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Unformatted text preview: Chapter P3: Energies of Chemical Reactions
In addition to its geometry, the internal energy of a molecule is its most
fundamental property. Differences in internal energies between the products
and reactants of a reaction (reaction enthalpies) indicate whether the reaction
will be favorable (exothermic) or unfavorable (endothermic). Combined with
the entropy, which may be calculated from knowledge of equilibrium
geometry and vibrational frequencies, reaction enthalpy may be used to
provide the Gibbs energy.
Our first objective in this chapter will be to establish the ability of each of
the three classes of models that have now been introduced: Hartree-Fock,
B3LYP density functional and MP2 models, to reproduce the energies for
different types of chemical reactions. We will employ a very-large basis set
to explore the limits of each model and to attempt to separate the LCAO
approximation from the underlying treatment of electron correlation. While
we cannot hope to actually achieve these limits, by repeating the calculations
with a somewhat smaller basis set we can hope to put bounds on the
magnitude of the error caused by the LCAO approximation. Our second
objective will be to establish the performance of the same three classes of
methods with smaller, more practical basis sets. This will allow us to
provide much greater numbers and diversity of examples and to construct
meaningful problems. The third objective will be to define combinations of
theoretical models (“recipes”) that are able to provide accurate heats of
formation to be used in thermochemical comparisons. We will consider both
recipes with and without empirical parameters. The final objective of the
chapter will be to address the calculation of entropy and subsequent
determination of the Gibbs energy.
Problems throughout this chapter are intended to exemplify the types of
questions that may be posed to modern quantum chemical methods and the
types and quality of responses that can be expected. Some will need only a
few seconds or minutes of computer time, while others may require several
hours. 1 Total Energy vs. Heat of Formation
The heat of formation typically reported in an experimental study differs in
three ways from the total energy obtained from a quantum chemical
calculation. The obvious difference is that enthalpy and energy are not the
same, but are related through a pressure-volume term.
∆H = ∆E + ∆(PV) or at constant pressure ∆H = ∆E+ P∆V
However, except where comparisons involve experiments that have been
carried out under very high pressures, the energy and enthalpy of a reaction
can be safely assumed to be identical.
Reactions can be (and have been) carried out as a function of pressure and provide
information about differences in the volumes of reactants and products. A more
interesting involves extracting the volumes of transition states (relative to those of
reactants) from the pressure dependence of activation energies. This will be discussed in
Pressure Affects Reaction Enthalpy: Ring opening of cyclobutene to 1,3-butadiene is
exothermic by ~56 kJ/mol. Use the HF/6-31G* model to calculate equilibrium geometries for 1,3-butadiene and
cyclobutene. Based on space-filling models as a measure, which molecule takes up less
volume? If it is cyclobutene, what pressure would be required to reverse the direction of
the isomerism. Need information on units.
Instead of space-filling models, use surfaces that enclose 99% of the total electron density
as a measure of volume. Do your results change significantly? Another difference is that experimental measurements are always carried out
at finite temperature (typically at or near room temperature) whereas
quantum chemical calculations refer to systems at 0K. The change in
enthalpy from 0K to a finite temperature (T) is given by:
ΔH(T) = ΔHtr(T) + ΔHrot(T) +ΔHvib(T) + RT
ΔHtr(T) = 3/2 RT
ΔHrot(T) = 3/2 RT (RT for a linear molecule) 2 This requires the vibrational frequencies, νi. R is the gas constant, k is
Boltzmann’s constant, h is Planck’s constant and N is Avogadro’s number.
Were it actually possible to measure the enthalpy at 0K, it would still not
correspond to the enthalpy (energy) obtained from a quantum chemical
calculation. This is because the calculated energy calculated from a quantum
chemical model refers to a molecule resting at the bottom of a potential
energy well, whereas a measured enthalpy refers to a molecule in its lowestenergy (“zeroeth”) vibrational state. The difference, referred to as the zero-point-vibrational energy, is given by. The summation is carried out over the vibrational frequencies, υ. h is
Planck’s constant. Zero-point vibrational energies are quite large, but largely
cancel in a chemical reaction.
Zero-Point Energies Affect Reaction Energy: Use the HF/6-31G* model to calculate
equilibrium geometry and vibrational frequencies for acetonitrile, CH3CN, and its isomer,
methyl isocyanide, CH3NC. Evaluate the zero-point energy for each. Does its explicit
inclusion change in the energy of isomerism by more than 10%?
Repeat your calculations and analysis for propene and its isomer cyclopropane.
Temperature Affects Reaction Energy: Using the results from the calculations you
performed in the previous problem, evaluate the change in the energy of acetonitrile and
methyl isocyanide with change in temperature from 0 K to 298 K. Combine this with the
difference in zero-point energies for the two isomers (see previous problem). Does the
full correction (zero-point energy + temperature) change in the energy of isomerism by
more than 10%?
Repeat your calculations and analysis for propene and its isomer cyclopropane.
Mass Affects Reaction Energy: As detailed in Chapter xx, the Born-Oppenheimer
approximation eliminates nuclear mass from the Schrödinger equation. However, because
of differences in zero-point energies, reaction energies actually change with change in
nuclear mass (isotope).
The most common mass substitution is deuterium for hydrogen. Use the B3LYP/6-31G* 3 model to determine the difference in zero-point energy (and the difference in bond
energies) between H2 and D2. Is the change in bond energy likely to be noticed? Assume
that a change in bond energy of 5% is detectable.
Repeat the calculations and the analysis for the change from
molecule. 14 N to 15 N in nitrogen A trivial difference is one of units. Heats of formation are most commonly
reported in kJ/mol, whereas total energies are most commonly reported in
atomic units (hartrees).
1 hartree = 2625 kJ/mol
kJ/mol will be used throughout this text, replacing the more familiar but now defunct
kcal/mol (1 kcal/mol = 4.184 kJ/mol). The reason for the difference in units is simply a matter of convention and to
some extent convenience. Heats of formation are typically only a few tenths
of a hartree while total energies are typically tens of thousands to several
million kJ/mol. In their respective units, both lie in a more convenient range
from a few tens to a few thousands.
Finally, and perhaps most important, while both total energy and heat of
formation refer to the energies (heats) of specific chemical reactions, the
reactions are different. Heat of formation refers to a balanced chemical
reaction in which a molecule is converted to a set of standard products, each
corresponding to the most stable form of the element at room temperature.
The heat of formation of each standard is defined as zero. For example, the
heat of formation of ethylene is defined by the reaction.
C2H4 → 2C (graphite) + 2H2 (gas) Graphite and hydrogen molecule are the carbon and hydrogen standards,
respectively. Of course, the experimental measurement is not actually
carried out for this reaction, but more typically (but not necessarily) for a
combustion reaction (reaction with O2), for example, for ethylene:
C2H4 + 3O2 → 2CO2 + 2H2O
In some cases, combustion leads to products that cannot be fully characterized. The most
conspicuous case involves combustion of molecules incorporating silicon where
polymeric silicon oxide polymers (sand) are formed. The clever solution is to “burn” the
molecule in fluorine rather than oxygen, leading to gaseous SiF4 as a product. 4 Heats of formation may be either positive or negative quantities and their
(absolute) values will generally span a range of only a few hundred kJ/mol.
Molecules with positive heats of formation much greater than this are likely
to the thermodynamically unstable and not likely to “stick around” long
enough to be detected let alone characterized.
Heats of formation may not be obtained directly from quantum chemical
calculations simply because some of the standards are not isolated species on
which calculations may be performed. A suitable alternative is to use a
hypothetical reaction that splits a molecule into isolated nuclei (not atoms)
and electrons, for example, for ethylene:
C2H4 → 2C+6 + 4H+ + 16e – Each of the products (H+, C+6 and e-) contains but a single particle, meaning
that its energy is zero. Total energies, as the energies of such reactions are
termed, are very large negative numbers, (several tens of thousands to
several million kJ/mol), but only a few tens to a few thousands of hartrees.
There are two important points to be made. First, is it is straightforward to
obtain heats of formation indirectly from total energies (or vice versa), either
by using experimental atomic data or a mixture of experimental and
theoretical atomic data. In fact, combinations of theoretical models
(“recipes”) have been developed in order to provide accurate heats of
formation. These will be discussed later in this chapter. The second point is
more important. Either heats of formation or total energies are suitable for
calculations of the energies (heats) of mass-balanced chemical reactions.
Here, the “standards” cancel. 5 Sources and Quality of Experimental Thermochemical Data
Experimental heats of formation have been reported for approximately three
thousand compounds, a large fraction of which are hydrocarbons and
oxycarbons. Among the most extensive compilations is the NIST database,
freely available on line (http://webbook.nist.gov).
A snapshot of part of the NIST database for closed-shell neutral organic molecules has
been provided with Spartan Student, and heats of formation where available are
displayed in the Molecule Properties dialog. No attempt has been made to identify
suspect experimental values. While care has been taken to ensure the integrity of this collection, because
the data derive from a variety of experimental techniques and has been
assembled over many decades, heats of formation for individual entries vary
widely in quality. The most egregious source of error is that the structure is
incorrect, meaning that the reported heat does not correspond to the reported
structure. Because most of the compounds in the NIST database are fairly
simple and readily available commercially, it is likely that only a very few
structures are incorrectly assigned. More likely sources of error include
incomplete combustion and poorly characterized combustion products.
Hydrocarbons and oxycarbons present fewest problems as their combustion
leads only to carbon dioxide and water, the amounts of which may easily be
determined. However, combustion of molecules with other elements may
give rise to a complex mixture of products and greater uncertainty.
Despite their importance, heats of formation are not routinely determined (or
at least are not routinely reported) for new compounds. While combustion
experiments are straightforward and the results easily interpreted, accurate
measurements require considerable diligence and may need to be repeated
several times to establish useful error limits. More relevant, combustion
experiments may require (and destroy) significant quantities of compound.
Very few synthetic chemists are willing to part with hundreds of mg (a huge
amount by modern standards) of a compound that they have just spent weeks
or months preparing.
One alternative source of experimental thermochemical data follows from
measurement of equilibrium constants. The best examples of this are for ionmolecule reactions carried out in the gas phase using ion cyclotron
resonance spectroscopy and related techniques. Most important among these
6 are protonation/deprotonation reactions, leading to gas-phase basicities and
acidities. For example, it is possible to accurately establish the energy of
protonation of a base, B, relative to that of a standard base, Bstandard, by
determining the amounts of the two protonated bases present at equilibrium.
BH+ + Bstandard B + BstandardH+
Equilibrium measurements require that the proton affinities of the two bases
be similar (within ~10 kJ/mol). In practice, a proton affinity “ladder” of base
strengths (involving hundreds of individual compounds) has been
constructed, allowing an appropriate standard to be selected over a range of
several hundred kJ/mol. The advantages to this kind of approach are that it is
quite accurate and requires only miniscule amounts of materials. The
obvious disadvantage is that it applies to ion-molecule reactions only. The
NIST database contains a large collection of heats of formation for both
negative and positive ions based on such measurements.
Another source of experimental thermochemical data on positive ions derives from
measurements of ionization potentials and electron affinities.
M + e- M+ + 2eM + e- MThe former are widely available whereas the latter are less common. 7 Reaction Energies and Boltzmann Distributions
As commented previously, the energy (heat) of a balanced chemical reaction
may be obtained using either heats of formation or total energies.
∆E(reaction) = Eproduct 1 + Eproduct 2 + … - Ereactant 1 - Ereactant 2 - … A negative ∆E indicates an exothermic (energetically favorable) reaction,
while a positive ∆E indicates an endothermic (unfavorable) reaction.
An important special case is where there is only one reactant molecule and
one product molecule. Here, the reactants and products are isomers, and the
reaction energy accounts directly for their difference in their energies.
∆E(reaction) = ∆E(isomer) = Eisomer 2 – Eisomer 1 A negative ∆E(reaction) means that isomer 2 is more stable than isomer 1,
and that the reaction will proceed as written (it is exothermic).
The equilibrium composition of a mixture of isomers is given by the
Isomer 1 Isomer 2 % Isomer i = Isomer 3 ... 100 exp (-EIsomer i /kT) ! exp (-1060 E Isomer j ) j ΔEisomer i is the energy of isomer i in hartrees relative to the energy of the
lowest-energy isomer. T is the temperature (in K) and k is the Boltzmann
constant. In the case where the equilibrium is between two isomers,
Isomer 1 Isomer 2 and assuming room temperature (298 K) the expression becomes.
