201. Examine the following claims.
Claim I: Glucose and fructose are functional isomers.
Claim II: Both can reduce Fehling solution, but their reasons for doing so are not identical.
Claim III: Both produce only one alditol on reduction.
The acceptable claims are:
ⓐ. I and II only
ⓑ. II and III only
ⓒ. I and III only
ⓓ. I, II and III
Correct Answer: I and II only
Explanation: Glucose and fructose have the same molecular formula but different carbonyl groups, so they are functional isomers. Glucose reduces Fehling solution through its open-chain aldehydic form. Fructose first rearranges through an enediol intermediate to aldose forms in alkaline medium. Glucose reduction produces sorbitol as a single principal alditol. Fructose reduction creates two configurations at \(\mathrm{C_2}\), giving sorbitol and mannitol, so Claim III is unacceptable.
202. A reaction record for two monosaccharides gives the following observations.
Sugar P forms a cyclic hemiacetal, has its anomeric carbon at \(\mathrm{C_1}\), and gives sorbitol on reduction.
Sugar Q forms a cyclic hemiketal, has its anomeric carbon at \(\mathrm{C_2}\), and forms the same osazone as P.
The best interpretation is:
ⓐ. P and Q are enantiomers
ⓑ. P is fructose and Q is glucose
ⓒ. P is glucose and Q is fructose
ⓓ. P and Q are identical cyclic structures
Correct Answer: P is glucose and Q is fructose
Explanation: A cyclic hemiacetal with anomeric carbon \(\mathrm{C_1}\) is characteristic of glucose. Reduction of glucose gives sorbitol, supporting this identification of P. A cyclic hemiketal with anomeric carbon \(\mathrm{C_2}\) is characteristic of fructose. Glucose and fructose form the same osazone because differences involving their first two carbon atoms are removed during derivative formation. Their common osazone does not make their original carbonyl structures identical.
203. A mixture contains \(0.100\,mol\) glucose and \(0.150\,mol\) fructose. Complete catalytic reduction consumes one mole of \(\mathrm{H_2}\) per mole of monosaccharide. The volume of hydrogen consumed at a molar gas volume of \(22.4\,L\,mol^{-1}\) is:
ⓐ. \(2.24\,L\)
ⓑ. \(3.36\,L\)
ⓒ. \(4.48\,L\)
ⓓ. \(5.60\,L\)
Correct Answer: \(5.60\,L\)
Explanation: \( \textbf{Amount of glucose:} \)
\[
n_{\text{glucose}}=0.100\,mol
\]
\( \textbf{Amount of fructose:} \)
\[
n_{\text{fructose}}=0.150\,mol
\]
\( \textbf{Total monosaccharide amount:} \)
\[
n_{\text{total}}=0.100+0.150
\]
\[
n_{\text{total}}=0.250\,mol
\]
\( \textbf{Hydrogenation ratio:} \)
\[
1\,mol\text{ monosaccharide}:1\,mol\mathrm{H_2}
\]
\( \textbf{Moles of hydrogen consumed:} \)
\[
n_{\mathrm{H_2}}=0.250\,mol
\]
\( \textbf{Molar gas volume supplied:} \)
\[
V_m=22.4\,L\,mol^{-1}
\]
\( \textbf{Volume relation:} \)
\[
V=nV_m
\]
\( \textbf{Substitution:} \)
\[
V_{\mathrm{H_2}}
=0.250\,mol\times22.4\,L\,mol^{-1}
\]
\( \textbf{Calculation:} \)
\[
V_{\mathrm{H_2}}=5.60\,L
\]
\( \textbf{Product interpretation:} \)
Glucose forms sorbitol, while fructose forms a sorbitol–mannitol mixture, but each monosaccharide consumes one mole of hydrogen.
\( \textbf{Final answer:} \)
The mixture consumes \(5.60\,L\) of hydrogen.
204. Assertion: Glucose and fructose form the same osazone but give different products on reduction.
Reason: Osazone formation removes differences involving \(\mathrm{C_1}\) and \(\mathrm{C_2}\), whereas reduction preserves the consequences of their different carbonyl positions.
