Consider the following reaction sequence

${\mathrm{CH}}_2={\mathrm{CH}}_2\underset{ \mathrm{cold}}{\stackrel{ \mathrm{dil}. {\mathrm{KMnO}}_4 }{⟶}}\mathrm{A}\underset{ \mathrm{excess} }{\stackrel{{\mathrm{PCl}}_5}{⟶}}\mathrm{B}$

The products $\left(\mathrm{A}\right)$ and $\left(\mathrm{B}\right)$ are, respectively :

Consider the following reaction.

The products formed in the reaction are

Consider the following reaction sequence,

The product $\left(\mathrm{B}\right)$ is-

Propene is allowed to react with $\mathrm{m}$-chloroperoxybenzoic acid. The product $\left(\mathrm{A}\right)$ is then reduced with ${\mathrm{LiAlH}}_4$ in dry ether to give $\left(\mathrm{B}\right).$

${\mathrm{CH}}_3\mathrm{CH}={\mathrm{CH}}_2\stackrel{\mathrm{mCPBA}}{⟶}\mathrm{A}\underset{2·{\mathrm{H}}_3{\mathrm{O}}^{+}}{\overset{1.{\mathrm{LiAlH}}_4}{\to }}\mathrm{B}$

The structure of the product $\left(\mathrm{B}\right)$ is:

In the reaction

the product $\left(\mathrm{B}\right)$ is :

The final product in the reaction is

$\mathrm{Ph}-\mathrm{OH}+{\mathrm{CCl}}_4\underset{\left(2\right){\mathrm{H}}^{\oplus }}{\overset{\left(1\right) \mathrm{KOH},△}{\to }}$ product

about above reaction the incorrect statement is