101. Complete hydrolysis of starch ultimately gives:
ⓐ. amino acids
ⓑ. glucose
ⓒ. fructose only
ⓓ. nucleotides
Correct Answer: glucose
Explanation: Starch is a polysaccharide built from many \(\alpha\)-D-glucose residues. Hydrolysis progressively shortens its chains through smaller carbohydrate fragments. Complete hydrolysis cleaves all glycosidic linkages and releases glucose molecules. Amino acids arise from protein hydrolysis, while nucleotides are associated with nucleic acids. The final product reflects the monomeric identity of the repeating units in starch.
102. Consider the following statements about glucose.
Statement I: It is a sweet, water-soluble monosaccharide.
Statement II: Its open-chain form is classified as an aldohexose.
Statement III: It can be obtained by hydrolysis of sucrose or starch.
The acceptable statements are:
ⓐ. I and II only
ⓑ. II and III only
ⓒ. I and III only
ⓓ. I, II and III
Correct Answer: I, II and III
Explanation: Glucose is a sweet monosaccharide that dissolves well in water because of its multiple hydroxyl groups. Its six-carbon open-chain structure contains a terminal aldehydic group, making it an aldohexose. Hydrolysis of sucrose releases glucose along with fructose. Complete hydrolysis of starch ultimately produces glucose because starch is a glucose polymer. The three statements describe compatible physical, structural and preparative features of the same biomolecule.
103. Formation of an oxime when glucose reacts with hydroxylamine provides evidence for the presence of:
ⓐ. a phosphate group
ⓑ. a peptide linkage
ⓒ. a carboxyl group
ⓓ. a carbonyl group
Correct Answer: a carbonyl group
Explanation: Hydroxylamine reacts with aldehydes and ketones to form oximes. Glucose produces an oxime because a small fraction of its molecules exists in an open-chain carbonyl form in solution. This reaction establishes the presence of a carbonyl group somewhere in the molecule. Oxime formation alone does not distinguish an aldehyde from a ketone because both carbonyl classes can react with hydroxylamine. Additional oxidation evidence is needed to identify the carbonyl group of glucose specifically as aldehydic.
104. A laboratory record states that glucose adds hydrogen cyanide to form a cyanohydrin. The observation most directly supports the conclusion that glucose:
ⓐ. contains a carboxyl group that undergoes cyanide addition
ⓑ. contains an open-chain carbonyl group
ⓒ. contains six ether linkages
ⓓ. behaves only as a cyclic acetal
Correct Answer: contains an open-chain carbonyl group
Explanation: Hydrogen cyanide undergoes nucleophilic addition across the carbon–oxygen double bond of aldehydes and ketones. Cyanohydrin formation therefore shows that glucose can provide a reactive carbonyl group. Although glucose exists mainly in cyclic forms in solution, equilibrium continuously produces a small amount of open-chain glucose. That open-chain fraction undergoes the addition reaction. The result does not show that glucose remains entirely open-chain, nor does it by itself identify whether the carbonyl is aldehydic or ketonic.
105. Mild oxidation of glucose with bromine water produces gluconic acid. This transformation indicates that the open-chain form of glucose contains:
ⓐ. a terminal aldehyde group
ⓑ. an internal ketone group
ⓒ. two terminal carboxyl groups
ⓓ. an amide group at \(\mathrm{C_1}\)
Correct Answer: a terminal aldehyde group
Explanation: Bromine water is a mild oxidising agent that converts an aldehyde group into a carboxylic acid under suitable conditions. Glucose gives gluconic acid because its \(\mathrm{C_1}\) aldehyde group is oxidised to \(\mathrm{-COOH}\). The primary alcohol group at the opposite end remains unchanged during this mild oxidation. An ordinary ketone would not be converted into a monocarboxylic acid by this simple aldehyde oxidation route. The product therefore identifies the carbonyl group of open-chain glucose as terminal and aldehydic.
106. Formation of glucose pentaacetate on treatment with excess acetic anhydride demonstrates that one glucose molecule contains:
ⓐ. one hydroxyl group
ⓑ. three hydroxyl groups
ⓒ. five hydroxyl groups
ⓓ. six hydroxyl groups
Correct Answer: five hydroxyl groups
Explanation: Acetic anhydride acetylates free hydroxyl groups by converting each \(\mathrm{-OH}\) group into an acetate ester group. The prefix “penta” in glucose pentaacetate indicates incorporation of five acetyl groups. Glucose must therefore contain five hydroxyl groups capable of acetylation. Its sixth oxygen atom belongs to the carbonyl-related part of the open-chain structure rather than to a sixth alcoholic hydroxyl group. Counting total oxygen atoms instead of acetylatable hydroxyl groups would incorrectly suggest six.
