Question

In $\underset{Specificrotaion=+{50}^o}{Ph-CH\left(OH\right){CH}_3}\underset{}{\overset{SO{Cl}_2}{\to }}\underset{|Cl}{PhCl}-{CH}_3$
the reaction is carried out in the presence of pyridine. Which of the following mechanisms does it follow?

• A. ${S_N}^1$
• B. $S_{N}2$
• C. $S_{N}i$
• D. $E_2$

Answer 1

The correct option is B $S_{N}2$

The reaction proceeds through $S{N}^2$ mechanism when it is carried out in presence of pyridine. The $-OH$ group of optically active substrate is replaced with $-Cl$ atom with inversion in configuration.