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1 Chemistry Dr. Jean M. Standard Homework Problem Set 3 Solutions 1. The part of a particular MM3-like force field that describes stretching energy for an O-H single bond is given by the following expression: U s = 380( r r eq 2 75( r r eq 3, where r is the O-H bond distance, r eq= Å, and the stretching energy U s is expressed in kcal/mol. Notice that in many force fields based on MM3, the harmonic stretching energy is corrected by an anharmonic term like the cubic one shown above. A molecule containing an O-H group is built on the computer with an initial O-H bond distance of Å. a. What is the stretching energy of the O-H bond with r = Å? Substituting into the expression for the stretching energy yields: ( 2 75( U s = U s = U s = kcal/mol. b. What is the gradient (or first derivative of the energy with respect to the O-H bond at r = Å? The gradient can be obtained by taking the derivative of the stretching energy with respect to the bond distance r: U " s = d U s = 760( r r eq 225( r r eq 2. By substituting the specific values for the O-H bond given above, the gradient can be calculated: d U s d U s d U s = 760( ( = = kcal mol 1 Å 1.

2 1. Continued 2 c. What is the curvature (or second derivative at r = Å? The curvature is the second derivative of the energy with respect to the bond distance r: U s "" = d 2 U s 2 = ( r r eq. By substituting the specific values for the O-H bond given above, the curvature can be calculated: d 2 U s = ( d 2 U s = d 2 U s 2 = kcal mol 1 Å 2. d. Using the Newton-Raphson method for geometry optimization, carry out one step to determine the new value of r. The Newton-Raphson formula for updating the bond length r is r new = r old ( (. U # r old U # r old Substituting r old = Å and the values of the gradient and curvature from parts 1b and c, the new coordinate can be calculated. ( U # r r new = r old old U #( r old r new = r new = Å.

3 1. Continued 3 e. Calculate the gradient at the new value of r determined in part (d. If the tolerance for reaching the minimum is that the absolute value of the gradient be less than or equal to 0.1 kcal/mol-å, is the optimization complete? The gradient at r = Å can be calculated using the same formula given in part 1b. d U s d U s d U s = 760( ( = = kcal mol -1 Å -1. Since the absolute value of the gradient is kcal/mol-å, this is not less than or equal to the tolerance of 0.1 kcal/mol-å. Therefore, the optimization is not complete. At least one more Newton-Raphson step will be required to reach the minimum. 2. A typical molecular mechanics force field expression for the energy of torsional motion is given by the equation = A + Bcosω + C cos3ω, where is the energy in kcal/mol, ω is the torsional angle in degrees, and A, B, and C are constants. For a torsional angle such as the C-C-C-C torsion in n-butane, the parameters are given by A = 4.67 kcal mol 1 B = 2.52 kcal mol 1 C = 1.79 kcal mol 1. A scientist constructs the n-butane molecule using a molecule-building software package on a personal computer. The initial C-C-C-C torsional angle is ω init = 67.0º. Using the steepest descent energy minimization method, determine the next three values of the torsional angle in degrees and the energy in kcal/mol. Assume that the parameter γ in the steepest descent equation equals Calculate the gradient for the third iteration to determine whether or not convergence has been reached. Using the constants given above, the torsional energy expression is given by = cosω cos3ω, where the energy is in kcal/mol. To carry out the steepest descents method to find the optimized geometry, the equation to be used is ω new = ω old γ ", where! is the first derivative of the energy (evaluated at ω = ω old and γ is a constant.

4 4 2. Continued Step 1 Using the expression given for the energy, the first derivative is! = 2.52 sinω 5.37 sin3ω. Substituting ω = 67.0º, the first derivative is! = 2.52 sinω 5.37 sin3ω = 2.52 sin 67.0! ( 5.37 sin( !! = kcalmol 1 deg 1. Substituting these numerical values into the steepest descent equation yields ω new = ω old γ " = ω new = deg. ( ( Step 2 Now, to do this a second time, we use the angle that we just calculated and call that ω old calculation. So, the first derivative evaluated at ω = 67.36!! = 2.52 sinω 5.37 sin3ω = 2.52 sin 67.36! is ( 5.37 sin( !! = kcalmol 1 deg 1. anepeat the Substituting these numerical values into the steepest descent equation yields ω new = ω old γ " = ω new = deg. ( ( Step 3 Finally for the third step, we use the new angle from step 2, rename it ω old, nepeat the calculation. So, the first derivative evaluated at ω = 67.63! is! = 2.52 sinω 5.37 sin3ω = 2.52 sin 67.63! ( 5.37 sin( !! = kcalmol 1 deg 1. Substituting these numerical values into the steepest descent equation yields ω new = ω old γ " = ω new = deg. ( ( 0.241

