Page 279 - Arrow Pushing in Inorganic Chemistry A Logical Approach to the Chemistry of the Main Group Elements
P. 279

7.2 ALKALI-INDUCED DISPROPORTIONATION OF MOLECULAR HALOGENS  259
                                              −       −     −
                                          3XO → XO   3  + 2X                      (7.12)

               This is not an easy mechanism to envision. Multiple negatively charged hypohalite anions
               must somehow come together to generate the products. The simplest mechanism involves
               an S 2-type attack on a hypohalite oxygen, with halide as the leaving group:
                   N
                                     +
                                   Na                          −
                                      −                      O     −
                             −        O    X           −        + X  Na +
                             O   X                     O   X
                                                             +
                                                                                  (7.13)
                                                            −
                                     −    +               O
                                   O   Na
                            −    +     −              −    X  2+  + X  −  Na  +
                             O   X     O   X           O      O −

               A negatively charged oxygen atom as an electrophile is admittedly a rather awkward pro-
               position. Certainly, alternative pathways are conceivable. To make a long story short,
               however, DFT calculations appear to favor the above path: hydrogen bonding by water and
                     +
               the Na counterions provide the electrostatic stabilization necessary for various anionic
               reactants to come together.
                  A milder version of the above disproportionation can lead to dichlorine monoxide
               (Cl O). The reaction involves the interaction of Cl with moist Na CO :
                  2                                    2            2   3

                            2Cl + 2Na CO + H O → 2NaHCO + 2NaCl + Cl O            (7.14)
                               2      2   3   2            3            2
               As usual, it’s useful to consider the product structures. The carbonate ion (CO 3 2− ) ends
                                    −
               up as bicarbonate (HCO ). That strongly suggests that the oxygen in Cl O derives from
                                   3                                      2
               water, there being no other oxygen source. We therefore begin by using carbonate to strip
               a proton off water:

                              −    −          O             −
                             O      O     H      H        O       OH   −
                                 C                            C      +  OH        (7.15)

                                O                             O

                      −
               The OH so produced reacts with Cl exactly as described above to produce HOCl. A
                                              2
                                                                       −
               second carbonate ion may then deprotonate the HOCl, producing ClO .
                                                   −
                                  −             − Cl       O
                                 HO     Cl   Cl
                                                        H      Cl
                            −    −          O            −                        (7.16)
                           O      O     H      Cl       O      OH     −
                               C                            C      +  O   Cl
                               O                            O
   274   275   276   277   278   279   280   281   282   283   284