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The three vibrational frequencies in \(\mathrm{H}_{2} \mathrm{O}\left(1595,3657, \text { and } 3756 \mathrm{cm}^{-1}\right)\) are all much larger than the corresponding frequencies in \(\mathrm{D}_{2} \mathrm{O}(1178,1571,\) and \(2788 \mathrm{cm}^{-1}\) ). This follows from the fact that vibrational frequency is given by the square root of a (mass-independent) quantity, which relates to the curvature of the energy surface at the minima, divided by a quantity that depends on the masses of the atoms involved in the motion. As discussed in Section \(26.8 .4,\) vibrational frequencies enter into both terms required to relate the energy obtained from a quantum chemical calculation (stationary nuclei at 0 K) to the enthalpy obtained experimentally (vibrating nuclei at finite temperature), as well as the entropy required to relate enthalpies to free energies. For the present purpose, focus is entirely on the so-called zero point energy term, that is, the energy required to account for the latent vibrational energy of a molecule at \(0 \mathrm{K}\) The zero point energy is given simply as the sum over individual vibrational energies (frequencies). Thus, the zero point energy for a molecule in which isotopic substitution has resulted in an increase in mass will be reduced from that in the unsubstituted molecule: A direct consequence of this is that enthalpies of bond dissociation for isotopically substituted molecules (light to heavy) are smaller than those for unsubstituted molecules. a. Perform B3LYP/6-31G* calculations on HCl and on its dissociation products, chlorine atom and hydrogen atom. Following geometry optimization on HCl, calculate the vibrational frequency for both HCl and DCl and evaluate the zero point energy for each. In terms of a percentage of the total bond dissociation energy, what is the change noted in going from HCl to DCl? \(\mathrm{d}_{1}\) -Methylene chloride can react with chlorine atoms in either of two ways: by hydrogen abstraction (producing HCl) or by deuterium abstraction (producing DCl): Which pathway is favored on the basis of thermodynamics and which is favored on the basis of kinetics? b. Obtain the equilibrium geometry for dichloromethyl radical using the B3LYP/6-31G* model. Also obtain vibrational frequencies for both the unsubstituted and the deuterium-substituted radical and calculate zero point energies for the two abstraction pathways (you already have zero point energies for HCl and DCl). Which pathway is favored on the basis of thermodynamics? What would you expect the (thermodynamic) product ratio to be at room temperature? c. Obtain the transition state for hydrogen abstraction from methylene chloride using the B3LYP/6-31G* model. A reasonable guess is shown here: Calculate vibrational frequencies for the two possible structures with one deuterium and evaluate the zero point energies for these two structures. (For the purpose of zero point energy calculation, ignore the imaginary frequency corresponding to the reaction coordinate.) Which pathway is favored on the basis of kinetics? Is it the same or different from the thermodynamic pathway? What would you expect the (kinetic) product ratio to be at room temperature?

Short Answer

Expert verified
In this exercise, we performed B3LYP/6-31G* calculations on HCl, DCl, and the dichloromethyl radical to obtain optimized geometries, vibrational frequencies, and zero-point energies. We compared the bond dissociation energies between HCl and DCl and evaluated the thermodynamic and kinetic favorability of different reaction pathways involving hydrogen and deuterium abstraction. The thermodynamically favored pathway and product ratio at room temperature were determined using the obtained zero point energies, and the favored pathway based on kinetics was determined by analyzing the transition state structures and their zero-point energies.

Step by step solution

01

Perform B3LYP/6-31G* Calculations for HCl and DCl

To begin, perform B3LYP/6-31G* calculations on HCl and its dissociation products. This will allow you to obtain optimized geometries and frequencies for both HCl and DCl.
02

Evaluate Zero-Point Energies

Using the vibrational frequencies obtained from the B3LYP/6-31G* calculations, evaluate the zero-point energies for HCl and DCl. This can be done by summing over the individual vibrational energies.
03

Calculate Bond Dissociation Energies and Compare

Determine the bond dissociation energies for HCl and DCl, and compare them as a percentage of the total bond dissociation energy. This will show the change in energy when going from HCl to DCl. b. Determine Favored Pathway based on Thermodynamics
04

