Optically pure $S (+) - 2 -$ Bromoctane with specific rotation $+ 36^{0}$ reacts with aq. $N a O H$ in acetone to give pure $R (-) - 2 -$ octanol with specific rotation $- {10.3}^{0}$ . Partially racemized 2 – Bromoctane with specific rotation $+ 30^{0}$ on reaction with aq. $N a O H$ produces 2 – octanol with specific rotation $- 6^{0}$ . Among the following correct statement(s) regarding second reaction is / are

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Optical purity of 2 – Octanol formed is nearly 58.25%

Percentage of back side attack is nearly 50%

Optical purity of 2– Bromoctane taken is nearly 83.3%

Percentage of $S_{N} 2$ reaction is nearly 100%

answer is A, B.

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