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In aqueous solution glucose remains as
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In all three forms in equilibrium
Hints: \(\begin{aligned} & \beta-\mathrm{D}-\text { glucose } \rightleftharpoons \mathrm{D}-\text { glucose } \rightleftharpoons \alpha-\mathrm{D}-\text { glucose } \\ & (\approx 64 \%) \quad\qquad(\text {open chain } \approx 0.02 \%) \qquad~~(\approx 34 \%)\end{aligned}\)
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