Carbohydrates Questions in English

(C) When $D-(+)$-glucose is dissolved in water,its specific rotation gradually changes from $+112^{\circ}$ to $+52.7^{\circ}$.
This phenomenon is known as mutarotation.
It occurs due to the equilibrium established between the $\alpha-D$-glucose and $\beta-D$-glucose anomers through the open-chain form in the solution.

(C) An aldose is a monosaccharide containing an aldehyde group $(-CHO)$ at the end of the carbon chain.
Glucose,Ribose,and Mannose are all aldoses as they contain an aldehyde group.
Fructose is a ketohexose,meaning it contains a ketone group $(>C=O)$ at the $C-2$ position.
Therefore,Fructose is not an aldose.

(C) Cellulose is a polysaccharide that is insoluble in water and common organic solvents.
However,it is soluble in Schweizer's reagent,which is a solution of ammoniacal cupric hydroxide $[Cu(NH_3)_4(OH)_2]$.

(A) Glucose and mannose differ in configuration at the $C_2$ carbon atom.
Therefore,they are known as epimers.

(B) The 'Tollens' test' is a chemical test used to distinguish between an aldehyde and a ketone. It is given by compounds containing a free aldehyde group $(-CHO)$ or those that can tautomerize to form an aldehyde group in an alkaline medium.
Glucose is an aldose and contains a free aldehyde group,so it gives a positive 'Tollens' test'.
Fructose is a ketose,but in an alkaline medium (used in the 'Tollens' reagent),it undergoes tautomerization to form glucose and mannose,which contain aldehyde groups,thus it also gives a positive 'Tollens' test'.
Sucrose is a non-reducing sugar because its glycosidic linkage involves the anomeric carbons of both glucose and fructose,preventing the formation of a free aldehyde group. Therefore,it does not give the 'Tollens' test'.

(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 at the hemiacetal or hemiketal carbon (anomeric carbon),are known as anomers.

(A) Glucose $(C_6H_{12}O_6)$ contains an aldehyde group $(-CHO)$ and six carbon atoms. Therefore,it is classified as an aldohexose.

(C) Benedict's solution contains $CuSO_4$,$Na_2CO_3$,and sodium citrate.
Benedict's solution is more stable than Fehling's solution and is not affected by substances like uric acid present in urine.
Therefore,it is used to measure the presence of glucose in urine to detect diabetes.

(B) Ribose is a $5$-carbon sugar containing an aldehyde group,therefore it is classified as an $Aldopentose$.

(A) Glucose $(C_6H_{12}O_6)$ is a dextrorotatory sugar,meaning it rotates the plane of plane-polarized light to the right.
Fructose is levorotatory.
Cellulose and starch are polysaccharides composed of glucose units,but the term 'dextrorotatory sugar' specifically refers to the monosaccharide glucose.

(C) The open-chain structure of glucose fails to explain why glucose pentaacetate does not react with hydroxylamine $(NH_2OH)$.
This is because the cyclic structure of glucose involves the formation of a hemiacetal,which prevents the presence of a free aldehyde group in the pentaacetate form.
Therefore,the inability of glucose pentaacetate to react with hydroxylamine is evidence for the cyclic structure of glucose.

(D) Glucose,fructose,and maltose are all reducing sugars that contain a free or potentially free carbonyl group (aldehyde or ketone).
When these sugars react with excess phenylhydrazine $(C_6H_5NHNH_2)$,they form characteristic crystalline derivatives known as osazones.
Therefore,all three sugars listed form osazones.

(D) Glucose contains a carbonyl group (aldehyde) but it does not form a bisulphite addition product with $NaHSO_3$ because the equilibrium in aqueous solution exists primarily in the cyclic hemiacetal form,which is not reactive towards $NaHSO_3$.
$1.$ Glucose reacts with $Br_2/H_2O$ to form gluconic acid.
$2.$ Glucose reacts with $H_2NOH$ to form an oxime.
$3.$ Glucose reacts with $HI$ to form $n$-hexane.

(A) Starch is a polysaccharide composed of $\alpha$-$D$-glucose units linked by glycosidic bonds. It consists of two components: amylose and amylopectin. Therefore, statement $A$ is correct.
Amylose is a component of starch, not cellulose. Thus, statement $B$ is incorrect.
Proteins are polymers of various amino acids, not just one type. Thus, statement $C$ is incorrect.
The cyclic structure of fructose (fructofuranose) contains a five-membered ring consisting of four carbon atoms and one oxygen atom. Thus, statement $D$ is incorrect.

