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Chapter 26: Problem 35

A surface for which the electrostatic potential is negative delineates regions in a molecule that are subject to electrophilic attack. It can help you to rationalize the widely different chemistry of molecules that are structurally similar. Optimize the geometries of benzene and pyridine using the HF/3-21G model and examine electrostatic potential surfaces corresponding to \(-100 \mathrm{kJ} / \mathrm{mol}\). Describe the potential surface for each molecule. Use it to rationalize the following experimental observations: (1) Benzene and its derivatives undergo electrophilic aromatic substitution far more readily than do pyridine and its derivatives; (2) protonation of perdeuterobenzene \(\left(\mathrm{C}_{6} \mathrm{D}_{6}\right)\) leads to loss of deuterium, whereas protonation of perdeuteropyridine \(\left(\mathrm{C}_{5} \mathrm{D}_{5} \mathrm{N}\right)\) does not lead to loss of deuterium; and (3) benzene typically forms \(\pi\) -type complexes with transition models, whereas pyridine typically forms \(\sigma\) -type complexes.

Short Answer

Expert verified
In summary, the electrostatic potential surfaces of benzene and pyridine show that benzene has a more delocalized negative potential, making it more susceptible to electrophilic attack and promoting 蟺-type complex formation with transition metals. Conversely, pyridine has a localized negative charge on the nitrogen atom, which explains its reduced reactivity for electrophilic aromatic substitutions, lack of deuterium loss upon protonation, and preference for 蟽-type complex formation. These differences in electrostatic potential surfaces help rationalize the distinctive chemistry of these structurally similar molecules.

Step by step solution

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1. Optimize molecular geometries using the HF/3-21G model

First, you need to use software like Gaussian or other quantum chemistry software to optimize the molecular geometries of benzene and pyridine by employing the Hartree-Fock (HF) method and the 3-21G basis set. This process will yield optimized geometries for both benzene and pyridine with the lowest possible potential energy.
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2. Examine the electrostatic potential surfaces

Using the optimized geometries obtained in the previous step, visualize the electrostatic potential surfaces of the molecules corresponding to -100 kJ/mol. This can be done using visualization software like GaussView, Chemcraft, or other molecular visualization tools. This will display the areas on the molecules with negative potentials that are susceptible to electrophilic attack.
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3. Compare the electrostatic potential surfaces of benzene and pyridine

Analyze and compare the distribution of negative electrostatic potentials on the surfaces of benzene and pyridine. You will notice that benzene has a more delocalized negative electrostatic potential, whereas pyridine has a localized negative charge on the nitrogen atom.
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4. Rationalize the experimental observations

Based on the comparison of the electrostatic potential surfaces, we can provide explanations for the experimental observations: (1) Benzene and its derivatives undergo electrophilic aromatic substitution more readily than pyridine and its derivatives: This is because the negative electrostatic potential in benzene is more uniformly distributed across its 蟺-electron cloud, making it more susceptible to electrophilic attack. Pyridine's negative charge is localized on the nitrogen atom, which makes electrophilic aromatic substitution less favorable. (2) Protonation of perdeuterobenzene (C6D6) leads to loss of deuterium, while protonation of perdeuteropyridine (C5D5N) does not: In the case of benzene, upon protonation, the deuterium atom can be readily exchanged with other hydrogens, leading to the loss of deuterium. In pyridine, the negative charge is localized on the nitrogen atom, thus protonation will not lead to deuteron exchange easily. (3) Benzene typically forms 蟺-type complexes with transition metals, whereas pyridine typically forms 蟽-type complexes: Due to the delocalized negative electrostatic potential in benzene, it is more likely to form 蟺-type complexes with transition metals by interacting with the entire 蟺-electron cloud. In pyridine, the negative charge is localized on the nitrogen atom, making it preferentially form a 蟽-type complex through the lone electron pair on the nitrogen atom.

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Key Concepts

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

Molecular Geometry Optimization
In the realm of chemistry, the shape or geometry of a molecule significantly influences its reactivity and interactions. Molecular geometry optimization is a computational technique used to find the most stable or lowest energy arrangement of atoms in a molecule. This process eliminates any unnecessary strain or repulsion between atoms and finds an arrangement that allows the molecule to reside in a minimum energy state.