Obtain Equilibrium Geometry for Dichloromethyl Radical

Perform B3LYP/6-31G* calculations to obtain the equilibrium geometry for the unsubstituted and deuterium-substituted dichloromethyl radical. This will be used for further analysis of the abstraction pathways.
05

Calculate Zero-Point Energies of Abstraction Pathways

Using the vibrational frequencies obtained for both the unsubstituted and deuterium-substituted radicals, calculate the zero-point energies. Alongside the previously calculated zero-point energies for HCl and DCl, this will help determine the favored pathway based on thermodynamics.
06

Thermodynamic Product Ratio

Using the calculated values, determine which reaction pathway is thermodynamically favored and calculate the expected thermodynamic product ratio at room temperature. c. Determine Favored Pathway based on Kinetics
07

Obtain Transition State

Perform B3LYP/6-31G* calculations to obtain the transition state for hydrogen abstraction from methylene chloride. You might want to use a reasonable guess shown in the exercise.
08

Calculate Zero-Point Energies for Transition State Structures

Calculate the vibrational frequencies for each of the two possible structures (with one deuterium) using the B3LYP/6-31G* calculations. Evaluate their zero-point energies, ignoring the imaginary frequency corresponding to the reaction coordinate.
09

Kinetic Product Ratio

Determine which reaction pathway is favored based on kinetics using the zero-point energies and compare it to the thermodynamically favored pathway. Calculate the expected kinetic product ratio at room temperature.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Zero Point Energy
Zero Point Energy (ZPE) refers to the minimum amount of energy that a quantum mechanical system will possess, even when at rest (0 K). Think of it as the lowest rung on an energy ladder that molecules can't step off. In the case of vibrations in molecules, ZPE is the energy due to vibrational motion that persists even when a molecule is cooled to absolute zero.

To calculate it, one sums the energies of all the individual vibrational modes of a molecule. The formula for a single vibrational mode is \( \frac{1}{2}hv \) where \( h \) is Planck's constant and \( v \) is the vibrational frequency. Thus, heavier isotopes tend to lower the ZPE because their vibrational frequencies are less due to greater mass, as seen with the comparison of \( \mathrm{H}_2\mathrm{O} \) and \( \mathrm{D}_2\mathrm{O} \).
Bond Dissociation Energy
Bond Dissociation Energy (BDE) is the energy required to break a bond in a molecule, producing isolated atoms. It's a critical concept for understanding the stability of molecules and the energy involved in chemical reactions.

Since the ZPE is part of the total energy of a molecule, when calculating BDE, the ZPE must be considered. As isotopic substitution increases the mass and lowers the ZPE, the BDE is also affected—typically reduced, because less energy is required to dissociate the heavier isotope-laden molecule. This is particularly relevant in reactions involving isotopes, as seen in the example of \( \mathrm{HCl} \) versus \( \mathrm{DCl} \) where the change of vibrational frequencies due to isotopic substitution affects the BDE.
Isotopic Substitution Thermodynamics
Isotopic Substitution Thermodynamics involves understanding how the replacement of an atom by its isotope affects the properties of a molecule. This substitution can change the mass, vibrational frequencies, and thus the ZPE.

The effects are quite significant when it comes to thermodynamic properties like enthalpy and entropy. Since vibrational frequencies are lower for heavier isotopes, ZPE decreases, and the substance becomes more thermodynamically stable. This concept can help predict the outcome of reactions—whether a reaction favors the lighter or heavier isotope, like in the thermodynamics of \( \mathrm{HCl} \) and \( \mathrm{DCl} \) where bond energies differ due to isotopic substitution.
B3LYP/6-31G* Calculations
B3LYP/6-31G* calculations are a computational method used in quantum chemistry to predict molecular structures, energies, and other properties. B3LYP refers to a hybrid functional, which combines Hartree-Fock methods with density functional theory (DFT), while 6-31G* represents a particular basis set describing the electrons in the system. This level of theory is commonly used for its balance between computational cost and accuracy.

In practice, B3LYP/6-31G* calculations optimize molecular geometries and predict vibrational frequencies—essential in determining the zero-point energies of molecules, as evidenced by the problems involving ZPE calculations for \( \mathrm{HCl} \) and \( \mathrm{DCl} \).
Transition State Analysis
Transition State Analysis focuses on identifying the configuration of a molecule at the highest energy point along the reaction pathway—the transition state. This transient state cannot be isolated, but its properties can be inferred through computational simulations such as B3LYP/6-31G* calculations.