5 2. Continued 5 Energies Note that the energy does not have to be calculated to carry out the steepest descent method; however, since it was also requested in the problem, the energy at each point may be calculated using the expression, = cosω cos3ω. Substituting the initial angle, ω = 67.0º, the torsional energy is = cos( 67.0! cos( ! = 0.64 kcal/mol. Substituting the second angle, ω = 67.36º, the torsional energy is = cos( 67.36! cos( ! = 0.66 kcal/mol. Substituting the third angle, ω = 67.63º, the torsional energy is = cos( 67.63! cos( ! = 0.68 kcal/mol. Substituting the final angle, ω = 67.85º, the torsional energy is = cos( 67.85! cos( ! = 0.70 kcal/mol. Summary and Convergence Check To summarize, the table below shows the three steps of the steepest descent method. Step ω new ω old! (deg. (deg. (kcal mol 1 deg 1 (kcal/mol To check convergence, we also need to calculated the gradient at the final coordinate value. This result is! = 2.52 sinω 5.37 sin3ω = 2.52 sin 67.85! ( 5.37 sin( !! = kcalmol 1 deg 1. Since the absolute value of the gradient is not less than 0.1 kcal/mol-deg, the minimization is not yet converged.

6 6 3. The potential energy from a particular molecular mechanics force field for bond stretching is given by U = 1 2 k s x2 1 3 k 3 x3, where the x coordinate corresponds to the bond displacement, x = r r eq. The parameters for C-H bond stretching are given by k s = 780 kcal mol 1 Å 2 k 3 = 50 kcal mol 1 Å 3. A molecule is built so that the intial bond displacement is x=0.20 Å. Carry out energy minimization on this system using the steepest descent method using the following conditions: a. Assume that the parameter γ in the steepest descent equation equals b. Assume that the parameter γ in the steepest descent equation equals Continue the iterations until the gradient reaches a value of 0.1 kcal mol 1 Å 1 or less; Excel might be helpful for this because many iterations may be required. For each minimization, report the number of steps required to reach convergence, the optimized value of the bond displacement x, the final energy, and the final gradient. From your results, discuss how the parameter γ impacts the energy minimization process. Part a To carry out the steepest descents method to find the optimized geometry, the equation to be used is x new = x old γ U ", where U! is the first derivative of the stretching energy (evaluated at x = x old and γ is a constant (0.001 in this case. For the stretching potential given above, the first derivative is U! = k s x k 3 x 2, or U! = 780x 50x 2. For the first step of the steepest descent method, substituting x=0.2 Å, the first derivative is U! = 780x 50x 2 = 780( ( U! = 154 kcal mol -1 Å -1. Substituting this into the steepest descent equation yields x new = x old γ U" ( ( 154 = x new = Å.

7 3. Continued 7 The stretching energy U at the initial point is U = 1 2 k s x2 1 3 k 3 x3 = 1 2 (780x ( x 3 = 1 ( U = kcal/mol. ( ( ( For the remainder of the steps, Microsoft Excel was used to generate the energy and first derivative as well as the new position predicted by the steepest descent method. The results are shown in the table below. Table 1. Results from steepest descent method with γ= step x old (Å U (kcal/mol U' (kcal mol 1 Å Here we see that the gradient (or first derivative has dropped to below 0.1 kcal mol 1 Å 1 (actual value 0.09 kcal mol 1 Å 1 after the 5th step of the steepest descent method. At that point, the calculated stretching energy is 0.00 kcal/mol and the new position is Å. x new (Å Part b The same equation for the steepest descent method is used here, x new = x old γ U ", where U! is the first derivative of the stretching energy (evaluated at x = x old and γ is a constant ( this time. For the first step of the steepest descent method, substituting x=0.2 Å, the first derivative is again U! = 780x 50x 2 = 780( ( U! = 154 kcal mol -1 Å -1. Substituting this into the steepest descent equation yields x new = x old γ U" ( ( 154 = x new = Å. The stretching energy U at the initial point is again kcal/mol.

8 3. Continued 8 For the remainder of the steps, Microsoft Excel was used to generate the energy and first derivative as well as the new position predicted by the steepest descent method. The results are shown in the table below. Table 2. Results from steepest descent method with γ= step x old (Å U (kcal/mol U' (kcal mol 1 Å x new (Å