(B) The formation of osazone involves the reaction of glucose with phenylhydrazine.
Initially,the carbonyl group at $C-1$ reacts with phenylhydrazine to form a phenylhydrazone.
Subsequently,the $C-2$ carbon atom undergoes oxidation to a carbonyl group,which then reacts with another molecule of phenylhydrazine to form the osazone.
Thus,the $C-1$ and $C-2$ carbon atoms are involved in the process due to the oxidation of the $C-2$ hydroxyl group.

(C) Cellulose is a linear polysaccharide composed of $D$-glucose units joined by $\beta$-glycosidic linkages.
Upon complete hydrolysis,cellulose yields only $D$-glucose.
The reaction is: $(C_6H_{10}O_5)_n + nH_2O \rightarrow nC_6H_{12}O_6$ (Glucose).

(D) The phenomenon where the specific rotation of an optically active compound changes over time until it reaches a constant equilibrium value is known as mutarotation. $\alpha-D-$glucose and $\beta-D-$glucose undergo this process in aqueous solution due to the interconversion between their anomeric forms through an open-chain aldehyde intermediate.

(C) Lactose,also known as milk sugar,is a disaccharide.
It is composed of two monosaccharide units: $\beta-D$-galactose and $\beta-D$-glucose.
These two units are linked by a $\beta-1,4$-glycosidic linkage.

(D) The structure shown is a six-membered ring containing an oxygen atom,which is characteristic of a pyranose form of a sugar.
Specifically,the configuration at the anomeric carbon (the carbon attached to two oxygen atoms) shows the $-OH$ group pointing downwards,which corresponds to the $\alpha-$anomer.
Therefore,the structure is $\alpha-$pyranose.

(B) Glucose is an aldohexose with the formula $C_6H_{12}O_6$.
Its open-chain structure is $CHO-(CHOH)_4-CH_2OH$.
From this structure,we can observe:
$1$. It contains one aldehyde $(-CHO)$ group.
$2$. It contains four secondary alcoholic $(-CHOH)$ groups.
$3$. It contains one primary alcoholic $(-CH_2OH)$ group.
Therefore,the statement that it has a ketone group is incorrect,as glucose is an aldose,not a ketose.

(C) The Molisch test is a sensitive chemical test for the presence of carbohydrates.
It involves the reaction of carbohydrates with concentrated $H_2SO_4$ to form furfural or its derivatives,which then react with $\alpha$-naphthol to produce a purple or violet colored ring at the interface of the two liquids.

(D) $\alpha-D$-glucose and $\beta-D$-glucose are anomers of each other.
Anomers are a type of stereoisomer that differ in configuration only at the anomeric carbon atom,which is $C-1$ in the case of glucose.
In $\alpha-D$-glucose,the $-OH$ group at $C-1$ is below the plane of the ring,while in $\beta-D$-glucose,the $-OH$ group at $C-1$ is above the plane of the ring.

(D) Monosaccharides with $5$ or $6$ carbon atoms exist in cyclic forms in aqueous solution.
These cyclic structures are either $5$-membered rings,known as furanose structures,or $6$-membered rings,known as pyranose structures.
Therefore,they exist in either pyranose or furanose structures.

(D) The $\alpha-D-(+)-$glucose and $\beta-D-(+)-$glucose are isomers that differ in configuration only at the $C-1$ carbon atom,which is the anomeric carbon.
Such isomers,which differ in configuration at the hemiacetal or hemiketal carbon,are known as anomers.
Therefore,they are classified as anomers.

(D) Reducing sugars are carbohydrates that can reduce Fehling's solution and Tollens' reagent.
All monosaccharides (like $Glucose$ and $Fructose$) and most disaccharides (like $Maltose$) are reducing sugars because they contain a free hemiacetal or hemiketal group that can open to form an aldehyde or ketone group.
Therefore,all the given options are reducing sugars.

(A) Glucose $(C_6H_{12}O_6)$ is a primary source of energy for living organisms. It undergoes cellular respiration to produce $ATP$,which powers various biological processes,including growth and maintenance of cellular functions. Therefore,it is essential for providing energy.

(A) In the given structure,ring $(a)$ is a six-membered ring containing oxygen,which is a pyranose ring. The glycosidic linkage connects the anomeric carbon of ring $(a)$ to the other unit. Since the oxygen atom at the anomeric carbon of ring $(a)$ is pointing downwards relative to the ring plane,it represents an $\alpha$-glycosidic linkage. Ring $(b)$ is a five-membered ring,which is a furanose ring. Therefore,the correct statement is that ring $(a)$ is pyranose with an $\alpha$-glycosidic linkage.

