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Standard XII

Chemistry

Mutarotation

The optical r...

Question

The optical rotation of the $α -$ form of a pyranose is $+ {150.7}^{\circ},$ that of the $β$ form is $+ {52.8}^{\circ}$ . In solution an equilibrium mixture of the anomers has an optical rotation of $+ {80.2}^{\circ}$ . The percentage of the $α -$ form at
equilibrium is

28 %

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32 %

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68 %

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72 %

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Solution

The correct option is A $28 %$
$x ({150.7}^{\circ}) + (100 - x) ({52.8}^{\circ}) = 100 \times 80.2 \Rightarrow x = 28 %$

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Similar questions

Q. The optical rotation of the

α

- form of a pyranose is

+ {150.7}^{\circ}

, that of the

β

-form is

+ {52.8}^{\circ}

. In solution an equilibrium mixture of these anomers has an optical rotation of

+ {80.2}^{\circ}

. The percentage of the

α

- form in equilibrium mixture is:

Q. The optical rotation of the

α - f o r m

of a pyramose is

+ {150.7}^{\circ}

, that of the

β - f o r m

+ {52.8}^{\circ}

. In solution an equilibrium mixture of these anomers has an optical rotation of

+ {80.2}^{\circ}

. The percentage of the

α

form in equilibrium mixture is:

Q. The optical rotation of the

α

form of pyranose is

+ {150.7}^{0}

, that of the

β

form is

+ {52.8}^{0}

. In solution an equilibrium mixture of these anomers has an optical rotation of

+ {80.2}^{0}

. The percentage of the

α

- form in equilibrium mixture is:

Q. The optical rotation for the

α

-form of a carbohydrate is

+ {107.6}^{o}

, that of the

β

-form is

+ {52.8}^{o}

. The solution at equilibrium of these anomers has an optical rotaion of

+ {80.2}^{o}

.
The value of the

\frac{α}{β}

is:

α

anomer of an aldohexose E has a specific rotation of

120^{\circ}

while

β

anomer of E has a specific rotation of

80^{\circ}

. If an equilibrium mixture of

α, β

anomers of E has the specific rotation of

96^{\circ}

, then what will be percentage of the

α

-anomer in the equilibrium mixture assuming the concentration of the open structure is almost negligible?

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The optical rotation of the α− form of a pyranose is +150.7∘, that of the β form is +52.8∘. In solution an equilibrium mixture of the anomers has an optical rotation of +80.2∘. The percentage of the α− form at equilibrium is

The correct option is A 28%x(150.7∘)+(100−x)(52.8∘)=100×80.2⇒x=28%

The optical rotation of the $α -$ form of a pyranose is $+ {150.7}^{\circ},$ that of the $β$ form is $+ {52.8}^{\circ}$ . In solution an equilibrium mixture of the anomers has an optical rotation of $+ {80.2}^{\circ}$ . The percentage of the $α -$ form at
equilibrium is

The correct option is A $28 %$
$x ({150.7}^{\circ}) + (100 - x) ({52.8}^{\circ}) = 100 \times 80.2 \Rightarrow x = 28 %$