107. Prolonged heating of glucose with excess hydrogen iodide gives \(n\)-hexane. The most useful structural inference is that glucose has:
ⓐ. a branched five-carbon skeleton
ⓑ. a cyclic chain containing only five carbon atoms
ⓒ. two separate three-carbon fragments
ⓓ. an unbranched chain of six carbon atoms
Correct Answer: an unbranched chain of six carbon atoms
Explanation: Under vigorous conditions, hydrogen iodide removes oxygen-containing functional groups and replaces them ultimately with hydrogen. The carbon framework of glucose is retained during this conversion. Formation of \(n\)-hexane therefore shows that the six carbon atoms of glucose are connected in one continuous unbranched chain. A branched glucose skeleton would be expected to produce a branched hydrocarbon instead. This experiment establishes carbon-chain arrangement but does not locate the individual hydroxyl groups.
108. Consider the following statements about structural evidence for glucose.
Statement I: Oxime formation supports the presence of a carbonyl group.
Statement II: Glucose pentaacetate formation supports the presence of five hydroxyl groups.
Statement III: Formation of \(n\)-hexane with prolonged hydrogen iodide treatment supports a straight six-carbon chain.
The acceptable statements are:
ⓐ. I and II only
ⓑ. I, II and III
ⓒ. II and III only
ⓓ. I and III only
Correct Answer: I, II and III
Explanation: Oxime formation is a characteristic reaction of a carbonyl-containing compound, so Statement I is acceptable. Acetylation replaces the hydrogen of each free hydroxyl group with an acetyl group, and the pentaacetate therefore reveals five hydroxyl groups. Vigorous reduction with hydrogen iodide removes the oxygen functions while preserving the carbon skeleton, producing straight-chain \(n\)-hexane. The three observations provide different and complementary structural information. No single one of them alone establishes the entire open-chain structure of glucose.
109. Match each reaction of glucose in Column I with the structural conclusion in Column II.
| Column I | Column II |
|---|
| P. Oxime formation | 1. Five hydroxyl groups |
| Q. Oxidation with bromine water | 2. Carbonyl group |
| R. Pentaacetate formation | 3. Straight six-carbon chain |
| S. Prolonged heating with \(\mathrm{HI}\) | 4. Aldehyde group |
ⓐ. P-2, Q-4, R-1, S-3
ⓑ. P-4, Q-2, R-3, S-1
ⓒ. P-2, Q-1, R-4, S-3
ⓓ. P-3, Q-4, R-1, S-2
Correct Answer: P-2, Q-4, R-1, S-3
Explanation: Oxime formation confirms a carbonyl group, so P matches 2. Bromine water oxidises an aldehyde to a carboxylic acid, linking Q with 4. Formation of a pentaacetate reveals five acetylatable hydroxyl groups, so R matches 1. Conversion into \(n\)-hexane preserves the unbranched six-carbon skeleton, linking S with 3. These reactions must be interpreted together because each probes a different feature of glucose.
110. The statement “glucose must be an aldehyde because it forms an oxime” is incomplete because:
ⓐ. only aldehydes form oximes, so the aldehyde assignment is complete
ⓑ. oxime formation identifies an aldehyde but excludes a ketone
ⓒ. ketones also form oximes, so oxime formation alone is inconclusive
ⓓ. oxime formation identifies carbonyl position without another test
Correct Answer: ketones also form oximes, so oxime formation alone is inconclusive
Explanation: Both aldehydes and ketones react with hydroxylamine to form oximes. Oxime formation therefore confirms a carbonyl group but does not independently identify its type. Bromine-water oxidation supplies the stronger distinction because glucose is converted into the monocarboxylic acid gluconic acid. That behaviour is consistent with oxidation of a terminal aldehyde group. Structural conclusions should not assign more information to a reaction than the reaction can actually provide.
111. Among the following observations, the one that most specifically distinguishes the aldehydic nature of open-chain glucose from a ketonic carbonyl is:
ⓐ. cyanohydrin formation by carbonyl addition
ⓑ. oxime formation by carbonyl condensation
ⓒ. reduction of the carbonyl group to an alcohol
ⓓ. bromine-water oxidation to gluconic acid
Correct Answer: bromine-water oxidation to gluconic acid
Explanation: Cyanohydrin and oxime formation are shown by both aldehydes and ketones, so neither observation alone distinguishes the two carbonyl classes. Water solubility mainly reflects the presence of several polar hydroxyl groups. Bromine water, however, mildly oxidises an aldehydic group to a carboxylic acid. Formation of gluconic acid therefore identifies the terminal carbonyl of open-chain glucose as an aldehyde. The unchanged primary alcohol group in gluconic acid also shows that the oxidation is selective under these conditions.