(C) Reducing sugars react with phenylhydrazine to form characteristic crystalline compounds known as osazones.
These osazones have distinct melting points and crystalline structures,which are used to identify specific reducing sugars like glucose,fructose,and mannose.
Therefore,the formation of osazones is a standard method for the identification of reducing sugars.

(D) Glucose is an aldose sugar containing a free aldehyde group $(-CHO)$.
Due to the presence of this aldehyde group,glucose acts as a reducing sugar.
It can reduce Tollens' reagent to metallic silver $(Ag)$,Fehling's solution to red cuprous oxide $(Cu_2O)$,and Benedict's solution to red cuprous oxide $(Cu_2O)$.
Therefore,glucose reduces all of the given reagents.

(D) Monosaccharides are the simplest form of carbohydrates that cannot be hydrolyzed further into smaller carbohydrate units.
They are classified based on the number of carbon atoms they contain,typically ranging from $3$ to $7$ carbon atoms.
For example,trioses ($3$ carbons),tetroses ($4$ carbons),pentoses ($5$ carbons),hexoses ($6$ carbons),and heptoses ($7$ carbons).

(D) Amylopectin is a branched-chain polymer of $\alpha$-$D$-glucose units.
It is a major component of starch (about $80-85\%$).
Due to its branched structure and high molecular weight,it is insoluble in water.
However,it can form a colloidal solution when dispersed in water.

(C) $\alpha - D - $glucose and $\beta - D - $glucose are anomers of each other.
Anomers are a type of stereoisomers that differ in configuration only at the anomeric carbon (the $C-1$ carbon in glucose).
Therefore,they differ in their configuration at the $C-1$ carbon atom.

(B) Ninhydrin reacts with amino groups in proteins and amino acids to produce a blue-violet color.
Monosaccharides (like glucose) contain a free aldehyde or ketone group,which reduces Benedict's solution (alkaline $CuSO_4$ and citrate ions) to form a red precipitate of $Cu_2O$.
Since monosaccharides do not contain free amino groups,they do not react with ninhydrin,making them the correct answer.

(A) The $\alpha-$ and $\beta-$ forms of glucose are anomers of each other.
Anomers are cyclic monosaccharides that differ in configuration only at the anomeric carbon.
For glucose,the anomeric carbon is $C_1$.
In $\alpha-D-$glucose,the $-OH$ group at $C_1$ is below the plane of the ring,while in $\beta-D-$glucose,the $-OH$ group at $C_1$ is above the plane of the ring.

(D) Anomers are a specific type of diastereomers (epimers) that differ in configuration only at the hemiacetal or hemiketal carbon,which is known as the anomeric carbon.
For glucose,the anomeric carbon is $C_1$.
Therefore,$\alpha-D$-glucose and $\beta-D$-glucose are anomers because they differ in the configuration of the hydroxyl group at the $C_1$ position.

(B) Glycogen is a branched-chain polymer of $\alpha-D-glucose$ units. It is structurally similar to amylopectin but with more extensive branching. While amylopectin is branched at every $24-30$ glucose units,glycogen is more highly branched,occurring at every $8-12$ glucose units.

(A) Mutarotation is the change in the optical rotation of a solution of a sugar due to the equilibrium between its $\alpha$ and $\beta$ anomeric forms.
This phenomenon is observed in reducing sugars that possess a free hemiacetal or hemiketal group.
Sucrose is a non-reducing disaccharide composed of $\alpha$-$D$-glucose and $\beta$-$D$-fructose linked by a glycosidic bond between $C1$ of glucose and $C2$ of fructose.
Since both anomeric carbons are involved in the glycosidic linkage,sucrose does not have a free hemiacetal or hemiketal group and therefore does not exhibit mutarotation.
Both $D$-glucose and $L$-glucose are monosaccharides with free anomeric carbons,so they exhibit mutarotation.

(D) ketohexose is a monosaccharide containing six carbon atoms and a ketone group.
$Fructose$ $(C_6H_{12}O_6)$ is a classic example of a ketohexose because it contains a ketone functional group at the $C-2$ position and has a total of six carbon atoms.
$Mannose$ and $Galactose$ are aldohexoses,and $Maltose$ is a disaccharide.

(A) non-reducing sugar is a carbohydrate that does not have a free aldehyde or ketone group,meaning it cannot reduce Tollens' reagent or Fehling's solution.
Sucrose is a disaccharide composed of glucose and fructose linked by a glycosidic bond between their anomeric carbons ($C_1$ of glucose and $C_2$ of fructose).
Since both anomeric carbons are involved in the glycosidic linkage,there is no free functional group available to act as a reducing agent.
Therefore,sucrose is a non-reducing sugar.