ⓐ. Both Assertion and Reason are true, and Reason explains Assertion
ⓑ. Both Assertion and Reason are true, but Reason does not explain Assertion
ⓒ. Assertion is true, but Reason is false
ⓓ. Assertion is false, but Reason is true
Correct Answer: Both Assertion and Reason are true, and Reason explains Assertion
Explanation: Osazone formation modifies the first two carbon atoms of both sugars. The original aldehyde–ketone distinction and the relevant configuration at those positions are therefore lost in the common derivative. Reduction acts differently because glucose has a terminal aldehyde, while fructose has an internal ketone. Glucose gives sorbitol, whereas fructose forms two alditols because a new stereogenic centre is generated. The Reason accounts for why the same pair behaves alike in one reaction and differently in another.
205. A structural comparison states that glucose has an anomeric carbon at \(\mathrm{C_1}\), while fructose has one at \(\mathrm{C_2}\). This difference arises because the anomeric carbon is:
ⓐ. always the carbon bearing the terminal \(\mathrm{-CH_2OH}\) group
ⓑ. the highest-numbered stereogenic carbon
ⓒ. the former carbonyl carbon in the open-chain structure
ⓓ. the carbon that supplies the ring oxygen
Correct Answer: the former carbonyl carbon in the open-chain structure
Explanation: The anomeric carbon is created when the planar carbonyl carbon becomes tetrahedral during cyclisation. Glucose is an aldose with its carbonyl at \(\mathrm{C_1}\), so \(\mathrm{C_1}\) becomes anomeric. Fructose is a ketose with its carbonyl at \(\mathrm{C_2}\), making \(\mathrm{C_2}\) the anomeric centre. The oxygen that enters the ring comes from another hydroxyl group and does not determine the carbon label. The rule follows carbonyl origin rather than a fixed carbon number for every sugar.
206. A reagent is required to distinguish glucose from fructose without using alkaline conditions that promote fructose rearrangement. The most suitable choice is:
ⓐ. Tollens reagent
ⓑ. Fehling solution
ⓒ. phenylhydrazine
ⓓ. bromine water
Correct Answer: bromine water
Explanation: Tollens and Fehling reagents are alkaline, so both glucose and fructose give positive results. Fructose responds after enediol rearrangement to reducing aldose forms. Phenylhydrazine is also unsuitable because glucose and fructose form the same osazone. Neutral bromine water mildly oxidises the aldehydic group of glucose to gluconic acid. Fructose, being a ketose, does not undergo the same selective oxidation under these conditions.
207. A glycosidic linkage is formed when:
ⓐ. two anomeric hydroxyl groups always condense with each other
ⓑ. an anomeric hydroxyl condenses with another hydroxyl group
ⓒ. an anomeric hydroxyl condenses with a carboxyl group
ⓓ. two ordinary hydroxyl groups condense without an anomeric carbon
Correct Answer: an anomeric hydroxyl condenses with another hydroxyl group
Explanation: A glycosidic linkage involves the anomeric carbon of a monosaccharide. Its anomeric hydroxyl group reacts with a hydroxyl group of another molecule during condensation. Water is eliminated, and an acetal- or ketal-like carbon–oxygen–carbon linkage is produced. The second hydroxyl group may belong to another monosaccharide or, in some glycosides, to a non-sugar alcohol. Carboxyl and amino-group condensation instead produces linkages belonging to other biomolecule classes.
208. The notation \(\alpha(1\rightarrow4)\) indicates that:
ⓐ. \(\mathrm{C_1}\) of an \(\alpha\)-configured sugar is linked to \(\mathrm{C_4}\) of another unit
ⓑ. \(\mathrm{C_4}\) of an \(\alpha\)-configured sugar is linked to \(\mathrm{C_1}\) of the same molecule
ⓒ. four anomeric carbons are joined to one hydroxyl group
ⓓ. the linkage contains one oxygen and four carbon atoms
Correct Answer: \(\mathrm{C_1}\) of an \(\alpha\)-configured sugar is linked to \(\mathrm{C_4}\) of another unit
Explanation: The first number identifies the anomeric carbon that supplies the glycosidic bond. For an aldose unit such as glucose, this is commonly \(\mathrm{C_1}\). The arrow points to the carbon bearing the hydroxyl group on the accepting monosaccharide, here \(\mathrm{C_4}\). The \(\alpha\) symbol describes the configuration of the donor anomeric centre involved in the linkage. The notation records carbon positions and stereochemistry rather than the total number of atoms in the bond.
209. Use the arrangement described below.
The anomeric hydroxyl group at \(\mathrm{C_1}\) of an \(\alpha\)-D-glucose unit reacts with the hydroxyl group at \(\mathrm{C_6}\) of another glucose unit.
