Counting the Number of Isomers of Chain Alkanes — Application of Pólya Enumeration Theorem

Fa Yongwang, Li Enshu, Xu Wenqi

This paper discusses the number of isomers (not considering enantiomers) of chain alkanes with chemical formulas. First, the Pólya generating function theorem is introduced by coloring the vertices of a cube in space; then the generating function of the number of carbon alkane bases is obtained; finally, the generating function of the number of carbon alkanes is derived.

I. Introduction

Firstly, consider the following question: Use four colors A, B, C, D (not necessarily all of them) to color the eight vertices of a cube that can move freely in space.(1) How many coloring patterns are there?(2) How many ways can you use 1 of A, 2 of B, 3 of C and 2 of D to color?

If the cube is immobile, the answer is ___, clearly one type and another. However, the rotation of the cube makes some coloring methods essentially the same as each other. If we label the eight vertices in the order of the eight trigrams as 1, 2, …, 8, then (in the order of 1, 2, …, 8, same below) ___ and ___ are clearly of the same type of coloring method.

We define a movement of the cube that maps it onto itself (via rotation) as a permutation. If the cube occupies the same position before and after the permutation, it is represented by a rotation. For example, if the cube is rotated 90° along the line passing through the centers of planes 1234 and 5678 from 1 to 2, the vertices will change in the order 1 → 2 → 3 → 4 → 1, 5 → 6 → 7 → 8 → 5. This rotation is represented by the permutation (1234)(5678), which consists of two disjoint cycles.

If the cube is rotated 180° along the line passing through the centers of planes 1234 and 5678 from 1 to 3, 2 to 4, 5 to 7, 6 to 8, it will change in the order 1 → 3 → 1, 2 → 4 → 2, 5 → 7 → 5, 6 → 8 → 6. This permutation is denoted as (13)(24)(57)(68), with 4 disjoint cycles.

If the cube is rotated 180° along the line passing through the midpoint of segments 12 and 78 from 1 to 2, the vertices will change in the order 1 → 2 → 1, 3 → 5 → 3, 4 → 6 → 4, 7 → 8 → 7. This permutation is denoted as (12)(35)(46)(78), with 4 disjoint cycles.

If the straight line 17 of the cube is rotated 120° from 5 to 2, the vertices will change in the order 1 → 1, 7 → 7, 2 → 4 → 5, 3 → 8 → 6. This permutation is denoted as (245)(386), with 4 disjoint cycles. Notably, the identity permutation (where all vertices remain unchanged) is denoted as (1), with 8 disjoint cycles.

If a coloring method can be transformed into another through one substitution, then these two coloring methods are essentially the same. For example, in the first coloring method mentioned earlier, after substitutions (16), (25), (38), (47), the second coloring method can be obtained, so these two methods are the same coloring method.

We note that there are a total of 24 possible substitutions for a cube, which include: 1 identity transformation (1), 6 transformations around a second axis such as (12)(35)(46)(78), 8 transformations around a third axis such as (245)(386), 6 transformations around a fourth axis by 90° such as (1234)(5678), and 3 transformations around a fourth axis by 180° such as (13)(24)(57)(68). Among these, the axis that can be restored during one full rotation around a certain axis is called the primary axis.

Thus, the original problem is equivalent to finding the number of essentially different coloring methods under these 24 permutations, that is, the number of equivalence classes of all coloring methods with respect to these 24 permutations (all essentially the same coloring methods are placed in the same equivalence class).

2. Preliminary Knowledge

This section briefly introduces some preparatory knowledge; see relevant books on group theory for details. Without considering the movement of the cube, all 65,536 ways to color the vertices are placed into one set X, and all 24 possible rotations of the cube are placed into another set S.

We note that the multiplication of transformations forms a group of permutations. For any (x \in X) and (a \in S), we denote (a(x)) as the permutation acting on the colored configuration to produce another colored configuration. We point out that this gives an action of group S on the set X.