(C) The process of mutarotation involves the interconversion of $\alpha-D-$glucose and $\beta-D-$glucose in an aqueous solution until an equilibrium is reached.
$\alpha-D-$glucose has a specific rotation of $+112^o$.
$\beta-D-$glucose has a specific rotation of $+19^o$.
At equilibrium,the mixture contains approximately $36\%$ of $\alpha-D-$glucose and $64\%$ of $\beta-D-$glucose.
The specific rotation of the equilibrium mixture is calculated as: $(0.36 \times 112^o) + (0.64 \times 19^o) \approx 40.32^o + 12.16^o = 52.48^o \approx +52^o$.

(D) pentose is a monosaccharide containing $5$ carbon atoms.
Galactose,glucose,and fructose are hexoses ($6$ carbon atoms).
Arabinose is an aldopentose with the formula $HOCH_2-(CHOH)_3CHO$.

(D) pentose sugar is a monosaccharide containing $5$ carbon atoms.
Among the given options,$Arabinose$,$Lyxose$,and $Ribose$ are all pentose sugars.
However,in the context of standard chemistry multiple-choice questions where only one answer is expected,$Ribose$ is the most common and fundamental example of a pentose sugar found in $RNA$.
If the question implies selecting the most representative pentose,$Ribose$ is the standard answer.

(A) disaccharide is a carbohydrate composed of two monosaccharide units linked by a glycosidic bond.
$Cellobiose$ is a disaccharide formed by two $D-glucose$ units linked by a $\beta(1 \to 4)$ glycosidic bond.
$Starch$ and $Cellulose$ are polysaccharides.
$Fructose$ is a monosaccharide.

(B) non-reducing sugar is a carbohydrate that does not have a free hemiacetal or hemiketal group,meaning it cannot act as a reducing agent.
In disaccharides,if the glycosidic linkage involves the anomeric carbons of both monosaccharide units,the sugar is non-reducing.
Sucrose is composed of $\alpha-D-glucose$ and $\beta-D-fructose$ linked by a glycosidic bond between $C1$ of glucose and $C2$ of fructose. Since both anomeric carbons are involved in the linkage,it has no free hemiacetal or hemiketal group,making it a non-reducing sugar.
Cellobiose,lactose,and maltose all contain at least one free anomeric carbon,making them reducing sugars.

(B) Monosaccharides are the simplest form of carbohydrates and cannot be further hydrolyzed into smaller polyhydroxy aldehyde or ketone units.
Disaccharides,oligosaccharides,and polysaccharides undergo hydrolysis to yield smaller units,which are ultimately monosaccharides (aldoses or ketoses).
Therefore,monosaccharides do not undergo hydrolysis.

(C) The reaction of glucose with methanol in the presence of dry $HCl$ is a Fischer glycosidation reaction.
This reaction involves the formation of an acetal from the hemiacetal form of glucose.
Since glucose exists in a cyclic hemiacetal form (pyranose structure) in solution,it reacts with methanol to form $\alpha$ and $\beta$ methyl glucosides.
This formation of glycosides is a characteristic property of the cyclic structure of glucose.

(D) Glucose exists in two anomeric forms: $\alpha$-$D$-glucose and $\beta$-$D$-glucose.
When glucose is crystallized from a concentrated aqueous solution at $30 \ ^\circ C$,$\alpha$-$D$-glucose is obtained,which has a specific rotation of $+111^\circ$.
When glucose is crystallized from hot glacial acetic acid,$\beta$-$D$-glucose is obtained,which has a specific rotation of $+19^\circ$.
Therefore,the specific rotation of glucose crystals obtained from acetic acid is $19^\circ$.

(D) Disaccharides are carbohydrates that yield two molecules of monosaccharides on hydrolysis. These two molecules may be the same or different. For example,sucrose on hydrolysis gives one molecule of glucose and one molecule of fructose (different),while maltose on hydrolysis gives two molecules of glucose (same).

(B) The structures of methyl $-\alpha-D$-glucoside and methyl $-\beta-D$-glucoside differ only in the configuration at the $C-1$ carbon atom (the anomeric carbon).
Such isomers,which differ in configuration only at the anomeric carbon,are known as anomers.
Therefore,they are anomers.

(D) The reaction of glucose with phenylhydrazine involves the formation of an osazone derivative.
One molecule of glucose reacts with $3$ molecules of phenylhydrazine $(C_6H_5NHNH_2)$.
The first molecule forms a phenylhydrazone,the second molecule acts as an oxidizing agent to convert the $C-2$ hydroxyl group into a carbonyl group,and the third molecule forms the final osazone structure.
Thus,the value of $X$ is $3$.