The linkage formed is:
ⓐ. \(\beta(1\rightarrow4)\)
ⓑ. \(\alpha(1\rightarrow2)\)
ⓒ. \(\alpha(1\rightarrow6)\)
ⓓ. \(\beta(1\rightarrow6)\)
Correct Answer: \(\alpha(1\rightarrow6)\)
Explanation: The donor monosaccharide contributes its anomeric carbon \(\mathrm{C_1}\). The accepting glucose unit contributes the hydroxyl group attached to \(\mathrm{C_6}\). These carbon numbers give the notation \((1\rightarrow6)\). The donor anomeric configuration is specified as \(\alpha\), so the full notation is \(\alpha(1\rightarrow6)\). Such linkages commonly produce branch points in amylopectin and glycogen.
210. Evaluate the following statements.
Statement I: Formation of a glycosidic bond is a condensation process.
Statement II: Hydrolysis of a glycosidic bond consumes water.
Statement III: Every molecule containing a glycosidic bond is necessarily non-reducing.
The acceptable statements are:
ⓐ. I only
ⓑ. I and II only
ⓒ. II and III only
ⓓ. I, II and III
Correct Answer: I and II only
Explanation: Glycosidic-bond formation commonly joins two hydroxyl-containing units with elimination of water, so Statement I is acceptable. Hydrolysis reverses this process by adding water across the linkage, making Statement II acceptable. Statement III is too broad because a disaccharide may retain a free anomeric carbon. Maltose and lactose contain glycosidic bonds yet remain reducing sugars. Non-reducing behaviour occurs when no anomeric centre remains available to generate an open-chain carbonyl form.
211. Two hexose molecules form a disaccharide by eliminating one molecule of water. The balanced molecular-formula equation is:
ⓐ. \(\mathrm{2C_6H_{12}O_6\rightarrow C_{12}H_{24}O_{12}}\)
ⓑ. \(\mathrm{2C_6H_{12}O_6+H_2O\rightarrow C_{12}H_{26}O_{13}}\)
ⓒ. \(\mathrm{2C_6H_{12}O_6\rightarrow C_{12}H_{22}O_{11}+H_2O}\)
ⓓ. \(\mathrm{2C_6H_{12}O_6\rightarrow C_{12}H_{20}O_{10}+2H_2O}\)
Correct Answer: \(\mathrm{2C_6H_{12}O_6\rightarrow C_{12}H_{22}O_{11}+H_2O}\)
Explanation: \( \textbf{Formula of one hexose molecule:} \)
\[
\mathrm{C_6H_{12}O_6}
\]
\( \textbf{Formula total for two hexoses:} \)
\[
\mathrm{2C_6H_{12}O_6=C_{12}H_{24}O_{12}}
\]
\( \textbf{Condensation change:} \)
One molecule of water is eliminated.
\[
\mathrm{H_2O}
\]
\( \textbf{Hydrogen atoms remaining:} \)
\[
24-2=22
\]
\( \textbf{Oxygen atoms remaining:} \)
\[
12-1=11
\]
\( \textbf{Carbon atoms remaining:} \)
\[
12
\]
\( \textbf{Disaccharide formula:} \)
\[
\mathrm{C_{12}H_{22}O_{11}}
\]
\( \textbf{Balanced condensation equation:} \)
\[
\mathrm{2C_6H_{12}O_6\rightarrow C_{12}H_{22}O_{11}+H_2O}
\]
\( \textbf{Final answer:} \)
Formation of one glycosidic linkage gives \(\mathrm{C_{12}H_{22}O_{11}}\) and one molecule of water.
212. A branched oligosaccharide contains seven monosaccharide residues connected in one acyclic molecule. Five linkages are \(\alpha(1\rightarrow4)\), and one linkage is \(\alpha(1\rightarrow6)\). The number of water molecules released during its formation from free monosaccharides is:
ⓐ. \(5\)
ⓑ. \(6\)
ⓒ. \(7\)
ⓓ. \(12\)
Correct Answer: \(6\)
Explanation: \( \textbf{Number of monosaccharide residues:} \)
\[
n=7
\]
\( \textbf{Connected acyclic structure rule:} \)
A single connected acyclic molecule containing \(n\) residues has \(n-1\) inter-residue linkages.
