A pure sample of 2 -chlorobutane shows rotation of PPL by 3 0 ∘ in standard conditions. When above sample is made impure by mixing its opposite form, so that the composition of the mixture becomes 75%

Question

A pure sample of 2 -chlorobutane shows rotation of PPL by 3 0 ∘ in standard conditions. When above sample is made impure by mixing its opposite form, so that the composition of the mixture becomes 75% d -form and 25% l-form, then what will be the observed rotation for mixture.

TestHub · Accepted Answer

The observed rotation is calculated using the enantiomeric excess (ee).

Given: Pure sample rotation = $3 0^{\circ}$ .

Mixture: 75% d-form, 25% l-form.

Enantiomeric excess (ee) = % d-form - % l-form = $75% - 25% = 50%$ .

Observed rotation = ee $\times$ rotation of pure enantiomer = $0.50 \times 3 0^{\circ} = + 1 5^{\circ}$ .

The final answer is $+ 1 5^{\circ}$ .

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