112. One mole of glucose is completely acetylated with excess acetic anhydride to form glucose pentaacetate. Assuming one mole of acetic anhydride is consumed per hydroxyl group acetylated, the required amount of acetic anhydride is:
ⓐ. \(4\,mol\)
ⓑ. \(5\,mol\)
ⓒ. \(6\,mol\)
ⓓ. \(10\,mol\)
Correct Answer: \(5\,mol\)
Explanation: \( \textbf{Given amount:} \)
Glucose taken \(=1\,mol\).
\( \textbf{Structural information:} \)
One glucose molecule contains \(5\) acetylatable hydroxyl groups.
\( \textbf{Reaction meaning:} \)
Each hydroxyl group forms one acetate ester group.
\( \textbf{Stoichiometric assumption:} \)
One mole of acetic anhydride is consumed per mole of hydroxyl groups acetylated.
\( \textbf{Hydroxyl-group amount:} \)
\[
1\,mol\text{ glucose}\times5=5\,mol\text{ hydroxyl groups}
\]
\( \textbf{Acetic anhydride requirement:} \)
\[
5\,mol\text{ hydroxyl groups}\times
\frac{1\,mol\text{ acetic anhydride}}{1\,mol\text{ hydroxyl groups}}
=5\,mol
\]
\( \textbf{Product check:} \)
Five acetylations give glucose pentaacetate rather than a mono-, tetra- or hexaacetate.
\( \textbf{Final answer:} \)
Complete acetylation of \(1\,mol\) glucose requires \(5\,mol\) of acetic anhydride under the stated assumption.
113. Glucose gives \(n\)-hexane rather than a branched hexane on prolonged treatment with \(\mathrm{HI}\). This result is inconsistent with a glucose structure containing:
ⓐ. branching in its carbon chain
ⓑ. five hydroxyl groups
ⓒ. a terminal aldehyde group in its open-chain form
ⓓ. six carbon atoms
Correct Answer: branching in its carbon chain
Explanation: Vigorous hydrogen iodide treatment removes oxygen-containing groups but does not rearrange the carbon framework in the structural inference being made. A branched carbon skeleton would therefore be expected to produce a branched hydrocarbon. The observed product is straight-chain \(n\)-hexane, so the carbon atoms of glucose must be connected without branching. The experiment remains compatible with six carbon atoms, a terminal aldehyde in the open-chain form and several hydroxyl groups before reduction. It reveals connectivity of the carbon skeleton rather than the original oxidation state of each carbon.
114. A compound with molecular formula \(\mathrm{C_6H_{12}O_6}\) gives an oxime, is oxidised by bromine water to a monocarboxylic acid, forms a pentaacetate and yields \(n\)-hexane with prolonged \(\mathrm{HI}\). The open-chain structure most consistent with all observations is:
ⓐ. \(\mathrm{CH_3CO(CHOH)_3CH_2OH}\)
ⓑ. \(\mathrm{HOOC(CHOH)_4CH_3}\)
ⓒ. \(\mathrm{CHO(CH_2)_4CH_2OH}\)
ⓓ. \(\mathrm{CHO-(CHOH)_4-CH_2OH}\)
Correct Answer: \(\mathrm{CHO-(CHOH)_4-CH_2OH}\)
Explanation: Oxime formation requires a carbonyl group, while bromine-water oxidation to a monocarboxylic acid identifies that group as an aldehyde. The aldehyde must therefore occur at one end of the chain as \(\mathrm{-CHO}\). Pentaacetate formation requires five hydroxyl groups, supplied by four \(\mathrm{-CHOH}\) units and one terminal \(\mathrm{-CH_2OH}\) group. Formation of \(n\)-hexane establishes a continuous unbranched chain containing six carbon atoms. The structure \(\mathrm{CHO-(CHOH)_4-CH_2OH}\) satisfies all four independent observations simultaneously.
115. In glucose pentaacetate, the prefix “penta” is directly related to the number of ______ groups present in glucose before acetylation.
ⓐ. aldehyde
ⓑ. carbonyl
ⓒ. hydroxyl
ⓓ. carboxyl
Correct Answer: hydroxyl
Explanation: Acetylation converts each free hydroxyl group into an acetate ester. Formation of the pentaacetate means that five such groups have reacted. The aldehydic carbonyl is not counted as an alcoholic hydroxyl group in this evidence. A carboxyl group is absent from unoxidised glucose. The prefix therefore records the number of acetylatable \(\mathrm{-OH}\) groups rather than the total number of oxygen atoms.
116. Assertion: Both oxime formation and cyanohydrin formation support the existence of a carbonyl-containing form of glucose.
Reason: Hydroxylamine and hydrogen cyanide can react by addition or condensation at a carbonyl carbon.
ⓐ. 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: Hydroxylamine reacts at a carbonyl group and ultimately forms an oxime. Hydrogen cyanide adds across the carbonyl group to form a cyanohydrin. Glucose undergoes both reactions because its cyclic forms remain in equilibrium with a small open-chain carbonyl form. The Reason describes the shared chemical basis of the two observations. Neither reaction by itself proves that the carbonyl is aldehydic, because ketones can also undergo corresponding reactions.