By using quantum chemistry software such as Gaussian, researchers can simulate how different molecular structures would behave without the need for experimental synthesis. During this process, parameters like bond angles and bond lengths are adjusted iteratively until the optimal configuration is achieved. This is essential in predicting how molecules will behave during chemical reactions, and it provides insights into molecular properties.

In the case of benzene and pyridine, geometry optimization ensures that the arrangement of atoms is as close as possible to their natural equilibrium, aiding in subsequent analyses like electrostatic potential mapping.
HF/3-21G Model
The HF/3-21G model stands for Hartree-Fock method with the 3-21G basis set, which is a widely used computational approach in molecular simulations. At its core, Hartree-Fock is a quantum mechanical method that approximates the behavior of electrons in a molecule. It simplifies these interactions by assuming that each electron moves independently in an average field created by the other electrons.

The 3-21G basis set is a part of the model that specifies the level of detail included in the calculation. It divides orbitals into groups (or basis functions) to better represent the electron cloud around an atom. The '3-21G' terminology reflects how this division is made. While more detailed basis sets like 6-31G** exist, 3-21G is often used for initial, less computationally intensive calculations.

This model can efficiently predict molecular properties and behaviors, such as optimized geometries and energies, which are crucial for understanding the chemistry of complex systems like benzene and pyridine. It balances computational efficiency with the accuracy of results.
Electrophilic Aromatic Substitution
Electrophilic Aromatic Substitution (EAS) is a fundamental reaction mechanism for aromatic compounds, wherein an electrophile replaces a hydrogen atom on the aromatic ring. Aromatic systems like benzene are rich in electrons due to their conjugated 蟺-electron clouds, making them prime targets for electrophiles, which are electron-poor species.

The tendency of a compound to undergo EAS is influenced by its electronic density distribution. Benzene's 蟺-electrons are delocalized across the ring, providing a uniform site for electrophiles to attack. This makes benzene highly reactive in EAS.

In contrast, pyridine features a nitrogen atom within its ring, which holds a pair of non-bonded electrons, consequently altering its electron distribution. This electron pair contributes to the ring's basicity but causes the rest of the aromatic system to be less electron-rich compared to benzene. As a result, pyridine is less susceptible to EAS than benzene. This understanding is crucial for predicting reaction outcomes in organic synthesis.
蟺-type and 蟽-type Complexes
蟺-type and 蟽-type complexes refer to different manners by which transition metals can interact with organic molecules. This distinction is essential in coordination chemistry, where transition metals form complexes with various ligands.

蟺-type complexes are characterized by the interaction between a metal and the delocalized 蟺-electron cloud of an aromatic ring like benzene. These complexes are formed because the extended electron cloud provides a broad area for the metal to interact with. Common examples involve transition metals binding to the aromatic rings in compounds like benzene, where the 蟺-bonding offers stability.

In contrast, 蟽-type complexes, often seen with pyridine, involve direct bonding to a metal via sigma bonds. This is due to the presence of localized electron pairs, such as those on the nitrogen atom in pyridine. The 蟽-type interaction occurs through the lone pair, providing a strong, directional bond between the ligand and the metal.

Understanding these interactions enables chemists to manipulate bond formation and predict the structures of metal complexes in synthetic and catalytic processes.

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Most popular questions from this chapter

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.

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.

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.

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?

Hydrocarbons are generally considered to be nonpolar or weakly polar at best, characterized by dipole moments that are typically only a few tenths of a debye. For comparison, dipole moments for molecules of comparable size with heteroatoms are commonly several debyes. One recognizable exception is azulene, which has a dipole moment of 0.8 debye: Optimize the geometry of azulene using the HF/6-31G* model and calculate an electrostatic potential map. For reference, perform the same calculations on naphthalene, a nonpolar isomer of azulene. Display the two electrostatic potential maps side by side and on the same (color) scale. According to its electrostatic potential map, is one ring in azulene more negative (relative to naphthalene as a standard) and one ring more positive? If so, which is which? Is this result consistent with the direction of the dipole moment in azulene? Rationalize your result. (Hint: Count the number of \(\pi\) electrons.)
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