Rajasthan BoardEnglish MediumSTD 12 ScienceChemistryBiomolecules2 Marks
Question
Enumerate the reactions of $D -$ glucose which cannot be explained by its open chain structure.
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Answer
$D (+) -$ glucose does not undergo certain characteristic reactions of aldehydes, e.g., glucose does not form $NaHSO_{3 }$ addition product.
Glucose reacts with $NH_2OH$ to form an oxime but glucose pentaacetate does not. This implies that the aldehydic group is absent in glucose pentaacetate.
$D - (+) -$ glucose exists in two stereoisomeric forms, i.e., $\alpha -$ glucose and $\beta -$ glucose.
Both $ \alpha - D -$ glucose and $\beta - D -$ glucose undergo mutarotation in aqueous solution. Although the crystalline forms of $\alpha -$ and $ \beta - D (+) -$ glucose are quite stable in aqueous solution but each form slowly changes into an equilibrium mixture of both.
$D (+) -$ glucose forms two isomeric methyl glucosides. Aldehydes normally react with two moles of methanol per mole of the aldehyde to form an acetal but $D (+) -$ glucose when treated with methanol in presence of dry $HCI$ gas, reacts with only one mole of methanol per mole of glucose to form a mixture of two methyl $D - $ glucosides i.e., methyl $- \alpha - D -$ glucoside $($melting point $438 K$, specific rotation $+ 158°)$ and methyl $- \beta - D -$ glucoside $($melting point $308 K$, specific rotation $- 33°)$.
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