117. The equation representing mild oxidation of glucose to gluconic acid is:
ⓐ. \(\mathrm{C_6H_{12}O_6+2[O]\rightarrow C_6H_{12}O_8}\)
ⓑ. \(\mathrm{C_6H_{12}O_6+[O]\rightarrow C_6H_{12}O_7}\)
ⓒ. \(\mathrm{C_6H_{12}O_6+[H]\rightarrow C_6H_{14}O_6}\)
ⓓ. \(\mathrm{C_6H_{12}O_6-H_2O\rightarrow C_6H_{10}O_5}\)
Correct Answer: \(\mathrm{C_6H_{12}O_6+[O]\rightarrow C_6H_{12}O_7}\)
Explanation: \( \textbf{Starting compound:} \)
Glucose has the molecular formula \(\mathrm{C_6H_{12}O_6}\).
\( \textbf{Functional-group change:} \)
The terminal aldehyde group is oxidised to a carboxyl group.
\[
\mathrm{-CHO+[O]\rightarrow -COOH}
\]
\( \textbf{Atom change:} \)
Conversion of \(\mathrm{-CHO}\) into \(\mathrm{-COOH}\) adds one oxygen atom.
\( \textbf{Carbon and hydrogen check:} \)
The numbers of carbon and hydrogen atoms remain \(\mathrm{C_6}\) and \(\mathrm{H_{12}}\).
\( \textbf{Oxygen count:} \)
\[
6+1=7
\]
\( \textbf{Product formula:} \)
\[
\mathrm{C_6H_{12}O_7}
\]
\( \textbf{Overall equation:} \)
\[
\mathrm{C_6H_{12}O_6+[O]\rightarrow C_6H_{12}O_7}
\]
\( \textbf{Chemical interpretation:} \)
Mild oxidation affects the aldehyde end without oxidising the terminal primary alcohol group.
\( \textbf{Final answer:} \)
Glucose is converted into gluconic acid according to \(\mathrm{C_6H_{12}O_6+[O]\rightarrow C_6H_{12}O_7}\).
118. Although the reactions of glucose support the open-chain formula \(\mathrm{CHO-(CHOH)_4-CH_2OH}\), that formula alone cannot satisfactorily explain:
ⓐ. the presence of six carbon atoms
ⓑ. formation of a pentaacetate derivative containing five acetyl groups
ⓒ. oxidation to gluconic acid
ⓓ. two crystalline forms and mutarotation
Correct Answer: two crystalline forms and mutarotation
Explanation: The open-chain structure explains the aldehydic reactions, five hydroxyl groups and straight six-carbon skeleton of glucose. It does not, however, account for the existence of distinct \(\alpha\) and \(\beta\) crystalline forms. It also cannot explain the gradual change in optical rotation observed when either crystalline form is dissolved in water. These observations require cyclic hemiacetal structures and interconversion through a ring–chain equilibrium. The open-chain form remains chemically important, but it represents only a small part of the equilibrium mixture in solution.
119. The open-chain structure of glucose is considered incomplete mainly because it cannot explain why glucose:
ⓐ. contains six carbon atoms
ⓑ. forms a pentaacetate
ⓒ. shows mutarotation and has two crystalline forms
ⓓ. yields \(n\)-hexane on prolonged heating with \(\mathrm{HI}\)
Correct Answer: shows mutarotation and has two crystalline forms
Explanation: The open-chain formula successfully accounts for the straight six-carbon skeleton, five hydroxyl groups and aldehydic reactions of glucose. A single open-chain structure, however, does not provide two distinct configurations that could correspond to separate crystalline forms. It also does not explain why the optical rotation of a freshly prepared glucose solution changes gradually before becoming constant. Both observations require interconversion between cyclic forms through a small open-chain fraction. The open-chain formula is therefore chemically useful but does not represent the predominant structure of glucose in solution.
120. A freshly prepared solution of one crystalline form of glucose gradually changes its optical rotation without undergoing decomposition. This observation most strongly indicates:
ⓐ. reversible interconversion of glucose forms
ⓑ. irreversible oxidation of glucose by dissolved oxygen
ⓒ. complete conversion of glucose into fructose
ⓓ. cleavage of glucose into smaller carbohydrates
Correct Answer: reversible interconversion of glucose forms
Explanation: A gradual change followed by a constant optical rotation is characteristic of mutarotation. The process occurs because one cyclic form opens to a small amount of open-chain glucose and then recloses to form either cyclic configuration. No carbon–carbon bond is broken, and the molecular formula remains unchanged. Chemical decomposition would generally produce new substances rather than a stable equilibrium rotation. The observed change therefore reflects a reversible structural equilibrium rather than hydrolysis or oxidation.
