(+)-2 -butanol has [θ ]λ 25=+13.9° . A sample of 2 -butanol containing both the enantiomers was found to have a specific rotation value of -3.5° under similar condition. The percentage of the (+) and (-) enantiomer present in the sample are, respectively:

Question

(+)-2 -butanol has [θ ]λ 25=+13.9° . A sample of 2 -butanol containing both the enantiomers was found to have a specific rotation value of -3.5° under similar condition. The percentage of the (+) and (-) enantiomer present in the sample are, respectively: (A) 37.4 % and 62.6 % (B) 62.6 % and 37.4 % (C) 42.2 % and 57.8 % (D) 35.5 % and 64.5 %

Tardigrade Admin · Accepted Answer

Enantiomeric excess =((d - l/ d + l) × 100)

Q. $(+) - 2$ -butanol has $[θ]_{λ}^{25} = + 13. 9^{\circ}$ . A sample of $2$ -butanol containing both the enantiomers was found to have a specific rotation value of $- 3. 5^{\circ}$ under similar condition. The percentage of the $(+)$ and $(-)$ enantiomer present in the sample are, respectively:

$37.4%$ and $62.6%$

$62.6%$ and $37.4%$

$42.2%$ and $57.8%$

$35.5%$ and $64.5%$

Enantiomeric excess $= (\frac{d - l}{d + l} \times 100)$

Q. (+)−2 -butanol has [θ]λ25​=+13.9∘ . A sample of 2 -butanol containing both the enantiomers was found to have a specific rotation value of −3.5∘ under similar condition. The percentage of the (+) and (−) enantiomer present in the sample are, respectively:

37.4% and 62.6%

62.6% and 37.4%

42.2% and 57.8%

35.5% and 64.5%