the dextrorotation of glucose.

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

(B) The reactions of glucose provide evidence for its structure:
$a$. Acetic anhydride reacts with glucose to form glucose pentaacetate,indicating the presence of $5$ hydroxyl groups $(iii)$.
$b$. Bromine water is a mild oxidizing agent that oxidizes the aldehyde group of glucose to gluconic acid,indicating the presence of an aldehyde group $(i)$.
$c$. Hydroiodic acid $(HI)$ reduces glucose to $n$-hexane,confirming that glucose has a straight chain of $6$ carbon atoms $(ii)$.
$d$. Hydrogen cyanide $(HCN)$ reacts with the carbonyl group of glucose to form a cyanohydrin,indicating the presence of a carbonyl group $(iv)$.
Therefore,the correct matching is $a-iii, b-i, c-ii, d-iv$.

(B) Glucose gives a positive Fehling's solution test.
This is because glucose is a reducing sugar,as it contains a free aldehyde group $(-CHO)$ in its open-chain structure,which can reduce $Cu^{2+}$ ions to $Cu^+$ ions.

(A) When glucose is heated with $HI$ and red phosphorus,it undergoes complete reduction to form $n$-hexane. The reaction is as follows:
$CHO-(CHOH)_4-CH_2OH \xrightarrow{HI/\text{red } P} CH_3-(CH_2)_4-CH_3$ ($n$-hexane).
This reaction confirms that glucose contains a straight chain of six carbon atoms.

(D) Sometimes the blood sugar level of diabetic patients decreases suddenly.
Glucose is a simple sugar that is absorbed directly into the bloodstream.
Therefore,diabetic patients generally carry a packet of glucose,which can increase the blood sugar level almost instantaneously to prevent hypoglycemia.

(B) The $\alpha$-glycosidic linkage is formed when the hydroxyl group on the $C1$ carbon of an $\alpha$-glucose molecule is involved in a glycosidic bond.
Starch is a polymer of $\alpha$-glucose and consists of two components: amylose and amylopectin.
Amylose is a linear chain of $\alpha$-$D$-glucose units linked by $\alpha(1 \to 4)$-glycosidic bonds.
Amylopectin is a branched chain of $\alpha$-$D$-glucose units with $\alpha(1 \to 4)$-glycosidic bonds in the linear chain and $\alpha(1 \to 6)$-glycosidic bonds at the branching points.
Therefore,both amylose and amylopectin exhibit $\alpha$-glycosidic linkages.

(B) Sucrose does not contain a $C_{1}-C_{4}$ glycosidic linkage.
In sucrose,the glycosidic linkage is formed between $C_{1}$ of $\alpha$-glucose and $C_{2}$ of $\beta$-fructose.
Therefore,it is a $C_{1}-C_{2}$ linkage.

(D) Amylose is a linear polymer of $\alpha-D-(+)$-glucose units.
These units are connected by $C_{1}-C_{4}$ glycosidic linkages.
Therefore,the correct option is $D$.

(D) Glycogen is often referred to as animal starch because its structure is similar to amylopectin.
Both are polymers of $\alpha-D$-glucose units.
The primary difference is that glycogen is more highly branched than amylopectin.
While amylopectin has branching every $20-25$ glucose units,glycogen has branching every $10-14$ glucose units,making it more compact and extensively branched.

(A) Carbohydrates that can reduce Tollen's reagent or Fehling's solution are called reducing sugars. All monosaccharides and most disaccharides (except sucrose) are reducing sugars.
In sucrose,the glycosidic linkage is formed between the $C1$ of glucose and $C2$ of fructose. Since both the anomeric carbons are involved in the linkage,there is no free aldehyde or keto group available to act as a reducing agent. Therefore,it fails to react with Tollen's reagent and Fehling's solution.

(D) Maltose is a disaccharide composed of two $\alpha$-$D$-glucopyranose units.
These two units are joined together by an $\alpha$-glycosidic linkage between the $C1$ of one glucose unit and the $C4$ of the other glucose unit,known as a $1-4$ glycosidic linkage.

