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The reagents with which glucose does not react to give the corresponding tests/products are A. Tollen's reagent B. Schiff's reagent C. HCN D. NH₂OH E. NaHSO₃ Choose the correct options from the given below:
Glucose forms an oxime (NH2OH) and cyanohydrin (HCN), reduces Tollen’s, but FAILS Schiff’s test and does NOT form NaHSO3 adduct because the -CHO group is locked in a cyclic hemiacetal ring!
Although glucose contains an aldehyde carbonyl group in its Fischer projection, it fails to exhibit several characteristic reactions of aliphatic aldehydes: 1. It does not react with Schiff’s reagent to restore the pink/magenta color. 2. It does not form a sodium bisulphite addition product with NaHSO3. 3. The pentaacetate of glucose does not react with hydroxylamine (NH2OH), proving the absence of a free -CHO group.
The -OH at C-5 adds intramolecularly to the C-1 carbonyl carbon, locking glucose into a stable six-membered cyclic hemiacetal (glucopyranose). The concentration of free open-chain aldehyde is less than 0.02%. Bulky bisulphite ions (HSO3-) and Schiff’s reagent cannot drive ring opening because their addition products are thermodynamically less stable than the closed hemiacetal ring.
In contrast: - Tollen’s reagent is alkaline and drives tautomerization (Lobry de Bruyn-Alberda van Ekenstein rearrangement) and oxidation to gluconic acid. - HCN and NH2OH are powerful, unhindered nucleophiles that irreversibly trap the minute open-chain form, pulling the equilibrium forward to completion.
Diagnose every choice: Know exactly why the right option works and why distractors fail.
Verbatim NCERT evidence demonstrating that the -CHO group is tied up in a cyclic hemiacetal.
Glucose reacts readily with NH2OH to form an oxime.
Glucose reacts with HCN to yield a cyanohydrin.
Glucose reduces Tollen’s reagent (silver mirror test) and forms an oxime with NH2OH.
Students who only memorize glucose as an "aldohexose" assume all classic aldehyde tests work. Glucose is a reducing sugar (reduces Tollen’s and Fehling’s), but FAILS Schiff’s and NaHSO3 tests!
"Glucose does not give Schiff’s test and does not form an addition product with NaHSO3 because it exists predominantly as a cyclic hemiacetal."
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