So I know that $\ce{KMnO4}$ converts any side chain irrespective of chain length to -$\ce{COOH}$ group and Soda lime converts that into the respective alkane. But in benzyl chloride, the alkyl group side chain also has a chlorine atom attached. How would the mechanism proceed?

It is given that Benzene is the product.

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    $\begingroup$ You have water present, benzyl chloride is a reactive species. Consider the possibility that the benzyl chloride hydrolyses to benzyl alcohol early in the reaction $\endgroup$ – Waylander Oct 23 '19 at 15:56
  • $\begingroup$ @Waylander Ok, but then how would KMnO4 react with a side chain containing alcohol? $\endgroup$ – Techie5879 Oct 23 '19 at 16:00
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    $\begingroup$ Alcohol to acid is a std permangate oxidation $\endgroup$ – Waylander Oct 23 '19 at 16:37
  • $\begingroup$ @Waylander Ah i haven't gotten to that part i guess $\endgroup$ – Techie5879 Oct 23 '19 at 16:39

This is not a magical reaction. Yet, it is a two step reaction as depicted in diagram below:


As @Waylander's comment elsewhere, benzyl chloride might have converted to benzyl alcohol before get oxidized to benzoic acid in acidic $\ce{KMnO4}$ solution (the first step). This oxidation is well known (see here). The second step is heating (may be $\gt \pu{200 ^\circ C}$) of solid benzoic acid in the presence of solid soda lime to cause the decarboxylation. This reaction is well described by Dr. Jim Clark (here) and also here.


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