Solutions to Assignment #4 Getting Started with HyperChem

Solutions to Assignment #4 Getting Started with HyperChem 1. This first exercise is meant to familiarize you with the different methods for visualizing molecules available in HyperChem. (a) Create a molecule of n-butane in the trans conformation and geometry optimize it using the MM+ force field. Explore the depictions of the butane model provided by all of the Atom Rendering options shown at the right. Turn in versions of Sticks, Balls, Balls and Cylinders, and Tubes representations of butane (using program default parameters) with your solutions (b) Which of these representations comes closest to your own mental picture of a butane molecule. Why? (a) The four representations are illustrated below. (b) Different rendering methods are useful for different purposes. None of them can claim any hold on reality. The balls representation is closest to my mental picture of molecules, because I do not like thinking of bonds as lines connecting atoms but rather as continuous regions of electron density. 2. A handy feature of HyperChem is that it provides an automated way of exploring 1-dimensional scans over potential energy surfaces. Here we ll use this feature to calculate the interaction between two Xe atoms as computed by the MM+ force field. (a) Create a Xe atom, then select molecular mechanics and the MM+ force field as the type of calculation. Report the values of the non-bonded parameters used by the MM+ force field in the case of Xe. (b) Create a second Xe atom at some reasonable distance from the first. Run a potential scan by selecting both Xe atoms and using the command /Compute/Potential. (Note that the Potential option will not be operative if both atoms are not properly selected.) A reasonable set of parameters for the scan is 3.5-8 Å in.1 Å steps with an energy limit of 2 kcal/mol. HyperChem will produce a convenient graph of the results for you. Label the graph appropriately and submit a copy with your report. 1

(c) Deselect the atoms and then run an energy minimization (i.e. geometry optimization) calculation on the Xe pair. Look at the bottom of the HyperChem window to find the energy after optimization. Make sure that the calculation says it has converged and that the gradient is smaller than.1 kcal/mol. Also find the separation at the optimized geometry by selecting both atoms and reading the result at the bottom banner of the HyperChem window. Record the energy and separation. Do these values coincide with what you would read from the graph of part (b)? (d) Rationalize the energy and separation obtained from part (c) with what is expected based on the treatment of nonbonding interactions by the MM+ FF (see Molecular Mechanics excerpt) and the potential parameters recorded in part (a). (a) The parameters are: Atom Type: XE Epsilon:.495 R: 2.28 (b) The potential scan is shown at the right. (c) The optimized separation and energy are: R = 4.56 Å E = -.5788 kcal/mol (d) According to the HyperChem manual Ch. 11, the parameters of the MM+ van der Waals term are the van der Waals radius, r * i, and the hardness parameter, ε i. These parameters are used via the equations: r = r + r (37) * * i * j 1/ 2 ε = ( ε iε j ) (38) { 5 * * ) 2.9 1 exp[ 12.5( R / r )] 2.25( R / r 6 } E ( R ) ε (39) vdw = Either numerical calculations or analytic manipulation of Eq. 39 shows that the minimum of E vdw * occurs when R = r. Unfortunately, the energy at this minimum point is not -ε as it would be in a * sensible description, but it is instead Emin = Evdw( R = r ) = 1.169ε. So although ε is the parameter describing the depth of the potential well, its value does not provide this depth directly. For two identical Xe atoms the parameters entering into Eq. 39 are r * 2 r * = 4.56Å and = Xe ε = ε Xe =.495 kcal/mol so one expects the well minimum to occur at 4.56 Å and the energy at this point should be -.5787 kcal/mol. These are indeed the values reported by the program. 3. One criterion for realism in the depiction of molecules is to require the sizes of atoms to indicate how closely atoms and molecules tend to approach one another. This goal can be achieved by depicting each atom using a radius equal to ½ the distance of closest approach of a like atom when 2

