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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
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B
Retained in path I and Inverted in path II
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C
Inverted in path I and Retained in path I
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D
Inverted in both paths
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Solution

The correct option is B Retained in path I and Inverted in path II
in path I, the nucleophile is $$C_3H_7\overset{\overset{\displaystyle O^\ominus}|}CHCH_3$$ 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 $$(C_2H_5O^{\circleddash})$$ attacks chiral centre back-side in an $$SN^2$$ reaction with inversion of configuration in ether.

Chemistry

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