/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 36 Hydrocarbons are generally consi... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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.

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 91Ó°ÊÓ!

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.

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

Discussion of the VSEPR model in Section 25.1 suggested a number of failures, in particular, in \(\mathrm{CaF}_{2}\) and \(\mathrm{SrCl}_{2},\) which (according to the VSEPR) should be linear but which are apparently bent, and in \(\operatorname{Se} \mathrm{F}_{6}^{2-}\) and \(\mathrm{TeCl}_{6}^{2-},\) which should not be octahedral but apparently are. Are these really failures or does the discrepancy lie with the fact that the experimental structures correspond to the solid rather than the gas phase (isolated molecules)? a. Obtain equilibrium geometries for linear \(\mathrm{CaF}_{2}\) and \(\mathrm{SrCl}_{2}\) and also calculate vibrational frequencies (infrared spectra). Use the HF/3-21G model, which has actually proven to be quite successful in describing the structures of main-group inorganic molecules. Are the linear structures for \(\mathrm{CaF}_{2}\) and \(\mathrm{SrCl}_{2}\) actually energy minima? Elaborate. If one or both are not, repeat your optimization starting with a bent geometry. b. Obtain equilibrium geometries for octahedral \(\mathrm{SeF}_{6}^{2-}\) and \(\mathrm{TeCl}_{6}^{2-}\) and also calculate vibrational frequencies. Use the HF/3-21G model. Are the octahedral structures for \(\operatorname{Se} \mathrm{F}_{6}^{2-}\) and \(\mathrm{TeCl}_{6}^{2-}\) actually energy minima? Elaborate. If one or both are not, repeat your optimization starting with distorted structures (preferably with \(C_{1}\) symmetry).

Aromatic molecules such as benzene typically undergo substitution when reacted with an electrophile such as \(\mathrm{Br}_{2},\) whereas alkenes such as cyclohexene most commonly undergo addition: What is the reason for the change in preferred reaction in moving from the alkene to the arene? Use the Hartree- Fock \(6-31 G^{*}\) model to obtain equilibrium geometries and energies for reactants and products of both addition and substitution reactions of both cyclohexene and benzene (four reactions in total). Assume trans addition products (1,2-dibromocyclohexane and 5,6-dibromo-1,3-cyclohexadiene). Is your result consistent with what is actually observed? Are all four reactions exothermic? If one or more are not exothermic, provide a rationale as to why.

Further information about the mechanism of the ethyl formate pyrolysis reaction can be obtained by replacing the static picture with a movie, that is, an animation along the reaction coordinate. Bring up "ethyl formate pyrolysis" (on the Spartan download) and examine the change in electron density as the reaction proceeds. Do hydrogen migration and CO bond cleavage appear to occur in concert or is one leading the other?

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.

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.

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.