The specific rotation of (R) (-) 2- bromooctane is - 40 . What is the percentage composition of mixture of enantiomers of 2- bromooctane whose rotation is +20 ?

Question

The specific rotation of (R) (-) 2- bromooctane is - 40 . What is the percentage composition of mixture of enantiomers of 2- bromooctane whose rotation is +20 ? (A) The mixture has 25% R and 75% S (B) The mixture has 20% R and 80% S (C) The mixture has 50% R and 50% S (D) The mixture has 75% R and 25% S

Tardigrade Admin · Accepted Answer

Specific rotation of (R) (-) 2-bromooctane is =- 40 Specific rotation of (S) (+) 2-bromooctane is =+40 Let the mixture contains x %R and (100-x) % S form. (x× (-40)+(100-x)(+40)/100)=20 x=25 and S=100-x=75 Hence, the mixture has 25 % R and 75 % S.

Q. The specific rotation of (R)(−)2− bromooctane is −40 . What is the percentage composition of mixture of enantiomers of 2− bromooctane whose rotation is +20 ?

The mixture has 25% R and 75% S

The mixture has 20% R and 80% S

The mixture has 50% R and 50% S

The mixture has 75% R and 25% S

Specific rotation of (R)(−)2-bromooctane is =−40 Specific rotation of (S)(+)2-bromooctane is =+40 Let the mixture contains x%R and (100−x) form. 100x×(−40)+(100−x)(+40)​=20 x=25 and S=100−x=75 Hence, the mixture has 25%R and 75%S.

Q. The specific rotation of $(R) (-) 2 -$ bromooctane is $- 40$ . What is the percentage composition of mixture of enantiomers of $2 -$ bromooctane whose rotation is $+ 20$ ?