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My textbook mentions that chlorination of alkanes by $\ce{SO2Cl2}$ is preferred over $\ce{SOCl2}$, but I didn't understand why.
Could someone explain which of the two reagents I should use in what situations, and briefly tell me why?
There is a similar question here: How does SO2Cl2 react with alcohols?
However the answers there did not completely clarify my doubts.
The reactivity patterns of $\ce{SO2Cl2}$ and $\ce{SOCl2}$ are quite different. $\ce{SOCl2}$ is a good electrophile, and can be thought of as a source of $\ce{Cl-}$ ions. These ions can go on to react in their typical nucleophilic fashion. $\ce{SO2Cl2}$ however is often a $\ce{Cl2}$ source, as it readily decomposes giving off sulfur dioxide. Usually much easier/safer to use this than measuring out (and getting into solution) chlorine gas. The chlorination of simple alkanes by $\ce{Cl2}$ gas (or something that makes it in solution) happens by a radical mechanism i.e. $\ce{Cl.}$ not $\ce{Cl-}$.
Chlorination of alkanes by $\ce{SO2Cl2}$ is preferred over $\ce{SOCl2}$. Just pay little more attention on the word alkanes. So, the statement is absolutely correct. Just go through it (link)! Hence, For chlorination in alkanes, if you use $\ce{SO2Cl2}$, It undergoes a free radical substitution reaction and will get the product.
If we use $\ce{SOCl2}$, it undergoes a nucleophilic substitution reaction. Here, this reaction mechanism is explained.
So, You can see here that $\ce{SOCl2}$ doesn't produce a chlorine free radical. Hence, It is not possible to chlorinate alkanes using $\ce{SOCl2}$, which $\ce{SO2Cl2}$ does.
