Chemistry 380.37
Dr. Jean M. Standard
Homework Problem Set 2 Solutions
1.
Indicate whether it would be appropriate to carry out a molecular mechanics calculation using the
MMFF94 force field to determine the minimized energy for the molecular systems listed below. If not,
suggest a more appropriate force field for the system.
a.
t-butylcyclohexane
t-butylcyclohexane is a hydrocarbon similar to those that we have studied using MMFF94, so this force
field would be appropriate for study of the molecule. In fact, any force field designed for organic
molecules would likely perform well for t-butylcyclohexane.
b.
ferrocene
Ferrocene has the formula Fe(C5H5)2. Since this is an organometallic molecule containing a transition metal
element, Fe, most typical organic/biological force fields like MMFF94 will likely not be able to handle it.
A general purpose force field such as UFF that includes parameters for metal atoms, or one specially
constructed to treat organometallic systems of this type, would be required.
c.
myoglobin
Myoglobin is an oxygen-binding protein which contains a heme unit (a porphyrin-type system with a bound
Fe2+ in the center). While the MMFF94 force field was constructed primarily for the study of organic and
biological molecules, one has to question whether or not it would be able to handle myoglobin due to the
iron center that it includes. A quick perusal of the atom types defined in the original MMFF94 paper
includes Fe2+ [Halgren, T. A. J. Comp. Chem. 1996, 17, 490-519], so perhaps MMFF94 could deal with
myoglobin. Some further investigation of the ability of MMFF94 to model biological systems containing
Fe2+ would be required. Alternately, a force field constructed with proteins in mind, such as Amber or
CHARMM, might be more appropriate for treating myoglobin.
d.
IF5
The molecule IF5 is a main group compound with somewhat unusual hypervalent bonding (i.e., one of
those compounds that in general chemistry provided an exception to the octet rule). Thus, it is not likely
that a organic/biological force field such as MMFF94 would be able to successfully model IF5. A
specialized force field would be required; UFF might be one such choice, if the training and validation sets
included compounds of a similar nature and UFF performed well for them.
2
1.
CONTINUED
e.
the water pentamer, (H2O)5
Since the water pentamer is formed through hydrogen bonding, the water pentamer would be an
appropriate molecular system to treat with molecular mechanics only if the force field is adequately
parameterized to deal well with hydrogen bonding. As we saw in Assignment 1 Part 2, MMFF94 describes
the water dimer reasonably well, it does not perform very well for the water trimer, and it does a reasonable
job at representing the water 20-mer. Most force fields developed for biological systems are able to
simulate hydrogen bonding because it is so important in biological systems. In addition, some force fields
even have explicit energy terms included to deal with hydrogen bonding directly. Thus, a force field such
as Amber might be a better choice for simulating the water pentamer.
f.
porphyrin
Porphyrin is a planar aromatic organic molecule containing only C, N, and H atoms. Thus, any typical
force field for organic/biological molecules, including MMFF94, likely could be used to perform molecular
mechanics calculations on porphyrin.
g.
NH(SiH3)2
** NOTE: I botched the molecular formula for this part in the copy of the homework that I
distributed. You could still answer the question because the formula involved the correct atoms.
You can download a PDF of the homework with the correct formula from the course web site.
The compound NH(SiH3)2 is disilylamine, the silicon analog of dimethylamine. While dimethylamine
clearly would be able to be modeled using the MMFF94 force field, disilylamine could only be modeled
with MMFF94 if suitable force field parameters are available. In class, we saw that MMFF94 was able to
model disilylether, (SiH3)2O, and thus we would expect that MMFF94 would be able to model disilylamine.
In addition, review of the atom types specified in the original MMFF94 paper [Halgren, T. A. J. Comp.
Chem. 1996, 17, 490-519] finds the silicon atom defined as atom type #19, with Si-H and Si-N bond
parameters definitely available.
h.
NH(GeH3)2
** NOTE: I botched the molecular formula for this part in the copy of the homework that I
distributed. You could still answer the question because the formula involved the correct atoms.
You can download a PDF of the homework with the correct formula from the course web site.