interacting in a non-bonding situation. Loosely speaking, this is what is meant by the van der Waals radius of an atom and these are the atomic radii used in CPK models. (a) Examine the Balls rendering option with respect to this criterion using Xe+Xe as a test case. With the default settings as shown at the right, are van der Waals radii being depicted? If the answer is no (which it should be) use the distances of closest approach you find in geometry optimizations of Xe+Xe and Ne+Ne to determine approximately what value of the ball radius slider is required so that the program displays van der Waals radii. (Use MM+ parameters.) Report what value of the slider radius you find (say in terms of tick marks from one end of the range) (b) Use this value to depict MM+ geometry optimized representations of (trans) n-butane and benzene. Capture and include these images with your solutions. (a) The figures below show the optimized pair distances in Ne+Ne and Xe+Xe. Using the default value of the ball radius setting gives lots of space between the atoms. I adjusted the slide to the 4 th position from the maximum end to achieve the bottom pair of snapshots. I take this setting to be about right. Ne+Ne Default Radius Setting Xe+Xe Default Radius Setting Ne+Ne Optimal Radius Setting Xe+Xe Optimal Radius Setting (b) The representations of butane and benzene made with this setting are shown on the following page. for fun I ve included a figure of the minimum energy structure of benzene + Xe to show that this representation provides a correct looking picture of this interaction as well. 3

t-butane - two views Benzene and Benzene+Xe 4. Use the Potential feature discussed in problem #2 to examine the MM+ representation of the torsional potentials of n-butane. (a) Create trans n-butane and optimize its geometry using the MM+ force field. Record the optimized energy. (b) Select the CH 3 torsional angle as indicated in the figure below. Use /Setup/Edit_Parameters to find the parameters that the MM+ force field uses to describe this torsion. (Record your results.) Then run a potential scan on this coordinate from -36. Include a plot of your scan in your solutions. (c) Quantitatively explain the features of this scan based on the form of the MM+ torsion term and parameters associated with this torsion. (d) Repeat parts (b) and (c) for the central C-C torsion angle. (e) Now geometry optimize the gauche conformation of n-butane. To obtain a starting point for this optimization select the central torsion and use the /Edit/Set_Bond_Torsion command. Note: Make sure you deselect all atoms before you perform the geometry optimization. Record the optimized energy and compute the gauche-trans energy difference using this value and the value obtained from part (a). (f) Scan the central C-C torsion again starting with the optimized gauche conformation. Record your scan and compare it to the scan of the trans form. Why are the two scans different? (g) Read the gauche-trans energy differences from the two potential scans and compare these values to the value from part (e). Explain any differences you find. (a) The optimized energy I find for trans-butane is 2.171 kcal/mol. The energy is greater than zero mainly because of van der Waals interactions between 1-4+ atoms are repulsive. (b,c) The MM+ force field classifies the CH 3 torsion in butane of type C4-C4-C4-H where C4 refers to an sp 3 hybridized carbon atom and H a hydrogen atom not attached to N or O. Aside: Note that the classification of atom types is described, at least in first pass, in the file mmptyp.txt which can be found in the course web site under parameters on the bottom of the Worksheets & Stuff web page. The relevant lines for butane from this file are: TYPE MASS REMARK 4

c4 12. 1. C CSP3 h 1.8 5. H EXCEPT ON N/O The parameters listed for this torsional interaction are: V 1 =, V 2 =, V 3 =.267 kcal/mol In this case the form of the torsional potential (Eq. 36 of the manual) specializes to: 1 ω ) = V {1 + cos(3ω )} E tor ( 2 3 where ω is the torsion angle. What is expected based on this interaction alone is a sinusoid with three minima at angles of 6, 18, and 3 where E tor = and three maxima at, 12, and 24 where E tor = V 3 =.267 kcal/mol. The CH 3 torsional scan is shown at the right. The curve has the qualitative form expected, but the energies at the minima are not at zero and the amplitude of the sinusoid is about 3 kcal/mol rather than.3 kcal/mol. The offset in energy is due to the other terms in the potential that are not zero in the optimized structure. The minimum energy as the CH 3 group is rotated is the global minimum energy of 2.17 kcal/mol found in part (a). The amplitude of the sinusoid is much larger than what is calculated for the particular C4-C4-C4-H term highlighted because this is not the only interaction that changes when the CH 3 group is rotated. Using the numbering scheme at the right, when the C(1) CH 3 group rotates, the following 1-4 torsional interactions all change: CCCH: 5123, 6123, 7213 HCCH: 5128, 5129, 6128, 6129, 7128, 7129 The H-C4-C4-H torsional interaction terms in MM+ are nearly identical to the C4-C4-C4-H terms. They are 3-fold symmetric with V 3 just slightly smaller, V 3 =.237 kcal/mol. Taking all of these torsional changes into account the amplitude of the sinusoid is expected to be: ΔE tors / kcal mol -1 = 3.267 + 6.237 = 2.223 These changes account for most of the energy change observed. But 1-4 and higher van der Waals interactions also change when the CH 3 group rotates and these variations apparently also add about.9 kcal/mol to the net torsional barrier. (d) The parameters listed for the C4-C4-C4-C4 torsional interaction are: V 1 =.2, V 2 =.27, V 3 =.93 kcal/mol In this case all three terms in the torsional potential are included. 1 1 E ω ) = 1 V {1 + cos( ω)} + V {1 cos(3ω )} + V {1 + cos(3 ω tor )} (36) ( 2 3 2 2 2 3 5

