Cyclohexane Chair & Boat

Cyclohexane The applet shows boat & chair conformations of cyclohexane. Radio buttons are provided to switch between two conformations. Angle θ provides facility to rotate molecule about y axis and α is about x axis thus one can study full details of geometry. The angles between the bonds of cyclohexane are close to 109o 28', hence it is quite stable and nonreactive. Two most most important conformations for cyclohexane are chair & boat forms. Out of these two chair form is more stable than boat. The positions of Hydrogen atoms in chair form are at a greater distance than those in boat form. The flag pole interaction (repulsion) between the Hydrogen atoms in boat form leads to changing boat to chair form. Therefore cyclohexane exists mainly in chair form conformation. This is a Java Applet created using GeoGebra from www.geogebra.org - it looks like you don't have Java installed, please go to www.java.com Coordinate system – x axis is taken to right, y axis up and z coming out of screen towards viewer. Thus it is right hand coordinate system. Rotations about two axes are sufficient to view all the details of the 3D body. α is rotation about x axis such that y axis move towards z axis. And θ is rotation about y axis so that z moves towards x axis. Thus rotations follow right hand rule. Let there be a point P(X, Y, Z) in 3D. If this is viewed in 2D with both alpha & Theta to be zero, then point will appear at (X, Y) Due to θ rotation about y axis x = X*cos(θ) + Z*sin(θ) and y = Y, And then by θ & α rotations together x = X*cos(θ) +Z*sin(θ) & y = Y*cos(α) - Z*cos(θ)* sin(α). Let us plot the three +Ve axes. Let L be the length of the vector of axes. The Tip of x axis XT(L, 0, 0) YT(0, L, 0) & ZT = (0, 0, L) then xT is (L*cos(θ), 0) & yT is (0, L*cos(α) and zT is (L*sin(θ), - L*cos(θ)*sin(α). For the tetrahedron formed by Methane molecule: let bond length be B, Then the height of the tetrahedron H = 4*B/3. The angle made by Height with two sides appearing to be coincident in elevation γ = asin(1/3) = 19.47° and median of triangle in plan is M = (3/2)B*cos(γ) = 1.41 B Side of tetrahedron S = M/cos(30°) = 1.63 B. The angle made by top vertex with the vertex at right at base with center is obtained as δ = 2*asin(S/(2B)) = 109.47°. The angle at top vertex made by other two vertexes coinciding with each other in elevation and center is ϕ = (360-2δ)/2 = 125.26° Let rectangle formed by C2, C3, C5 & C6 lie in xz plane with origin at its center. Then x of C2 & C6 will be – B/2 and x of C3 & C4 will be B/2. C2, C3, C5 & C6 will have y to be zero. C2 & C3 will have z = -S/2 and C5 & C6 will have z = S/2. C1 & C4 have z = 0. Now to get x & y coordinates of C1, consider a tetrahedron with C2 at center and C3 other vertex to right in in elevation with C1 as one of the two vertexes coincident with each other. Hence C1 will have X = - B/2 - H/4, Y = M*sin(γ) & Z = 0. C4 is mirror reflection of C1.Thus matrix C(3 x 6) is generated. These are 3D coordinates of 6 carbon atoms. Now to get 2D coordinates one has to multiply T & C(Transpose). Hydrogen atoms H1 to H4 lie in x-y plane and hence their locations are found by considering a tetrahedron with C1 at center and C2, C6, H1 & H2 at vertexes. XH1 = xC1+ B cos(2ϕ – π) & yH1= yC1 + B sin(2ϕ – π). zH1 is obviously zero. H2 is aligned with x axis so xH2 = xC1 – B & yH2 = yC1. Obtaining coordinates of other Hydrogen atoms is bit complicated and is achieved as follows: Pass a plane through three known atoms draw two vectors of bond length from the central atom. The Hydrogen atoms lie in a plane which bisects this plane. The bisecting plane is located with the help of unit vector along two bond. Their resultant lies in plane and cross product lies ⊥ to plane. Now the Hydrogen atoms are at angle ϕ = 125.26° in bisecting plane. Dr Vasant Barve, 24 August 2014, Created with GeoGebra