9 3. Continued 9 This time we see that the gradient (or first derivative has dropped to below 0.1 kcal mol 1 Å 1 (actual value 0.09 kcal mol 1 Å 1 after the 44th step of the steepest descent method. At that point, the calculated stretching energy is 0.00 kcal/mol and the new position is Å. Clearly, selecting an appropriate value for the steepest descent parameter γ is an important factor governing how fast convergence to an energy minimum is achieved. If γ is selected to be too small, convergence will require many more steps. 4. The potential energy from a particular molecular mechanics force field for bond stretching and angle bending is defined as U = 1 2 k s x k b y2 k sb xy, where the x coordinate corresponds to the bond displacement, x = r r eq, and y is the angle displacement, y = θ θ eq. The last term in the force field is a stretch-bend term that accounts for interactions between bond stretching and angle bending; terms like this are included in the MMFF94 force field. The parameters are given by k s = 1000 kcal mol 1 Å 2 k b = 500 kcal mol 1 radian 2 k sb = 100 kcal mol 1 Å 1 radian 1. a. Determine the locations of any extrema on the potential energy surface. The criterion for determining an extremum (either a minimum, maximum, or saddle point is that the first derivative is equal to zero (in both the x and y directions in this case. Taking the first derivatives, we have! U $ # & " x y! U $ # & " y x = k s x k sb y = k b y k sb x. Setting these equal to zero and solving gives the positins of the extrema on the potential energy surface, k s x k sb y = 0 k b y k sb x = 0. Solving the first equation for x yields, x = k sb k s y.

10 4 a. Continued 10 Substituting this result into the second equation allows us to solve for y, Using y=0 in the equation we obtained for x gives, k b y k sb x = 0 k b y k sb k sb k s y = 0 # y k b k 2 & sb ( = 0, $ ' k s or y = 0. Therefore, there is one extremum for this function: x = k sb k s y = k sb k s 0 x = 0. ( x, y = ( 0, 0. b. Classify any extrema found in part (a as a minimum, maximum, or saddle point by determination of the curvature at each point. To determine whether an extremum is a minina, maxima, or saddle point, we must determine the curvature (second derivatives and evaluate the curvature at the extremum. Taking the second derivatives,! 2 U $ # & " x 2 y! 2 U $ # & " y 2 x = k s = k b. Since both second derivatives are positive, the point at (0,0 corresponds to a minimum.

11 5. A potential energy surface has a form given by V( x, y = 1 3 x 3 + xy 2 + 2x 2 2y a. Determine the locations of all of the extrema on the potential energy surface. The criterion for determining an extremum (either a minimum, maximum, or saddle point is that the first derivative is equal to zero (in both the x and y directions in this case. Taking the first derivatives, we have # V & ( $ x ' y = x 2 + y x # V & ( $ y ' x = 2xy 4 y. Setting these equal to zero and solving gives the positions of the extrema on the potential energy surface, The second equation has a common factor of y, x 2 + y 2 + 4x = 0 2xy 4 y = 0. y( 2x 4 = 0. This means that there is a solution of the second equation for y=0 or for 2x 4 = 0, x = 2. Going back to the first equation, x 2 + y 2 + 4x = 0, we can substitute y=0 and simplify: This equation has a solution when x=0 or when x x = 0, ( = 0. x x + 4 x + 4 = 0 x = 4. Thus, the solutions of this equation are x=0 and x = 4. Along with the result that y=0 from the second equation, we have so far obtained the following extrema: ( x, y = ( 0,0 ( 4,0.

12 5 a. Continued 12 Now, going back to the other solution from the second equation, x=2, substitution of this into the first equation, x 2 + y 2 + 4x = 0, yields, 4 + y = 0, y 2 = 12. Since this solution for y is imaginary, it is not a valid solution for a real potential energy surface. Thus, the only physically valid extrema for this potential energy surface are the ones found previously: ( x, y = ( 0,0 ( 4,0. b. Classify each of the extrema found in part (a as a minimum, maximum, or saddle point by determination of the curvature at each point. To determine whether the extrema are minina, maxima, or saddle points, we must determine the curvature (second derivatives and evaluate the curvature at the extrema. Taking the second derivatives, # 2 V & $ x 2 ( ' y = 2x + 4 # 2 V & $ y 2 ( ' x = 2x 4. Now, we need to evaluate these second derivatives at each of the two extrema in order to determine whether or not they are minima, maxima, or saddle points. ** For the point at ( x, y = ( 0,0: # 2 f & $ x 2 ( = 2x + 4 = = 4 ' y x =0,y =0 # 2 f & $ y 2 ( = 2x 4 = 0 4 = 4. ' x x =0,y =0 Since one second derivative is negative and the other is positive, the point at (0,0 corresponds to a saddle point.

13 5 b. Continued 13 ** For the point at ( x, y = ( 4,0 : # 2 f & $ x 2 ( = 2x + 4 = = 4 ' y x = 4,y =0 # 2 f & $ y 2 ( ' x x = 4,y =0 = 2x 4 = 8 4 = 12. Since both second derivatives are negative, the point at (-4,0 corresponds to a maximum.