For each (x), let (Sx) be the orbit of x under S; let (S_x) be the stabilizer subgroup of x under S; and let (X^a) be the invariant set under a. Next, we have the following theorem:

Theorem. Denote the set of equivalence classes of X under S as X/S (where the elements are all orbits of S, and X/S is a family of sets). Then, the number of equivalence classes of all colorings is given by[|X/S|=\frac{1}{|S|}\sum_{a\in S}|X^a|.]

Proof (sketch). This follows from Burnside’s Lemma and the orbit–stabilizer relationship. Two colorings are essentially the same if and only if they belong to the same orbit under the group action. Distinct orbits partition X, which yields the stated counting formula.

Burnside’s Lemma is a “symmetry averaging” method: it counts distinct patterns by averaging how many patterns stay unchanged under each symmetry operation (like rotations).

Verifying Burnside’s Lemma: a Triangle Example

We verify Burnside’s Lemma with a simpler example: color the vertices of an equilateral triangle in space with two colors, A and B. Mark each vertex counterclockwise 1, 2, 3. Then the permutation group of the equilateral triangle is[S={(1),(123),(132),(12),(23),(13)},]and the set of all colorings is[X={(AAA),(AAB),(ABA),(ABB),(BAA),(BAB),(BBA),(BBB)}.]

It can be shown there are exactly four orbits, corresponding to four essentially different coloring configurations:[{(AAA)},\ {(AAB),(ABA),(BAA)},\ {(BBA),(BAB),(ABB)},\ {(BBB)}.]Thus (|S|=6) and (|X/S|=4).

Each element in S fixes a set of colorings; counting these fixed colorings gives[|X/S|=\frac{1}{|S|}\sum_{a\in S}|X^a|=\frac{1}{6}(8+2+2+4+4+4)=4,]matching the orbit count.

Pólya’s Theorem and the Cube Coloring Problem

Pólya’s theorem states that if there are m available colors and a permutation has (i(a)) disjoint cycles, then the number of colorings fixed under that permutation is[|X^a|=m^{i(a)}.]

Reason: within each disjoint cycle, all positions must share the same color to remain unchanged. Each cycle contributes one independent color choice.

Example (triangle swap (12)): the cycle structure has two cycles (1↔2) and (3 fixed), so (i(a)=2). With two colors, the number of unchanged patterns is (2^{2}=4): AAA, AAB, BBA, BBB.

Returning to the cube question (1), with 4 colors and 24 rotations: there is 1 identity with (i(a)=8), 14 rotations with (i(a)=4), and 9 rotations with (i(a)=2). Therefore the number of essentially distinct colorings is[\frac{1}{24}\left(4^8+14\cdot 4^4+9\cdot 4^2\right).]

Generating Function Form of Pólya’s Theorem

To solve question (2), we use the generating-function version of Pólya’s theorem. The generating function form is[P(S)=\frac{1}{|S|}\sum_{a\in S}\prod_{j\ge1}\left(x_1^{,j}+x_2^{,j}+\cdots +x_m^{,j}\right)^{c_j(a)},]where (c_j(a)) is the number of j-cycles in permutation a, and (x_1,\dots,x_m) represent colors.

A term (x_1^{n_1}x_2^{n_2}\cdots x_m^{n_m}) in the expansion corresponds to colorings using exactly (n_1,n_2,\dots,n_m) of each color, and its coefficient counts how many essentially distinct colorings exist with that composition.

For the triangle example with colors A and B, this reproduces the correct result. Returning to the cube question (2), applying the cycle index for the cube and extracting the coefficient of (a^1b^2c^3d^2) yields:

There are 70 distinct coloring patterns using exactly 1 A, 2 B, 3 C, and 2 D.

Counting the Number of Alkyl Isomers

First, consider the number of structural isomers of n-carbon alkyl groups (carbon chains with one free end), denoted (A_n). Treat the hydrogen atom at the free end as a zero-carbon alkyl group and define (A_0=1). Using the sequence ({A_0,A_1,A_2,\dots}), we construct the alkyl generating function[A(x)=\sum_{n\ge0}A_n x^n.]