The compound NH(GeH3)2 is digermylamine, the germanium analog of dimethylamine. We saw in
problem 1g that the silicon analog is able to be modeled using MMFF94. In class, we saw that MMFF94
was not able to model disgermylether, (GeH3)2O, because force field parameters were not available, and
thus we would expect that MMFF94 would not be able to model digermylamine. In addition, review of the
atom types specified in the original MMFF94 paper [Halgren, T. A. J. Comp. Chem. 1996, 17, 490-519]
finds no parameters for germanium listed (and indeed, none of the subsequent papers refer to any
extensions which include germanium). Thus, we would have to turn to a general force field such as UFF in
order to model NH(GeH3)2, contigent upon the force field having been parameterized and tested for similar
germanium compounds.
2.
A researcher is interested in using molecular mechanics to study molecules containing amine functional
groups. The force field that the researcher is using has no energy term to describe the torsional motion of
amines, so the researcher constructs a new set of parameters to describe the amine torsional motion using
experimental data from the molecule CH3NH2. The researcher uses the following functional form for the
energy of torsional rotation of the C-N bond:
(
U torsion = k torsion ω − ω eq
)
2
,
where ktorsion is a constant, ω eq is the equilibrium value of the torsional angle, and ω is the torsional
angle. Is the form that the researcher
chose for the torsional energy a reasonable one? Explain. If not,
€
sketch a form that might be more appropriate.
€
€
The form for the torsional energy given in the equation above is a parabola (i.e., it is a harmonic potential). This
means that the energy would vary with angle according to the shape sketched in the figure below.
E
torsional angle
As can be seen from the parabolic shape of the graph, with this type of functional form there is only one
minimum possible. However, when considering the methylamine molecule and its Newman projections as
shown below, there is more than one local minimum (or stable structure) possible.
H
H
H
N
N
H
H
H
staggered
HH
HH
eclipsed
The staggered form of methylamine is lowest in energy and is degenerate for torsional angles of 60 (shown),
180, and 300 degrees. The eclipsed form is highest in energy and is degenerate for torsional angles of 0, 120
(shown), and 240 degrees. The simple parabolic functional form given in the question does not account for these
minima and maxima. A sketch of a function that does is given below.
3
4
2. CONTINUED
1.2
1
Energy
0.8
0.6
0.4
0.2
0
-0.2
0
50
100
150
200
250
300
350
Angle (degrees)
Notice that this plot is a simple cosine function with a three-fold angle, such as
U torsion =
V3
(1+ cos 3ω) ,
2
where V3 is a constant and ω is the torsional angle. This is very similar to the torsional potential for ethane.
€
It should be noted, though, that the original function proposed would be okay if the torsional barrier (that is, the
€ difference in torsional energy between the maxima and minima) is large. For rotation about single bonds,
though, the torsional barriers tend to be on the order of 1-3 kcal/mol, and thus a potential like the cosine function
that accounts for the three-fold rotation of methylamine would be required to properly describe the torsional
behavior of the molecule.
In most force fields, the interaction between two nonbonded atoms is described using a function called
the Lennard-Jones 6-12 potential. It has the general form
U nb = −
a
b
+ 12 ,
6
r
r
where a and a are constants which depend on the two interacting atoms and r is the distance between
them. For example, for two carbon
€ atoms interacting, a = 745 and b = 2524840 (these parameters yield
energy in kcal/mol if the distance is expressed in angstroms).
a.
Sketch the nonbonded energy as a function of distance for the case of two carbon atoms.
A sketch of the function is shown below. It was generated by calculating a few values of the energy for
different values of the distance between the atoms. Notice that for large distances, the energy approaches
zero as the interaction between the atoms dies off. This can be seen in the function by taking the limit as r
goes to infinity, then 1/r6 and 1/r12 both go to zero. Also, for short distances, the energy grows rapidly as
the atoms repel one another.
0.8
0.6
Energy (kcal/mol)
3.
0.4
0.2
0
2
b.
4
6
8
r (angstroms)
10
12
At approximately what distance does the minimum energy occur for this particular nonbonded
interaction?
From the graph, the energy is a minimum at a distance of about 4.3-4.5 angstroms. The approximate
answer is acceptable. To get the answer more accurately, the minimum occurs when the derivative equals
zero. Using the functional form for the nonbonded energy, this can be calculated exactly.
