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Q.Explain mutarotation in terms of glucose. (Structure is not necessary). OR What is peptide bond? Clarify peptide bond with example of dipeptide.

Gujarat GsebGSEB Higher Secondary Certificate (HSC) Examination 2019Subjective· 2mImportance★★★★★
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Freshly dissolved glucose slowly changes its optical rotation as its cyclic alpha- and beta-anomeric forms interconvert via the open-chain form, eventually settling at a constant equilibrium value.

Glucose exists predominantly in a cyclic (six-membered, pyranose ring) form, and this ring closure creates a NEW stereocentre at C1 (the anomeric carbon), giving two distinct cyclic isomers: alpha-D-glucose (specific rotation +112 degrees) and beta-D-glucose (specific rotation +18.7 degrees).

When either pure anomer (say, pure alpha-D-glucose, freshly crystallised) is dissolved in water, the ring can open up momentarily to the open-chain aldehyde form and then re-close, which allows it to close again as EITHER anomer (alpha or beta) at random. Over time, this interconversion continues until the alpha and beta forms reach a fixed equilibrium ratio (roughly 36% alpha : 64% beta at equilibrium for glucose). …

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