[ Isomer 1 ]
= exp [-1060 (Eisomer1 – Eisomer2 )]
[ Isomer 2 ] HCN vs. HNC: Under “normal” (laboratory) conditions, hydrogen isocyanide (HNC) is
in equilibrium with its more stable isomer, hydrogen cyanide (HCN). According to the
HF/6-31G* model, what is the room-temperature Boltzmann distribution of isomers?
Assuming that 5% as the lower bound for detection, what is the lowest temperature that
would be needed to see the minor isomer? 8 Radioastronomy confirms that both hydrogen cyanide and hydrogen isocyanide are
present in interstellar clouds. The interesting observation is that they occur in similar
amounts. Speculate why.
Protonation of Dimethylamine and Aziridine: Molecules contained the high vacuum
environment of an ion cyclotron resonance spectrometer can be protonated and ionmolecule equilibria established. The relative abundance of the ions involved in the
equilibrium can be measured which, together with knowledge of the relative amounts of
the neutral molecules, allows highly accurate determination of relative proton affinities.
Use the HF/6-31G* model to calculate the equilibrium geometries of the four molecules
involved in the proton-transfer equilibrium between dimethylamine and aziridine.
dimethylamine-H+ + aziridine = dimethylamine + aziridine-H+
Which molecule has the higher proton affinity and by how much? Assuming a 1:1
mixture of the two amines, and that the limit of detection of the “minor” ion is 5%, is it
possible to directly establish the relative proton affinities of dimethylamine and
aziridine? 9 Classifying Chemical Reactions
Chemical reactions may be conveniently be divided into one of three
categories depending on the extent that overall bonding is maintained.
Reactions that Change the Total Number of Electron Pairs
The most common reactions of this type are homolytic bond dissociation
reactions. Hartree-Fock models are expected to provide bond energies
that are too large. To see why this is so, consider bond dissociation in H2.
H–H → H• +H• While the Hartree-Fock energy for the product (two hydrogen atoms) is
exact (each contains only a single electron), that for the reactant
(hydrogen molecule) is too high. This means that the bond dissociation
reaction will not be sufficiently endothermic. In reality, homolytic bond
dissociation reactions are not very important as they are only infrequently
encountered. However, they serve as “absolute standards” with which to
judge the performance of different theoretical models.
An important possible “exception” involves comparisons between reactants and
transition states (activation energies), whic may also lead to a change in the number
of electron pairs. These will be discussed in Chapter P4. Reactions that Conserve the Total Number of Electron Pairs
This category separates reaction into two classes, depending on whether
or not total bond count in addition to total electron-pair count is
maintained. Protonation of trimethylamine (defining its absolute proton
affinity) and association of trifluoroborane and carbon monoxide (leading
to trifluoroborane carbonyl) are examples of the latter.
(CH3)3N: + H+ (CH3)3NH+
BF3 + :CO BF3CO
Here the product contains one more bond than the reactants but the same
number of electron pairs. Examples of reactions that conserve bonds as
well as electron pairs are more common. Among the most important are
comparisons of structural isomers, for example, comparison of allene and
propyne. Here, the reactant incorporates two double bonds and the
product one single and one triple bond.
CH2=C=CH2 → CH3CH≡CH2
10 Some structural isomer comparisons, for example, that between 1-butyne and 2butyne, not only maintain overall bond count, but also maintain the numbers of
individual bond types involving (in this case, two single bonds and one triple bond).
HC CCH2CH3 → CH3C≡CCH3
In this case, only the nature of the “atomic hybrids” (environments) are altered. One
of the carbon-carbon single bonds in 1-butyne made from sp and sp3 hybrids and the
other from two sp3 hybrids, whereas both of the carbon-carbon bonds in 2-butyne are
made from sp and sp3 hybrids. This and other reactions like it will be put into a third
class (see next section). Addition reactions such as cycloaddition of 1,3-butadiene and ethylene to
form cyclohexene (a Diels-Alder reaction) and of addition of molecular
bromine to cyclohexene to yield trans-1,2-dibromocyclohexane, also
conserve total bond count (but not individual bond counts). In the first
case, there is a loss of two π bonds but a gain of two new carbon-carbon
σ bonds, while in the second case a π bond and a Br-Br bond are traded
for two C-Br bonds.
+ Reactions that conserve the total number of electron pairs would be
expected to benefit from error cancellation to a greater extent than
reactions that do not. Because of this, it might be anticipated that their
energies will be described even by quantum chemical models that do not
properly take electron correlation into account, specifically Hartree-Fock
models. We will explore this later in the chapter.
Reactions that Conserve Individual Bond Counts
Processes that conserve not only the total number of bonds and electron
pairs but also the numbers of each kind of chemical bond (and each kind
of non-bonded lone pair) represent another very important and very
diverse class of chemical reactions. Some structural isomer comparisons,
for example, that between 1-butyne and 2-butyne mentioned above,
formally fit this category, as do all comparisons of regio and
stereoisomers. For example, comparison of isobutene and cis and trans2-butene. The latter two are stereoisomers and both are regioisomers of
isobutene. 11 H3 C H
C H3 C H3C C CH3
C H H H3 C C H
C H H C
CH3 Also of note are reactions that compare the energies of protonation
(proton affinities) of closely-related molecules, for example, that
compare the proton affinities of ammonia and trimethylamine. This is a
quantity that can be accurately measured although not necessarily
directly (see previous discussion).
(CH3)3NH+ + NH3 → (CH3)3N + NH4+ In all of these cases, comparisons are between molecules with the same
number of each kind of chemical bond and of each kind of electron pair.
Only the local environment changes. It might be expected that these
kinds of reaction would benefit from error cancellation to a greater
extent than the previous types of reactions.
The obvious suspicion is that it will be “easier” to calculate relative
quantites, for example, the relative proton affinities of ammonia and
methylamine, than it is to calculate absolute quantities, for example, the
absolute proton affinity of ammonia. This will be tested in the next
section. At the outset it should be pointed out that while absolute
energies will certainly be required in some instances, in many others
relative energies will suffice. 12 Limiting Behavior of Hartree-Fock, Density Functional and MP2
Models for Reaction Energies
We first set out to establish or at least to estimate the limiting behavior of
Hartree-Fock, B3LYP and MP2 models with regard to reaction energies
using examples of each of the types processes discussed in the previous
section. This will allow us to separate the effects of the LCAO
approximation from effects arising from replacement of the exact manyelectron wavefunction by an approximate Hartree-Fock, density functional
or MP2 wavefunction. As commented in the introduction to this chapter and
in Chapter P2, it is not possible to actually reach the limit for either
Hartree-Fock and B3LYP models, nor is it possible to establish the MP2
limit. However, it should be possible to use a sufficiently large basis set such
that the addition of further functions to the basis set will have only a small
effect on calculated reaction energy. As for geometry comparisons, the ccpVQZ basis set will be employed. This should be sufficiently large to be
able to reflect the properties of the limit, but small enough to be applied to
the molecules in the comparisons provided. Of more practical concern, ccpVQZ is about as large a basis set that can be applied to the molecules used
in the comparisons that follow.
The results provided in this section based on use of the cc-pVQZ basis set
will be paralleled by those obtained using the slightly smaller cc-pVTZ basis
set. This will allow us to quantify the extent to which reactions energies
based on the cc-pVQZ basis set actually approach limiting values.
The first comparison involves AH bond energies in molecular hydrogen and
in one-heavy-atom hydrides and AB bond energies in two-heavy-atom
AH A +H AB A +B Reference bond energies have been obtained from the G3(MP2) recipe (see
discussion later in this chapter) rather than from experiment. This provides a
more uniform baseline and allows comparisons to be extended to molecules
where experimental bond energies are not accurately known (if known at
all). G3(MP2) heats of formation have already been corrected for zero-point
vibrational energy and for finite temperature (298 K). That is, they
correspond to the experimental data most commonly found in the chemical
literature (including that in the NIST database). 13 These corrections are based on vibrational frequencies obtained from the HF/6-31G*
model scaled by 0.9 to take account of a known systematic error in these frequencies. This means that energies from the Hartree-Fock, B3LYP and MP2
calculations also need to be corrected. Signed differences between
B3LYP/cc-pVQZ and MP2/cc-pVQZ and G3(MP2) AH bond energies are
shown in Figure P3-1 and between G3(MP2) AB bond energies are shown
in Figure P3-2.
Excel spreadsheets containing AH and AB bond dissociation energies discussed in this
section (including Hartree-Fock bond energies) are provided on the CD-ROM
accompanying this text (limiting AH bond energies and limiting AB bond energies,
respectively). The Hartree-Fock cc-pVQZ model provides a very poor account of both AH
and AB bond energies and individual errors have not been included in the
figures. As expected, Hartree-Fock bond energies are much larger than
G3(MP2) values, with mean absolute deviations of 125 and 160 kJ/mol for
AH and AB bond energies, respectively. Bond energies calculated using the
smaller cc-pVTZ basis set (not provided in the figures but available in the
Excel spreadsheet) are nearly identical with those, confirming that the
problem is the Hartree-Fock approximation and not the use of a finite basis
The B3LYP/cc-pVQZ model provides a far better account. Mean absolute
errors are 8 and 17 kJ/mol for AH and AB bond energies, respectively.
Individual bond energies are always smaller than G3(MP2) values, and
usually within 10 kJ/mol of G3(MP2) values for AH bond energies and 20
kJ/mol for AB bond energies. The B3LYP/cc-pVTZ model yields bond
energies that differ from the corresponding cc-pVQZ values by only 1-2
kJ/mol, although the difference is 6 kJ/mol for the OH bond energy in water
and 3-4 kJ/mol for some bonds involving two heteroatoms. This suggests
that B3LYP models are more sensitive to basis set than Hartree-Fock
models. Still, most of the discrepancy between B3LYP and G3(MP2) bond
energies does not appear to be due to limitations in the basis set.
Individual bond energies from the MP2/cc-pVQZ model show much wider
variation from G3(MP2) values than those from the corresponding B3LYP
model, although mean absolute errors are virtually identical (9 kJ/mol for
AH bond energies and 17 kJ/mol for AB bond energies). These range from
25 kJ/mol too large for the bond energy in hydrogen, to 35 kJ/mol too small 14 Figure P3-1: Signed Deviations Between B3LYP/cc-pVQZ (left) and MP2/cc-pvQZ
(right) and G3(MP2) AH Bond Energies (kJ/mol) Figure P3-2: Signed Deviations Between B3LYP/cc-pVQZ (left) and MP2/cc-pvQZ
(right) and G3(MP2) AB Bond Energies (kJ/mol) 15 for the OO bond energy in hydrogen peroxide. In part, the problem may
reside with the basis set. Bond energies from the MP2/cc-pVQZ model
differ by as much as 18 kJ/mol (in Cl2) from those obtained from the
model, and differences on the order of 10 kJ/mol are common. This result,
while disturbing, is not unexpected. MP2 models (unlike Hartree-Fock and
density functional models) directly use “excited-state” wavefunctions which
involve both higher-order and more diffuse component than required in the
In summary, “limiting” Hartree-Fock models fail to provide an acceptable
account of bond dissociation energies. The corresponding B3LYP and MP2
models perform much better, although sizable deviations are noted for
individual molecules, in particular, for MP2 models. Bond energies obtained
from both Hartree-Fock and B3LYP models do not change significantly in
moving from the cc-pVQZ to the smaller cc-pVTZ basis set. On the other
hand, large changes are noted for some molecules between the
corresponding MP2 models.
The second comparison (Table P3-1) involves energy differences among
structural isomers. Here, the total number of electron pairs is conserved but
the numbers of individual bond types are not maintained. Both experimental
data and data from the G3(MP2) thermochemical recipe have been used as
references. “Limiting” Hartree-Fock, B3LYP and MP2 isomer energies have
been corrected for zero-point energy and finite temperature in the same way
as G3(MP2) energy differences.
Both the mean absolute error of Hartree-Fock isomer energies from
experimental enthalpies and the mean absolute deviation from G3(MP2)
values are 11 kJ/mol, much less than the errors seen previously for bond
energy comparisons. Note, however, that differences in isomer energies are
much smaller than bond energies. The largest individual error from
experiment is 33 kJ/mol, and involves comparison of 1,3-butadiene with its
highly strained isomer, bicyclo[1.1.0]butane. The corresponding deviation
from G3(MP2) is 21 kJ/mol. Other large errors (from experiment) involve
comparisons of propyne and cyclopropene and 1,3-butadiene and
cyclobutene (both 20 kJ/mol). Other large deviations (from G(MP2)) involve
comparisons of acetaldehyde and oxirane (17 kJ/mol) and acetonitrile and
methyl isocyanide (16/kJ/mol) and methyl ethyl ketone and 2,3-dihydrofuran
(16 kJ/mol). As with bond dissociation energies, isomer energy differences 16 from Hartree-Fock calculations with the cc-pVTZ basis set (not provided in
the table) are nearly identical to those with the cc-pVQZ basis set.