\[
\text{Total linkages}=7-1
\]
\[
\text{Total linkages}=6
\]
\( \textbf{Linkage information supplied:} \)
\[
5\text{ linkages are }\alpha(1\rightarrow4)
\]
\[
1\text{ linkage is }\alpha(1\rightarrow6)
\]
\( \textbf{Total from the given types:} \)
\[
5+1=6
\]
\( \textbf{Condensation relation:} \)
Formation of each glycosidic linkage eliminates one water molecule.
\[
6\text{ linkages}\rightarrow6\mathrm{H_2O}
\]
\( \textbf{Branching interpretation:} \)
Branching changes linkage positions but does not change the \(n-1\) count for one connected acyclic structure.
\( \textbf{Final answer:} \)
Formation of the oligosaccharide releases \(6\) water molecules.
213. Match each notation in Column I with the suitable interpretation in Column II.
| Column I | Column II |
|---|
| P. \(\alpha(1\rightarrow4)\) | 1. Common branch-point linkage |
| Q. \(\beta(1\rightarrow4)\) | 2. Anomeric \(\mathrm{C_1}\) joined to \(\mathrm{C_4}\) with \(\beta\) configuration |
| R. \(\alpha(1\rightarrow6)\) | 3. Anomeric \(\mathrm{C_1}\) joined to \(\mathrm{C_4}\) with \(\alpha\) configuration |
| S. Hydrolysis | 4. Cleavage of the linkage using water |
ⓐ. P-2, Q-3, R-1, S-4
ⓑ. P-3, Q-1, R-4, S-2
ⓒ. P-1, Q-2, R-3, S-4
ⓓ. P-3, Q-2, R-1, S-4
Correct Answer: P-3, Q-2, R-1, S-4
Explanation: The notation \(\alpha(1\rightarrow4)\) describes an \(\alpha\)-configured anomeric \(\mathrm{C_1}\) linked to \(\mathrm{C_4}\), so P matches 3. The corresponding \(\beta\) linkage matches Q with 2. An \(\alpha(1\rightarrow6)\) bond commonly forms a branch point, linking R with 1. Hydrolysis uses water to break a glycosidic bond and restore hydroxyl groups, so S matches 4. The carbon numbers and \(\alpha\)–\(\beta\) designation must both be read to interpret a linkage completely.
214. A disaccharide contains one glycosidic linkage but retains a free anomeric hydroxyl group on one monosaccharide unit. It should be expected to:
ⓐ. be non-reducing and resistant to hydrolysis
ⓑ. be reducing and capable of hydrolysis
ⓒ. be reducing but incapable of ring opening
ⓓ. contain no cyclic sugar units
Correct Answer: be reducing and capable of hydrolysis
Explanation: The glycosidic bond can be cleaved by hydrolysis, so the disaccharide is hydrolysable. The remaining free anomeric hydroxyl group belongs to a hemiacetal or hemiketal centre. That centre can undergo ring opening and generate a reactive carbonyl form. The molecule can therefore reduce Tollens or Fehling reagent. The presence of one glycosidic linkage does not automatically make a disaccharide non-reducing; the decisive feature is whether any anomeric carbon remains free.
215. Sucrose is formed from:
ⓐ. two \(\alpha\)-D-glucopyranose units
ⓑ. \(\beta\)-D-galactopyranose and \(\alpha\)-D-glucopyranose
ⓒ. \(\alpha\)-D-glucopyranose and \(\beta\)-D-fructofuranose
ⓓ. two \(\beta\)-D-fructofuranose units
Correct Answer: \(\alpha\)-D-glucopyranose and \(\beta\)-D-fructofuranose
Explanation: Sucrose contains one glucose residue and one fructose residue. The glucose component is present as \(\alpha\)-D-glucopyranose, while the fructose component is present as \(\beta\)-D-fructofuranose. Their anomeric carbons participate directly in the glycosidic bond. Sucrose therefore differs from maltose, which contains two glucose units, and lactose, which contains galactose and glucose. Correct component identification is essential for predicting its hydrolysis products and reducing behaviour.
