Question

Increasing tendency for $S_{N}1$ and $S_{N}2$ reaction is:

$\left(I\right)C{H}_{3}C{H}_{2}C{H}_{2}C{H}_{2}Cl$

$\left(II\right)C{H}_{3}C{H}_2\underset{\underset{Cl}{|}}{C}HC{H}_3$

$\left(III\right)C{H}_3\underset{\underset{C{H}_3}{|}}{C}{H}_2-Cl$

$\left(IV\right)C{H}_3\underset{\underset{C{H}_3}{|}}{\stackrel{\stackrel{C{H}_3}{|}}{C}}-Cl$

• A. $S_{N}1$ ; I < III < II < IV
• B. $S_{N}2$ ; IV < II < III < I
• C. (A) and (B) both are correct
• D. both incorrect

Answer 1

The correct option is C (A) and (B) both are correct

In both $S_{N}1$ and $S_{N}2$ reactions, the nucleophile competes with the leaving group. The $S{N}_1$ reaction is feasible with $3^o$ haloalkane because tertiary carbocations are relatively stable compared with secondary or primary ones.

A primary carbocation would be much more energetically unstable than the tertiary one formed from a tertiary alkyl halide, and therefore much more difficult to produce and shows $S_{N}2$ reaction.

Therefore the increasing tendency of $S_{N}1$ is $I<III<II<IV$ and that for $S_{N}2$ is $IV<II<III<I$

So, both option $A$ and $B$ are correct.