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Chapter 26: Problem 14

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.)

Short Answer

Expert verified
In this exercise, we optimize the structures of singlet and triplet states of methylene, cyanomethylene, methoxymethylene, and cyclopentadienylidene using computational chemistry software and either HF/6-31G* or B3LYP/6-31G* model. We apply an energy correction factor to estimate singlet-triplet energy separation in each carbene and assign their ground state as singlet or triplet. The preference for singlet or triplet ground states is rationalized based on the characteristics of substituents like 蟺 donor or acceptor properties and incorporation into a cyclopentadienyl ring.

Step by step solution

01

Identify the computational chemistry software

Before solving this problem, it is necessary to have access to computational chemistry software capable of performing calculations with Hartree-Fock and B3LYP density functional models. Examples of such software include Gaussian, ORCA, and Psi4.
02

Optimize the structures of singlet and triplet methylene

Once you have access to the appropriate software, set up and perform geometry optimizations for both singlet and triplet states of methylene using the Hartree-Fock and B3LYP density functional models with the 6-31G* basis set.
03

Compare calculated energies and apply correction

After the optimization of singlet and triplet methylene, find the singlet-triplet energy separation with each model and compare the calculated energies to experimental values. Determine which model leads to a better agreement and calculate the energy correction factor for singlet-triplet energy separations.
04

Optimize structures of singlet and triplet states for cyanomethylene, methoxymethylene, and cyclopentadienylidene

Using the chosen model (either HF/6-31G* or B3LYP/6-31G*), perform geometry optimizations for singlet and triplet states of cyanomethylene, methoxymethylene, and cyclopentadienylidene.
05

Estimate singlet-triplet energy separation in each carbene

Apply the energy correction factor obtained earlier to estimate the singlet-triplet energy separation for each carbene. Assign their ground states as either singlet or triplet.
06

Analyze the effect of substituents on singlet and triplet stability

Analyze the results and explain the favoring of singlet or triplet states in cyanomethylene and methoxymethylene by characterizing cyano and methoxy substituents as 蟺 donors or 蟺 acceptors. Investigate how a donor or acceptor stabilizes or destabilizes singlet and triplet methylene. For cyclopentadienylidene, discuss the influence of incorporation into the cyclopentadienyl ring on singlet or triplet stability relative to methylene. Consider the number of 蟺 electrons associated with the rings in both singlet and triplet states to explain the preference.

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Key Concepts

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

Singlet and Triplet States
Carbenes, which are molecules that contain a neutral carbon atom with two unpaired electrons, can exist in two electronic states known as singlet and triplet states. This distinction is crucial in understanding carbene chemistry. - **Singlet State:** In a singlet state carbene, both electrons are paired and occupy the same orbital. This results in a molecule that is relatively stable and less reactive compared to the triplet state. The singlet state typically participates in reactions involving additions to double bonds. - **Triplet State:** A triplet state carbene contains two unpaired electrons residing in different orbitals, making it a biradical or diradical. This typically results in higher reactivity and is characterized by radical-type reactions. In the triplet state, carbenes can easily interact with other radical species. Understanding these states helps in predicting the chemical behavior of carbenes鈥攚hich can aid in designing reactions that selectively yield products such as cyclopropanes.
Computational Chemistry
Computational chemistry uses computer simulations to solve chemical problems, which allows scientists to explore the properties of molecules that might be difficult to examine experimentally. Software like Gaussian, ORCA, and Psi4 are commonly employed to perform these calculations. - **Geometry Optimization:** This refers to finding the arrangement of atoms in a molecule where the energy is minimized. For carbenes like methylene, optimizations are crucial for both its singlet and triplet states to understand their relative energies and predict reactivity. - **Energy Calculations and Corrections:** Computational methods help calculate the energy levels of different states and apply corrections based on experimental data. For example, calculating the singlet-triplet energy separation in methylene allows for more accurate predictions of carbene behavior in various chemical environments.
Density Functional Theory
Density Functional Theory (DFT) is an essential method in computational chemistry. It is used for calculating the electronic structure of atoms, molecules, and solids. The B3LYP method is a commonly used exchange-correlation function in DFT for systems like carbenes. - **Hartree-Fock vs. B3LYP:** These models are employed to understand which state is lower in energy for methylene. - **Hartree-Fock Approximation:** Provides a basic level of calculation that considers the interaction of electrons within the molecule. - **B3LYP Method:** Adds an approximation of electron correlation, offering an often more accurate and reliable result compared to Hartree-Fock alone. For carbenes, employing these methods allows chemists to not only predict but also rationalize the observed reactivity and stability by examining their electronic structure.
Singlet-Triplet Energy Separation
The energy gap between the singlet and triplet states of a carbene is a critical factor influencing its stability and reactivity. For methylene, the triplet state is experimentally known to be approximately 42 kJ/mol lower in energy than the singlet state. - **Calculating Energy Separation:** The energy difference can be determined using computational methods to optimize both singlet and triplet states. Adjustments are then made by applying a correction factor derived from experimental data. - **Influence of Substituents and Structure:** Different substituents can stabilize or destabilize the singlet or triplet state, altering the energy separation. - **Substituents like Cyano and Methoxy:** By acting as 蟺 donors or acceptors, these substituents shift the energy preference towards one state over the other. - **Effect of Ring Structures:** The integration of a carbene into a structure like a cyclopentadienyl ring can also affect the singlet-triplet energy separation by introducing additional 蟺 electron systems, influencing the chemical properties significantly. Understanding these influences aids chemists in the design of carbenes with specific properties and reactivities.

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Most popular questions from this chapter

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.

For many years, a controversy raged concerning the structures of so-called "electron-deficient" molecules, that is, molecules with insufficient electrons to make normal two-atom, two- electron bonds. Typical is ethyl cation, \(\mathrm{C}_{2} \mathrm{H}_{5}^{+}\) formed from protonation of ethene. Is it best represented as an open Lewis structure with a full positive charge on one of the carbons, or as a hydrogenbridged structure in which the charge is dispersed onto several atoms? Build both open and hydrogen-bridged structures for ethyl cation. Optimize the geometry of each using the B3LYP/6-31G* model and calculate vibrational frequencies. Which structure is lower in energy, the open or hydrogenbridged structure? Is the higher energy structure an energy minimum? Explain your answer.

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.

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?

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?
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