Calculation of Coordinates
Scott MacKenzie
University of Rhode Island Kingston, Rhode Island
For Substituent Atoms on Carbon
In calculations involving carbon systems, it is generally the practice to utilize the locations of atoms in the form of coordinates in three-dimensional space. This is, after all, the most economical way to present all interatomic distances. For example, let us suppose that calculations are being done for a three-carbon system which is bonded a t the middle atom to two other groups, F and G (see figure). Most often, F and O are either carbons or hydrogens and only these two cases need be considered. If one knows the coordinates of carbon atoms A , B , and C, the coordinates of F and G can be calculated. Many of the solutions to this problem are trivial and one does well to choose the coordinate system so that elementary trigonometry suffices. However, in dealing with rings and larger carbon systems, one is often forced to choose a coordinate system which is skew to the system -4-B-C. I n this situation, solving for the coordinates of F and G becomes more difficult. It is this case which we shall consider here. Let the system ABCFG occupy any position in threedimensional space and let the known coordinates be Z A , Y A , X A , ZB, Y B , X B , ZC, YC, and XC. It will be assumed that arbo on B is tetrahedral, that A B and BC are each 1.54 A, and that F and G are either hydrogen atoms (BF is 1.09 A) or carbon atoms (BF is 1.54 A). B F = 1.09 ZD = (ZC YD = (YC XD = (XC
alternately, B F +(or, ZA)/2 ++ XA)I2 YA)/2
=
1.54 A)
The solution is given here in Fortran language which is
suitable for presentation to electronic computers. I n this problem, these Fortran statements differ very little from the usual mathematical notation. The distance BD is constant (BD = 1.54 cos 54' 44' = 0.88916) while RE depends upon the choice made for BF(BE = BF cos 54"44').
(In Fortran, * is "multiply.") By definition, one next calculates the divection cosines of the line BD. = (ZB - ZD)/BD E = (YB - YD)/BD F = (XR - XD)/BD
D
After calculnting the length of line AC(AC = 2 X 1.54 X sin 54'44' = 2.51532), one can calculate the direction cosines of this line. .4C = 2.51532 A = (ZC - ZA)/AC B = (YC - YA)IAC C = (xC - XA)/AC
The heart of the problem has now been reached. There must be applied two laws from solid geometry. The first states that "the sum of the squares of the direction cosines of any straight line is equal to unity" and the second says that "the sum of the products of the corresponding direction cosines of two lines which are perpendicular is equal to zero." Let the unknown direction cosines of the line FG be Z, Y, and X. Then Z2+Y2+XZ= l,Z.A+Y.B+X.C=O,andZ.D Y .E X . F = 0, for line FG, is, i n direction, perpendimdar to both BD and AC. Solution of the three equations for the three unknowns gives rather complex statements which can be simplified by introducing four new quantities, P, Q, R, and S.
+
+
P Q R S
= = = =
C*E - B*F A*F - C*D BID - A*E SQRT(P*P
+ Q*Q + R=R)
(In Fortran, SQRT is "square root".) Each of the unknown direction cosines can then be simply presented.
z
A three-carbon system (A-B-C) with rvbrtitventr IF, G). The coordinoterof A, B, and C are known. The coordinates of all other points (in paren-
theses) are
solculoted.
Y
= =
X
=
PIS Q/S RIS
It will be recalled that con~puters,in taking square roots, always take the positive sign. The signs of the unknown direction cosines are therefore determined only by the signs of the constants P, Q, and R, since S is always positive. The final few statements convert the calculated direction cosines into the desired information. The length Volume 43, Number I , Januory 1966
/
27
of line EF is equal to BF sin 54"44'. The coordinates for both F and G should always be calculated, even when only one set is desired. It is necessary to use models to ascertain which of the points F or G corresponds to each set of coordinates obtained.
28 / Journal o f Chemical Educofion