By analyzing the transition state, chemists can predict the rate of a chemical reaction—higher energy barriers correspond to slower reactions. Transition states are crucial for understanding reaction kinetics, including predicting which pathway a reaction may prefer at room temperature, as in the determination of either hydrogen or deuterium abstraction from methylene chloride.
Enthalpy and Entropy in Physical Chemistry
Enthalpy and entropy are fundamental concepts in thermodynamics, representing heat content and the measure of disorder, respectively, in a physical system. Enthalpy changes indicate the heat released or absorbed during a reaction, while entropy measures the dispersal of energy. Both properties are influenced by vibrational frequencies.

When a reaction occurs, there is a change in enthalpy (\(\Delta H\)) and entropy (\(\Delta S\)), which interact to determine the free energy change (\(\Delta G\)). This determines whether a reaction is thermodynamically favorable. Calculating these properties can guide the prediction of reaction outcomes, including whether a reaction is endothermic or exothermic and the balance between order and chaos in a system's energy distribution.
Quantum Chemical Calculations
Quantum Chemical Calculations involve the use of quantum mechanics to determine the electronic structure of molecules. These calculations predict a wide range of properties, such as energy levels, molecular geometries, and reaction pathways. By solving Schrödinger's equation for a system, quantum chemistry offers insights into both the stationary and dynamic aspects of molecular behavior.

Methods such as the aformentioned B3LYP/6-31G* involve approximations because exact solutions are often not feasible for complex molecules. Nevertheless, these calculations provide invaluable information on the fundamental forces governing molecular interactions, helping to explain observations and predict new phenomena.
Kinetic vs Thermodynamic Control
Kinetic versus Thermodynamic Control refers to the divergent paths a reaction can take based on different reaction conditions. Kinetic control often involves lower temperatures and rapid reaction times, steering the reaction towards products that form faster, even if they are not the most stable. In contrast, thermodynamic control favors the most stable products, given sufficient time and higher temperatures, allowing the system to reach the lowest energy state.

Understanding this can predict which product will prevail under certain conditions. For instance, in reactions like those involving methylene chloride and chlorine atoms, the choice between the formation of HCl or DCl is influenced by whether the reaction is kinetically or thermodynamically controlled, and this choice determines the final product distribution at room temperature.

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Most popular questions from this chapter

Hydrazine would be expected to adopt a conformation in which the NH bonds stagger. There are two likely candidates, one with the lone pairs on nitrogen anti to each other and the other with the lone pairs gauche: On the basis of the same arguments made in VSEPR theory (electron pairs take up more space than bonds) you might expect that anti hydrazine would be the preferred structure. a. Obtain energies for the anti and gauche conformers of hydrazine using the HF/6-31G* model. Which is the more stable conformer? Is your result in line with what you expect from VSEPR theory? You can rationalize your result by recognizing that when electron pairs interact they form combinations, one of which is stabilized (relative to the original electron pairs) and one of which is destabilized. The extent of destabilization is greater than that of stabilization, meaning that overall interaction of two electron pairs is unfavorable energetically: b. Measure the energy of the highest occupied molecular orbital (the HOMO) for each of the two hydrazine conformers. This corresponds to the higher energy (destabilized) combination of electron pairs. Which hydrazine conformer (anti or gauche) has the higher HOMO energy? Is this also the higher energy conformer? If so, is the difference in HOMO energies comparable to the difference in total energies between the conformers?

