Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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 π-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 σ-type complex formation. These differences in electrostatic potential surfaces help rationalize the distinctive chemistry of these structurally similar molecules.

Step by step solution

01

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.
02

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.
03

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.
04

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 π-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 π-type complexes with transition metals, whereas pyridine typically forms σ-type complexes: Due to the delocalized negative electrostatic potential in benzene, it is more likely to form π-type complexes with transition metals by interacting with the entire π-electron cloud. In pyridine, the negative charge is localized on the nitrogen atom, making it preferentially form a σ-type complex through the lone electron pair on the nitrogen atom.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

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 π-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 π-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.
π-type and σ-type Complexes
π-type and σ-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.

π-type complexes are characterized by the interaction between a metal and the delocalized π-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 π-bonding offers stability.

In contrast, σ-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 σ-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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

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.)

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.

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?

It is well known that cyanide acts as a "carbon" and not a "nitrogen" nucleophile in \(\mathrm{S}_{\mathrm{N}} 2\) reactions, for example, How can this behavior be rationalized with the notion that nitrogen is in fact more electronegative than carbon and, therefore, would be expected to hold any excess electrons? a. Optimize the geometry of cyanide using the HF/3-21G model and examine the HOMO. Describe the shape of the HOMO of cyanide. Is it more concentrated on carbon or nitrogen? Does it support the picture of cyanide acting as a carbon nucleophile? If so, explain why your result is not at odds with the relative electronegativities of carbon and nitrogen. Why does iodide leave following nucleophilic attack by cyanide on methyl iodide? b. Optimize the geometry of methyl iodide using the HF/3-21G model and examine the LUMO. Describe the shape of the LUMO of methyl iodide. Does it anticipate the loss of iodide following attack by cyanide? Explain.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free