Table P3-1: Errors in “Limiting” Hartree-Fock, B3LYP and MP2 Energies of
Structural Isomers (kJ/mol)
propyne B3LYP MP2 G3(MP2) Expt. isomer
allene 7 -9 18 1 7 cyclopropene 113 99 99 100 93 cyclopropane 43 40 19 38 29 2-butyne 32 37 21 39 36 cyclobutene 68 67 40 56 48 bicyclo[1.1.0]butane 141 131 90 120 108 cyclobutane 54 53 35 47 46 1,4-pentadiene 55 56 83 67 70 methyl isocyanide 84 99 113 100 88 vinyl alcohol 52 43 44 41 43 oxirane 132 121 110 115 118 ethanol dimethyl ether 44 44 54 50 51 acetic acid methyl formate 70 68 75 70 75 methyl vinyl ketone cyclobutanone 32 27 11 22 23 2-hydroxy-1,3-butadiene 52 41 38 38 2,3-dihydrofuran 51 45 35 43 divinyl ether 104 91 105 95 102 mean absolute error 11 9 (10) 5 - mean absolute deviation from G3(MP2) 11 5 (12) - 5 propene
acetaldehyde 17 An Excel spreadsheet containing structural isomer energies for Hartree-Fock, B3LYP and
MP2 models with both cc-pVQZ and cc-pVTZ basis sets is provided on the CD-ROM
accompanying this text (limiting structural isomer energies). The B3LYP model offers some improvement over the corresponding
Hartree-Fock model. The mean absolute error from experiment is reduced to
9 kJ/mol and the mean absolute deviation from G3(MP2) is reduced to 5
kJ/mol. The largest individual deviation error from experiment is reduced to
9 kJ/mol (for comparison of 1,3-butadiene and bicyclo[1.1.0]butane). The
largest individual deviations from G3(MP2) are 11 kJ/mol (for ). Consistent
with previous results for bond energies, isomer energies change only slightly
in moving to the smaller cc-pVTZ basis set (data are not provided in the
The “limiting” MP2 model shows a mean absolute error with experimental
results of 10 kJ/mol, similar to that for the corresponding B3LYP and MP2
models. However, the largest individual error (comparison of 1,3-butadiene
and bicycle[1.1.0] butane) is larger (30 kJ/mol). The mean absolute
deviation with G3(MP2) is 12 kJ/mol, somewhat greater than that for the
other two models.
The final comparison (Table P3-2) involves regio and stereoisomers and
exemplifes reactions in which both the total number of electron pairs and the
numbers of individual bond types are conserved. As with the previous
comparisons, the reference is to G3(MP2) energy differences, and the
“limiting” Hartree-Fock, B3LYP and MP2 isomer energies have been
corrected for zero-point energy and finite temperature in the same way as
G3(MP2) energy differences.
An Excel spreadsheet containing region and stereoisomer energies for Hartree-Fock,
B3LYP and MP2 models with both cc-pVQZ and cc-pVTZ basis sets is provided on the
CD-ROM accompanying this text (limiting regio and stereoisomer energies). The “limiting” Hartree-Fock model provides a good account of differences
in isomer energies. The mean deviation from G3(MP2) energy differences is
5 kJ/mol, and with a single exception (comparison of cyclobutene and
methylenecyclopropane) individual deviations are 7 kJ/mol or less. This is
not unexpected. Regio and stereoisomers are more closely related to each
other than structural isomers, and energy comparisons should benefit from 18 and cancellation of errors. Results from Hartree-Fock calculations with the
smaller cc-pVTZ basis set (not provided in the table) are nearly the same.
Table P3-2: Errors in “Limiting” Hartree-Fock, B3LYP and MP2 Energies of Regio and
1,3-butadiene B3LYP MP2 G3(MP2) Expt. isomer
1,2-butadiene 55 44 51 51 53 1-butyne 24 27 21 19 20 methylenecyclopropane 26 19 33 40 44 trans-2-butene 2 1 6 6 7 cis-2-butene 9 7 11 11 10 1-butene 13 15 18 16 17 methylenecyclobutane 87 80 90 87 86 1,4-pentadiene 21 29 31 30 30 1,1-dimethylallene 53 44 52 53 53 1,3-dimethylallene 55 45 58 57 1,2-pentadiene 62 55 66 65 65 2-methyl-2-propenal 7 4 1 3 9 trans-2-butenal 13 10 13 6 5 mean absolute error 4 8 (3) 2 - mean absolute deviation from G3(MP2) 4 7 (3) - 2 2-butyne
2-methyl-1,3-butadiene methyl vinyl ketone 19 In terms of mean absolute deviation from G3, the B3LYP/cc-pVQZ model is
slightly inferior to the corresponding Hartree-Fock model for region and
stereoisomer energy comparisons, although the difference is not large. This
suggests that errors inherent to Hartree-Fock theory in large part cancel
where reactants and products have the same number of bonds and electron
pairs, and differ only in detailed environment.
MP2 20 Practical Hartree-Fock and Density Functional Models for Reaction
Except for very small molecules, Hartree-Fock, B3LYP and especially MP2
models with very large basis sets such as cc-pVTZ and cc-pVQZ are not
practical for calculation of the energies for any but reactions involving very
small molecules. While this will slowly change with improvements in
computer speed and storage capabilities, at the present time these basis sets
are primarily of value in judging the limits and ultimately the quality of the
underlying (Hartree-Fock and B3LYP and MP2) models. Here we examine
the performance of two smaller basis sets, 6-31G* and 6-311+G**, both of
which are presently easily applicable to much larger molecules (with
molecular weights approaching 500 amu). The primary focus will be to
establish the ability of the models to reproduce G3(MP2) relative energies to
“chemical accuracy” that is, within 8 kJ/mol.
A summary of mean absolute errors for collections of AH and AB bond
dissociation energies are provided in Table P3-3. These include HartreeFock, B3LYP and MP2 models with the 6-31G* and 6-311+G** basis sets
as well as with the cc-pVTZ and cc-pVQZ basis sets previously examined.
An Excel spreadsheet containing AH and AB bond dissociation energies for HartreeFock, B3LYP and MP2 models with 6-31G*, 6-311+G**, cc-pVQZ and cc-pVTZ basis
sets is provided on the CD-ROM accompanying this text (AH and AB bond dissociation
energies). Hartree-Fock models lead to unacceptable results for AH and AB bond
energies, irrespective of basis set. They should not be employed for this
The B3LYP/6-311+G** model yields similar AH bond energies (and a
similar mean absolute error) to that of the corresponding model with the ccpVQZ basis set. The MP2 model is more sensitive to basis set. The mean
absolute error in AH bond energies obtained from the MP2/6-311+G**
model is three times larger than that from the corresponding model with the
cc-pVQZ basis set. This result could have been anticipated by previously
noted differences between MP2 models with the cc-pVQZ and cc-pVTZ
basis sets (see Table P3-1). The 6-31G* basis set provides a much poorer
account of AH bond energies for both B3LYP and MP2 models.
In terms of mean absolute error, bond energies from the B3LYP model with
the 6-31G* and cc-pVQZ basis sets are comparable, while those obtained
using the 6-311+G** basis set are not as good.
21 Table P3-3: Summary of Deviations from G3(MP2) of Hartree-Fock, B3LYP and
MP2 AH and AB Bond Dissociation Energies (kJ/mol)
Basis Set AH AB B3LYP MP2 AH AB AH AB 6-31G* 20 15 54 20 6-311+G** 10 27 24 21 8 17 cc-pVTZ
cc-pVQZ 125 160 22 9 17 There are situations where the 6-311+G** basis set may not be adequate to provide an
accurate account of isomer energies. A particularly simple example is the identity of the
molecule resulting from dimerization of chlorine oxide, ClO. It is an important example
because the observed (high) concentration of ClO in the stratosphere above Antarctica as
a function of time of year corrlelates with the observed (low) concentrations of ozone, O3.
Current thinking involves a catalytic photochemical mechanism that starts with (ClO)
dimer formation and eventually loss of atomic chlorine. This in turn reacts with ozone
leading to molecular oxygen and chlorine oxide.
ClO + ClO (ClO)2
(ClO)2 + hυ Cl + ClOO
ClOO Cl + O2
Cl + O3 ClO + O2
The obvious choice for the structure of ClO dimer is chlorine peroxide, ClOOCl, but
“limiting” (cc-pVQZ basis set) B3LYP calculations show that chloryl chloride, ClClO2, is
only 19 kJ/mol higher in energy. The fact that analogous calculations with smaller basis
sets show progressively larger isomer differences (47 kJ/mol from the cc-pVTZ basis set
and 132 kJ/mol from the 6-311+G** basis set) strongly suggest that the separation is
even smaller (and perhaps actually favoring chlorine perchlorate. This would have
serious impact on the proposed ozone destruction mechanism.
Relative CH Bond Dissociation Energies in Hydrocarbons: CH bond energies in
acetylene, benzene and ethylene are known to be 113, 25 and 17 kJ/mol smaller than that
in ethane, whereas that in ethane is known to be 25 kJ/mol larger. Given this knowledge,
which bond in propyne is more likely to break, that on C1 (the alkyne) or on C3 (the
methyl group)? Which bond in propene is most likely to break, those on C1 or C2 (the
alkene) or on C3 (the methyl group)? Use calculations from the B3LYP/6-31G* model to
back up your answers.
Use the B3LYP/6-31G* model to obtain equilibrium geometries for allene, H2C=C=CH2,
and the radical resulting from CH bond dissociation. Calculate the bond dissociation
energy relative to that of methane (you will need to do calculations on methane and
methyl radical). Is this the result you expect based on the experimental result for bond
dissociation in ethylene? (You can confirm this result by performing calculations on
ethylene and vinyl radical.) If it is not, provide an explanation as to why not.
Chlorine Nitrate: Chlorine nitrate, ClONO2, may play an active role in the balance
ozone in the upper atmosphere. Specifically, the molecule may serve as a sink for NO2
and ClO free radicals that are known to react with ozone and destroy ozone.
ClO + NO2 ClONO2 23 Use the B3LYP/6-311+G** model to obtain equilibrium geometries for the two free
radicals as well as chlorine nitrate. Is radical recombination sufficiently exothermic to
make chlorine nitrate an effective radical sink? Elaborate.
Chlorine peroxynitrite (ClOONO) is a plausible isomer of chlorine nitrate. Were it in
equilibrium with chlorine nitrite, it might undergo cleavage of the OO bond leading back
to the initial radicals. Obtain the equilibrium geometry of chlorine peroxynitrite using the
B3LYP/6-311+G** model, and calculate the room-temperature Boltzmann distribution
between the two isomers. Is chlorine peroxynitrite likely to be involved? Elaborate.
Carbon-Fluorine Bond Stengths in Fluoromethanes: Carbon-fluorine bond lengths in
fluoromethanes, CFnH4-n (n=1-4), decrease with increasing number of fluorines, from
1.38Ǻ in fluoromethane to 1.32Ǻ in tetrafluoromethane. Does increase in the number of
fluorines also lead to increase in CF bond energies? Use the B3LYP/6-31G* model to
evaluate energies for each of the of the reactions below.
CH3F2 → CH3. + F.
CH2F2 → CH2F. + F.
CF3H → CHF2. + F.
CF4 → CF3. + F. Do CF bond energies in the fluoromethanes parallel CF bond distances?
Repeat your calculations and analysis for the analogous fluorosilanes, SiFnH4-n (n=1-4).
Repeat your calculations and analyses on both carbon and silicon compounds substituted
with chlorine instead of fluorine, CClnH4-n and SiClnH4-n (n=1-4).
Singlet and Triplet Carbenes: Molecules incorporating divalent carbon are referred to
as carbenes or methylenes. The parent compound, CH2 (methylene), is known to possess
a triplet ground state, with one unpaired electron residing in an in-plane σ orbital and the
other in an out-of-plane π orbital. The lowest-energy singlet state (with both electrons in
the σ orbital) is known experimentally to be approximately 42 kJ/mol higher in energy.
Because triplet methylene and other triplet carbenes have one fewer electron pair than the
corresponding singlets, Hartree-Fock models will always bias in favor the former. Thus,
the (estimated) Hartree-Fock limiting (cc-pVQZ basis set) energy difference for triplet
and singlet methylene (triplet favored) is 118 kJ/mol. B3LYP and MP2 models show no
such bias. Limiting (cc-pVQZ basis set) singlet-triplet energy splittings are 48 and 60
kJ/mol (in favor of the triplet). These two models appear to correctly assign the ground
state, and (in the very few cases for which experimental data exist) provide a good
account of singlet-triplet energy differences.
