201. Why do glucose and fructose form the same osazone?
ⓐ. They are both polysaccharides with identical linkages.
ⓑ. They contain no carbonyl group in any form.
ⓒ. Osazone formation involves changes at \(C_1\) and \(C_2\).
ⓓ. Fructose first becomes cellulose before reacting.
Correct Answer: Osazone formation involves changes at \(C_1\) and \(C_2\).
Explanation: Glucose and fructose differ mainly in arrangement around the first two carbon atoms in their open-chain forms. During osazone formation, the reaction changes the groups at \(C_1\) and \(C_2\). Because the rest of the carbon chain remains effectively the same, both sugars form the same osazone. This is a common comparison point between glucose and fructose.
202. Which pair correctly matches reagent and main reaction of glucose?
ⓐ. \(Br_2/H_2O\) — gluconic acid formation
ⓑ. \(NH_2OH\) — oxidation to saccharic acid
ⓒ. \(H_2/Ni\) — oxidation to gluconic acid
ⓓ. \(HCN\) — reduction to sorbitol
Correct Answer: \(Br_2/H_2O\) — gluconic acid formation
Explanation: Bromine water is a mild oxidising agent and converts glucose into gluconic acid. Hydroxylamine, \(NH_2OH\), forms an oxime rather than reducing glucose. \(H_2/Ni\) reduces glucose to sorbitol and does not acetylate hydroxyl groups. \(HCN\) adds to the carbonyl group to form a cyanohydrin rather than oxidising glucose.
203. Which statement best distinguishes sorbitol formation from gluconic acid formation?
ⓐ. Both are hydrolysis reactions of glycosidic bonds.
ⓑ. Sorbitol forms by oxidation, while gluconic acid forms by reduction.
ⓒ. Both require phenylhydrazine and give the same osazone.
ⓓ. Sorbitol forms by reduction, while gluconic acid forms by oxidation.
Correct Answer: Sorbitol forms by reduction, while gluconic acid forms by oxidation.
Explanation: Sorbitol is formed when the carbonyl group of glucose is reduced to an alcohol group. Gluconic acid is formed when the aldehyde group of glucose is oxidised to a carboxylic acid group. These two reactions move in opposite redox directions. They are not hydrolysis reactions and do not involve phenylhydrazine osazone formation.
204. Which assertion-reason pair is correctly evaluated?
Assertion: Glucose pentaacetate formation supports the presence of five hydroxyl groups in glucose.
Reason: Acetylation occurs at hydroxyl groups capable of reacting with an acetylating agent.
ⓐ. Assertion is true, but Reason is false.
ⓑ. Both are true, and Reason explains Assertion.
ⓒ. Assertion is false, but Reason is true.
ⓓ. Both are true, but Reason is unrelated to Assertion.
Correct Answer: Both are true, and Reason explains Assertion.
Explanation: Glucose pentaacetate contains five acetylated hydroxyl positions. Acetylation is a reaction shown by hydroxyl groups, so forming a pentaacetate means five such groups are present. The reason directly explains why this reaction gives structural evidence for glucose. It is separate from reactions that prove carbonyl character.
205. Which description best represents a glycosidic bond?
ⓐ. A bond joining amino acids through \(-CO-NH-\)
ⓑ. A bond joining nucleotides through phosphate groups
ⓒ. An oxygen bridge joining sugar units
ⓓ. A bond joining fatty acids to glycerol units
Correct Answer: An oxygen bridge joining sugar units
Explanation: A glycosidic bond is the characteristic linkage between sugar units in disaccharides, oligosaccharides, and polysaccharides. It is commonly formed when a hydroxyl group at the anomeric carbon of one sugar reacts with a hydroxyl group of another molecule. This produces an oxygen bridge between the sugar units. It is different from a peptide bond in proteins and a phosphodiester bond in nucleic acids.
206. Which carbon of a monosaccharide is most important in deciding whether its glycosidic linkage blocks reducing behaviour?
ⓐ. Anomeric carbon
ⓑ. Terminal methyl carbon
ⓒ. Penultimate carbon only
ⓓ. Lowest-numbered alcohol carbon
Correct Answer: Anomeric carbon
Explanation: The anomeric carbon is the carbon derived from the carbonyl carbon during ring formation. If this carbon remains as a free hemiacetal or hemiketal centre, the sugar can open into a carbonyl form. If it is locked in a glycosidic linkage, ring opening becomes difficult or impossible at that centre. This is why the anomeric carbon is central to reducing and non-reducing sugar behaviour.
207. Which process breaks a glycosidic bond in carbohydrates?
ⓐ. Sublimation
ⓑ. Neutralisation
ⓒ. Polymerisation only
ⓓ. Hydrolysis
Correct Answer: Hydrolysis
Explanation: Hydrolysis uses water to break bonds in larger molecules. In carbohydrates, hydrolysis breaks glycosidic linkages between sugar units. This converts disaccharides into monosaccharides and polysaccharides into many smaller sugar units. The process is the reverse of condensation bond formation between sugar units.