For two carbon atoms,
U nb = −
€
745
2524840
+
,
6
r
r12
5
6
3b. CONTINUED
Taking the derivative with respect to r yields
d U nb
4470
30298080
=
−
.
dr
r7
r13
We next have to set this equation equal to zero and solve for r.
€
4470
r
7
−
30298080
r 13
= 0
13
As long as r is not equal to zero, we can multiply both sides of the equation by r to get r into the
numerator:
4470 r 6 − 30298080 = 0.
6
Next, we can solve for r :
r 6 = 6778.09 .
Finally, taking the 6th root of both sides yields r at the minimum:
r = 4.350 Å.
c.
Which of the two terms in the expression for the nonbonded energy is responsible for the attractive
part of the interaction? Which term is responsible for the repulsive part of the interaction?
6
The attractive part of the interaction is the r term since it is negative and leads to the energy becoming
12
smaller as r is decreased. The repulsive part of the interaction is the r term since it leads to the energy
getting larger and larger as r is decreased.
4.
A computational chemist working at DrugsRUs, Inc. performed an energy minimization using the Amber
force field on the lactic acid molecule and found that the lowest energy conformer had an energy of
19.856 kcal/mol. Upon carrying out a literature search, the chemist found a journal article from 1998 in
which a study using the MMFF94 force field determined the global minimum for lactic acid to be 3.898
kcal/mol. Is there something wrong with the DrugsRUs chemist's work? Explain.
There is nothing necessarily wrong with the DrugsRUs chemist’s work. Notice that the literature study employs
the MMFF94 force field, while the DrugsRUs chemist used the Amber force field. Therefore, the total energies
of the lactic acid molecule calculated with the two different force fields cannot be directly compared.
Remember that only relative energies should be compared when different force fields are used.
What the DrugsRUs chemist should probably do is to obtain the energies of the first few lowest energy
conformations of lactic acid and make a more appropriate comparison of the relative energy differences and
structures of these results to the literature study.
7
5.
Consider the energies of the axial and equatorial forms of t-butylcyclohexane.
a.
Which form would you expect to be higher in energy, the axial or equatorial? Why?
The axial form of t-butylcyclohexane would be expected to have the highest energy. This is due to large
steric interactions between the t-butyl group and other hydrogens in axial positions in cyclohexane. In the
equatorial position, the steric interactions are much lower. This is illustrated in the figures below.
H3C
C H3
H3C
C7
H
H
C1
H
C5
C6
H
H
C3
H
H
C2
C4
H
H
H
H
axial t-butylcyclohexane
H
C1
H3C
H
H
H
C6
C7
H3C
C H3
C5
H
C3
H
C2
H
H
C4
H
H
H
equatorial t-butylcyclohexane
In the figure showing the axial form, the unfavorable steric interactions between the t-butyl group in the
axial position and the axial hydrogens in the molecule are indicated by the double arrows. In the equatorial
form, these are replaced only by lower energy interactions between hydrogens only in the axial positions.
8
5. CONTINUED
b.
Do molecular mechanics calculations agree with your qualitative expectations? Use the data below to
verify the relative energies of the two forms.
Energy of t-butylcyclohexane calculated using MMFF94 and Amber force fields (kcal/mol)
Force Field
Bonds
Angles
Torsions
Nonbonded
Total
Axial
MMFF94
Amber
Equatorial
MMFF94
Amber
1.517
0.732
4.281
4.434
5.458
1.837
6.625
4.303
18.313
11.305
1.394
0.584
1.969
1.465
3.755
0.311
5.892
3.989
13.312
6.350
The molecular mechanics calculations agree with the qualitative argument that t-butyl cyclohexane is more
stable in the equatorial form than the axial form. From the molecular mechanics calculations using the
MMFF94 force field, we see that the equatorial form has a total energy of 13.312 kcal/mol, 5.001 kcal/mol
lower in energy than the axial form with a total energy of 18.313 kcal/mol.