Use the B3LYP/6-31G* model to obtain equilibrium geometries of both singlet and
triplet states of methylene, difluoromethylene (CF2), dichloromethylene (CCl2) and
dibromomethylene (CBr2). Adjust the calculated singlet-triplet energy differences for
CF2, CCl2 and CBr2 to account for the error in the singlet-triplet difference for CH2 and
assign the ground state for each. Rationalize any change in preferred ground state
(relative to the parent compound) that you uncover. 24 Singlet and Triplet Cyclobutadiene: While cyclobutadiene, C4H4, is a very short-lived
molecule, the cyclobutadienyl ligand is common throughout organometallic chemistry.
Here, the coordinated metal may change the number of π electrons from 4 (meaning that
cyclobutadiene is formally an antiaromatic molecule) to a lesser or greater number. Does
cyclobutadiene itself possess a singlet or triplet ground state? To tell, obtain equilibrium
geometries for both singlet and triplet cyclobutadiene using the B3LYP/6-311+G**
model, starting from structures that are not square. Describe the geometries. Is either or
both square? If not, provide a rational as to why not?
Calculate the singlet-triplet energy difference in cyclobutadiene, and correct it to account
for the error in the corresponding difference in methylene (experimentally the triplet is
favored by 42 kJ/mol). Which state is favored and by how much?
Cyclopropylidene and Tropylidene Use the B3LYP/6-31G* model to obtain equilibrium geometries for both singlet and
triplet state of cyclopropylidene. Correct the calculated singlet-triplet energy difference to
account for the error in the corresponding difference in methylene (experimentally the
triplet is favored by 42 kJ/mol). Is the ground state different from that for methylene?
Examine both the carbon-carbon bond distances in singlet cyclopropylidene as well as its
highest-occupied molecular orbital (use cyclopropene as a reference). Speculate on the
cause behind the state preference
Repeat the calculations and analysis for tropylidene. Use cycloheptatriene as a reference
and locate the molecular orbitals in the carbene that correspond to the three occupied π
orbitals in cycloheptatriene.
Dissociation of Krypton Difluoride: While numerous compounds of xenon are known,
the only neutral krypton compound to be reported is the difluoride, XeF2. Is such a
species thermodynamically stable with regard to dissociation into xenon atom and
fluorine molecule? To tell, use the B3LYP/6-31G* model to obtain equilibrium structures
of XeF2, F2 and (the energy of) Xe.
Sulfur-Sulfur Linkages in Proteins: Nearby cysteine residues in proteins may form
sulfur-sulfur linkages. In so doing they impose geometrical constraints on the protein
which in turn affects its secondary structure. How strong are sulfur-sulfur bonds? To tell,
consider the model reaction.
H3CS-SCH3 2 CH3S
Use the B3LYP/6-311+G* model to calculate equilibrium geometries for dimethyl
disulfide and thiomethoxy radical, and calculate the energy of SS bond cleavage. How
does it compare with other bond energies (see Excel spreadsheet AH and AB bond
dissociation energies for typical examples)? 25 The next comparison involves energy differences among structural isomers.
Here, the total number of electron pairs is conserved but the numbers of
individual bond types are not maintained. G3(MP2) serves as a reference. As
with the previous bond energy comparisons, only a summary of mean
absolute errors is provided (Table P3-4).
An Excel spreadsheet containing structural isomer energies for Hartree-Fock, B3LYP and
MP2 models with 6-31G*, 6-311+G**, cc-pVTZ and cc-pVQZ basis sets is provided on
the CD-ROM accompanying this text (structural isomer energies). Hartree-Fock models with the 6-31G* and 6-311+G** basis sets actually
yield slightly lower overall errors for this particular set of comparisons than
the corresponding models with larger “limiting” basis sets. Overall errors for
B3LYP models do not change significantly with basis set, while those for
MP2 models with the two smaller basis sets are larger than those for the
larger sets. Individual errors show wider deviations (see Excel spreadsheet).
The B3LYP/6-31G* and B3LYP/6-311+G** models show the best overall
performance, although those of the corresponding Hartree-Fock models are
quite similar. 26 Table P3-4:
Basis Set Summary of Deviations from G3 of Hartree-Fock, B3LYP and MP2
Energies of Structural Isomers (kJ/mol)
Hartree-Fock B3LYP MP2 6-31G* 9 8 17 6-311+G** 10 7 12 16 7 cc-pVTZ
cc-pVQZ 27 Acidity of Propyne: Loss of hydride anion from propyne can either occur from the sp
hybridized carbon or from the sp3 hybridized carbon. The former might be expected as
acetylene is a much stronger acid than ethane, while the latter might be expected as it
would lead to a delocalized anion. Which deprotonation is thermodynamically favored?
H3 C C C– H3C C CH H2C C C H2 – Use the B3LYP/6-311+G** model to obtain equilibrium geometries for the two anions.
Which is lower in energy? Is the less stable anion likely to be detectable in an equilibrium
mixture at room temperature? (Assume a threshold of 5%.)
Isomers of Carboranes: Carboranes are compounds of carbon, boron and hydrogen.
Depending on stoichiometry, they exhibit caged (closo), partially caged (nido from the
Latin word for nest) or open (arachno from the Greek word for spider) structures. are
B4C2H6 and B10C2H12 simple examples of closo structures. The former can exist in one of
two isomers, and the latter in one of three isomers. H
BH HB BH
H HB CH HB BH
H ChemDraw B10C2H12for isomers
Use the B3LYP/6-31G* model to obtain equilibrium geometries for the two isomers of
B4C2H6 and the three isomers of B10C2H12. Examine carbon-carbon bond lengths in the
appropriate isomer of each. Do these appear to be “normal” single bonds or are the
significantly shorter or longer? If the latter provide a rationalization? Identify the lowestenergy isomer for B4C2H6 and B10C2H12 and provide a rationale for the observed
Performance of Practical Models for Hydrogenation Reactions: A hydrogenation
reaction maintains overall bond count but does not maintain individual bond counts. For
example, the products of hydrogenation of ethane have the same number (eight) of σ
bonds as the reactants but two new CH bonds have replaced a C-C bond and H-H bond.
CH3-CH3 + H-H → 2CH4 Calculate hydrogenation energies for ethane, hydrazine, hydrogen peroxide and fluorine
using the Hartree-Fock, B3LYP and MP2 models with both the 6-31G* and 6-311+G** 28 basis sets (six models in total). Correct these for zero-point energy and finite temperature
using the data in Appendix X. Compare these to the results from the G3(MP2) recipe.
F- F -60 kJ/mol
-548 kJ/mol Which (if any) models yield hydrogenation energies that are within +/-12 kJ/mol of the
G3(MP2) values? Which model provides the best results overall?
Energy Content of Hydrazine Fuels: According to the B3LYP/6-31G* model, which
fuel delivers the greater energy on a per gram basis, hydrazine or tetramethylhydrazine?
The products of oxidation are N2 and water for hydrazine and N2, water and CO2 for
tetramethylhydrazine. Make certain to include the mass of the oxidizer (O2) in your
calculations. How does the better of the two fuels compare with molecular hydrogen on a
per gram basis?
Combustion of Hydrocarbons and Fluorocarbons: Hydrocarbons are commonly used
as fuels but fluorocarbons are not. Of course, fluorocarbons cannot be mined or drilled
from the earth, but is there also a fundamental reason? Use the B3LYP/6-31G* model to
calculate the energy of complete combustion of methane, ethane and propane (to CO2 and
water) and tetrafluoromethane, hexafluoroethane and decafluoropropane (to CO2 and
OF2). Point out differences between energies of combustion for the hydrocarbon and
analogous fluorocarbon and relate this to their use as fuels.
“Combustion” of Silanes: Silicon-oxygen polymers (“sand”) result from combustion (in
oxygen) of silanes and other silicon-containing compounds. This makes it difficult to
assign heats of formation. One clever solution is to “burn” such compounds in fluorine
(F2) rather than in oxygen. This leads only to gaseous products (SiF4 and HF) the amounts
of which may easily be determined, for example, combustion of silane.
SiH4 + 4F2 SiF4 + 4HF
Use the B3LYP/6-31G* model to determine energies of complete “combustion” (in F2) of
silane, disilane, trisilane, 2-silyltrisilane and 2,2-disilyltrisilane (resulting in only SiF4 and
HF as products). On a per gram basis, combustion of which of these produces the greates
amount of heat?
SiH4 H3 S i SiH3 H3 S i Si S i H2
H3 S i S i H3 SiH3 H3 S i SiH3
Si SiH3 H3 S i SiH3 Compare silane combustion energies with those of the corresponding hydrocarbons (see
previous problem; you need to perform B3LYP/6-31G* calculations on 2-methylpropane
and 2,2-dimethylpropane). On a per gram basis, are alkanes or silanes better fuels?
2,4-Cyclohexadiene vs. Phenol: As a rule, enols (unsaturated alcohols) are less stable
than their keto isomers (aldehydes and ketones). For example, vinyl alcohol is estimated
to be 43 kJ/mol less stable than its “keto” tautomer, acetaldehyde.
29 ∆E ≈ 43 kJ/mol Use the B3LYP/6-31G* model to obtain the equilibrium geometry and energy of 2,4cyclohexadienone and its enol tautomer, phenol.
O OH Which is more stable? If this is an exception to the “rule”, provide a rationale as to why.
What temperature would be needed in order for the higher-energy structure to be present
as 10% of the equilibrium mixture?
Addition vs. Substitution: Alkenes typically undergo addition reactions whereas
aromatic compounds typically undergo substitution reactions. For example, reaction of
bromine and cyclohexane yields trans-1,2-dibromocyclohexane not 1-bromocyclohexene,
whereas bromination of benzene yields bromobenzene not trans-5,6-dibromo-1,3cyclohexadiene.
+ Br2 Br
vs. + HBr Br
+ Br2 Br
vs. + HBr Br What is the reason for the change in preferred reaction in moving from the alkene to the
arene? Use the HF/6-31G* model to obtain equilibrium geometries and energies for
reactants and the products of both addition and substitution reactions of both cyclohexene
and benzene (four reactions in total). Is your result consistent with what is actually
observed? Are all four reactions exothermic? If one or more are not, provide a rationale
as to why.
Bromonium Ions: Addition of Br2 to an alkene, for example, cyclopentene, is usually
represented as a two-step process. In the first step, the electrophile Br+ adds to the double
– bond giving rise to a “bromonium ion” intermediate. In the second step, Br reacts with
the intermediate to give a trans brominated product. Br2 Br–
Br The 13C NMR spectrum of the intermediate cyclopentenebromonium ion has been
recorded and shows resonances at 19, 32 and 115 ppm (relative to tetramethylsilane).
This is consistent with one of two possible structures. Either the bromine is bonded to
both carbons, leaving it with the (formal) positive charge, or it bonded to only a single
30 carbon, leaving the positive charge on the other carbon. In the latter case, the bromine
needs to transfer between the two carbons with a very low energy barrier.
Br+ + H
H H Br +
H H H Br Use the B3LYP/6-31G* model to obtain equilibrium structures, relative energies and 13C
NMR chemical shifts for the two alternative structures. Is the cyclic structure better
represented as a three-membered ring (like cyclopropane or oxacyclopropane) or as a
weak complex between cyclopentene and bromine cation? Examine the carbon-carbon
bond distance to tell. Which structure is lower in energy? Which better fits the observed
C NMR spectrum, the bridged structure or an equal mixture of the two open structures?
Would you expect the NMR spectrum to change were the measurement carried out at
160K? Elaborate. Are the energetic and NMR results consistent with each other?
Hydrogenation of Acetylene: Both steps in the hydrogenation of acetylene to ethane
involve formation of two new CH bonds from the loss of an HH bond and a π bond. Use
the HF/6-31G* model to obtain equilibrium geometries and energies for all molecules
involved in the hydrogenation reaction. Which step is the more exothermic? Rationalize
H C C H H2 H H
C H2 C H CH3CH3 H Ketene Dimer: There are six possible structures for the dimer of ketene, H2C=C=O.
