Is this how #"CHCl"_2^(+)# acts as an electrophile in the electrophilic aromatic substitution? I feel like I have my mechanism arrows in the wrong direction but I don't know how they should go.

Question

Brooklyn Barber · Accepted Answer

Nope. You can see that #"CHCl"_2^(+)# is positively charged. In fact, it is similar to a carbocation if you draw it out:

And so, we call #"CHCl"_2^+# an electrophile, and a Lewis acid (an electron pair acceptor). It has an empty central #p# orbital, and so, it has no electrons to readily donate, just like #"BH"_3# or #"BF"_3#:

In electrophilic aromatic substitution reactions like these, benzene acts as a nucleophile (a Lewis base), and then restores its aromaticity afterwards.

The electrons should come from benzene first.

(The way you've drawn it suggests you have lost track of two electrons (they ended up on the top middle carbon!), since on the final structure they are not there.)

There is only one arrow. It donates away #pi# electrons from one carbon, and leaves the other one #(+)#.

Then, some base in solution can come to pluck off the #"H"^(+)# to perform a #beta# elimination, restoring the aromaticity of benzene.

Jacob Anderson · Answer

undefined Yes, that is correct. "CHCl"_2^(+) can act as an electrophile in electrophilic aromatic substitution reactions. In the mechanism, the aromatic ring's electron density attacks the electrophile, leading to the formation of a sigma complex intermediate. The arrows should show the movement of electrons from the aromatic ring to the electrophile, indicating the formation of the sigma complex.