Most students of organic chemistry, when they start learning about the theories of chemical structure, are told that the set of structural (viz. constitutional) isomers of saturated acyclic hydrocarbons (alkanes) may be found by means of a simple graphical method (see, for example, ref 1). Actually this problem, which in the parlance of graph theory is the task of enumerating free trees with degree no greater than 4, is rather complicated. Historically, it has been the object of much effort and ingenuity.Eventually, students may be taught, or may find for themselves, a simple and primitive algorithm. For the alkanes with molecular formula CnH2"+2, this algorithm consists of drawing, in turn, the following structures:(1) a linear chain of n carbon atoms;(2) a chain of re -1 carbon atoms to which a methyl group is attached at one of the carbons numbered 2, 3,... , [re/2j, where UJ means the integer part of x;(3) a chain of re -2 carbon atoms with either an ethyl group attached to one of the carbons numbered 3, 4,. .., [(rel)/2j, or two methyl groups attached to carbon atoms numbered 2,3,..., re -3 in all possible different arrangements; (4) a chain of re -3 carbons with an re-propyl group at one of the carbons numbered 4, 5, . .., L(n -2)/2_[, an isopropyl group at one of the carbons numbered 3, 4,.. ., L(re -2)/2j, or one ethyl and one methyl in all possible distinct arrangements, or three methyl groups in all possible distinct arrangements; and so on.
