Question

Which of the following reactions of alcohol (A), proceeds by $S_{N}2$ mechanism?

(I) $\left(A\right)+SOC{l}_2⟶$

(II) $\left(A\right)+COC{l}_2⟶$

(III) $\left(A\right)+SOC{l}_2\stackrel{Pyridine}{⟶}$

(IV) $\left(A\right)+COC{l}_2\stackrel{Pyridine}{⟶}$

• 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

Answer: (C)

$\left(III\right),\left(IV\right)$

$SOC{l}_2$ and $COC{l}_2$ both replace $OH$ of alcohols by $S_{N}1$ mechanism. They coordinate to the alcohol, with loss of $HCl$ and formation of a good leaving group (“chlorosulfite”). But when pyridine (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. So chloride attacks the carbon from the backside, leading to inversion of configuration and formation of a $C-Cl$ bond. This proceeds in the $S{N}^2$ reaction.