$\mathrm{S}_{\mathrm{N}} 2$ reaction involving inversion of configuration takes place with an optically…

$\mathrm{S}_{\mathrm{N}} 2$ reaction involving inversion of configuration takes place with an optically active compound $Z$. The compound $Z$ is
  1. $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{X}$
  2. $\left(\mathrm{CH}_3\right)_2 \mathrm{CHX}$
  3. $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}\left(\mathrm{CH}_3\right) \mathrm{X}$
  4. $\left(\mathrm{CH}_3\right)_3 \mathrm{CX}$

Solution

In $\mathrm{S}_{\mathrm{N}} 2$ reaction the nucleophile attacks the electrophilic carbon of the electrophile. The bond between nucleophile and carbon forms at the same time that the bond between carbon and leaving group breaks.

Asked in: AP EAMCET 2018 (23 Apr Shift 2)

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