How is the electrophile generated in the nitration of benzene?

1 Answer
May 5, 2016

The nitration of benzene involves sulfuric acid and nitric acid, whose pKas are about #-3# and #-1.3#, respectively. Thus, sulfuric acid is the stronger acid and can protonate nitric acid.

The #"H"_2"O"# leaving group just dissociates on its own. The full reaction is:

#"C"_6"H"_6 + "H"_2"SO"_4 + "HNO"_3 stackrel(k_"obs"" ")(->) "C"_6"H"_5"NO"_2 + "HSO"_4^(-) + "H"_3"O"^(+)#

Water can form because water has a pKa of about #15.7#, and thus its formation in equilibrium is heavily favored in this case.

The electrophile is the bent #\mathbf( :stackrel(..)"O"=stackrel((+))("N")=stackrel(..)"O": )#. Your book may call it the nitronium ion.