Glucose,when heated with CH_3OH in the presen | vedclass

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Question

(C) Glucose exists in a cyclic hemiacetal form in equilibrium with its open-chain structure. When glucose is treated with methanol $(CH_3OH)$ in the p

Vedclass · Accepted Answer

$A$ ring structure

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(C) Glucose exists in a cyclic hemiacetal form in equilibrium with its open-chain structure. When glucose is treated with methanol $(CH_3OH)$ in the presence of dry $HCl$ gas,the hemiacetal hydroxyl group (at $C_1$) reacts with methanol to form an acetal,known as a methyl glucoside. The formation of two isomeric products,$\alpha$-methyl glucoside and $\beta$-methyl glucoside,is possible only because the $C_1$ carbon is a chiral center in the cyclic (ring) structure of glucose. Therefore,the presence of a ring structure is responsible for the formation of these two glucosides.

Glucose,when heated with $CH_3OH$ in the presence of dry $HCl$ gas,gives $\alpha$ and $\beta$-methyl glucosides because it contains:

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