(C) The statement that aldose or ketose sugars in alkaline medium do not isomerize is incorrect.
In an alkaline medium,aldose and ketose sugars undergo a reversible isomerization process known as the $Lobry \ de \ Bruyn-van \ Ekenstein \ rearrangement$.
For example,in a dilute solution of $NaOH$,glucose (an aldose) undergoes isomerization to form an equilibrium mixture of $D$-glucose,$D$-mannose,and $D$-fructose.
Therefore,option $C$ is the incorrect statement.

(D) $\alpha-D-(+)$-glucose and $\beta-D-(+)$-glucose are called anomers.
These are the two cyclic hemiacetal forms of glucose that differ only in the configuration of the hydroxyl group at $C1$.

(A) Sucrose is a disaccharide composed of one molecule of $\alpha$-$D$-glucose and one molecule of $\beta$-$D$-fructose.
These two monosaccharide units are joined together by a glycosidic linkage between $C-1$ of $\alpha$-$D$-glucose and $C-2$ of $\beta$-$D$-fructose.

(A) Ammoniacal $AgNO_3$ solution is known as Tollen's reagent,which is used to test for reducing sugars.
Reducing sugars contain a free aldehydic or ketonic group that can reduce $Ag^+$ to metallic silver.
$Sucrose$ is a non-reducing sugar because the glycosidic linkage involves the anomeric carbons of both glucose and fructose,leaving no free aldehydic or ketonic group.
$Maltose$,$Lactose$,and $Fructose$ are reducing sugars as they possess at least one free functional group capable of oxidation.

(A) The structure of $\beta-D-(-)-$Fructofuranose is a five-membered ring (furanose form) where the $-CH_2OH$ group at $C-2$ and the $-OH$ group at $C-2$ are in the $\beta$-configuration (the $-OH$ group is on the same side as the $-CH_2OH$ group at $C-5$).
In the Haworth projection of $\beta-D-(-)-$Fructofuranose:
$1$. The $C-2$ anomeric carbon has the $-OH$ group pointing upwards.
$2$. The $-CH_2OH$ group at $C-5$ also points upwards.
$3$. The $-OH$ groups at $C-3$ and $C-4$ point downwards and upwards respectively.
Comparing the given structures,option $A$ correctly represents the $\beta-D-(-)-$Fructofuranose configuration.

(B) In the open-chain structure of $D$-glucose $(CHO-(CHOH)_4-CH_2OH)$,there are $5$ hydroxyl $(-OH)$ groups.
In the cyclic (ring) structure of $D$-glucose (glucopyranose),one $-OH$ group is involved in the formation of the hemiacetal linkage,but it remains as an $-OH$ group (the anomeric hydroxyl group). Thus,there are still $5$ hydroxyl groups present in the ring structure.

(D) Carbohydrates are stored in plants in the form of starch.
In animals,carbohydrates are stored in the form of glycogen.
Both starch and glycogen serve as energy reserves and are broken down during metabolism to release energy.

(B) $(i)$ Monosaccharides are the simplest carbohydrates and cannot be further hydrolysed into smaller units. Thus,statement $(i)$ is false.
$(ii)$ Disaccharides yield two monosaccharide units upon hydrolysis,which can be identical or different. For example,sucrose yields glucose and fructose,while maltose yields two glucose units. Thus,statement $(ii)$ is true.
$(iii)$ Polysaccharides are complex polymers of many monosaccharide units and are generally tasteless (non-sweet). Thus,statement $(iii)$ is true.
$(iv)$ All monosaccharides are reducing sugars because they contain a free aldehyde or ketone group. Thus,statement $(iv)$ is false.
Therefore,statements $(ii)$ and $(iii)$ are correct.

(C) Tollen's reagent is $(AgNO_3 + NH_4OH)$ and it provides a basic medium. Fructose is a keto-hexose,yet it can reduce Tollen's reagent. This is because,in a basic medium,it undergoes enolisation and rearrangement (Lobry de-Bruyn-van Ekenstein transformation) to form an aldose (like $D$-glucose or $D$-mannose) which contains an aldehyde group. The equilibrium involves the formation of an enediol intermediate.

(B) Oligosaccharides are formed by the linkage of two or more monosaccharide units through $O$-glycosidic bonds.
This linkage involves the formation of a $C-O-C$ bond between two sugar units,which is chemically classified as an ether linkage.
Thus,the correct option is $(B)$.