The results from the Amber force field are very similar when comparing the relative energies of the
equatorial and axial forms. In this case, the equatorial form has an energy of 6.350 kcal/mol, 4.955
kcal/mol lower than the axial form with an energy of 11.305 kcal/mol. From this data, both force fields
predict that the energy difference between the axial and equatorial forms of t-butylcyclohexane is about 5.0
kcal/mol, with the equatorial form being lower in energy as expected.
c.
From the force field data given in part b, what element of the force field shows the largest change
when comparing the axial and equatorial conformations? Why do you suppose this is?
For the MMFF94 and Amber force fields, the energy differences of specific components of the force field
between the high enery axial form and the lower energy equatorial form are as follows (in kcal/mol):
bonds
angles
torsions
nonbonded
MMFF94
0.123
2.312
1.703
0.733
Amber
0.148
2.969
1.526
0.314
The two force fields exhibit some variations in the components that are emphasized, but both show the
largest difference in the axial and equatorial forms comes from the angle bending term, followed by
torsions, and then nonbonded interactions. It makes sense that the bond angles show the greatest variation
because in order to alleviate some of the steric repulsion between the axial t-butyl group and the axial
hydrogens in the axial form of t-butyl cyclohexane, the C-C-C angles must expand. This means that they
are further from their optimal tetrahedral values of 109.47 degrees and this increases the energy in these
angles relative to the equatorial form.
For example, the C7-C1-C2 and C7-C1-C6 angles shown in the figures are 117.4 degrees in the axial form
compared to 113.4 degrees in the equatorial form (using the Amber force field). Similarly, the C1-C2-C3
and C1-C6-C5 angles are 114.9 degrees in the axial form compared to 111.1 degrees in the equatorial form
(again using the Amber force field).
9
6.
In class, we discussed the parameterization and testing of a simple force field for alkanes. The training
set of molecules included methane, ethane, and n-butane. In this problem, you will use the force field to
calculate the energy of propane. The force field parameters are given in the handout from Lewars.
The structure of propane is shown below. For the rest of the problem, the atom numbering scheme shown in the
figure will be employed.
The simple alkane force field that we constructed has the form
U = ∑
Us + bonds
∑
angles
U b + ∑
torsions
Ut + ∑
U nb .
nonbond
interactions
Note that we only considered uncharged alkanes, so there is no term included for electrostatic interactions. The
bond stretching terms include C-C and C-H single bonds only (note the factors of 1/2 missing in the definition
in Lewars):
2
UC−Cstretch = kC−C ( rC−C − rC−C,eq ) ,
2
UC−Hstretch = kC−H ( rC−H − rC−H,eq ) .
The angle bending terms include C-C-C angles and H-C-C angles (note the factors of 1/2 missing in the
definition of Lewars):
UC−C−Cbend = kC−C−C (θC−C−C − θC−C−C,eq )
2
U H −C−Cbend = kH −C−C (θ H −C−C − θ H −C−C,eq )
2
The only torsional motion considered in the alkane force field is for rotation of methyl groups about a central CC single bond:
UC−C−C−Ctorsion = k0 + k1 [1+ cosω ] + k2 [1+ cos2ω ] + k3 [1+ cos3ω ] + k4 [1+ cos 4ω ] ,
where ω is the C-C-C-C torsional angle.
Finally, the non-bonded interactions were treated with a united atom approximation for the methyl groups in
which only CH3/CH3 non-bonded interactions are considered. The form employed is a typical Lennard-Jones 612 potential:
6.
10
continued
(! σ $12 ! σ $6 +
UCH 3 /CH 3 = knb *# & − # & - ,
" r % -,
*)" r %
where knb and σ are parameters, and r is the distance between the carbons of the methyl groups.
The parameters for the stretching and bending terms are listed in Table 1. The torsional and non-bonded
parameters are listed in Table 2.
Table 1. Stretching and bending force constants and equilibrium bond lengths
req (Å)
ks (kJ mol–1Å–2)
C-C bond
C-H bond
1735
1934
1.538
1.083
kb (kJ mol–1deg–2)
θ eq (deg.)
