Which one of the following statements is incorrect regarding the chemical reactivity of Glucose?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Acetylation of glucose with excess of acetic anhydride gives glucose tetraacetate.
The reactions of glucose reveal its open-chain structure, \( \ce{CHO-(CHOH)4-CH2OH} \): a straight chain of six carbons, one aldehyde group and five –OH groups.
Acetylation with acetic anhydride converts every –OH into –\( \ce{OCOCH3} \). Glucose has five –OH groups, so it forms glucose pentaacetate, not a tetraacetate:
\( \displaystyle \ce{CHO-(CHOH)4-CH2OH} \xrightarrow{\ce{(CH3CO)2O}} \ce{CHO-(CHOCOCH3)4-CH2OCOCH3} \)
Option (1) Oxidation of glucose with bromine water gives gluconic acid. — A correct statement about glucose. The mild oxidant oxidises only the –CHO group to –COOH: \( \ce{CHO-(CHOH)4-CH2OH} \xrightarrow{\ce{Br2}\text{ water}} \ce{COOH-(CHOH)4-CH2OH} \)
Option (2) Acetylation of glucose with excess of acetic anhydride gives glucose tetraacetate. — The incorrect statement. Glucose has five –OH groups, so it gives glucose pentaacetate.
Option (3) Oxidation of glucose with nitric acid gives saccharic acid. — A correct statement about glucose. \( \ce{HNO3} \) oxidises both the –CHO and the primary –\( \ce{CH2OH} \) to –COOH: \( \ce{CHO-(CHOH)4-CH2OH} \xrightarrow{\ce{HNO3}} \ce{COOH-(CHOH)4-COOH} \)
Option (4) On prolonged heating with hydrogen iodide (HI), glucose forms n-hexane. — A correct statement about glucose. This shows that the six carbon atoms are in a straight chain: \( \ce{CHO-(CHOH)4-CH2OH} \xrightarrow{\ce{HI},\ \Delta} \ce{CH3-(CH2)4-CH3} \)
NCERT (2026-27 edition):
Glucose – “… 2. On prolonged heating with HI, it forms n-hexane, suggesting that all the six carbon atoms are linked in a straight chain. …”
“4. Glucose gets oxidised to six carbon carboxylic acid (gluconic acid) on reaction with a mild oxidising agent like bromine water. This indicates that the carbonyl group is present as an aldehydic group.”
“5. Acetylation of glucose with acetic anhydride gives glucose pentaacetate which confirms the presence of five –OH groups. Since it exists as a stable compound, five –OH groups should be attached to different carbon atoms.”
“6. On oxidation with nitric acid, glucose as well as gluconic acid both yield a dicarboxylic acid, saccharic acid. This indicates the presence of a primary alcoholic (–OH) group in glucose.”