Question

In which of the following reactions the inversion of configuration takes place?

(I) $pentan-2-ol$ + $SOC{l}_2\to$
(II) $pentan-2-ol$ + $COC{l}_2⟶$

(III) $pentan-2-ol$ + $\left(A\right)+SOC{l}_2$ $\stackrel{Pyridine}{\to }$

(IV) $pentan-2-ol$ + $COC{l}_2$ $\stackrel{Pyridine}{\to }$

• A. $\left(I\right),\left(III\right)$
• B. $\left(I\right),\left(II\right)$
• C. $\left(III\right),\left(IV\right)$
• D. $\left(I\right),\left(IV\right)$

Answer 1

The correct option is C $\left(III\right),\left(IV\right)$

Retention of stereochemistry with $SOC{l}_2$ or $COC{l}_2$ alone, inversion with $SOC{l}_2$ and pyridine, both reactions form the “chlorosulfite” intermediate. But when pyridine (a nucleophile) is present, it can attack the chlorosulfite, displacing chloride ion and forming a charged intermediate.

Now, if the leaving group departs, forming a carbocation, there’s no lone pair nearby on the same face that can attack. Chloride attacks the carbon from the backside, leading to inversion of configuration and formation of a $C-Cl$ bond through $S{N}^2$ mechanism. Reaction $III$ and $IV$ both proceed by $S{N}^2$ mechanism and lead to inversion.