(B) $(X) \longrightarrow$ Sucrose on hydrolysis gives glucose and fructose.
Sucrose $\longrightarrow$ glucose + fructose
$(Y) \longrightarrow$ Lactose on hydrolysis gives $D$-glucose and $D$-galactose.
Lactose $\longrightarrow$ $D$-glucose + $D$-galactose
$(Z) \longrightarrow$ Maltose on hydrolysis gives $2$ units of glucose.
Maltose $\longrightarrow$ glucose + glucose
Thus,the correct option is $(B)$.

(C) The structure of $\beta-D-(-)-\text{fructofuranose}$ is characterized by a five-membered furanose ring.
In the $\beta$-anomer,the hydroxyl group $(-OH)$ at the anomeric carbon $(C-2)$ is on the same side as the $CH_2OH$ group at $C-5$.
Specifically,for $D$-fructose,the $CH_2OH$ group at $C-5$ is pointing upwards.
Therefore,in the $\beta$-anomer,the $-OH$ group at $C-2$ also points upwards.
Comparing the given options,the structure in option $C$ correctly represents this configuration.

(D) 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 glucose).
However,the question asks for the enantiomer of $D-(+)-glucose$,which is $L-(-)-glucose$.
In $D-(+)-glucose$,the $-OH$ groups are at positions: $C-2$ (right),$C-3$ (left),$C-4$ (right),and $C-5$ (right).
In $L-(-)-glucose$,the configuration at every chiral center is inverted compared to $D-glucose$.
Therefore,in $L-(-)-glucose$,the $-OH$ groups are at positions: $C-2$ (left),$C-3$ (right),$C-4$ (left),and $C-5$ (left).
This corresponds to the structure shown in option $D$.

(B) -fructose is a ketohexose. Its open-chain structure has a ketone group at $C-2$. The configuration at the chiral centers $C-3, C-4,$ and $C-5$ is as follows:
At $C-3$,the $-OH$ group is on the left.
At $C-4$,the $-OH$ group is on the right.
At $C-5$,the $-OH$ group is on the right (which determines the $D$-configuration).
Comparing this with the given options,the structure in option $B$ matches the correct Fischer projection of $D$-fructose.

(A) The structure of $\beta-D-(+)-$glucopyranose in its Haworth or cyclic form is characterized by the orientation of the hydroxyl group at the anomeric carbon $(C_1)$.
In the $\beta$-anomer,the $-OH$ group at $C_1$ is on the same side as the $-CH_2OH$ group (upwards in the Haworth projection,or in the same relative configuration in the Fischer-like cyclic projection).
Specifically,for $D$-glucose,the configuration at $C_2, C_3, C_4,$ and $C_5$ follows the standard $D$-glucose pattern.
Comparing the given options,structure $A$ represents $\beta-D-(+)-$glucopyranose where the $-OH$ group at the anomeric carbon is oriented such that it corresponds to the $\beta$-configuration.

(D) The correct answer is $D$.
Glucose is an aldohexose containing a free aldehyde group,but it does not form an addition product with $NaHSO_3$ because the aldehyde group is involved in the formation of a cyclic hemiacetal structure.
Option $A$ is correct because glucose does not restore the pink color of Schiff's reagent.
Option $B$ is correct as glucose exists in $\alpha-$ and $\beta-$ anomeric forms.
Option $C$ is correct because the pentaacetate of glucose lacks a free $-OH$ group at the $C-1$ position,preventing the formation of the open-chain aldehyde required to react with $NH_2OH$.

(C) When $D-(+)-$glucose is oxidized with strong oxidizing agents like 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 a dicarboxylic acid known as saccharic acid (also called glucaric acid).
The reaction is as follows:
$CHO-(CHOH)_4-CH_2OH + [O] \xrightarrow{HNO_3} COOH-(CHOH)_4-COOH$
Thus,the correct option is $C$.

(B) Glucose does not react with Schiff's reagent because the aldehyde group in glucose is not free; it is involved in the formation of a hemiacetal structure.
Glucose exists in two different crystalline anomeric forms,$\alpha-D$-glucose and $\beta-D$-glucose.
Thus,both statements $I$ and $II$ are correct.

