Question

The best method to prepare neopentyl chloride is :-
$A){\text{ }}{(C{H_3})_3}CC{H_2}OH\xrightarrow{{PC{l_5},\Delta }}$
$B){\text{ }}{(C{H_3})_3}CC{H_2}OH\xrightarrow{{HCl,\Delta }}$
$C){\text{ }}{(C{H_3})_3}CC{H_2}OH\xrightarrow{{SoC{l_2},pyridine}}$
$D){\text{ }}{(C{H_3})_3}CC{H_3}\xrightarrow{{C{l_2},h\nu {\text{ }}\Delta }}$

Answer 1

Hint: In present studies, it is found that neopentyl chloride (1-Chloro-2,2-dimethylpropane) formed by photo-chemical free –radical chlorination of neopentane (${(C{H_3})_3}CC{H_3}$) is stable even at ${200^0}C$ for long periods of time and is extraordinary inactive.

Complete step by step answer:
It has been founded in laboratory that neopentyl chloride cannot be prepared from alcohol because we get ${1^0}$ neopentyl cation as an intermediate here, which has a strong tendency to rearrange itself from less stable ${1^0}$carbocation to a more stable ${3^0}$ carbocation ($\because $ order of stability of carbocations is: tertiary carbocation (${3^0}$) > secondary carbocation (${2^0}$ ) > primary carbocation (${1^0}$) ).
 Consequently, here we get 2-chloro-2-methylbutane instead of neopentyl chloride.
When we prepare neopentyl chloride from the photo-chemical free –radical chlorination (i.e. $C{l_2}{\text{ in the presence of }}h\nu $) of neopentane (${(C{H_3})_3}CC{H_3}$) , we get neopentyl radical as an intermediate here. Since free radicals usually do not rearrange, it is the best method to prepare neopentyl chloride.
So, the correct answer is “Option D”.

Note: The reactions of neopentyl alcohols with thionyl chloride($SOC{l_2}$ ) and phosphorus pentachloride $(PC{l_5})$ will give chloroalkanes. Moreover, thionyl chloride method is preferred over hydrogen chloride or phosphorus chloride method for the preparation of chloroalkanes because when we use thionyl chloride as reagent ,we get ${\text{S}}{{\text{O}}_{\text{2}}}{\text{ and HCl}}$ as the by-products. Both these by-products are escapable gases, leaving behind the chloroalkanes in almost pure state.