The Valence Angle of the Carbon Atom GORDON L. GOMBERT Kent State University, Kent, Ohio
I
N EXPLAINING optical isomerism a model of the carbon atom, as in Figure 1, is practically essential. It is mentioned that the four atoms or groups united with the carbon atom are equally distant from each other because the angles between the carbon valences are equal. In such discussions the statement is usually made that this angle is 109°28', but few organic texts, if any, offer a proof for this value, leaving this to the mathematical initiative of the student.
In triangle ABM
AT=AM~+BM~ and letting each of the edges, that is AB, A V , VC, and BC, of the tetrahedron equal S and since - S BM=-
2
we have
s z = m + s2 -4 which reduces to
Since VF is the altitude of the regular tetrahedron V-ABC, i t meets the base triangle ABC a t the point F which is two-thirds distant from the vertices of this triangle and one-third distant from the midpoints of its bases. Therefore
or by substitution
T FS = ~ & 0
FIGURE1
and I+ The tetrahedral conception of the carbon atom is F N =. attributed to van't Hoff (1)and Le Be1 (2) who published papers on this geometrical spatial arrangement S = 6dF of atoms in 1874. It is ~ossiblethat their ideas mew out of those already advanced by Wollaston (3) in 1808, In the triangle VFA Pasteur (4) in 1860, and Kekule (5) in 1867. The tetrahedral carbon atom is today substantiated by -A = +,TA X-ray investigation of the diamogd by Bragg (6), or and more recentlv . bv. means of quantum mechanics by Pauling (7). In solving this problem I found that a model, shown in Figure 2, of a regular tetrahedron with specially opening sides was particularly helpful. The model was prepared from four identical equilateral triangles, joined so as to form a regular tetrahedron, V-ABC. Then in order to study the center of the figure, where the valence angles are located, a vertical plane was passed through the vertex V, bisecting the base and one face in the lines AM and VM,respectively. At the same In triangle AOF time one of the face triangles and half the bisected face r'=m+zF were hinged along the base and allowed to open. The radii, OA and OV, were drawn from the center 0, which and substituting values of 7 and S for OF and AF is the point of intersection of the four altitudes of the tetrahedron, to the two vertices lying on the bisecting plane. The radius OV was continued to F. With these lines drawn, the value of the valence angle or may be calculated.
Angle a equals 180' - 8
or Solving for r we get
180' ,=
S -
Also in this same triangle Sin 6 =
m= r
S
-
3
S dz + -
- 70"31336"= 109%8'24"
In the preparation of this paper the author is grateful for the constructive criticism given by Professors E. M.
Collins, M. B. Palmer, and F. L. Brooks of Kent State University. LITERATURESITED
..
Then by canceling the S's Sin 8 = 0.94280 and 8
=
iOS1'36*
(1) VAN'THOP;. Bull. Soc. Chzm., 23,295 (1875). (2) LE BEL, ibid., 22, 337 (1874). Phil. Trans., 98, 96 (1808). (3) WOLLASTON, (4) PASTEUR,Alembic Clzrb Reprints, 14 (1897). (5) KEKULB,2. Chem., 3, 217 (1867). . Soc., A, 89, 277 (1913). (6) BRAGGAND BRADD,P ~ cRoy. (7) PAULING, PWC.NaI. Acnd. Sri.,14 (1928).