The computed energy as a function of central C-C angle is shown below (left). Also shown is the form of the single E tor (ω) term of Eq. 36. I computed this using the torsion_play.xls worksheet supplied on the course web. The black line a normalized (scaled by 1 / Σ i Vi ) representation of E tor (ω) and the colored curves are the individual components. (Converting the relative amplitudes shown here to kcal / mol one requires multiplication by.563 kcal/mol.) 1..9.8.7 C4-C4-C4-C4 Term amplitude.6.5.4.3.2.1. 6 12 18 24 3 36 angle / deg In this case, even the form of the potential is not the same as what is calculated based on the single term associated with the CCCC torsion. Again, the energy change resulting from rotating about the 2-3 bond results from a summation of a number of terms. Here there are four H-C4-C4-H torsion terms that change: 8-2-3-1, 8-2-3-11, 9-2-3-1, and 9-2-3-11. These terms would add to the 3-fold part of the potential. But more important in this case are the changes to the van der Waals interactions that occur between the end methyl groups. These terms cause the much larger energy of the eclipsed form at. (e) The optimized energy of the gauche form of n- butane is 3.35 kcal/mol which gives a g-t energy difference of: Eg t Δ =.863 kcal/mol (f) The scan of the potential for rotation about the central C-C bond after optimizing the gauche form of n-butane is shown at the right. This scan differs from the scan initiated from the trans-optimized form because the other geometric variables (bond lengths, angles, other torsions) of the two forms differ. When a scan is performed, these other variables are fixed at their starting values and thus the scans represent distinct slices of the potential energy surface. They are qualitatively similar, but not quantitatively the same. 6

(g) The results of various calculations are tabulated below. From this table one finds that the partially optimized structures yield gauche-trans energy differences that are upper and lower bounds to the true difference between the fully optimized structures. Calculation E(t-min) E(g-min) ΔE kcal/mol kcal/mol kcal/mol trans optimized 2.171 3.814 1.643 gauche optimized 2.365 3.35.67 individually optimized 2.171 3.35.863 5. Compare the bending potentials of water using all four of the force fields available in HyperChem. (a) First create a water molecule and select the bond angle. Then choose each of the four force fields available (the default versions) using the /Setup/Molecular_Mechanics command and record the atom types and parameters associated with each FF using the /Setup/Edit_Parameters command. (b) Use the parameters obtained from (a) and a program such as Excel to compute and plot the bending potentials of the different force fields on the same graph. (A description of the various force fields is provided in the HpyerChem Molecular Mechanics.) Make plots that show the potentials over a wide range of angles (say ±5 ) and over the range relevant to modeling motion near room temperature. Use units of kcal/mol and Å. (You can verify the accuracy of your calculations using potential scans within HyperChem. (c) Comment on the similarities and differences among these potentials. (d) What would you expect the relative bending frequencies of these potentials to be? (Assuming the equilibrium geometries are all identical, the relative bending frequencies should depend only on the relative force constants.) (e) If one assumes that the average energy in the bending vibration is equal to k B T, what is the range of angles expected to be sampled by a typical molecule at room temperature using each of the models? (a) The parameters read from the /Setup/Edit_Parameters command are as follows: Table 5-1: Parameters of the HOH Bend as Reported in HyperChem Force Field Atom Types Force Constant Eq. Bond Angle MM+ HO, O2, HO.4 14.5 AMBER HW, OW, HW 1 14.52 BIO+(CHARMM) HT, OT, HT 55 14.52 OPLS HT, OT, HT 6 14.52 According to the HyperChem manual, the AMBER, BIO+(CHARMM), and OPLS force fields all use the simple quadratic form of the angle bending term: 2 E bnd ( θ ) = Kθ ( θ θ (16) ) where K θ is the bending force constant (provided in units of kcal mol -1 radian -2 ) and θ the natural bond angle provided in degrees. The MM+ force field uses the more complicated form: 7