208. A disaccharide has both anomeric carbons locked in the glycosidic bond. Which property is expected?
ⓐ. It must form a peptide bond easily.
ⓑ. It is expected to be non-reducing.
ⓒ. It must contain no oxygen atoms.
ⓓ. It is always a protein catalyst.
Correct Answer: It is expected to be non-reducing.
Explanation: Reducing behaviour requires a free hemiacetal or hemiketal centre that can open into a carbonyl-containing form. If both anomeric carbons are involved in the glycosidic linkage, neither unit can open freely at that centre. The disaccharide therefore cannot reduce Tollens’ or Fehling’s reagent in the usual way. This logic explains the non-reducing nature of sucrose.
209. Which feature is usually required for a sugar to behave as a reducing sugar?
ⓐ. A free hemiacetal or hemiketal carbon
ⓑ. A fully blocked anomeric carbon only
ⓒ. A peptide linkage between sugar units
ⓓ. A phosphate group on every carbon
Correct Answer: A free hemiacetal or hemiketal carbon
Explanation: A reducing sugar must be able to open into a form containing a reactive carbonyl group. This is possible when the anomeric carbon is present as a free hemiacetal or hemiketal centre. The open-chain form can then reduce mild oxidising reagents. If the anomeric carbon is completely locked in an acetal-type glycosidic bond, reducing behaviour is lost.
210. Which statement best explains why some disaccharides are reducing and others are non-reducing?
ⓐ. Reducing disaccharides always contain nitrogen, while non-reducing ones do not.
ⓑ. Reducing disaccharides are always polysaccharides with many branches.
ⓒ. Reducing behaviour depends on whether a free anomeric carbon is available.
ⓓ. Non-reducing disaccharides have no covalent bonds between sugar units.
Correct Answer: Reducing behaviour depends on whether a free anomeric carbon is available.
Explanation: Disaccharides are classified as reducing or non-reducing mainly by the availability of a free anomeric carbon. If one anomeric carbon is free, the ring can open and the sugar can reduce Tollens’ or Fehling’s reagent. If both anomeric carbons are tied up in the glycosidic linkage, ring opening is blocked at both centres. This structural difference is more important than merely counting the number of sugar units.
211. Which statement about an acetal-type glycosidic linkage is most suitable?
ⓐ. It is the same as a peptide bond in proteins.
ⓑ. It is formed only by nitrogenous bases.
ⓒ. It always gives a free aldehyde group directly.
ⓓ. It can lock the anomeric carbon.
Correct Answer: It can lock the anomeric carbon.
Explanation: When the anomeric carbon becomes part of a full glycosidic acetal linkage, it is no longer a free hemiacetal centre. This blocks easy ring opening at that sugar unit. If all anomeric carbons in a sugar molecule are blocked, reducing behaviour is absent. This structural idea is important for understanding why sucrose is non-reducing.
212. Which observation is most consistent with a reducing sugar?
ⓐ. It cannot open into a carbonyl form under any condition.
ⓑ. It reduces Tollens’ or Fehling’s reagent.
ⓒ. It has both anomeric carbons fully blocked.
ⓓ. It is always a non-carbohydrate molecule.
Correct Answer: It reduces Tollens’ or Fehling’s reagent.
Explanation: A reducing sugar can reduce mild oxidising reagents such as Tollens’ or Fehling’s reagent. This behaviour comes from its ability to generate a reactive carbonyl form through ring opening. The presence of a free anomeric carbon is usually the key structural requirement. If both anomeric centres are blocked, the sugar generally becomes non-reducing.
213. Which statement correctly links mutarotation with reducing character in many sugars?
ⓐ. Both require absence of all hydroxyl groups.
ⓑ. Both occur only in peptide hormones.
ⓒ. Both depend on ring opening.
ⓓ. Both require a phosphate backbone in the molecule.
Correct Answer: Both depend on ring opening.
Explanation: Mutarotation requires opening of the cyclic sugar form followed by reclosure to different anomers. Reducing behaviour also depends on the ability to form an open-chain carbonyl structure. Therefore a free anomeric centre often supports both mutarotation and reducing character. If the anomeric carbon is locked in a glycosidic bond, both behaviours may be prevented for that centre.
214. Which option best describes a non-reducing sugar?
ⓐ. A sugar lacking a free anomeric centre
ⓑ. A sugar that contains no carbon, hydrogen, or oxygen
ⓒ. A sugar that must always be a single amino acid
ⓓ. A sugar that reduces Tollens’ reagent instantly
Correct Answer: A sugar lacking a free anomeric centre
Explanation: A non-reducing sugar lacks a free anomeric carbon capable of ring opening into a reactive carbonyl form. This usually happens when the anomeric carbon is locked in a glycosidic linkage. As a result, the sugar does not reduce common mild oxidising reagents in the usual test. This definition is structural, not based simply on whether the sugar is sweet or soluble.
