Question

The SN$_2$ reactivity of the following compounds will be in the order:
I) C$_6$H$_5$CH(CH$_3$)Br
II) (C$_6$H$_5$)$_2$CHBr
III) (C$_6$H$_5$)$_2$C(CH$_3$)Br
IV) C$_6$H$_5$CH$_2$Br

• A. III< II< I< IV
• B. III< I< II< IV
• C. II< III< IV< I
• D. II< IV< I< III

Hint:

In SN$_2$ reactions, the reactivity increases with decreasing steric hindrance. Primary halides react faster than secondary, and tertiary halides are the least reactive.

Answer 1

The Correct Option is A

SN$_2$ reactions proceed faster when steric hindrance is minimal, as the nucleophile attacks the electrophilic carbon directly. Let’s evaluate steric hindrance around the carbon attached to Br in each compound: \begin{itemize} \item III: (C$_6$H$_5$)$_2$C(CH$_3$)Br – A tertiary benzylic halide; highly hindered $\Rightarrow$ slowest. \item II: (C$_6$H$_5$)$_2$CHBr – Secondary benzylic halide; still hindered. \item I: C$_6$H$_5$CH(CH$_3$)Br – Secondary benzylic halide; less hindered than II. \item IV: C$_6$H$_5$CH$_2$Br – Primary benzylic halide; least hindered $\Rightarrow$ fastest. \end{itemize} Hence, SN$_2$ reactivity order: III< II< I< IV.