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Home/Practice questions/Chemistry
ChemistryOrganic: Biomolecules

Glucose on prolonged heating with HI forms

Please read the passage carefully for the following questions

Carbohydrates are primarily produced by plants and form a very large group of naturally occurring organic compounds. Some common examples of carbohydrates are cane sugar, glucose, starch, etc. Most of them have a general formula, Cx(H2O)y, and were considered hydrates of carbon from where the name carbohydrate was derived. For example, molecular formula of glucose (C6H12O6) fits into this general formula, C6(H2O)6. But all compounds which fit into this formula may not be classified as carbohydrates. For example, acetic acid (CH3COOH) fits into this general formula, C2(H2O)2 but is not a carbohydrate. Some of carbohydrates, which are sweet in taste, are also called sugars. Carbohydrates are classified on the basis of their behavior on hydrolysis. They have been broadly divided into following three groups: (i) Monosaccharides, (ii) Oligosaccharides, and (iii) Polysaccharides. The carbohydrates may also be classified as either reducing or non-reducing sugars. All those carbohydrates which reduce Fehling’s solution and Tollens’ reagent are referred to as reducing sugars. All monosaccharides, whether aldose or ketose, are reducing sugars. Glucose is an aldohexose and is also known as dextrose. It is the monomer of many of the larger carbohydrates, namely starch and cellulose. Glucose is found to exist in two different crystalline forms which are named α-D-Glucose and β-D-Glucose. The two monosaccharides are joined together by an oxide linkage formed by loss of a water molecule. Such a linkage between two monosaccharide units through oxygen atom is called glycosidic linkage. Polysaccharides contain a large number of monosaccharide units joined together by glycosidic linkages. Vitamins are classified into two groups depending upon their solubility in water or fat: (i) Fat-soluble vitamins, and (ii) Water-soluble vitamins.

Glucose on prolonged heating with HI forms

Answer & explanation+

Correct answer: option 3

The correct answer is Option (3) → n-Hexane

The open chain structure of glucose was worked out from a set of reactions, each of which reveals one feature of the molecule:

  • Prolonged heating with HI gives n-hexane: the six carbon atoms are in a straight chain.

  • Reaction with hydroxylamine (oxime) and HCN (cyanohydrin): a carbonyl group is present.

  • Oxidation by bromine water gives gluconic acid: the carbonyl group is an aldehyde group.

  • Acetylation with acetic anhydride gives a pentaacetate: five –OH groups are present.

  • Oxidation by nitric acid gives saccharic acid: one of the –OH groups is a primary alcohol group.

    \( \displaystyle \ce{C6H12O6} \xrightarrow[\Delta]{\ce{HI}} \ce{CH3-CH2-CH2-CH2-CH2-CH3} \)

HI removes all the oxygen atoms and reduces the chain completely. So glucose on prolonged heating with HI forms n-hexane.

Why the other options are incorrect

  • Option (1) Saccharic acid — Incorrect. It is formed by oxidation of glucose with nitric acid.

  • Option (2) Gluconic acid — Incorrect. It is formed by oxidation of glucose with bromine water.

  • Option (4) 3-Iodohexane — Incorrect. The final product of prolonged heating contains no iodine; HI reduces glucose all the way to the hydrocarbon.

NCERT (2026-27 edition):
Glucose – …

Textbook extract

Note: The same/similar question has been asked in different years by NTA as per the details given below:

20262025202420232022
18-05 S1––24-05 S3–

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