0.110
0.093
112.5
110.7
C-C-C angle
H-C-C angle
Table 2. Torsional and Non-bonded Interaction Parameters
C-C-C-C Torsion
k0 (kJ/mol)
20.1
k1
k2
k3
k4
(kJ/mol)
–4.7
(kJ/mol)
1.91
(kJ/mol)
–7.75
(kJ/mol)
0.58
CH3/CH3 Interaction
knb (kJ/mol)
4.7
σ (Å)
3.85
Note that propane has two C-C bonds, eight C-H bonds, one C-C-C angle, 10 H-C-C angles, no C-C-C-C
torsions, and no CH3/CH3 nonbonded interactions (because the CH3 groups involved in the nonbonded
interaction cannot be part of the same bond or bond angle).
a.
Assume that the C-C bond distances are 1.519 Å and the C-H bond distances are all 1.095 Å.
Compute the bond stretching energy of propane in kcal/mol.
With 2 C-C bonds of equal length and 8 C-H bonds of equal length, the stretching energy is:
U s = 2 UC−Cstretch + 8 UC−Hstretch
2
= 2 kC−C ( rC−C − rC−C,eq ) + 8 kC−H ( rC−H − rC−H,eq )
(
)(
)
2
(
2
)(
)
2
U s = 2 1735 kJ mol–1Å –2 rC−C − 1.538 Å + 8 1934 kJ mol–1Å –2 rC−H − 1.083 Å .
11
6
a.
continued
Substituting the C-C and C-H distances, the stretching energy is:
(
= 2 (1735 kJ mol
2
2
)
)
)(
(
)(
) (1.519 − 1.538 Å) + 8 (1934 kJ mol Å ) (1.095 − 1.083 Å)
U s = 2 1735 kJ mol–1Å –2 rC−C − 1.538 Å + 8 1934 kJ mol–1Å –2 rC−H − 1.083 Å
–1
2
Å –2
–1
–2
2
= 1.25 + 2.23 kJ/mol
U s = 3.48 kJ/mol .
b.
The C-C-C bond angle is 111.7 degrees, the H-C-C bond angles for the end methyl groups are 110.3,
111.0, and 110.0 degrees, and the H-C-C bond angles of the middle CH2 group are 109.4 degrees.
Compute the angle bending of propane in kcal/mol.
The angle bending energy is:
U b = UC1−C 2−C8 + U H 3−C1−C 2 + U H 4−C1−C 2 + U H 5−C1−C 2 + U H 9−C8−C 2 + U H10−C8−C 2 + U H11−C8−C 2 + U H 6−C 2−C1 + U H 7−C 2−C1 + U H 6−C 2−C8 + U H 7−C 2−C8 2
= kb,C−C−C (θC1−C 2−C8 − θC−C−C,eq ) 2
+ kb,H −C−C (θ H 3−C1−C 2 − θ H −C−C,eq ) + kb,H −C−C (θ H 4−C1−C 2 − θ H −C−C,eq )
2
2
+ kb,H −C−C (θ H 5−C1−C 2 − θ H −C−C,eq ) + kb,H −C−C ( θ H 9−C8−C 2 − θ H −C−C,eq )
2
2
2
+ kb,H −C−C (θ H10−C8−C 2 − θ H −C−C,eq ) + kb,H −C−C (θ H11−C8−C 2 − θ H −C−C,eq ) 2
2
2
2
+ kb,H −C−C (θ H 6−C 2−C1 − θ H −C−C,eq ) + kb,H −C−C (θ H 7−C 2−C1 − θ H −C−C,eq ) + kb,H −C−C (θ H 6−C 2−C8 − θ H −C−C,eq ) + kb,H −C−C (θ H 7−C 2−C8 − θ H −C−C,eq ) .
(
2
)
U b = 0.110 kJ mol–1deg –2 (θC1−C 2−C8 −112.5 deg) (
2
)
(
)
2
)
) (θ
) (θ
) (θ
2
+ 0.093 kJ mol–1deg –2 (θ H 3−C1−C 2 −110.7 deg) + 0.093 kJ mol–1deg –2 (θ H 4−C1−C 2 −110.7 deg)
(
+ ( 0.093 kJ mol
+ ( 0.093 kJ mol
+ ( 0.093 kJ mol
)
) (θ
) (θ
) (θ
2
(
−110.7 deg) + ( 0.093 kJ mol
−110.7 deg) + ( 0.093 kJ mol
−110.7 deg) + ( 0.093 kJ mol
+ 0.093 kJ mol–1deg –2 (θ H 5−C1−C 2 −110.7 deg) + 0.093 kJ mol–1deg –2 ( θ H 9−C8−C 2 −110.7 deg)
–1
deg –2
–1
deg –2
–1
deg –2
2
–1
deg –2
H10−C8−C 2
H 6−C 2−C1
H 6−C 2−C8
2
–1
2
–1
deg –2
deg –2
2
−110.7 deg) H11−C8−C 2
H 7−C 2−C1 −110.7 deg
)
2
2
H 7−C 2−C8
−110.7 deg) .