C O O C CH2 O O C C C O O H2 C CH2
C CH2 O
C H2 C C
H2 C CH2
C C C H2 C O O CH2 Before you do any calculations, predict which of these is likely to be the most stable
based on what you know about the relative stabilities of CC and CO π bonds (lost in the
dimerization) and CC and CC σ bonds (gained in the dimerization). Then, obtain
equilibrium geometries for all six using the HF/6-31G* model. Which structure is
actually preferred. Assuming thermodynamic control, are two or more structures likely to
be seen (assume 5% as the limit of detection of any structure). Is the preferred structure
in line with your prediction? Is the dimerization exothermic or endothermic? Is this
consistent with changes in bond strengths and the strain of the resulting four-membered
Molecular Phosphorous: The most stable arrangement of molecular phosphorus (white
phosphorus) is tetrahedral P4. Dissociation to two molecules of P2 is estimated to be
endothermic by >200 kJ/mol, and can only be detected upon heating to 1000 K. To the
31 contrary, the stable form of molecular nitrogen is N2, and N4 has never been detected let
Use the B3LYP/6-31G* model to obtain equilibrium geometries and energies for
tetrahedral P4 and N4 (as well as P2 and N2) and calculate energies for the two dissociation
reactions. Are your results consistent with what is known (or unknown) about the two
systems? Is tetrahedral N4 actually a stable molecule? Explain your reasoning.
Is tetrahedral As4 a stable molecule? Use the B3LYP/6-31G* model to obtain its
equilibrium geometry and vibrational frequencies. Is its dissociation to As2 endothermic
(like P4) or exothermic? What is the room temperature equilibrium distribution of As4 and
Tetramethylbutane: The central carbon-carbon bond in 2,2,3,3-tetramethylethane
(tetramethylbutane) is known to be significantly weaker than the CC bond in ethane. One
explanation is that bond cleavage relieves the crowding of the methyl groups. Another is
that is tert-butyl radical (the product of bond dissociation of tetramethylbutane) is much
more stable than the methyl radical (the product of dissociation of ethane). To find out
which explanation is correct or if both contribute, calculate energies for the following
four reactions. Use the B3LYP 6-31G* model. (Do not to start with a symmetrical
staggered structure for tetramethylbutane.)
Me3C-CMe3 + H3C· → Me3C· + H3C-CH3
Me3C-CMe3 + H2 → 2Me3CH
H3C-CH3 + H2 → 2CH4
Me3CH + H3C· → Me3C· + CH4 The first reaction directly compares carbon-carbon bond energies of tetramethylbutane
and ethane. The next two reactions provide the energies of hydrogenation of
tetramethylbutane and ethane (both giving rise to uncrowded products). The last reaction
compares bond energies of these (uncrowded) products (providing a measure of the
relative stabilities of the methyl and tert-butyl radicals).
Is the bond energy in tetramethylbutane less than that in ethane? If so, by how much? Is
there evidence for crowding in the geometry of tetramethylbutane? Is the central carboncarbon bond similar in length to the bond in ethane? Are all single bonds staggered? Does
the difference between the hydrogenation energies of tetramethylbutane and ethane
partially or fully account for the difference in bond energies? Does the difference in
difficulty of formation of tert-butyl and methyl radicals partially or fully account for the
difference in bond energies?
Dimerization of Alkylboranes: Borane (BH3) exists in an equilibrium with its dimer,
diborane (B2H6) that strongly favors the dimer (from B3LYP/6-31G* calculations).
ΔE = -164 kJ/mol
Trimethylborane also dimerizes but the equilibrium favors the monomer
32 ΔE = 113 kJ/mol
What is the reason for the change? Is a hydrogen bridge strongly favored over a methyl
bridge? To tell, consider the dimerization of methylborane which may lead to four
possible products: cis and trans isomers with both methyl groups terminal, with both
methyl groups bridged and with one methyl group terminal and the other bridged. Use the
B3LYP/6-31G* model to obtain equilibrium geometries for all four isomers in addition to
methylborane. Which dimer is favored? Is dimerization to one or both of the isomers with
the two methyl groups in terminal positions exothermic (as with dimerization of borane)?
Is dimerization to the isomer with both methyl groups in bridged positions endothermic
(as with dimerzation of hexamethyldiborane)? Speculate on the reason for the difference
in dimerization energies between borane and trimethylborane.
Repeat the calculations for dimerization of both fluoroborane and chloroborane, each
leading to four different products. Is bridging by hydrogen of fluorine (chlorine) favored?
Obtain dimerization energies for alane (AlH3) to dialane (Al2H6) and trimethylaluminum
(AlMe3) to hexamethyldialane (Al2Me6). Continue to use the B3LYP/6-31G* model.
Point out any significant differences from those of the corresponding boron compounds.
Hydrogen Azide: Hydrogen azide is purported to have a high energy cyclic isomer,
cyclotriazine. This is not unreasonable. While the cyclic isomer is likely to be highly
strained, its Lewis structure doe not involve separated positive and negative charges. Attempt to obtain equilibrium geometries for both using the B3LYP/6-311+G** model.
(Start from a non-planar geometry for the cyclic isomer.) Are both acyclic and cyclic
structures energy minima? Justify you answer. If both structures are minima, which is
more stable? What temperature would be required in order for both to be observed in an
equilibrium mixture? (Assume that the minor isomer must be at least 5% of the mixture
in order to be detected.)
Repeat your calculations and analysis for methyl azide and methyl substituted
cyclotriazene. Isomers of Adamantyl Cation: Tertiary carbocations, such as tert-butyl cation are more
stable than secondary carbocations, such as isopropyl cation. It is also important that the
33 carbocation center is planar or nearly planar. Which is more important when only one of
the two can be realized? The isomers of adamantly cation provide the opportunity to tell.
2-adamantyl cation incorporates a non-planar tertiary carbocation center, whereas the 1adamantyl cation incorporates a planar secondary carbocation center. Use the B3LYP/6-31G* model to obtain equilibrium geometries for the two isomers of
adamantyl cation. Are the CCC bond angles at the cation center in 2-adamantyl cation
significantly distorted from ideal (120o)? Which cation is lower in energy? Does it appear
that the need to accommodate a planar cation is more important that the benefit achieved
in going from a secondary to tertiary center? Elaborate. The final comparisons (Table P3-5) exemplify reactions in which both the
total number of electron pairs and the numbers of individual bond types are
conserved, specifically comparisons of regio and stereoisomers. As with
previous comparisons, data from G3(MP2) calculations serve as a reference. 34 Table P3-5: reference Deviations from G3(MP2) of Hartree-Fock, B3LYP and MP2
Energies of Regio and Stereoisomers (kJ/mol) isomer Hartree-Fock
6-31G* 6-311+G** 6-31G* 6-311+G** 6-31G* 6-311+G** G3 1,3-butadiene 1,2-butadiene 50 53 39 43 47 49 49 2-butyne 1-butyne 28 24 33 27 23 18 22 cyclobutene methylenecyclopropane 29 26 25 20 32 31 44 isobutene trans-2-butene
17 cyclopentene methylenecyclobutene 86 85 79 79 89 89 85 2-methyl-1,
62 methyl vinyl trans-2-butenal
4 mean absolute error 5 5 8 8 4 4 – 1,4-pentadiene
1,2-pentadiene 35 Isomers of Pentavalent Phosphorus Halides, PFnCl5-n: While pentavalent phosphorus
halides adopt trigonal bipyramidal geometries with distinct axial and equatorial
positions, these positions rapidly interconvert via pseudorotation (see discussion in
Chapter P5). Therefore, observed properties, for example, the dipole moment, will be
those of an equilbrium mixture of all possible isomers and depend on temperature.
Use the B3LYP/6-31G* model to obtain equilibrium geometries for the two isomers of
SF4Cl (with Cl in either an equatorial or axial position). Which isomer is lower in
energy? Why? Calculate the dipole moment of a sample at room temperature. Is it
different at 50 K? At 1000 K?
Repeat the calculations for SF3Cl2 (3 isomers), SF2Cl3 (3 isomers) and SFCl4 (2 isomers).
Are the results consistent with those for SF4Cl? Elaborate. Compute dipole moments at
50 K, room temperature and 1000 K.
Oxy Acids of Phosphorus: The oxy acids of phosphorus can exist in one of two isomeric
(tautomeric) forms, one in which the phosphorus is trivalent and the other in which it is
R OH R' P
R H The experimental evidence points strongly to the pentavalent species as the dominant
form. Specifically, the infrared spectra of phosphorus oxy acids contain lines
characteristic of P=O and PH bond stretches and no evidence of OH stretches (except
where R or R’ is OH). The one exception appears to be the bis(trifluoromethyl)
compound (R=R’=CF3). Despite the apparent preference for pentavalent forms, the
known “chemistry” of the phosphorus oxy acids demands involvement of the trivalent
structure. This suggests that the two must be in equilibrium, meaning that they two are
close in energy.
Use the B3LYP/6-31G* model to obtain equilibrium geometries for both trivalent and
pentavalent forms of dimethylphosphonate (R=R’=OMe). Precede your calculations by a
search of possible conformers using molecular mechanics (discussion of conformational
searching will be provided in Chapter P5). Which form is lower in energy? Is your result
consistent with the experiment? Elaborate. What temperature would be required for the
higher energy structure to be present as 5% of an equilibrium mixture?
Repeat your calculations with the bis(trifluoromethyl) compound (R=R’=CF3). If, as the
experimental data suggest, there is a change in preferred structure, propose a reason why.
Adamantene: The six atoms involved in a carbon-carbon double bond prefer to lie in the
same plane, for example, the two carbons and four hydrogens in ethylene lie in the same
plane. This does not appear to be possible for adamantene, the olefin formed from lss of
hydrogen from the stable hydrocarbon, adamantane.
- H2 36 Use the B3LYP/6-31G* model to obtain the equilibrium geometry for adamantene. Does
the molecule actually incorporate a double bond, that is, with a carbon-carbon distance in
the usual range of 1.30 to 1.34Ǻ? Is one or both of the carbons involved in the bond
puckered? Are they twisted relative to each other? Display the HOMO of adamantene,
and describe how it differs from the HOMO of a “normal” alkene.
Relate the hydrogenation energy of adamantene to that of 2-methyl-2-butene, a molecule
with similar “substituents” on the double bond. (Use the B3LYP/6-31G* model to obtain
geometries for 2-methylbutane and 2-methyl-2-butene.)
adamantene + 2-methylbutane adamantane + 2-methyl-2-butene
What does this tell you about the strengths of the two π bonds?
Bond Separation Reactions and Interaction of Substituents: A bond-separation
reaction relates any molecule that can be written in terms of a Lewis structure to the set
of “simplest molecules” that contain the same bonds. For example, the bond separation
reaction for acrolein, relates it to ethane, ethylene and formaldehyde, the simplest
molecules with carbon-carbon single and double bonds and a carbon-oxygen double
bond. Two molecules of methane need to be added to the left hand side to achieve
H2C=C-C(H)=O + 2CH4 → H2C=CH2 + H3C-CH3 + H2C=O Aside from the conformation of the reactant, the energy of a bond separation reaction is
well-defined and unique as long as the Lewis structure is well defined and unique.
Because it maintains both total and individual bond counts, it might be expected to be
reasonably well described even by models that provide poor account of electron
Bond separation reactions may be used to determine whether two substituents bonded to
carbon interact constructively (bond separation reaction is endothermic), destructively
(bond separation reaction is exothermic) or not at all.
XCH2Y + CH4 → CH3X + CH3Y Use the HF/6-31G* model to obtain equilibrium geometries and energies for all
molecules involved in bond separation reactions of molecules, CH2X2 (X=CH3, CMe3,
CN, F, SiH3). Interpret the calculated reaction energies both in terms of steric interactions
and the σ and π donor/acceptor properties of the individual substituents:
σ donor, π donor
σ acceptor, π acceptor
σ acceptor, π donor
σ donor, π acceptor Me, CMe3
SiH3 Which of the substituents interact favorably? Which interact unfavorably? Provide a
rationale for each.
Compare bond distances in CH3X and CH2X2 compounds. Do the changes parallel the
interaction energies? Elaborate. 37 Bond Stengths vs. Bond Separation Energies in Fluoromethanes: In an earlier
problem (see Chapter P2), you found that CF bond energies in fluoromethanes, CFnH4-n
(n=1-4), increase with increasing number of fluorines. Is this trend reflected in the bond
separation energies? Use the HF/6-31G* model to calculate energies for all reactants and
products for the three bond separation reactions below.
CH2F2 + CH4 → 2 CH3F
CF3H +2CH4 → 3CH3F
CF4 + 3CH4 → 4CH3F Do the bond energies of these reactions (normalized for the number of CF bonds) parallel
the bond dissociation energies calculated previously?
