368 Mr. 0. J. Lodge on Nodes and Loops 



benzol, C 6 H 6 , 12 + 2 — 2x = 6, or # = 4; and jj 



there are 4 cycles, viz. the three double bars and q 



the whole ring. TTP^ Vr 



(2) I now find that a general expression for M 



the number of loops * in a compound may be HC CH 

 established with great ease. First of all, the XQ,/ 



truth of Clifford's rule is evident; for draw H 



the graphic formula of C»H 2n+2 , and assume 

 for the present that it has no loops ; the H atoms can only be 

 removed in pairs, and for every pair removed the two liberated 

 bonds close up, " satisfying each other " and forming one loop. 

 Hence if 2x hydrogen atoms be removed, x loops are obtained, 

 which is the rule. 



It is also evident that, whether a pair of liberated bonds unite 

 with an atom of oxygen or with each other, the number of loops 

 is the same, that an atom of oxygen may be inserted anywhere 

 in a chain without making a loop, and that a chain of oxygen 

 atoms can only act as a single dyad radical. Hence the number 

 of loops in a compound is independent of the amount of oxygen 

 (or any other dyad element) which it may contain. This proves 

 that no paraffin can have any loops ; for H 2 has none, and they 

 can all be built up from H 2 by successive additions of CH 2 , 

 which is a dyad radical. 



Before proceeding to obtain a general expression for the 

 number of loops in a compound, it will be well to show by two 

 examples how the loops are to be counted. 



B^B has 2 loops, not 3 ; 

 and 



C^=C 



\ C / has 5 loops only. 



Let us now replace some of the hydrogen in a paraffin by one 

 atom of a A>ad (denoted by K), so as to form the compound 

 C n K~R 2 (n+i-x)+f(k), where f{k) has to be determined so that there 

 may be x loops. 



Now, if K is a dyad, we have seen that the term/(£) does not 

 appear, or/(2)=0. 



Also if K is a tetrad, it will act simply as another atom of car- 

 bon ; but another atom of carbon would add 2 to the suffix of 

 H ; therefore the tetrad must add 2, or/(4)=2. 



These two equations practically give us /(#)=£— 2; hence 

 there will be x loops in C n KU 2{n - X )+k- 



* I use the word loop because " cycle*' seems too good for the purpose. 



