Question

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:

• A. Specific rotation in beaker I will decrease while in II will increase upto a constant value
• B. The specific rotation of equilibrium mixture in two beakers will be different
• C. The equilibrium mixture in both beakers will be leavorotatory
• D. The equilibrium mixture in both beakers will contain only cyclic form of glucose

Answer 1

The correct option is A Specific rotation in beaker I will decrease while in II will increase upto a constant value

$\alpha -D-glucose$ and $\beta -D-glucose$ when dissolved in water and allowed to stand, the following equilibrium is established.

$\underset{(+{111}^o)}{\alpha -D-glucose}\rightleftharpoons \text{open chain form}\rightleftharpoons \underset{(+{19}^o)}{\beta -D-glucose}$

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.