216. The glycosidic linkage present in sucrose is represented as:
ⓐ. \(\alpha(1\rightarrow2)\beta\)
ⓑ. \(\alpha(1\rightarrow4)\alpha\)
ⓒ. \(\beta(1\rightarrow4)\beta\)
ⓓ. \(\alpha(1\rightarrow6)\alpha\)
Correct Answer: \(\alpha(1\rightarrow2)\beta\)
Explanation: The anomeric carbon \(\mathrm{C_1}\) of the glucose unit is connected to the anomeric carbon \(\mathrm{C_2}\) of the fructose unit. The glucose anomeric centre has the \(\alpha\)-configuration, while the fructose anomeric centre has the \(\beta\)-configuration. The linkage is therefore written as \(\alpha(1\rightarrow2)\beta\). An \(\alpha(1\rightarrow4)\) linkage is characteristic of maltose, while a \(\beta(1\rightarrow4)\) linkage occurs in lactose. The notation records both the participating carbon atoms and their anomeric configurations.
217. Sucrose is non-reducing because:
ⓐ. the glycosidic bond leaves the glucose anomeric carbon free
ⓑ. only the fructose anomeric carbon enters the linkage
ⓒ. the linkage opens directly to expose an aldehydic chain
ⓓ. both anomeric carbons form the glycosidic bond
Correct Answer: both anomeric carbons form the glycosidic bond
Explanation: The glycosidic linkage of sucrose connects the anomeric carbon of glucose with the anomeric carbon of fructose. Neither unit therefore retains a free hemiacetal or hemiketal hydroxyl group. Without a free anomeric centre, the rings cannot open to generate a reducing carbonyl form. Sucrose consequently gives negative Tollens and Fehling tests before hydrolysis. Its non-reducing nature does not imply that the glycosidic bond is resistant to acid- or enzyme-catalysed hydrolysis.
218. Consider the following statements about sucrose.
Statement I: It contains one glucose unit and one fructose unit.
Statement II: Its two anomeric carbons are joined to each other.
Statement III: It gives a positive Fehling test before hydrolysis.
The acceptable statements are:
ⓐ. I only
ⓑ. I and II only
ⓒ. II and III only
ⓓ. I, II and III
Correct Answer: I and II only
Explanation: Sucrose is composed of glucose and fructose, so Statement I is acceptable. Its linkage joins glucose \(\mathrm{C_1}\) with fructose \(\mathrm{C_2}\), and both are anomeric carbons. Statement II is therefore also acceptable. Because neither anomeric centre remains free, sucrose cannot readily form an open-chain reducing structure. Statement III is unacceptable because unhydrolysed sucrose gives a negative Fehling test.
219. Use the arrangement described below.
The \(\mathrm{C_1}\) anomeric carbon of an \(\alpha\)-D-glucopyranose unit is bonded through oxygen to the \(\mathrm{C_2}\) anomeric carbon of a \(\beta\)-D-fructofuranose unit.
The disaccharide is:
ⓐ. sucrose
ⓑ. maltose
ⓒ. lactose
ⓓ. cellobiose
Correct Answer: sucrose
Explanation: The description identifies both component monosaccharides and both linkage positions. Sucrose uniquely contains glucose joined through its anomeric \(\mathrm{C_1}\) to the anomeric \(\mathrm{C_2}\) of fructose. The stated configurations give the complete \(\alpha(1\rightarrow2)\beta\) linkage. Maltose contains two glucose residues, while lactose contains galactose and glucose. Participation of both anomeric centres also predicts that the identified disaccharide is non-reducing.
220. Match each sucrose feature in Column I with its suitable description in Column II.
| Column I | Column II |
|---|
| P. Monosaccharide components | 1. Negative before hydrolysis |
| Q. Glycosidic linkage | 2. Glucose and fructose |
| R. Fehling test | 3. \(\alpha(1\rightarrow2)\beta\) |
| S. Complete hydrolysis | 4. Equimolar monosaccharides |
ⓐ. P-3, Q-2, R-4, S-1
ⓑ. P-2, Q-4, R-1, S-3
ⓒ. P-2, Q-3, R-1, S-4
ⓓ. P-4, Q-1, R-3, S-2
Correct Answer: P-2, Q-3, R-1, S-4
Explanation: Sucrose contains glucose and fructose, so P matches 2. Their anomeric carbons are joined through an \(\alpha(1\rightarrow2)\beta\) linkage, linking Q with 3. Unhydrolysed sucrose gives a negative Fehling test because it has no free anomeric carbon, so R matches 1. Complete hydrolysis releases one mole each of glucose and fructose per mole of sucrose, linking S with 4. These properties follow consistently from the same anomeric-carbon arrangement.
