Question

2-Pentanol can be converted to 2-ethoxy pentane by two paths. In path I, $H^{\oplus }$ is abstracted from 2-pentanol to form the nucleophile while in path II, $H^{\oplus }$ is abstracted from ethanol. What is true about the configuration at chiral $C$ of 2-pentanol?

• A. Retained in both paths
• B. Retained in path I and Inverted in path II
• C. Inverted in path I and Retained in path I
• D. Inverted in both paths

Answer 1

The correct option is B Retained in path I and Inverted in path II

in path I, the nucleophile is $C_3{H}_7\stackrel{\stackrel{O^{\ominus }}{|}}{C}HC{H}_3$ ion. In path I, nucleophile has the chiral center and reaction takes place without breaking any bond, thus configuration in the final product is retained.
In path II, nucleophile $\left(C_2{H}_5{O}^{⊝}\right)$ attacks chiral centre back-side in an $S{N}^2$ reaction with inversion of configuration in ether.