VSEPR (valence state electron pair repulsion) theory was formulated to anticipate the local geometry about an atom in a molecule (see discussion in Section 25.1). All that is required is the number of electron pairs surrounding the atom, broken down into bonded pairs and nonbonded (lone) pairs. For example, the carbon in carbon tetrafluoride is surrounded by four electron pairs, all of them tied up in \(\mathrm{CF}\) bonds, whereas the sulfur in sulfur tetrafluoride is surrounded by five electron pairs, four of which are tied up in SF bonds with the fifth being a lone pair. VSEPR theory is based on two simple rules. The first is that electron pairs (either lone pairs or bonds) will seek to avoid each other as much as possible. Thus, two electron pairs will lead to a linear geometry, three pairs to a trigonal planar geometry, four pairs to a tetrahedral geometry, five pairs to a trigonal bipyramidal geometry, and six pairs to an octahedral geometry. Although this knowledge is sufficient to assign a geometry for a molecule such as carbon tetrafluoride (tetrahedral), it is not sufficient to specify the geometry of a molecule such as sulfur tetrafluoride. Does the lone pair assume an equatorial position on the trigonal bipyramid leading to a seesaw geometry, or an axial position leading to a trigonal pyramidal geometry? The second rule, that lone pairs take up more space than bonds, clarifies the situation. The seesaw geometry in which the lone pair is \(90^{\circ}\) to two of the SF bonds and \(120^{\circ}\) to the other two bonds is preferable to the trigonal pyramidal geometry in which three bonds are \(90^{\circ}\) to the lone pair. Although VSEPR theory is easy to apply, its results are strictly qualitative and often of limited value. For example, although the model tells us that sulfur tetrafluoride adopts a seesaw geometry, it does not reveal whether the trigonal pyramidal structure (or any other structure) is an energy minimum, and if it is, what its energy is relative to the seesaw form. Also it has little to say when more than six electron pairs are present. For example, VSEPR theory tells us that xenon hexafluoride is not octahedral, but it does not tell us what geometry the molecule actually assumes. Hartree-Fock molecular orbital calculations provide an alternative. a. Optimize the structure of \(\mathrm{SF}_{4}\) in a seesaw geometry \(\left(C_{2 v} \text { symmetry }\right)\) using the HF/3-21G model and calculate vibrational frequencies (the infrared spectrum). This calculation is necessary to verify that the energy is at a minimum. Next, optimize the geometry of \(\mathrm{SF}_{4}\) in a trigonal pyramidal geometry and calculate its vibrational frequencies. Is the seesaw structure an energy minimum? What leads you to your conclusion? Is it lower in energy than the corresponding trigonal pyramidal structure in accordance with VSEPR theory? What is the energy difference between the two forms? Is it small enough that both might actually be observed at room temperature? Is the trigonal pyramidal structure an energy minimum? b. Optimize the geometry of \(\mathrm{XeF}_{6}\) in an octahedral geometry \(\left(\mathrm{O}_{\mathrm{h}} \text { symmetry }\right)\) using the HF/3-21G model and calculate vibrational frequencies. Next, optimize \(\mathrm{XeF}_{6}\) in a geometry that is distorted from octahedral (preferably a geometry with \(\left.C_{1} \text { symmetry }\right)\) and calculate its vibrational frequencies. Is the octahedral form of \(\mathrm{XeF}_{6}\) an energy minimum? What leads you to your conclusion? Does distortion lead to a stable structure of lower energy?

Pyramidal inversion in the cyclic amine aziridine is significantly more difficult than inversion in an acyclic amine, for example, requiring \(80 \mathrm{kJ} / \mathrm{mol}\) versus \(23 \mathrm{kJ} / \mathrm{mol}\) in dimethylamine according to HF/6-31G* calculations. One plausible explanation is that the transition state for inversion needs to incorporate a planar trigonal nitrogen center, which is obviously more difficult to achieve in aziridine, where one bond angle is constrained to a value of around \(60^{\circ},\) than it is in dimethylamine. Such an interpretation suggests that the barriers to inversion in the corresponding four- and fivemembered ring amines (azetidine and pyrrolidine) should also be larger than normal and that the inversion barrier in the six-membered ring amine (piperidine) should be quite close to that for the acyclic. Optimize the geometries of aziridine, azetidine, pyrrolidine, and piperidine using the HF/6-31G* model. Starting from these optimized structures, provide guesses at the respective inversion transition states by replacing the tetrahedral nitrogen center with a trigonal center. Obtain transition states using the same Hartree-Fock model and calculate inversion barriers. Calculate vibrational frequencies to verify that you have actually located the appropriate inversion transition states. Do the calculated inversion barriers follow the order suggested in the preceding figure? If not, which molecule(s) appear to be anomalous? Rationalize your observations by considering other changes in geometry from the amine to the transition state.

