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What will be the product of the reaction between chlorobenzene and methyl chloride in the presence of anhydrous $\ce{AlCl3}$?

I think we should get 1-chloro-4-methylbenzene as the major product, as we'll have Friedel–Crafts alkylation. Is this correct? Or will we get toluene?

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    $\begingroup$ The answer is there in the mechanism of reaction itself.Friedel craft reaction is an electrophilic substitution reaction.Though,your doubt is common when the concept of electrophilic substitution and nucleophilic substitution is not clear.Think what is the purpose of anh AlCl3 as catalyst.Once you find it you will have the answer (If this still doesn't address your problem please notify.thanks). $\endgroup$ – Abhishek Pallippara gopakumar Feb 28 '16 at 11:13
  • $\begingroup$ @AbhishekPallipparagopakumar Actually I was quite sure of the product. It's the book which confused me. Thanks anyways :) $\endgroup$ – Gummy bears Feb 29 '16 at 4:38
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You will actually get a mixture of different products, with the product with the highest yield being p-chlorotoluene, or 1-chloro-4-methylbenzene. The other products resulting from the reaction include o-chlorotoluene and m-chlorotoluene, with a significantly higher amount of the former being produced*. In fact, there would also be products with more than one methyl substituent being produced in various proportions depending on the precise conditions being used.

Yes, the reaction will proceed by Friedel-Crafts alkylation: Firstly, the Lewis acid catalyst abstracts the chloride ion from methyl chloride, giving a reactive, electrophilic methenium ion. Then, electrons from the $\pi$ system of the benzene ring of chlorobenzene then attack the electrophilic ion. Finally, a proton is lost to restore aromaticity of the ring, giving the product.

*Refer to the following post to understand why the chlorine atom as a substituent would be ortho/para-directing: Why are halogens ortho para directing even though deactivating?

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1-chloro 4-methyl benzene because chlorine is an ortho para directing group so methyl group eithermust attack at ortho position or para position if chlorine. But chlorine shows negative inductive affect.so in order to increase stability methyl group must be attached at para position of chlorine

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    $\begingroup$ A para-substituted aromat is the product because the previous substituent is not para-directing? And I can’t understand your reasoning for why para should dominate over ortho … $\endgroup$ – Jan Dec 17 '17 at 4:31

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