Repeat your calculations and analysis for bond separation reactions of fluorosilanes,
Repeat your calculations and analysis for bond separation reactions on both carbon and
silicon compounds substituted by chlorine instead of fluorine, CClnH4-n and SiClnH4-n,
(n=1-4). 38 Ion Molecule Reactions
Quantum chemical calculations are not restricted to uncharged molecules,
but may also be applied to cations and anions. While there are virtually no
experimental data relating to the geometries of in the gas phase, there is a
wealth of data relating to their energies. The two most common sources are
proton transfer reactions, that is, acidities and basicities, and electron
transfer reactions, that is, ionization potentials and electron affinities.
There are an enormous number of X-ray crystal structures of ions (together with their
counterions). Discussion has already been provided in Chapter P2. As discussed earlier in this chapter, relative acidities and basicities are most
commonly determined by ion-cyclotron-resonance (ICR) spectroscopy.
What is actually measured is the equilibrium abundance of the ions involved
in a proton transfer reactions between two bases (or two acids) of similar
strength, for example, that between methylamine and ethylamine (bases) or
between acetic acid and propanoic acid (acids). The neutral molecules are
MeNH2 + EtNH3+ MeNH3+ + EtNH2
MeCO2H + EtCO2- MeCO2- + EtCO2H
Measurements are all finally related to a single standard base (ammonia) and
standard acid (???), giving rise to absolute basicities and acidities.
Relative Acidities of Propene and Propyne: Is propene or propyne the stronger acid in
the gas phase? Use the B3LYP/6-311+G** model to calculate the geometries for the
reactants and product of the reaction. Be certain to consider all possible anions resulting
from deprotonation of propene and propyne.
propene + propyne-H+ propene-H+ + propyne Rationalize your result. Is it anticipated by comparison of electrostatic potential maps for
propene and propyne? Elaborate.
Why is Cyclopentadiene a Strong Acid? Cyclopentadiene is a much stronger acid than
1,3-pentadiene in the gas phase, that is, the following reaction is highly exothermic.
cyclopentadiene + 1,3-pentadiene - H+ 39 cyclopentadiene-H+ + 1,3-pentadiene Use the B3LYP/6-31G* model to obtain equilibrium geometries for cyclopentadiene and
its deprotonated form (cyclopentadienyl anion). Note any structural changes that have
occurred to cyclopentadiene as a result of deprotonation, and use these to rationalize its
Gas and Aqueous-Phase Basicities of Amines: The relative base strengths (proton
affinities) of free amines are known to differ significantly from those in water. For
example, in the gas-phase, the proton affinity trimethylamine is 84 kJ/mol greater than
ammonia, whereas in water, the two are of nearly equal. Use the B3LYP/6-31G* model to
obtain equilibrium geometries for the ammonia and trimethylamine and their respective
protonated forms (ammonium and trimethylammonium ions). Is the energy that you
calculate for the transfer of a proton between the two in accord with their relative gasphase proton affinities?
Me3N + NH4+ Me3NH+ + NH3
For practical reasons, you can’t perform the equivalent B3LYP calculations in water. You
can, however, model the energy of the reaction in water by explicitly including a specific
number of water molecules in your calculation. The obvious thing to do is to attach the
minimum number of water molecules to account for all possible amine-water hydrogen
bonds. This number is four for ammonia and ammonium ion, but only one for
trimethylamine and trimethylammonium ion. Obtain equilibrium geometries for the four
amine/ammonium ion complexes involved in the reaction below and calculate the relative
“aqueous-phase” proton affinities of trimethylamine and ammonia.
Me3N…H2O + NH4+ …(H2O)4 Me3NH+…H2O + NH3…(H2O)4
Do the results show the observed trend in proton affinities in moving from the gas phase
into water? Elaborate.
Redo your calculations with trimethylamine and trimethylammonium ion “attached to”
four water molecules instead of only one, and calculate the energy of the reaction below.
Me3N…(H2O)4 + NH4+ …(H2O)4 Me3NH+…(H2O)4 + NH3…(H2O)4
Does this appear to be a better model? Elaborate. Try to identify a serious flaw in both
Boiling Points of Ethanol and Ethylamine: While ethanol and ethylamine have similar
molecular weights and molecular structures and while both can participate in three
hydrogen bonds, the two molecules have very different boiling points. Ethanol is a liquid
at STP whereas ethylamine is a gas. Is this because ethanol forms stronger intermolecular
hydrogen bonds than ethylamine? To tell, obtain equilibrium geometries for ethanol and
ethylamine and their respective hydrogen-bonded dimers, and compare hydrogen-bond
energies. Use the B3LYP/6-31G* model.
ethanol + ethanol ethanol … ethanol
ethylamine + ethylamine ethylamine … ethylamine
Which dimer is more tightly bound? Is your result consistent with the ordering of boiling
points? Elaborate. 40 Exploring Chemical Concepts with Quantum Chemical Models
The availability of reliable and practical quantum chemical models offers the
opportunity to critically examine a number of the qualitative models that
chemists have advanced over many decades. Among the most important are
conjugation, aromaticity and ring strain, the first leading to molecules that
are stablized and the second to molecules that are destabilized.
Molecules incorporating adjacent double bonds that are coplanar or nearly
coplanar are known to be preferred over non-coplanar arrangements or
arrangements in which the double bonds are separated by one or more sp3
“spacers”. Thus, the double bonds in both trans-1,3-pentadiene and in trans2-butenal are coplanar and each is more stable than its non-conjugated
isomer, 1,4-pentadiene and 3-butenal, respectively.
Hydrogenation of 1,3-Butadiene: Both steps in the hydrogenation of trans-1,3butadiene to 1-butene and then to n-butane involve formation of two new CH bonds
resulting from the loss of an HH bond and a π bond. Use the HF/6-31G* model to obtain
equilibrium geometries and energies for all molecules involved in both steps of the
hydrogenation. Which step is the more exothermic? Does this support the idea that
conjugated double bonds are stronger than isolated double bonds? Elaborate.
C H H C C H2 H
H CH2 CH3 H2 C H CH3 CH2 CH 2 CH3 H Hydrogenation Energy of 1,3-butadiene vs. But-1-yne-3-ene: Compare the energy of
the first step in the hydrogenation of 1,3-butadiene (see previous problem) with that of
hydrogenation of the double bond in but-1-yne-3-ene. Use the HF/6-31G* model to
obtain equilibrium geometries for all molecules in the two reactions.
C vs. H C H2 CH3C CH H Which reaction is more exothermic? What does this say about the conjugation energies of
two double bonds vs. a double bond and a triple bond? 41 Hydrogenation of 1,2-butadiene: Both steps in the hydrogenation of 1,2-butadiene first
to 1-butene and then to n-butane involve formation of two new CH bonds resulting from
the loss of an HH bond and a π bond. Use the Hartree-Fock 6-31G* model to obtain
equilibrium geometries and energies for all molecules involved in both steps of the
hydrogenation. Does the double bond in 1,2-butadiene appear to be stronger or weaker
than that in 1-butene? Is the ordering of bond strengths consistent with the ordering of
double bond lengths in the two molecules? Based on hydrogenation energies, would you
conclude that the double bond in 1,2-butadiene is stronger or weaker than that in 1,3butadiene (see first problem in this section)? Would you characterize 1,2-butadiene as a
conjugated molecule? Elaborate.
H C C CH2CH3 H
H H C
H Conjugated vs. Non-Conjugated Cyclic Dienes: 1,4-cyclohexadiene is known to be
slightly (4 kJ/mol) more stable than its isomer, 1,3-cyclohexadiene. This is a surprising
result as the former is a non-conjugated diene whereas the latter is a conjugated diene.
Use the HF/6-31G* model to obtain equilibrium geometries for the conjugated dienes,
1,3-cyclohexadiene, 1,3-cycloheptadiene and 1,3-cyclooctadiene and their nonconjugated isomers, 1,3-cyclohexadiene, 1,3-cycloheptadiene and 1,3-cyclooctadiene.
Assign the preferred isomer and calculate the energy difference for each pair. Do the
calculations reproduce the experimental result for cyclohexadiene? Do they show like
preferences for cycloheptadiene and cyclooctadiene? Aromaticity
Whereas coplanar arrangements of two or more connected π bonds show
modest energy stabilization, much larger effects are noted for some cyclic
arrangements of π bonds. Specifically, molecules with 4n+2 π electrons in a
planar or nearly planar arrangement are known to be unusually stable or
aromatic, and are likely to exhibit chemistry that is quite distinct from that
of unsaturated molecules (even conjugated unsaturated molecules). The
archetypical example is benzene with three carbon-carbon double bonds
(involving six π electrons) confined in a planar six-membered ring.
One way to “measure” the aromatic stabilization of benzene is to compare
the energy of adding H2 (leading to 1,3-cyclohexadiene) with the energy of
adding hydrogen to 1,3-cyclohexadiene (leading to cyclohexene), or the
energy of adding hydrogen to cyclohexene (leading to cyclohexane).
H2 H2 H2 2 4 k J /m o l -110 kJ/mol - 1 2 0 k J /m o l 42 The three steps are similar insofar as each trades a CC π bond and an HH
bond for two CH bonds. Normally, one would tend to think of such a trade
as energetically favorable as σ bonds are stronger than π bonds. However,
the first hydrogenation step is actually slightly endothermic, while the
remaining two steps are both strongly exothermic. The difference between
the first and second (or third) steps (~140 kJ/mol) reflects the additional
(aromatic) stabilization resulting from arrangement of three π bonds in a
planar six-membered ring.
The individual hydrogenation reactions do not conserve the numbers of each
kind of chemical bond (each reaction does maintain overall bond count), and
would not be expected to benefit fully from cancellation of errors. However,
the difference between the first and second reactions (or between the first
and third reactions) which actually relates to aromatic stabilization can be
expressed as a reaction that does maintain individual bond counts. Previous
experience with reactions of this type suggests that its energy should be
reasonably well described by quantum chemical methods that do not take
electron correlation into account.
+ 2 1,3,5-Cyclohexatriene: The six carbon-carbon bonds in benzene are all the same,
intermediate in length between “normal” single and double bonds. The hypothetical
molecule 1,3,5-cyclohexatriene is identical to benzene except that the CC bonds alternate
between single and double bonds. 1,3,5-cyclohexatriene is not an energy minimum and,
if given the chance, will collapse to benzene. Is the energy difference between benzene
and 1,3,5-cyclohexatriene roughly the same as the aromatic stabilization of benzene (~
140 kJ/mol; see discussion above), or is it significantly smaller? To decide, use the HF/631G* model to obtain an equilibrium geometry for benzene and an energy for 1,3,5cyclohexatriene, assuming a fixed geometry with alternating single and double CC bonds
set to 1.54Å and 1.32Å.
Aromaticity of Thiophene: Thiophene is a likely candidate for an aromatic molecule
with two double bonds and an out-of plane lone pair on sulfur making a total of six π
S Compare reaction energies for the individual steps in the hydrogenation of thiophene to
tetrahydrothiophene. As was the case for benzene, addition of the first equivalent of
hydrogen breaks one of the double bonds and leading to loss of aromaticity. Addition of
the second equivalent only breaks a double bond. Therefore, the difference in energy
between the two provides a measure of the aromatic stabilization.
43 X X = S thiophene
X = NH pyrrole
X = O furan X H2 X H2 X = S tetrahydrothiophene
X = NH tetrahydropyrrole
X = O tetrahydrofuran Use the HF/6-31G* model to obtain geometries and energies for thiophene,
dihydrothiophene and tetrahydrothiophene as well as hydrogen, and calculate
hydrogenation energies for the two steps. Is the difference in hydrogenation energies
comparable to that between the corresponding first and second hydrogenation energies
for benzene (see discussion above) or is it significantly smaller or greater? Comment on
the aromaticity of thiophene.
Perform analogous calculations and analyses on pyrrole (X=NH) and furan (X=O). Do
either or both of these molecules show aromaticity? Elaborate. Order the aromaticity of
thiophene, pyrrole and furan.
Aromaticiy of Borazine and Boraphosphazine: Borazine, B3N3H6, is isoelectronic with
benzene and like benzene is known to be a planar molecule with six π electrons
(originating from the three out-of-plane nitrogen lone pairs).
BH Does this mean that borazine is aromatic? Use the B3LYP/6-31G* model to calculate
energies for successive addition of one, two and three equivalents of hydrogen
(analogous to what has previously been done for benzene).
H NH H2 BH H2N
H NH H
2 H2 N BH H2B H2
H2 NH H2 H2N BH H2 B H2
BH Is the first hydrogenation step significantly more difficult (more endothermic) than the
second and third steps? If it is, estimate the “aromaticity” of borazine and compare this to
the value for benzene obtained from the same analysis.
Use the B3LYP/6-31G* model to calculate energies for the individual steps in the
hydrogenation of the phosphorous analogue of borazine (boraphosphazine).