singlet and triplet carbenes exhibit different properties and show markedly different chemistry. For example, a singlet carbene will add to a cis- disubstituted alkene to produce only \(c i s\) -disubstituted cyclopropane products (and to a trans-disubstituted alkene to produce only trans- disubstituted cyclopropane products), whereas a triplet carbene will add to produce a mixture of cis and trans products. The origin of the difference lies in the fact that triplet carbenes are biradicals (or diradicals) and exhibit chemistry similar to that exhibited by radicals, whereas singlet carbenes incorporate both a nucleophilic site (a low-energy unfilled molecular orbital) and an electrophilic site (a high- energy filled molecular orbital); for example, for singlet and triplet methylene: It should be possible to take advantage of what we know about stabilizing radical centers versus stabilizing empty orbitals and use that knowledge to design carbenes that will either be singlets or triplets. Additionally, it should be possible to say with confidence that a specific carbene of interest will either be a singlet or a triplet and, thus, to anticipate its chemistry. The first step is to pick a model and then to establish the error in the calculated singlet-triplet energy separation in methylene where the triplet is known experimentally to be approximately \(42 \mathrm{kJ} / \mathrm{mol}\) lower in energy than the singlet. This can then be applied as a correction for calculated singlet-triplet separations in other systems. a. Optimize the structures of both the singlet and triplet states of methylene using both Hartree-Fock and B3LYP density functional models with the \(6-31 G^{*}\) basis set. Which state (singlet or triplet) is found to be of lower energy according to the HF/6-31G* calculations? Is the singlet or the triplet unduly favored at this level of calculation? Rationalize your result. (Hint: Triplet methylene contains one fewer electron pair than singlet methylene.) What energy correction needs to be applied to calculated singlet-triplet energy separations? Which state (singlet or triplet) is found to be of lower energy according to the B3LYP/6-31G" calculations? What energy correction needs to be applied to calculated energy separations? b. Proceed with either the HF/6-31G* or B3LYP/6-31G* model, depending on which leads to better agreement for the singlet-triplet energy separation in methylene. Optimize singlet and triplet states for cyanomethylene, methoxymethylene, and cyclopentadienylidene: Apply the correction obtained in the previous step to estimate the singlet-triplet energy separation in each. For each of the three carbenes, assign the ground state as singlet or triplet. Relative to hydrogen (in methylene), has the cyano substituent in cyanomethylene and the methoxy substituent in methoxymethylene led to favoring of the singlet or the triplet? Rationalize your result by first characterizing cyano and methoxy substituents as \(\pi\) donors or \(\pi\) acceptors, and then speculating about how a donor or acceptor would stabilize or destabilize singlet and triplet methylene. Has incorporation into a cyclopentadienyl ring led to increased preference for a singlet or triplet ground state (relative to the preference in methylene)? Rationalize your result. (Hint: Count the number of \(\pi\) electrons associated with the rings in both singlet and triplet states.)

Ammonia provides a particularly simple example of the dependence of vibrational frequencies on the atomic masses and of the use of vibrational frequencies to distinguish between a stable molecule and a transition state. First examine the vibrational spectrum of pyramidal ammonia ("ammonia" on the precalculated Spartan file). a. How many vibrational frequencies are there? How does this number relate to the number of atoms? Are all fre- quencies real numbers or are one or more imaginary numbers? Describe the motion associated with each frequency and characterize each as being primarily bond stretching, angle bending, or a combination of the two. Is bond stretching or angle bending easier? Do the stretching motions each involve a single \(\mathrm{NH}\) bond or do they involve combinations of two or three bonds? b. Next, consider changes to the vibrational frequencies of ammonia as a result of substituting deuteriums for hydrogens ("perdeuteroammonia" on the precalculated Spartan file \() .\) Are the frequencies in \(\mathrm{ND}_{3}\) larger, smaller, or unchanged from those in \(\mathrm{NH}_{3}\) ? Are any changes greater for motions that are primarily bond stretching or motions that are primarily angle bending? c. Finally, examine the vibrational spectrum of an ammonia molecule that has been constrained to a planar geometry ("planar ammonia"' on the Spartan download). Are all the frequencies real numbers? If not, describe the motions associated with any imaginary frequencies and relate them to the corresponding motion(s) in the pyramidal equilibrium form.

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