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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.

Short Answer

Expert verified
The preferred reaction for benzene when reacted with Brâ‚‚ is a substitution reaction, forming bromobenzene, while cyclohexene undergoes an addition reaction, forming 1,2-dibromocyclohexane. Using Hartree-Fock 6-31G* calculations to obtain the energies and geometries of reactants and products allows for comparing these values to determine consistency with real observations. In general, the results obtained should be consistent with actual observations. All four reactions might not be exothermic, and if that's the case, the rationale behind this observation would be discussed. Overall, the type of reaction (substitution or addition) depends on the molecule's nature and the electrophile it reacts with.

Step by step solution

01

Understand the types of reactions involved

In this exercise, there are two types of reactions involved: 1. Addition reaction: When an electrophile reacts with a double bond, resulting in a saturated compound. 2. Substitution reaction: When an electrophile replaces an atom or a group of atoms in a molecule with another atom or group of atoms. Here, we need to analyze the reactions for both benzene and cyclohexene.
02

Analyze benzene reaction with Brâ‚‚

Benzene is an aromatic compound, and it typically undergoes substitution reactions when reacted with electrophiles. In the case of Brâ‚‚, benzene reacts with Brâ‚‚ and forms bromobenzene. We need to determine the equilibrium geometry and energy of benzene and the product molecules using Hartree-Fock 6-31G* calculations.
03

Analyze cyclohexene reaction with Brâ‚‚

Cyclohexene is an alkene and typically undergoes addition reactions when reacted with electrophiles. In the case of Brâ‚‚, cyclohexene reacts with Brâ‚‚ and forms 1,2-dibromocyclohexane. We need to determine the equilibrium geometry and energy of cyclohexene and the product molecules using Hartree-Fock 6-31G* calculations.
04

Compare the calculated energies and geometries

Once you have calculated the geometries and energies using the Hartree-Fock 6-31G* model, compare the results obtained for the addition and substitution reactions of both benzene and cyclohexene.
05

Determine the consistency of results and exothermicity

Based on the results obtained in step 4, analyze: 1. If the results are consistent with the actual observations. 2. If all four reactions are exothermic. If not, provide a rationale.
06

Write the conclusion

Summarize your findings regarding the preferred reactions for benzene and cyclohexene, the consistency of your results with actual observations, and whether the reactions are exothermic or not.

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Key Concepts

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

Aromatic Substitution Reactions
Understanding the fundamentals of aromatic substitution reactions is crucial for students of organic chemistry. These reactions involve an electrophile replacing a hydrogen atom on an aromatic ring such as benzene. Unlike alkenes, which readily participate in addition reactions, aromatic rings maintain their stability by undergoing substitution. This stability arises from the delocalized \textbf{pi} electrons across the ring, forming a resonance-stabilized structure.

The merit of substitution over addition for aromatics can be attributed to the preservation of the \textbf{aromaticity}, which is the key feature that provides the inherent stability of these compounds. So, when benzene reacts with bromine (\textbf{Br}\(_2\)), it undergoes an electrophilic aromatic substitution, yielding bromobenzene without disrupting the aromatic system. This process generally involves the formation of a sigma complex, where the aromaticity is temporarily lost, followed by a loss of a proton to restore aromaticity.

Key Points in Electrophilic Aromatic Substitution:

  • Aromatic rings retain stability through substitution, not addition.
  • Substitution maintains the aromatic character of the compound.
  • The reaction typically progresses via a sigma complex intermediate.
To support the understanding of this reaction’s preference, computational methods like Hartree-Fock calculations can help predict the energetics and feasibility of the reaction path.
Addition Reactions
Alkenes like cyclohexene are characterized by the presence of a carbon-carbon double bond, which is highly reactive towards electrophiles. Addition reactions occur when electrophiles and nucleophiles add to the two carbon atoms involved in the double bond. The hallmark of these reactions is the conversion of the double bond (\textbf{C=C}) into single bonds (\textbf{C-C}), leading to a saturated molecule.

In contrast to substitution reactions, addition reactions involve the breaking of the double bond and the formation of new single bonds. When cyclohexene reacts with bromine (\textbf{Br}\(_2\)), it results in 1,2-dibromocyclohexane through a trans addition, implying that the bromine atoms add on opposite sides of the double bond. This reaction is typically exothermic, as it leads to the formation of more stable sigma bonds from the higher-energy pi bond.

Key Aspects of Addition Reactions:

  • Double bonds are altered, creating new single bonds.
  • Addition is favored over substitution in non-aromatic alkenes.
  • The reaction often results in the release of energy (exothermic).
Computational simulations, such as those with Hartree-Fock 6-31G* models, enable deeper insight into the transition states and energy profiles of addition reactions.
Hartree-Fock 6-31G* Calculations
The Hartree-Fock 6-31G* approach offers a valuable computational tool for predicting the structures and energies of molecules undergoing chemical reactions. In our exercise, we use it to explore the equilibrium geometries and energies for the reactants and products of both addition and substitution reactions involving cyclohexene and benzene. This quantum chemical method provides a theoretical framework to determine the electronic structure of molecules through computational approximations.

Using the Hartree-Fock method with the 6-31G* basis set, which incorporates *polarization functions* improving accuracy, allows for the calculation of potential energy surfaces. This assists in understanding which reaction pathways are energetically favorable. To assess if a reaction is exothermic, one compares the total energy of the reactants with that of the products; a lower energy for products indicates an exothermic process.

Importance of Hartree-Fock Calculations:

  • They assist in predicting equilibrium geometries and electronic energies.
  • The 6-31G* basis set enhances the accuracy of these predictions.
  • Provides insight into reaction exothermicity and feasibility.
Using these computations to support experimental observations, one can explain the preference for certain reaction types and their energetic profiles. Resolving the reactions of benzene and cyclohexene illustrates the practical use of this chemistry-specific computational framework.

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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.)

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.

One of the most powerful attractions of quantum chemical calculations over experiments is their ability to deal with any molecular system, stable or unstable, real or imaginary. Take as an example the legendary (but imaginary) kryptonite molecule. Its very name gives us a formula, \(\mathrm{KrO}_{2}^{2-}\), and the fact that this species is isoelectronic with the known linear molecule, \(\mathrm{KrF}_{2}\), suggests that it too should be linear. a. Build \(\mathrm{KrF}_{2}\) as a linear molecule \((\mathrm{F}-\mathrm{Kr}-\mathrm{F}),\) optimize its geometry using the HF/6-31G* model, and calculate vibrational frequencies. Is the calculated KrF bond distance close to the experimental value \((1.89 \AA) ?\) Does the molecule prefer to be linear or does it want to bend? Explain how you reached this conclusion. b. Build \(\mathrm{KrO}_{2}^{2-}\) as a linear molecule (or as a bent molecule if the preceding analysis has shown that \(\mathrm{KrF}_{2}\) is not linear), optimize its structure using the HF/6-31G* model, and calculate vibrational frequencies. What is the structure of \(\mathrm{KrO}_{2}^{2-} ?\)

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?

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?

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