Note that in this equation, the atom numbering scheme is shown in the image of propane given above; the
bending energy accounts for 1 C-C-C angle and 10 H-C-C angles of propane.
6
b.
12
continued
Substituting the values of the C-C-C (111.7 deg) and H-C-C angles (2 @ 110.3 deg, 2 @ 111.0 deg., 2 @
110.0 deg, and 4 @ 109.4 deg), the bending energy is:
(
2
)
U b = 0.110 kJ mol–1deg –2 (111.7 −112.5 deg) (
+ 2 ( 0.093 kJ mol
2
)
(
) (110.0 −110.7 deg) + 4 (0.093 kJ mol
–1
deg –2
2
–1
deg –2
2
)
) (109.4 −110.7 deg)
+ 2 0.093 kJ mol–1deg –2 (110.3−110.7 deg) + 2 0.093 kJ mol–1deg –2 (111.0 −110.7 deg)
2
= 0.07 + 0.03 + 0.02 + 0.09 + 0.63 kJ/mol
U b = 0.84 kJ/mol .
c.
Assuming that the torsional angles correspond to a perfectly staggered configuration for propane,
what is the torsional energy in kcal/mol?
The torsional energy is zero since there are no C-C-C-C torsional angles in propane. Whether the H-C-C-H
or H-C-C-C torsional angles are perfectly staggered or not (i.e., 60/300 or 180 degrees) does not matter for
predicting the enegy of propane using the simple alkane force field. Only C-C-C-C torsions were
considered; this is of course a failing in the force field, and a more comprehensive model would need to
include H-C-C-H and H-C-C-C torsions as well as the C-C-C-C torsion.
d.
Finally, put all the components together and compute the total energy of propane using the alkane
force field. [Note that the non-bonded contribution is not required in this case.]
Since there is no contribution in the case of propane from the torsion or non-bonded energies, the total
energy of propane consists of the contributions from stretching and bending only,
U = U s + U b
= 3.48 + 0.84 kJ/mol
U = 4.32 kJ/mol .
e.
Repeat calculations (a)-(d), assumming all the bond distances remain the same, the H-C-C angles
remain the same, but the C-C-C bond angle increases to 130.0 degrees. Compare your results.
If the bond distances are the same, the stretching energy is the same as part (a), 3.48 kJ/mol.
If the H-C-C angles remain the same, then the contribution to bending energy from the H-C-C angles does
not change, but the total bending energy changes due to an increase in the C-C-C angle to 130.0 deg,
(
)
2
U b = 0.110 kJ mol–1deg –2 (130.0 −112.5 deg) (
+ 2 ( 0.093 kJ mol
2
)
(
) (110.0 −110.7 deg) + 4 (0.093 kJ mol
–1
deg –2
2
= 34.64 + 0.03 + 0.02 + 0.09 + 0.63 kJ/mol
U b = 35.41 kJ/mol .
–1
deg –2
2
)
) (109.4 −110.7 deg)
+ 2 0.093 kJ mol–1deg –2 (110.3−110.7 deg) + 2 0.093 kJ mol–1deg –2 (111.0 −110.7 deg)
2
6
e.
13
continued
With no torsional or non-bonded contribution for propane, the total energy is again the sum of the
stretching and bending energies,
U = U s + U b
= 3.48 + 35.41 kJ/mol
U = 38.89 kJ/mol .
So we see that when we stretch the C-C-C angle of propane by almost 20 degrees (from 111.7 to 130.0), the
total energy increases by almost 35 kJ/mol (from 4 to 39 kJ/mol).