215. Which statement best explains why a free anomeric carbon is important in maltose or lactose?
ⓐ. It converts the sugar into a protein.
ⓑ. It removes all oxygen atoms from the molecule.
ⓒ. It prevents any hydrolysis of the glycosidic bond.
ⓓ. It allows reducing behaviour and mutarotation.
Correct Answer: It allows reducing behaviour and mutarotation.
Explanation: In reducing disaccharides such as maltose and lactose, one anomeric carbon remains free. This free centre can open into a carbonyl-containing form. Because of this, these sugars can show reducing behaviour and may also undergo mutarotation. A blocked anomeric carbon would not allow the same ring-opening behaviour.
216. Which comparison between reducing and non-reducing sugars is most accurate?
ⓐ. Reducing sugars lack carbonyl possibilities, while non-reducing sugars always contain free aldehydes.
ⓑ. Reducing sugars can generate a carbonyl form, while non-reducing sugars cannot do so readily.
ⓒ. Reducing sugars are always proteins, while non-reducing sugars are always nucleic acids.
ⓓ. Reducing sugars contain no \(OH\) groups, while non-reducing sugars contain many \(OH\) groups.
Correct Answer: Reducing sugars can generate a carbonyl form, while non-reducing sugars cannot do so readily.
Explanation: Reducing sugars can open from a cyclic form to a carbonyl-containing structure. This carbonyl form is responsible for reducing mild oxidising reagents. Non-reducing sugars generally have their anomeric centres locked, preventing easy ring opening. The difference is based on structural availability of the anomeric carbon, not on protein or nucleic-acid identity.
217. Which assertion-reason pair is correctly evaluated?
Assertion: A sugar with a free hemiacetal group can behave as a reducing sugar.
Reason: A free hemiacetal group can open to a carbonyl-containing form.
ⓐ. Assertion is false, but Reason is true.
ⓑ. Assertion is true, but Reason is false.
ⓒ. Both are true, and Reason explains Assertion.
ⓓ. Both are true, but Reason is unrelated to Assertion.
Correct Answer: Both are true, and Reason explains Assertion.
Explanation: A free hemiacetal group at the anomeric carbon allows the cyclic sugar to open. The open-chain form contains a carbonyl group that can reduce suitable mild oxidising reagents. This is the structural basis of reducing sugar behaviour. The reason therefore directly explains the assertion.
218. Which situation would most likely prevent a sugar unit from showing reducing behaviour at that unit?
ⓐ. Its anomeric \(OH\) becomes a glycosidic acetal.
ⓑ. Its molecule contains several ordinary hydroxyl groups.
ⓒ. Its ring contains oxygen as part of the cyclic structure.
ⓓ. Its carbon chain contains carbon and hydrogen atoms.
Correct Answer: Its anomeric \(OH\) becomes a glycosidic acetal.
Explanation: The anomeric \(OH\) is the key group involved in ring opening. When it is converted into a glycosidic acetal linkage, the anomeric carbon becomes locked. This prevents formation of the open-chain carbonyl form from that unit. Ordinary hydroxyl groups elsewhere in the molecule do not have the same direct effect on reducing character.
219. Which statement is most suitable for the glycosidic bond in a disaccharide?
ⓐ. It is a metallic bond between two sugar ions.
ⓑ. It is a hydrogen bond between two nitrogenous bases.
ⓒ. It is a peptide bond formed by amino-acid condensation.
ⓓ. It is a covalent linkage between two sugar units.
Correct Answer: It is a covalent linkage between two sugar units.
Explanation: A disaccharide contains two monosaccharide units joined by a glycosidic bond. This bond is a covalent linkage, commonly involving an oxygen bridge. Hydrolysis of the disaccharide breaks this linkage and releases the sugar units. Metallic bonds, hydrogen bonds between bases, and peptide bonds belong to different chemical contexts.
220. Which statement best predicts the behaviour of a disaccharide with one free anomeric carbon?
ⓐ. It must be unable to hydrolyse.
ⓑ. It can behave as a reducing sugar.
ⓒ. It must contain a phosphate backbone.
ⓓ. It cannot contain a glycosidic bond.
Correct Answer: It can behave as a reducing sugar.
Explanation: If one anomeric carbon remains free in a disaccharide, that sugar unit can open into a carbonyl-containing form. This allows the disaccharide to reduce reagents such as Tollens’ or Fehling’s reagent. The molecule still contains a glycosidic bond joining the two monosaccharide units. A free anomeric carbon therefore supports reducing behaviour without making the molecule non-carbohydrate.
