Question

$C{l}^{-}C{H}_3{O}^{-}C{H}_3{S}^{-}{I}^{-}$
$\left(I\right)\left(II\right)\left(III\right)\left(IV\right)$
The correct order of increasing leaving group capability of above anions:

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

Answer 1

The correct option is B $II<III<I<IV$

The correct order of increasing leaving group capability of the anions is
$II\left(C{H}_3{O}^{-}\right)<III\left(C{H}_3{S}^{-}\right)<I\left(C{l}^{-}\right)<IV\left(I^{-}\right)$

Iodide ion is the best leaving group and methoxide ion is the worst leaving group.

The negative charge on large iodide atom is stabilized. In methoxide ion, the $+I$ (electron releasing) effect of methyl group intensifies the negative charge on $O$ atom and destablizes the ion.

A good leaving group should be
(a) electron-withdrawing to polarize the carbon
(b) stable once it has left (not a strong base)
(c) polarisable- to maintain partial bonding with the carbon in the transition state (both $S_{N}1$ and $S_{N}2$).

This bonding helps to stabilize the transition state and reduces the activation energy.