Enumerate the reactions of $D -$ glucose which cannot be explained by its open chain structure.

Q: Enumerate the reactions of $D -$ glucose which cannot be explained by its open chain structure.

1. $D (+) -$ glucose does not undergo certain characteristic reactions of aldehydes, e.g., glucose does not form $NaHSO_3$ addition product. 2. 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. 3. $D - (+) -$ glucose exists in two stereoisomeric forms, i.e., $\alpha$ - glucose and $\beta$ - glucose. 4. 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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$D (+) -$ 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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