BH H2 H2P
BH H2 H2 P
H2 P H H2 H2P BH H2 B H2
BH Rank the aromatic stabilization of benzene, borazine, and boraphosphazine.
Cyanuric Acid: Heating urea leads to loss of ammonia and production of hydrogen
cyanate. This in turn trimerizes to cyanuric acid, which may either take on a keto or enol
structure. Both are aromatic molecules insofar as both involve six π electrons in a planar
six-membered ring. Which is preferred and why? 44 O
H2N NH3 + HN C O NH 2 O HO NH
H O N N OH N
OH Use the B3LYP/6-31G* model to obtain equilibrium geometries for urea, ammonia,
hydrogen cyanate and the keto and enol forms of hydrogen cyanate trimer. Is loss of
ammonia from urea an endothermic or exothermic process? Is it is endothermic, what do
you think drives the reaction? Is trimerization to the more stable of the two forms of
cyanuric acid endothermic or exothermic? Is the overall reaction endothermic or
exothermic? What is the preferred structure of cyanuric acid?
Cyclobutadiene: Whereas a molecule with 4n+2 π electrons in a cycle is thought to be
unusually stable (aromatic), a cyclic arrangement of 4n electron has the potential of being
unusually unstable (antiaromatic). A good example of this is singlet cyclobutadiene. Compare the energy of hydrogenation of cyclobutadiene (to cyclobutene) with that of
hydrogenation of cyclobutene (to cyclobutane). Use the B3LYP/6-311+G** model. The
two reactions are similar in that each trades an HH bond and a CC π bond for two CH
bonds. However, only the first reaction removes interaction of the coplanar double bonds.
Aromaticity of Cyclopentadienyl and Cycloheptatrienyl Radicals: Cyclopentadienyl
radical assumes a planar geometry with a total of five π electrons. Cycloheptatrienyl
radical is also planar but has seven π electrons. Do these radicals benefit from aromatic
stabilization? To tell, use the same test previously employed for benzene and related
systems, that is, compare energies for the first step in an overall hydrogenation reaction
with that of subsequent steps. Use the B3LYP/6-31G* model to obtain equilibrium geometries for all the molecules in
the above two reactions. For each, calculate the sequence of hydrogenation energies.
Would you conclude that either cyclopentadienyl and/or cycloheptatrienyl radical is
Aromaticity of Cyclopentadienyl Anion and Cycloheptatrienyl (Tropylium) Cation:
The cyclopentadienyl anion is among the most common ligands found in transition-metal
organometallic compounds. Here it maintains a planar (or nearly planar) geometry and, is
generally thought to contribute six electrons to the valence shell of the metal to which it
is bonded. Is cyclopentadienyl anion aromatic? Certainly it fits the obvious criteria of
possessing a planar cyclic arrangement of 4n+2 π electrons (as does benzene). However,
these electrons are spread over only five carbons (and not six as in benzene). 45 The opposite situation exists for cycloheptatrienyl cation (commonly known as
tropylium). It is also planar with six π electrons, but these are spread over seven carbons.
Do these ions benefit from aromatic stabilization? To tell, use the same test previously
employed for benzene and related systems, that is, compare energies for the first step in
an overall hydrogenation reaction with that of subsequent steps.
– + H2 H2 – + H2 – H2 + + Use the B3LYP/6-31G* model to obtain equilibrium geometries for all the molecules in
the above two reactions. For each, calculate the sequence of hydrogenation energies.
Would you conclude that either cyclopentadienyl anion and/or cycloheptatrienyl cation is
aromatic? Elaborate. Ring Strain
A molecule with an sp3 center incorporated into three-membered rings
would be expected to be energetically disfavored. The main reason is that
the centers cannot adopt their ideal tetrahedral values (bond angles of
109.5o). Another reason is that planarity of the skeleton leads to eclipsing
interaction involving hydrogens on adjacent centers. This combination of
factors is commonly referred to as ring strain, and the associated “cost” in
energy is commonly referred to as the strain energy. For example,
cyclopropane, is known to be ~60 kJ/mol less stable than its isomer,
propene, while cyclohexane is ~80 kJ/mol more stable than 1-hexene. The
latter reflects the fact that, all other things being equal, σ bonds are stronger
than π bonds. The difference (~140 kJ/mol) might be considered as the strain
energy of cyclopropane.
cyclopropane propene cyclohexane 1-hexene -60 kJ/mol 80 kJ/mol Isomerization energy is but one way to assess strain energy. Hydrogenation
reactions may also be used to distinguish strained from unstrained
molecules. For example, hydrogenation of cyclopropane to propane is
exothermic by ~144 kJ/mol, while hydrogenation of cyclohexane to hexane
is exothermic by only ~44 kJ/mol, yielding a strain energy of ~100 kJ/mol. 46 Ring Strain in Oxirane and Aziridine: Which molecule is more strained oxirane or
cyclopropane? Use the HF/6-31G* model to evaluate the hydrogenation energy of
oxirane (leading to dimethyl ether) relative to that of cyclopropane (leading to propane).
(Hydrogen molecule appears on both sides of the equation and its energy is not needed.)
Offer an explanation for your result.
H2 C CH2 + CH2 H3 C O CH3 H3 C + CH3 CH2
H2 C CH2 Repeat the calculation for hydrogenation of aziridine (leading to dimethylamine) relative
to hydrogenation of cyclopropane.
H2 C CH2 + CH2
H3 C NH CH3 H3 C CH3 + CH2
H2 C CH2 Which is more strained, aziridine or oxirane? Suggest an explanation for your result.
Strain Energies of Small-Ring Cycloalkanes: As discussed above, cyclopropane is a
highly strained molecule whereas cyclohexane is not. Is the strain energy of cyclobutane
closer to that of cyclopropane or cyclohexane? Of cyclopentane? Is cycloheptane more
strained than cyclohexane? To assess the strain in cyclopropane, cyclobutane,
cyclopentane and cycloheptane relative to that in cyclohexane, obtain energies of
reactions with n = 1, 2, 3 and 5. Use the HF/6-31G* model. (Energies for cyclopropane
and propane are available if you completed the previous problem.)
H2C CH2 CH3 CH3 CH3 H2C CH3 + CH2 +
(CH2 )4 (CH2 )n (CH2 )n (CH2 )4 Cycloalkynes: Because alkynes exhibit a linear or nearly linear C-C≡C-C unit, it is
unlikely that they will be able to incorporate into small rings. Identify the smallest
cycloalkyne where both C-C≡C bond angles are within 10° of being linear (Start with
cyclohexyne and minimize with molecular mechanics. Use the B3LYP/6-31G* model to
calculate the equilibrium geometry of this cycloalkyne and cycloalkynes with one and
two fewer carbons. Also obtain geometries of the corresponding cycloalkenes as well as
hydrogen molecule. Calculate energies of the hydrogenation reaction.
C (CH2 )n H2 C C
H n = 4–9 C
H Is there a relationship between C-C≡C bond angle and hydrogenation energy for the three
Strain Energies in Polycyclic Alkanes: Polycyclic cycloalkanes are likely to be even
more strained than cycloalkanes. An extreme example of a strained molecule is
tetrahedrane. While the parent compound, C4H4, has yet to be experimentally
characterized, X-ray crystal structures exist for several simple derivatives, and appear to 47 be quite “normal”. For example, tetra-tert-butyltetrahedrane, shows carbon-carbon bond
lengths of 1.49Ǻ, virtually identical with those found in cylopropane (1.50Ǻ).
Use the B3LYP/6-31G* model to calculate equilibrium geometries for all molecules
involved in hydrogenation of tetrahedrane, first to bicyclo[1.1.0]butane and then to
H2 H2 Cubane: It may come as a surprise that despite what appears to be a highly-strained
carbon skeleton, derivatives of cubane are rather common (the crystal structures of well
over a hundred of them have been reported). Use the B3LYP/6-31G* model to obtain
equilibrium geometries for cubane and the first and second hydrogenation products (the
former is unambiguous (all twelve CC bonds are the same), but the second is not.
Assume the structure shown below.
H2 H2 Inter and Intramolecular Hydrogen Bonding
Discussion to be written
The Enol of 1,3-Malonaldehyde: The enols of most carbonyl compounds are higher in
energy than the keto forms. A possible exception arises in dicarbonyl compounds such as
1,3-malonaldehyde where there is the possibility of intramolecular hydrogen bonding.
O O O H O H O O Use the B3LYP/6-31G* model to obtain equilibrium geometries for malonaldehyde and
for cis-3-hydroxyacrolein. Start with a conformation of the latter that allows a hydrogen
bond to be formed. Also obtain the geometry of trans-3-hydroxyacrolein (a molecule that
cannot form an intramolecular hydrogen bond). Is cis-3-hydroxyacrolein lower in energy
than malonaldehyde? If so, does this appear to be due to the hydrogen bond (compare the
energies of cis and trans-3-hydroxyacrolein to tell)?
Problem (using canned files) on hydrogen-bonding in base pairs and related systems. 48 H3C CH3 N O O N H H3C H N O N H O H2N N
N CH3 O
N CH3 N
CH3 49 N N H H H HN O
N H N N H O
N N N CH 3 N NH H N N O
HN N CH3 NH2
N N N N H CH3
N O N N CH3
N H3 C O C H3 C H3 N N O Metal-Ligand Binding in Transition-Metal Organometallics
Discussion to be written
Metal-Ligand Binding Energies in Iron Carbonyls: Both ethylene and acetylene bind
to iron tetracarbonyl to form stable organometallics. Obtain equilibrium geometries for these two compounds as well as for their component
fragments (ethylene, acetylene and iron tetracarbonyl), and calculate metal-ligand
binding energies. Use the B3LYP/6-31G* model. Which is more tightly bound? Is your
result consistent with changes to the C=C and C≡C bonds in ethylene and acetylene that
result from binding to the metal carbonyl? Elaborate.
Perform analogous calculations on butadiene iron tricarbonyl and cyclobutadiene iron
tricarbonyl as well as 1,3-butadiene, cyclobutadiene and iron tricarbonyl. Which ligand is more tightly bond to iron tricarbonyl, butadiene iron or cyclobutadiene?
Is this result in line with the relative bond length changes in the two dienes? Elaborate.
Metal-Ligand Binding Energy in Benzene Chromium Tricarbonyl: Stable πcomplexes between chromium tricarbonyl and arenes are common and are sufficiently
stable to allow “chemistry” to be performed on the ligand. Cr
CO Use the B3LYP/6-31G* model to obtain equilibrium geometries for benzene chromium
tricarbonyl as well as for benzene and chromium tricarbonyl. Is the metal-ligand binding
energy of the same order of magnitude as a normal covalent bond (250-400 kJ/mol), or of
the same order of magnitude as a hydrogen bond (20-30 kJ/mol) or somewhere in
between? Compare the binding energy in benzene chromium tricarbonyl to those in
complexes considered in the previous problem. Does benzene appear to be a better (more
tightly bound) ligand than ethylene or acetylene? Than butadiene or cyclobutadiene?
Rationalize you results.
Tebbe Reagent: Early transition metal carbenes, LnM=CRR’, catalyze a variety of
related processes involving olefins, including metathesis and polymerization. The Tebbe
Me2AlCl, exhibits similar behavior suggesting that it is
nothing more than a weak complex between a titanium methylidene and a Lewis acid. 50 Use the B3LYP/6-31G* model to establish the equilibrium geometry for the Tebbe
reagent as well as for its methylidene and Lewis acid components, and calculate the
energy of complexation. is both an excellent olefin metatesis catalyst for olefin 51 Using Approximate Equilibrium Geometries
Is it actually necessary to obtain geometries with the same theoretical model
required to accurately calculate reaction energies, or are geometries from a
simpler (lower computation cost) model sufficient? This is an important
question because determining equilibrium geometry may easily require one
or two orders of magnitude more computation than calculating energy at a
fixed geometry, due to the need of multiple calculations at different
geometries. It is also straightforward to answer. Table P3-6 compares
energies for structural isomers obtained from B3LYP/6-311+G**
calculations using four different sets of geometries: 3-21G, 6-31G*,
B3LYP/6-31G* and “exact” (B3LYP/6-311+G**). Also provided are
reference isomer energies obtained from G3(MP2) calculations. The
conclusion is clear. The magnitude of the error introduced as a result of
using approximate geometries is insignificant in comparison with that
inherent to the underlying B3LYP/6-311+G** model. Even the HF/3-21G
model provides satisfactory geometries.