We interpret generating-function operations:

1. The generating function of atoms forming structure P is (P(x)).

2. The generating function of n atoms forming m identical structures P is (x^{n-m}P(x^m)).

3. The generating function of a union of structures is (P(x)+Q(x)).

4. The generating function of a structure made by combining P and Q is (P(x)Q(x)).

Treat the free end as the root carbon atom. Once the root is fixed, the alkyl group is composed of three sub-alkyl groups connected to it. Label these three positions 1, 2, 3 around the root. Since enantiomers are not considered, these positions behave like the vertices of an equilateral triangle with symmetry group ({(1),(123),(132),(12),(13),(23)}).

Applying Burnside’s Lemma gives the functional equation:[A(x)=1+\frac{x}{6}\big(A(x)^3+2A(x^3)+3A(x)A(x^2)\big),]which implicitly defines (A(x)). Iterating yields[A(x)=1+x+x^2+2x^3+4x^4+8x^5+17x^6+39x^7+89x^8+211x^9+507x^{10}+\cdots.]

Counting Isomers of Chain Alkanes

Define the diameter of an alkane as the longest carbon chain connecting two hydrogen atoms; its length is the diameter length. The center is the midpoint carbon atom (odd diameter) or midpoint carbon–carbon pair (even diameter). Thus, alkanes with odd diameter have 1 center; those with even diameter have 2 centers.

Let (N(x)) be the generating function for the number of alkane isomers, and use intermediate generating functions (P(x),Q(x),S(x)) derived from attaching alkyl groups in symmetry-aware ways:

• (P(x)): labeled-carbon alkanes (four alkyl groups attached to a labeled carbon)

• (Q(x)): labeled bond alkanes (two alkyl groups attached to a labeled C–C bond)

• (S(x)): symmetric C–C bond cases

• 
  

The relationship becomes:[N(x)=P(x)-Q(x)+S(x).]

With expansions:[P(x)=x+x^2+2x^3+4x^4+9x^5+18x^6+42x^7+96x^8+229x^9+549x^{10}+\cdots][Q(x)=x^2+x^3+3x^4+6x^5+15x^6+33x^7+82x^8+194x^9+482x^{10}+\cdots][S(x)=x^2+x^4+2x^6+4x^8+8x^{10}+\cdots]so[N(x)=x+x^2+x^3+2x^4+3x^5+5x^6+9x^7+18x^8+35x^9+75x^{10}+\cdots,]whose coefficients match the known isomer counts from methane to decane.

A second derivation based on diameter decomposition is also presented, splitting cases by whether the diameter length is even or odd and using controlled “height” constraints on attached alkyl groups. This approach yields the same coefficients for (N(x)).

References

[1] Ma, C., & Li, Q. (2022). Counting isomers of hydrocarbons. University Chemistry, 37(1), Article 2103067. https://doi.org/10.3866/PKU.DXHX202103067 Available at: https://www.dxhx.pku.edu.cn/article/2022/1000-8438/20220135.shtml

[2] Hongwu. (n.d.). Calculation of the number of isomers of alkanes. Zhihu. Retrieved from https://zhuanlan.zhihu.com/p/28032458

[3] Walk Away. (n.d.). In the answer to “How many isomers are there in pentyl, hexyl, and heptyl? What's the pattern?” Zhihu. Retrieved from https://www.zhihu.com/question/64046654/answer/661571148

[4] Yin, N. (n.d.). Read Abstract Algebra (1) in one article: Pólya counting principle. Zhihu. Retrieved from https://zhuanlan.zhihu.com/p/404310081

[5] The Distant Star of the Sword. (n.d.). Counting—Burnside with symmetry and Pólya’s theorem. Zhihu. Retrieved from https://zhuanlan.zhihu.com/p/80261375

[6] Pólya, G. (1937). Kombinatorische Anzahlbestimmungen für Gruppen, Graphen und chemische Verbindungen. Acta Mathematica, 68, 145–254. https://doi.org/10.1007/BF02546665