/*! 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 18 Hydrazine would be expected to a... [FREE SOLUTION] | 91Ó°ÊÓ

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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?

Short Answer

Expert verified
The more stable conformer of hydrazine is determined by comparing the energies obtained from HF/6-31G* calculations. In line with VSEPR theory, the anti conformation is expected to be more stable due to lone electron pairs having more space. HOMO energies can be used to further understand their stability; typically, a conformer with lower HOMO energy is more stable. By analyzing the differences in both total and HOMO energies between the conformers, we can assess the implications of electron pair interactions for their relative stability.

Step by step solution

01

Set up and perform calculations using HF/6-31G* model

In order to obtain the energies of the two conformers (anti and gauche), we first need to set up and perform quantum chemical calculations using an appropriate software package such as Gaussian, ORCA, or Psi4. Represent the molecular structure of hydrazine (N2H4) in both conformations (anti and gauche), and perform single-point energy calculations using the Hartree-Fock method with a 6-31G* basis set.
02

Obtain and compare energies of conformers

After running the calculations, extract the total energies of the anti and gauche conformers of hydrazine from the output files. Compare these energies to determine which conformer is more stable (i.e., has lower energy).
03

Compare the calculated energies to VSEPR theory expectations

Based on VSEPR theory, it is expected that the anti conformation is more stable than the gauche conformation, as lone electron pairs would have more space in the anti conformation. Compare the results from the energy calculations to this expectation.
04

Obtain HOMO energies and compare

From the output of the HF/6-31G* calculations, obtain the HOMO energy for each conformation (anti and gauche). Identify which hydrazine conformer has the higher HOMO energy.
05

Assess the relation between HOMO energies, conformer stability, and total energy differences

If the conformer with the higher HOMO energy is also the higher energy conformer, compare the difference in HOMO energies to the difference in total energies between the conformers. Evaluate whether the HOMO energies play a significant role in the stability of the two conformers of hydrazine based on the comparisons made in Steps 2 and 4. By following these steps, we will have a better understanding of the stability of the anti and gauche conformers of hydrazine, their relation to VSEPR theory expectations, and the roles played by HOMO energies in their stability.

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Key Concepts

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

VSEPR Theory
Valence Shell Electron Pair Repulsion (VSEPR) theory is a model used in chemistry to predict the geometry of individual molecules based on the number of electron pairs surrounding their central atoms. The principle behind VSEPR theory is simple: electron pairs around a central atom tend to orient themselves as far apart as possible to minimize repulsion between these negatively charged areas.

When applying VSEPR theory to hydrazine (N2H4), the theory suggests that the molecule might prefer an arrangement where the lone pairs of electrons on the nitrogen atoms are as far from each other as possible. This would lead to the assumption that the 'anti' conformation, where the lone pairs are opposite each other, is more stable than the 'gauche' conformation, where the lone pairs are closer to each other. However, quantum chemical calculations are needed to verify this prediction and to take into account other interactions that might influence stability.
Hartree-Fock Method
The Hartree-Fock method is a foundational algorithm in the field of quantum chemistry used to approximate the wave function and energy of a quantum many-body system. It is a self-consistent field method, meaning that the final solution is obtained through an iterative process where the wave function of each electron is calculated in the field of all the other electrons, under the simplifying assumption that each electron moves independently.

This method involves the use of basis sets like the 6-31G* to describe the orbitals of the electrons. In our exercise, the HF/6-31G* calculation assists in estimating the energies of different hydrazine conformers. By applying the Hartree-Fock method, we can deduce which conformer, anti or gauche, has lower energy and is thus more stable.
HOMO Energy
In molecular orbital theory, the HOMO, or Highest Occupied Molecular Orbital, is the most energetically favorable orbital occupied by electrons. HOMO energy is significant as it indicates the molecule's ability to donate electrons.

In the context of hydrazine conformers, comparing the HOMO energies can be revealing. A higher HOMO energy for one conformer might suggest greater electron pair repulsion, potentially corresponding to a less stable structure. By examining the HOMO energies, we can infer whether the conformer with a higher HOMO is the less stable one, as postulated by the destabilization of electron pair interactions outlined in the exercise.
Quantum Chemical Calculations
Quantum chemical calculations encompass a broad range of computational techniques used to derive the electronic and structural properties of molecules. The goal of these calculations is to solve the Schrödinger equation, or its approximations, for a molecular system to obtain energies, geometries, and other properties.