(D) Glucose contains an aldehyde group $(-CHO)$,but it does not give all the characteristic reactions of aldehydes due to the formation of a cyclic hemiacetal structure.
Glucose reacts with $NH_2OH$ to form an oxime.
Glucose reacts with $HCN$ to form a cyanohydrin.
Glucose reacts with $Br_2 / H_2O$ (mild oxidizing agent) to form gluconic acid.
However,glucose does not react with $NaHSO_3$ (sodium bisulphite) to form a hydrogen sulphite addition product,which is a characteristic reaction of most aldehydes and ketones.

(D) When glucose reacts with bromine water $(Br_2/H_2O)$,it undergoes mild oxidation of the aldehyde group to a carboxylic acid group,resulting in the formation of gluconic acid.
The reaction is as follows:
$CH_2OH(CHOH)_4CHO + [O] \xrightarrow{Br_2/H_2O} CH_2OH(CHOH)_4COOH$
Therefore,the correct option is $(D)$.

(A) When glucose $(CHO(CHOH)_4CH_2OH)$ is heated with concentrated $HI$ for a prolonged period,the reduction of all hydroxyl groups and the aldehyde group occurs.
This reaction results in the formation of $n-$hexane $(CH_3(CH_2)_4CH_3)$,which indicates that all six carbon atoms in glucose are linked in a straight chain.

(C) Amylose is a linear polymer of $\alpha-D-(+)$-glucose units linked by $\alpha-1,4$-glycosidic linkages.
It is water-soluble and constitutes about $15-20 \%$ of starch.
Amylopectin,on the other hand,is a highly branched polymer of $\alpha-D-(+)$-glucose units.
Therefore,the statement that amylose is a highly branched polymer is incorrect.

(D) Statement-$I$ is incorrect because lactose is composed of $\beta-D$-galactose and $\beta-D$-glucose linked by a $\beta(1 \to 4)$ glycosidic bond.
Statement-$II$ is correct because lactose contains a hemiacetal group in the glucose unit,which allows it to undergo mutarotation and act as a reducing sugar.

(C) Statement-$I$ is correct: Cane sugar (sucrose) is a disaccharide composed of $\alpha-D$-glucose and $\beta-D$-fructose linked by a glycosidic bond.
Statement-$II$ is incorrect: Milk sugar (lactose) is a disaccharide composed of $\beta-D$-galactose and $\beta-D$-glucose,not $\alpha-D$-glucose.
Therefore,Statement-$I$ is correct,but Statement-$II$ is not correct.

(D) . Amylose is a linear polymer of $\alpha-D-glucose$ units linked by $\alpha-1,4$-glycosidic bonds. Thus,$A-I$.
$B$. Cellulose is a linear polymer of $\beta-D-glucose$ units linked by $\beta-1,4$-glycosidic bonds. Thus,$B-III$.
$C$. Amylopectin is a branched polymer of $\alpha-D-glucose$ units with $\alpha-1,4$-glycosidic bonds in the chain and $\alpha-1,6$-glycosidic bonds at the branching points. Thus,$C-II$.
Therefore,the correct matching is $A-I, B-III, C-II$.

(B) Tollens' reagent is reduced by sugars that contain a free hemiacetal or hemiketal group,known as reducing sugars.
$a$) Fructose is a ketohexose that isomerizes to glucose and mannose in the presence of base,thus it acts as a reducing sugar.
$b$) Sucrose is a non-reducing sugar because the glycosidic linkage is between the anomeric carbons of glucose and fructose,leaving no free hemiacetal or hemiketal group.
$c$) Lactose is a reducing sugar as it contains a free hemiacetal group.
$d$) Cellulose is a polysaccharide consisting of $D$-glucose units linked by $\beta(1 \to 4)$ glycosidic bonds; it is a non-reducing sugar.
Therefore,sucrose and cellulose do not reduce Tollens' reagent.

(B) Monosaccharides: $A$ carbohydrate that cannot be hydrolysed further to give simpler units of polyhydroxy aldehyde or ketone is called a monosaccharide.
Examples: $Glucose$,$Fructose$,$Ribose$.
Disaccharides: The sugar formed when $2$ monosaccharides are joined by a glycosidic linkage.
Examples: $Sucrose$,$Lactose$,$Maltose$.
Since $Fructose$ is a monosaccharide,it is not a disaccharide.