686 The Philippine Journal of Science 1921 
of four electrons is immediately established between C (2) and 
C (3). The structure of the resulting nonionic dissociation prod- 
uct is shown in formula B, which is identical with the quinoid 
modification in Gomberg’s formula II. 
H 
8. (am neal 
When carbon atoms (1) and (2) separate it is not to be sup- 
posed that-C (1) continues to exist with only seven electrons in 
its shell, six of which are drawn away from the kernel by the 
chemically negative phenyl groups. The eighth electron is un- 
doubtedly supplied by a quinoid rearrangement in the left-hand 
part of the molecule (formula A) identical with the one de- 
scribed for the right-hand part and taking place practically at 
the same time as the dissociation or before it. That is, the dis- 
sociation of hexaphenylethane in nonionizing solvents is sym- 
metrical, giving two neutral molecules which cannot strictly be 
called triphenylmethyl and which have the structure shown in 
formula B. 
When hexaphenylethane is dissolved in a solvent, such as sul- 
phur dioxide, which contains unshared electrons (that is, an 
ionizing solvent) the dissociation occurs unsymmetrically. This 
is caused by an unstable union with a solvent molecule which 
supplies two additional electrons to the shell of carbon atom (6). 
A quinoid rearrangement takes place as before described except 
that two electrons are supplied by the ring to the union between 
‘C (2) and C (3). Carbon atom (2) now has momentarily ten 
electrons, so that C (1) separates, carrying with it both of the 
shared electrons. The left-hand part of the molecule (formula 
A) thus forms a colorless negative ion (C) and the right hand 
part forms a colored positive ion (D). The formula below shows 
only the ion proper without the solvent molecule, which undoubt- 
edly remains loosely attached to the para carbon atom, which 
otherwise would have two vacancies in the shell. 
Cc. ( C,H;);=C D. LG (emo 
Since C (1) has completed its octet and hence has a negative 
charge while C (6) has lost two electrons and therefore has 
a positive charge, it is quite conceivable that the two ions can 