In our exercise, we are specifically interested in calculating the total energies and HOMO energies of the anti and gauche hydrazine conformers. Performing these calculations allows us to determine the most stable conformation and understand how the theoretical VSEPR predictions correspond with the quantum mechanically calculated energies.
Conformational Analysis
Conformational analysis is the study of the different shapes (conformations) that a molecule can adopt due to the rotation around single bonds. Even small changes in conformation can significantly impact the molecule's stability and reactivity. In the case of hydrazine, the anti and gauche conformers represent two possible shapes that are analyzed to determine the more stable form.

Conformational analysis often involves computational methods such as the Hartree-Fock calculations mentioned to quantitatively compare the energies of different conformers. By relating the computational results to VSEPR theory and considering the HOMO energies, conformational analysis helps to construct a comprehensive understanding of molecular shape and stability.

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Most popular questions from this chapter

Benzyne has long been implicated as an intermediate in nucleophilic aromatic substitution, for example, Although the geometry of benzyne has yet to be conclusively established, the results of a \(^{13} \mathrm{C}\) labeling experiment leave little doubt that two (adjacent) positions on the ring are equivalent: There is a report, albeit controversial, that benzyne has been trapped in a low-temperature matrix and its infrared spectrum recorded. Furthermore, a line in the spectrum at \(2085 \mathrm{cm}^{-1}\) has been assigned to the stretching mode of the incorporated triple bond. Optimize the geometry of benzyne using the HF/6-31G* model and calculate vibrational frequencies. For reference, perform the same calculations on 2 -butyne. Locate the \(\mathrm{C} \equiv \mathrm{C}\) stretching frequency in 2 -butyne and determine an appropriate scaling factor to bring it into agreement with the corresponding experimental frequency \(\left(2240 \mathrm{cm}^{-1}\right) .\) Then identify the vibration corresponding to the triple-bond stretch in benzyne and apply the same scaling factor to this frequency. Finally, plot the calculated infrared spectra of both benzyne and 2-butyne. Does your calculated geometry for benzyne incorporate a fully formed triple bond? Compare with the bond in 2 -butyne as a standard. Locate the vibrational motion in benzyne corresponding to the triple bond stretch. Is the corresponding (scaled) frequency significantly different \(\left(>100 \mathrm{cm}^{-1}\right)\) from the frequency assigned in the experimental investigation? If it is, are you able to locate any frequencies from your calculation that would fit with the assignment of a benzyne mode at \(2085 \mathrm{cm}^{-1} ?\) Elaborate. Does the calculated infrared spectrum provide further evidence for or against the experimental observation? (Hint: Look at the intensity of the triple-bond stretch in 2-butyne.)

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?

Chemists know that nitric and sulfuric acids are strong acids and that acetic acid is a weak acid. They would also agree that ethanol is at best a very weak acid. Acid strength is given directly by the energetics of deprotonation (heterolytic bond dissociation); for example, for acetic acid: \\[ \mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H} \longrightarrow \mathrm{CH}_{3} \mathrm{CO}_{2}^{-}+\mathrm{H}^{+} \\] As written, this is a highly endothermic process, because not only is a bond broken but two charged molecules are created from the neutral acid. It occurs readily in solution only because the solvent acts to disperse charge. Acid strength can be calculated simply as the difference in energy between the acid and its conjugate base (the energy of the proton is 0 ). In fact, acid strength comparisons among closely related systems, for example, carboxylic acids, are quite well described with practical quantum chemical models. This is consistent with the ability of the same models to correctly account for relative base strengths (see discussion in Section \(26.8 .3)\) Another possible measure of acid strength is the degree of positive charge on the acidic hydrogen as measured by the electrostatic potential. It is reasonable to expect that the more positive the potential in the vicinity of the hydrogen, the more easily it will dissociate and the stronger the acid. This kind of measure, were it to prove successful, offers an advantage over the calculation of reaction energy, in that only the acid (and not the conjugate base ) needs to be considered. a. Obtain equilibrium geometries for nitric acid, sulfuric acid, acetic acid, and ethanol using the HF/3-21G model, and compare electrostatic potential maps. Be certain to choose the same (color) scale for the four acids. For which acid is the electrostatic potential in the vicinity of (the acidic) hydrogen most positive? For which is it least positive? Do electrostatic potential maps provide a qualitatively correct account of the relative acid strength of these four compounds? b. Obtain equilibrium geometries for several of the carboxylic acids found in the following table using the HF/3-21G model and display an electrostatic potential map for each. "Measure" the most positive value of the electrostatic potential associated with the acidic hydrogen in each of these compounds and plot this against experimental \(\mathrm{p} K_{\mathrm{a}}\) (given in the preceding table). Is there a reasonable correlation between acid strengths and electrostatic potential at hydrogen in this closely related series of acids?

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.)

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.

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