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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.)

Short Answer

Expert verified
Upon analyzing the electrostatic potential maps of azulene and naphthalene, it is observed that one ring of azulene is more negative and the other ring is more positive, relative to naphthalene. The differences in their electrostatic potential and dipole moments can be rationalized by considering the distribution of π electrons in both molecules. Azulene has 10 π electrons, leading to increased electron density around one ring, making it more negative, and lower electron density around the other ring, making it more positive. This result is consistent with the direction of the dipole moment in azulene (0.8 debye).

Step by step solution

01

Optimize the geometry of azulene and naphthalene

Using a computational chemistry software, like Gaussian, input the molecular structure of azulene and naphthalene and optimize their geometries using the HF/6-31G* method. Make sure to save the optimized structures for the next step.
02

Calculate electrostatic potential maps

Using the optimized structures, calculate the electrostatic potential maps for both azulene and naphthalene. Ensure that the calculation method remains consistent (HF/6-31G*). Save the generated electrostatic potential maps for comparison.
03

Analyze polarity based on electrostatic potential maps

Compare the electrostatic potential maps of azulene and naphthalene side by side, and on the same color scale. Determine if one ring in azulene is more negative and one ring is more positive relative to naphthalene, and identify which ring has which characteristic.
04

Compare the dipole moments

Compare the dipole moments of azulene (0.8 debye) and naphthalene (almost nonpolar). Check if the direction of the dipole moment in azulene is consistent with the electrostatic potential you observed in the previous step.
05

Rationalize the result

Rationalize the differences in electrostatic potential and dipole moments between azulene and naphthalene, considering the distribution of π electrons in both molecules. Count the number of π electrons in each molecule to help with your analysis. In summary, you need to perform geometry optimization, calculate electrostatic potential maps, compare the polarity of azulene and naphthalene, and rationalize the results based on the distribution of π electrons in both molecules.

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Key Concepts

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

Hydrocarbons
Hydrocarbons are organic compounds composed solely of carbon and hydrogen atoms. They are typically nonpolar due to the similar electronegativity values of carbon and hydrogen. This results in an even distribution of electric charge across the molecule. The lack of a strong dipole moment further supports their nonpolarity, usually recording only a few tenths of a debye if there is one.
\(\pi\) electron distribution in hydrocarbons is essential in understanding their chemical behavior. Azulene, a unique hydrocarbon, exhibits a relatively higher dipole moment compared to naphthalene, another hydrocarbon. This is due to the distinct arrangement of \(\pi\) electrons and molecular geometry features like rings that influence polarity.
This understanding helps in comparing and studying the properties of different hydrocarbons and predicting their behavior in chemical reactions or interactions.
Dipole Moments
Dipole moments are vectors that depict the polarity of a molecule's bonds. They represent the separation of positive and negative charges within a molecule. The unit of measurement for dipole moments is the debye (D).
When comparing molecules, like azulene and naphthalene, a significant difference in dipole moments can be noted. Azulene, with a dipole moment of 0.8 debye, is more polar than naphthalene, which is almost nonpolar. This difference stems from the structural and electronic differences between the two molecules.
Analyzing dipole moments gives insight into the electron distribution and geometry of molecules, and can often predict how molecules will interact with electrostatic fields or other polar molecules.
Geometry Optimization
Geometry optimization is a critical step in computational chemistry that configures the spatial arrangement of atoms in a molecule. This process minimizes the potential energy, ensuring a stable configuration for the molecule.
In the context of molecules like azulene and naphthalene, employing a given method, such as HF/6-31G*, for geometry optimization helps in accurate prediction of molecular properties. This involves inputting atomic coordinates into computational software, like Gaussian, and allowing the software to compute the optimal geometric structure.
The optimized geometry not only aids in understanding molecular stability but also in accurately determining properties like electrostatic potential maps and dipole moments, facilitating further analysis and comparison between molecules.
Computational Chemistry
Computational chemistry is a branch of chemistry that uses computer simulations to solve chemical problems. It involves calculating molecular properties and predicting behavior by applying the principles of quantum chemistry and classical mechanics.
Using computational tools such as Gaussian, researchers can perform tasks like geometry optimization and the generation of electrostatic potential maps. This allows for a deeper understanding of molecular interactions that are challenging to study empirically.
In the cases of azulene and naphthalene, computational chemistry provides a way to visualize and interpret the differences in electron distribution and polarity, without physically synthesizing the molecules. It opens doors to theoretical predictions, which can be vital in the fields of material science, pharmacology, and more.

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Most popular questions from this chapter

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.

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.

Lithium provides a very simple example of the effect of oxidation state on overall size. Perform HF/6-31G* calculations on lithium cation, lithium atom, and lithium anion, and compare the three electron density surfaces corresponding to enclosure of \(99 \%\) of the total electron density. Which is smallest? Which is largest? How does the size of lithium relate to the number of electrons? Which surface most closely resembles a conventional space-filling model? What, if anything does this tell you about the kinds of molecules that were used to establish the space-filling radius for lithium?

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.

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