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For methylbenzene (with the methyl group at position 1), is it easier for chlorine to substitute the hydrogen at position 2, 3 or 4?

I was thinking somewhere along the lines of different charge densities around the benzene ring, and that the charge density closest to the already existing methyl group has been diminished due to the methyl group itself. However, I am unable to explain it properly enough.

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  • $\begingroup$ I just corrected a little typo in the product names, but i left them wrong for didactic purposes ;) See below for an explanation :) $\endgroup$
    – Klaus-Dieter Warzecha
    Commented Feb 26, 2014 at 16:30

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We are talking about the electrophilic aromatic substitution of methylbenzene (A) with chlorine.

EAS

The possible products are 2-methylchlorobenzene (2), 3-methylchlorobenzene (3), and 4-methylchlorobenzene (4)

Chloromethylbenzene (B) is something different, also known as benzylchloride. Chloromethyl is the name for the $\ce{-CH2Cl}$ substituent :)

Please have a look at this site for an animated explanation of the mechanism and the selectivities.

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  • $\begingroup$ In order to emphasize the differences, you might want to consider using the systematic IUPAC names 1-chloro-2-methylbenzene (2), 1-chloro-3-methylbenzene (3), 1-chloro-4-methylbenzene (4), and (chloromethyl)benzene (B). $\endgroup$
    – user7951
    Commented Jul 8, 2015 at 9:26
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It very much depends on the reagents and reaction type you are using to halogenate the aromatic ring, however, the most common method taught during organic chemistry is electrophilic hologenation, which in this case the methyl will be para directing.

http://en.wikipedia.org/wiki/Electrophilic_halogenation#Scope

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