When $\alpha$ -D-glucose and $\beta$ -D-glucose are dissolved in water in two separate beakers I and II…

When $\alpha$ -D-glucose and $\beta$ -D-glucose are dissolved in water in two separate beakers I and II respectively and allowed to stand, then -
  1. specific rotation in beaker I will decrease while in II will increase upto a constant value
  2. the specific rotation of equilibrium mixture in two beakers will be different
  3. the equilibrium mixture in both beakers will be leavorotatory
  4. the equilibrium mixture in both beakers will contain only cyclic form of glucose

Solution

$\alpha$ -D-glucose or $\beta$ -D-glucose when dissolved in water and allowed to stand, following equilibrium is established.
$\alpha$ -D-glucose $ightleftharpoons$ Open chain form $ightleftharpoons \beta$ -D-glucose
$\left(+111^{\circ}ight)$ $\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$ $\left(+19^{\circ}ight)$
Specific rotation of $\alpha$ -form falls, while that of $\beta$ -form increases until a constant value of $+52.5^{\circ}$ is reached at equilibrium. This phenomenon is known as mutarotation.

Asked in: JEE-TOPICTESTS-CHEMISTRY

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