( ) (.43828) 1 2 8 4 E θ ( θ θ ) {1 7. 1 ( θ θ bnd = Kθ + ) } (34) 2 In this case K θ is provided in units of mdyne Å / rad 2 but the factor.43828 is used to convert it to kcal mol -1 deg -2. (The manual wrongly adds that these are the units of K θ used in the other FFs.) (b) The plots are shown below. (I verified that these are correct by using the Potential utility in HyperChem. 8 AMBER OPLS 2. AMBER OPLS Bio+ 6 1.5 E bnd / kcal mol -1 4 2 Bio+ MM+ E bnd / kcal mol -1 1..5 k B T MM+ 4 6 8 1 12 14 16 Bond Angle θ / deg. 9 95 1 15 11 115 12 Bond Angle θ / deg (c) All of the purely harmonic potentials are qualitatively the same but the different force constants lead to rather different widths in the bending potentials. The MM+ potential is unique by virtue of the 6 ( θ θ ) term, which makes the region of the minimum much broader than it would be otherwise. I am surprised at how large the difference are among these various bending potentials. Given that an H-O-H bend can only apply to water, and given water s importance in chemistry, I would have expected them to all be much more similar. The OPLS force field in particular appears to use rather unrealistic values for the bending force constant it is much too stiff. (d) Putting all of the (harmonic) force constants into the same units of kcal/deg 2 leads to the following comparison: Table 5-2: Force Constants, Relative Frequencies, and Amplitudes MM+ AMBER Bio+ OPLS K θ / kcal mol -1 deg -2.88.35.168.1828 ν/ν MM+ 1. 1.86 1.38 4.57 ( θ θ) range / deg 16.4 8.8 11.9 3.6 Vibrational frequencies are proportional to the square root of the force constants. The second row of the table shows frequencies relative to the MM+ (harmonic) frequency calculated according to 1/ 2 ν K FF θ, FF = ν MM + Kθ, MM + (e) For a harmonic potential, the range of angles sampled for an energy equal to RT is given by: 8

1/ 2 RT Δθ = 2 Kθ : where the factor of 2 is for ± θ θ ) angles. At room temperature (298 K) RT =.592 kcal /mol. ( This calculation gives the values listed in the last row of Table 5-2. (Use of the full MM+ potential gives approximately the same value of 16.) 6. (Extra Credit) Go through the same comparison as in problem #5 but for the O-H stretching term in the water potential. (a) The parameters read from the /Setup/Edit_Parameters command are as follows: Table 6-1: Parameters of the OH Stretch as Reported in HyperChem Force Field Atom Types Force Constant Eq. Bond Length Bond Dipole MM+ HO, O2 4.6.942 1.115 AMBER HW, OW 553.9572 Nil BIO+(CHARMM) HT, OT 45.9572 Nil OPLS HT, OT 6.9572 Nil As in the case of the bend potentials, the AMBER, BIO+(CHARMM), and OPLS force fields all use the simple quadratic form: E str ( r) r) 2 = Kr ( r (11) where K r is the stretching force constant (provided in units of kcal Å -2 ) and r is the natural length provided in Å. The MM+ force field uses the cubic function: 2 E r) = (143.88) 1 K ( r r ) {1 + 2.( r r ) switch } (34) str ( 2 r where switch is a function that switches off the cubic term before it can diverge to -. (For simplicity I ll ignore this term. It is described erroneously in the manual anyway.) (b,c) Plots are shown on the following page. Apart from the unrealistic behavior of the MM+ potential (when the switching function is neglected) all of the functions are qualitatively similar, again differing mainly quantitatively as a result of the differences in the bond stretching force constants. Again the OPLS force field seems to have a stretching force constant that is much too large, making the bond much stiffer than is realistic. 9

8 OPLS Bio+ 2. AMBER OPLS 6 1.5 E str / kcal mol -1 4 2 MM+ AMBER.4.6.8 1. 1.2 1.4 1.6 Bond Length r / Å E str / kcal mol -1 1. MM+ k B T.5 Bio+..85.9.95 1. 1.5 Bond Length r / Å (d,e) Using identical reasoning to that described in 5(d) and 5(e) I obtained the results listed in Table 6-2. Table 6-2: Force Constants, Relative Frequencies, and Amplitudes MM+ AMBER Bio+ OPLS K r / kcal mol -1 Å -2 331 553 45 6 ν/ν MM+ 1. 1.29 1.17 4.26 (r-r ) range / Å.8.7.7.2 Note that the total range of bond length variations at room T is less predicted to be than 1% in all force fields. 1