The correct option is A 67.4Specific rotation, [α]_λ^T = α(g/ml) × l (dm) Let, the ...

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Standard XII

Biology

Pyranose and Furanose

The

Question

The % of $α -$ anomer in the mixture is:

32.6

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67.4

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42.6

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57.4

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Solution

The correct option is A 67.4
Specific rotation, $[α]_{λ}^{T} = \frac{α}{(g / m l) \times l (d m)}$

Let, the % of $α$ form is $x$ .

Let, the % of $β$ form is $100 - x$ .

Therefore, at equilibrium, $\frac{29 \times x + (100 - x) \times (- 17)}{100} = 14$

On solving, we get $x = 67.4$

Therefore, % of $α$ anomer in the mixture is $67.4$ %

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The % of α−anomer in the mixture is:

The correct option is A 67.4Specific rotation, [α]Tλ=α(g/ml)×l(dm)Let, the % of α form is x.Let, the % of β form is 100−x.Therefore, at equilibrium, 29×x+(100−x)×(−17)100=14On solving, we get x=67.4Therefore, % of α anomer in the mixture is 67.4%

The % of $α -$ anomer in the mixture is: