Question

What will be the percentage amount of +2butanol and -2butanol in a mixture whose specific rotation is -9.72? (Given: specific rotation of pure recamic mixture is 13.5)

Topic=Optical Isomerism.

Answer 1

Dear student!

First of all we know that,

Optical purity %age = 100 x [α]_mixture / [α]_{pure sample}

Where, [α]_mixture = specific rotation of mixture = -9.72

and [α]_{pure sample} = specific rotation of pure sample = 13.5

Therefore optical purity %age = 100 x -9.72/ 13.5 = 72 %

Hence, there is 72 % excess R over S form.

Now the rest 28 % are optically inactive which must be equal to amount of R and S butanol . So if we assume the rest as 100 % then, there must be 82 % R form (as 72% excess over S) and 10 % S form of butanol.