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In Friedel Craft Alkylation of aromatic compounds using $\ce{BF3}$ :

  1. $\ce{BF3}$ acts as a lewis acid

  2. It is used as a catalyst

  3. Amount of $\ce{BF3}$ used in the reaction is almost equal to that of amount of benzene used.
  4. $\ce{BF3}$ acts as a nucleophile.

I am sure about options $1$ and $3$ being correct and $4$ th being incorrect but I am hesitant about option $2$.

Catalyst is a substance that lowers the activation energy and enhances the rate of the reaction, in the end remaining unconsumed.

Now, what Lewis Acids do is that they make halogens stronger electrophiles by forming complexes with them. Taking this further we can say that yes they serve as "catalysts" in the reaction because otherwise the reaction would have been lot slower.

However, the given answer is $1,3$. Is my reasoning correct and the answer wrong? Is it functioning as a catalyst in the reaction?

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On the formal definition, I think $\ce{BF3}$ may fail the remains unconsumed part. It will form $\ce{[BF3X]-}$, which would need to revert to $\ce{BF3 + X^-}$ in order to complete the catalytic cycle. I don't think the $K_\mathrm{d}$ of this complex is particularly high, so this is not likely to happen, hence it won't be a proper catalyst. This is also why you need a stoichiometric ammount of $\ce{BF3}$ for the reaction to proceed to completion.

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    $\begingroup$ Tbh, in school, we were always taught that this complex (or any such catalytic complex) dissociated completely during the reaction. So, unless we know the fact you stated, it is tough to arrive at the answer. $\endgroup$ – Gaurang Tandon Mar 22 '18 at 0:53
  • $\begingroup$ It's an assumption based on the fact that you need stoichiometric amounts of the compound (thus you don't get it back during catalysis), and the fact that $\ce{BF3}$ is quite a strong Lewis Acid since it lacks an octet. $\endgroup$ – ralk912 Mar 22 '18 at 0:57

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