Some properties will be more sensitive to small changes in geometry, for
example, the dipole moment. Also, the infrared spectrum of a molecule
(discussed in Chapter P2) requires that the geometry be calculated using the
same theoretical model. However, for reaction energy calculations, the
savings in computation cost achieved by the use of approximate geometries
more than offsets any small changes in results. 52 Table P3-6: Effect of Use of Approximate Geometries on the Energies of
Structural Isomers from B3LYP/6-311+G** (kJ/mol)
propyne 6-31G* B3LYP/6-31G* B3LYP/6-311+G** G3(MP2) isomer
allene -7 -7 -7 -7 3 cyclopropene 102 101 101 101 102 cyclopropane 38 38 38 38 38 2-butyne 36 37 37 37 3 bicyclo[1.1.0]butane 132 130 130 130 119 cyclobutane 47 46 47 47 48 1,4-pentadiene 63 63 63 63 65 methyl isocyanide 100 100 100 102 102 vinyl alcohol 45 43 43 43 43 oxirane 124 122 121 121 116 ethanol dimethyl ether 48 48 47 47 49 acetic acid methyl formate 69 69 68 68 70 methyl vinyl ketone cyclobutanone 24 23 24 24 23 divinyl ether 101 98 98 98 102 1 0 0 - - propene
acetaldehyde mean absolute deviation from “exact”
mean absolute error from G3(MP2) - 53 Silaolefins: Only a very few compounds incorporating a carbon-silicon double bond
(“silaolefins”) have been characterized, among them compounds 1-4. On the contrary,
stable compounds with silicon-carbon single bonds (“silanes”) are common. There is
indirect evidence that suggests that the lack of silaolefins is due at least in part to the high
reactivity of silicon-carbon double bonds. Specifically, all of the compounds that do exist
have large bulky groups shielding the double bond, thereby making it difficult for
reagents to approach.
SiMe(Cme3)2 Si M e3Si
M e3Si C Me OSiMe3
Si (1-adamantyl) 1 2
Si C (CMe3)Me2Si C C M e3Si
CMe3 3 C (CMe3)Me2Si
4 We examine the kinetic reactivity of silaolefins elsewhere in this text. For now we, probe
their thermochemical stability, specifically the thermochemistry of hydrogenation relative
to that of normal olefins. Obtain equilibrium geometries for 1,1-dimethylsilaethylene
(Me2Si=CH2) and isobutene, as well as their hydrogenation products, trimethylsilane and
isobutene. Use the B3LYP/6-31G* model. Evaluate the energy of the reaction comparing
hydrogenation energies for 1,1-dimethylsilaethylene and isobutene.
Me2Si=CH2 + Me3CH → Me3SiH + Me2C=CH2 What does it tell you about the strength of the SiC double bond in the silaolefin relative
to the CC double-bond strength in isobutene?
Repeat your calculations for the germanium analogue of isobutene, that is, evaluate the
energy of the reaction.
Me2Ge=CH2 + Me3CH → Me3GeH + Me2C=CH2 Do you conclude silicon or germanium forms the stronger π bond to carbon? Elaborate
any assumptions that you have made to reach this conclusion.
Finally, repeat both sets of calculations using geometries obtained from the HF/3-21G
model (in lieu of the B3LYP/6-31G* model), following these with energy calculations
with the B3LYP/model. Have the results changed significantly. Estimate the computation
cost of this approach relative to that employing “exact” geometries.
Incorporating Atoms and Molecules into BuckyBall: The cavity inside Buckyball,
aka, Buckmeister fullerene, fullerene of C60), is not large enough to incorporate anything
but atoms and very small molecules (atomic inclusion compounds are common but only a
few examples of molecular complexes have been reported). 54 Host-guest complexes need to be formed by “assembling” the fullerene in the presence
the atoms/molecules that are to be incorporated, for example, a helium complex needs to
be formed by assembling it helium gas. The fullerene “mesh” is very tight, and entrapped
atoms and molecules cannot escape. Therefore, it is not clear if complex formation is
thermodynamically driven force or whether complexes exist because fullerene does not
provide an exit for its guest.
Quantum chemical calculations may be employed to decide. These are large molecules
and “full” treatments with models capable of producing reliable binding energies may be
prohibitive in terms of computer time. An alternative is to obtain equilibrium geometries
with a simple model and follow these with energy calculations with a better (and
computationally more expensive) model. In this example, the HF/3-21G model serves the
first role and the B3LYP/6-31G* model the second role. Even so, calculations of this
magnitude are not suitable as “homework exercises”, and the results have been provided
to examine and ponder. No quantum chemical calculations are required.
Results for C60 and for helium, neon, hydrogen molecule, methane and water complexes
of C60 obtained form the B3LYP/6-31G* model based on HF/3-21G equilibrium
geometries are provided in C60 and C70 complexes. Also provided results for C70 and for
the C70-hydrogen molecule complex.
Calculate binding energies for the five C60 complexes.
fullerene (host) + guest host-guest complex
Which if any of the complexation reactions are thermodynamically driven? Are your
results consistent with electrostatic potential maps for the inside surface of fullerene
examined in Chapter P1? Elaborate.
Calculate the binding energy of hydrogen molecule into C70. Is the guest more or less
inclined thermodynamically to associate with the inside of the larger C70 host than it is
with the smaller C60 host?
Finally, calculate binding energies for hydrogen molecule inside both C60 and C70 cages
obtained from B3LYP/6-311+G** energies again assuming HF/3-21G equilibrium
geometries. Do you see a marked change either in the absolute numbers and/or in the
difference between binding energies for the smaller and larger hosts from you previous
Results for both C60 and C70 and together with their complexes with hydrogen molecule,
obtained form the B3LYP/6-311+G** model based on HF/3-21G equilibrium geometries
are provided in C60 and C70 hydrogen molecule complexes. 55 56 Thermochemical Recipes
This chapter has focused on the calculation of energies of diverse chemical
reactions. One lesson we have learned is that reactions that involve explicit
bond making or breaking require “better” quantum chemical models than
reactions that maintain bonding and only change local environment. Another
lesson is that even the “best” of the models that are presently routinely
applicable to molecules of moderate size may lead to errors of unacceptable
magnitude. It would certainly be desirable to have a single practical model
that would be able to provide energies for any reaction to within 4-8 kJ/mol
and be applicable to molecules with molecular weights up to 400 amu.
One possibility is the G3(MP2) method that we have used throughout this
chapter to supply reference data (in lieu of experimental data). In practice, it
is applicable only to very small molecules (with molecular weights below
150 amu). G3(MP2) heats of formation for several hundred small molecules
have been compared with experimental values contained in the NIST
thermochemical database. This leads to a mean absolute error of 6.2 kJ/mol
and an RMS error of 8.3 kJ/mol.
G3(MP2) actually involve a combination of several different quantum
chemical models, and perhaps is best referred to as a recipe. The first step is
to obtain an equilibrium geometry using the MP2/6-31G* model. Next, two
energy calculations are performed at this geometry, one an MP2 calculation
with the G3MP2large basis set an the other a QCISD(T) calculation with the
6-31G* basis set. The energy differences between the MP2 calculation with
the two basis sets and the QCISD(T) and MP2 calculations with the same
basis set are summed. This assumes that the changes in energy due to
increase in the size of the basis set from 6-31G* to the G3MP2large and
improvement in treatment of electron correlation from the MP2 to the
QCISD(T) are independent. This assumption is the reason that G3(MP2) is
as widely applicable as it is. Were it necessary to perform a QCISD(T)
calculation (which scales as the seventh power of the number of basis
functions) with the G3MP2large basis set, the procedure would be much
more limited. The fact that G3(MP2) heats of formation obtained are in
reasonable accord with experimental values validates the assumption. The
final step in the recipe involves a HF/6-31G* frequency calculation
(preceded by a HF/6-31G* geometry). This allows for calculation of the
zero-point energy and for correction for finite temperature. 57 A new recipe, designated T1, has been formulated that requires significantly
less computation than G3(MP2), but yields heats of formation that are nearly
identical. Both the QCISD(T) and Hartree-Fock frequency calculations have
been eliminated and the MP2/G3MP2large calculation has been replaced by
a RI-MP2 calculation in which the G3MP2large basis set is approximated
using so-called dual basis set techniques. A HF/6-31G* geometry replaces
the MP2/6-31G* geometry. Without further modification, these changes to
G3(MP2) result in heats of formation that are not sufficiently accurate to be
useful in thermochemical calculations. A successful recipe follows by
introducing a total of 66 linear regression parameters based on Mulliken
bond orders calculated from the Hartree-Fock wavefunction. These have
been determined using a training set of more than 1100 G3(MP2) heats of
formation. A plot of T1 vs. G3(MP2) heats of formation for this set is
provided in Figure P3-3.
Table P3-7 compares structural isomer energies obtained from T1 heats of
formation for a variety of simple systems with those obtained both from the
G3(MP2) recipe and with experimental heats. The mean absolute and rms
errors is 6.6 and 9.0 kJ/mol, roughly the same as those for the G3(MP2)
As discussed earlier in this chapter, while the energy of a molecule is one of its most
fundamental properties, it is only rarely determined. This means that quantum chemical
models (or combinations of quantum chemical models) that were able to provide routine
and reliable energies (heats of formation) are of significant value. 58 Figure P3-3: Comparison of Heats of Formation Obtained from the T1 and
G3(MP2) Recipes (kJ/mol) 59 Table P3-7: Comparison of Energies of Structural Isomers from T1 and
G3(MP2) Recipes with Experimental Values (kJ/mol)
formula (reference) isomer T1 G3(MP2) expt. C2H3 N (acetonitrile) methyl isocyanide 102 100 88 C2H4 O (acetaldehyde) vinyl alcohol
118 C2H4 O2 (acetic acid) methyl formate 68 70 75 C2H6 O (ethanol) dimethyl ether 49 50 51 C3H4 (propyne) allene
93 C3H6 (propene) cyclopropane 36 38 29 C4H6 (1,3-butadiene) 2-butyne
divinyl ether 7
102 C4H8 (2-methylpropene) trans-2-butene
46 C5H8 (cyclopentene) 2-methyl-1,3-butadiene
105 5 4 - C4H6 O (methyl vinyl ketone) trans-2-butenal mean absolute error 60 The version of Spartan provded with this textbook (Spartan Student) does not provide
for calculations using the T1 thermochemical recipe. However, the associated Spartan
Molecular Database includes T1 data for all the molecules required for the problems that
Double-Bond Disproportionation Reactions: A double-bond disproportionation
reaction relates the energy of a molecule incorporating a double bond with the average of
the energies of molecules incorporating the corresponding single and triple bonds. For
example, the double-bond disproportionation reaction for ethylene relates its energy to
the average of the energies of ethane and acetylene. Experimentally, this reaction is
endothermic by 38 kJ/mol, meaning that a CC double bond is weaker than the average of
carbon-carbon single and triple bonds.
2 H2C=CH2 → H3C-CH3 + HC≡CH Use the T1 recipe to calculate the energies of double-bond disproportionation reactions
for ethylene, methyleneimine (H2C=NH), formaldehyde (H2C=O) and thioformaldehyde
(H2C=S). (Build the molecules as you normally would and simply retrieve from the
Spartan Molecular Database.) Adjust your results for the last three to account for the
error in the reaction of ethylene. Note any significant differences among the four
disproportionation energies and try to rationalize them. 61 Calculation of Entropy and Gibbs Energy
The change in the Gibbs energy (ΔG) for a chemical reaction follows from
the change enthalpy (ΔH) by way of the usual thermodynamic relationship.
ΔG = ΔH - TΔS
T is the temperature in K and ΔS is the change in entropy. The entropy is a
sum of translational, rotational and vibrational parts.
S = Str + Srot + Svib
The translational part, which depends on the total molecular mass, M, and
pressure, P (n is the number of moles, and R and k are the gas and
Boltzmann constants), cancels in a mass-balanced equation.
3 Str = nR 2 2!MkT + ln 3/ 2 2 nRT
P The rotational part depends on the principal moments of inertia, IA, IB and IC,
which in turn depend on the geometry. s is termed the symmetry number
which is usually unity.
3 Srot = nR 2 ( !vA vBvC )1/2 + ln s vA = h2/8!IA kT, vB = h2 /8!IB kT, vC = h2/8!IC kT The vibrational part depends on the frequencies. It is based on the linear
harmonic oscillator approximation, and incorrectly goes to ∞ and not ½ RT
as the frequency goes to 0. It needs to be adjusted to approach the proper
Svib = nR !
i (uieui – 1)–1 – ln (1 – e –ui ) µi = h!i/kT 62 ...
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This note was uploaded on 02/22/2010 for the course CHEM N/A taught by Professor Head-gordon during the Spring '09 term at Berkeley.
- Spring '09