In the following the reaction ,
According to me the product should be compound (A) due to substitution reaction.
But the product formed is the compound (B) , what could be the mechanism for the reaction for that product.
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1$\begingroup$ Part of product B is missing. I assume it is meant to be just a methyl group. $\endgroup$– bonMay 24 '16 at 17:46
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$\begingroup$ @bon yes you are correct , it is methyl group only . $\endgroup$– KoolmanMay 24 '16 at 17:47
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To reflect on the comments below: In literature this reaction seems to actually produce product A and synthesis of product B seems to require much more exotic conditions than described in the question. As such I have to conclude that the correct answer would have be A.
Old answer:
Due to the polar aprotic solvent (DMF) one would think Sn2 is favoured as the reaction mechanism, however if the product is indeed B it would suggest an Sn1 mechanism where the aromatic ring stabilises the cation. One of the resulting resonance structures would indeed react with fluoride. Due to the stability of aromatics the product is formed by re-aromatisation.
answered May 25 '16 at 11:24
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2$\begingroup$ This is what I suspected must be the case but I can't give a good reason why the fluoride attacks at the ortho position rather than the benzyl position. $\endgroup$– bonMay 25 '16 at 11:37
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$\begingroup$ @bon If the primary carbocation is formed it will easily resonate to the secondary, much more stable, carbocation. Therefore the ortho position doesn't surprise me as much as the Sn1 mechanism does. $\endgroup$ May 25 '16 at 11:42
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2$\begingroup$ The secondary, non-aromatic resonance form, is more stable than the primary, aromatic resonance form...? Benzyl carbocations don't react on the ring $\endgroup$– orthocresol ♦May 25 '16 at 11:44
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$\begingroup$ @orthocresol It struck me as weird the first time I saw similar mechanisms as well, but yes the aromaticity can be broken for such stabilization. However, I cannot find any mention of a synthesis ortho-fluorotoluene from benzyl bromide in literature. bon, are you sure B is the correct answer? $\endgroup$ May 25 '16 at 11:53
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2$\begingroup$ I have no idea what the correct answer is. I didn't ask the question. I too was unable to find any reference to this in the literature though so I suspect it may be wrong. $\endgroup$– bonMay 25 '16 at 11:56