/*! 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 26 It is well known that cyanide ac... [FREE SOLUTION] | 魅影直播

魅影直播

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.

Short Answer

Expert verified
The optimized geometry of cyanide reveals that the HOMO is concentrated on the carbon atom, supporting its behavior as a carbon nucleophile in S_N2 reactions. This result does not contradict the relative electronegativities of carbon and nitrogen because the triple bond between C and N delocalizes the electrons, allowing carbon to act as a nucleophile. The optimized geometry of methyl iodide shows that the LUMO is concentrated around the carbon-iodine bond. This suggests that the iodide's leaving group tendency can be anticipated, as the LUMO can accept electron density from cyanide's HOMO, leading to the breaking of the C-I bond and iodide's departure in S_N2 reactions.

Step by step solution

01

Optimize the geometry of cyanide

The first step is to optimize the geometry of cyanide using the given Hartree-Fock (HF) method with the 3-21G basis set. After performing the optimization, analyze the resulting structure and pay special attention to the bonding between the carbon and nitrogen atoms.
02

Examine the HOMO of cyanide

Once the geometry is optimized, examine the HOMO of the cyanide molecule. Identify the shape of the HOMO and where the electron density is concentrated. Is the electron density more concentrated on the carbon or the nitrogen atom?
03

Relate the HOMO of cyanide to its behavior as a carbon nucleophile

Based on your observation from Step 2, determine whether the HOMO of cyanide supports its behavior as a carbon nucleophile. Explain how the location of the electron density in the HOMO is consistent with cyanide's behavior in S_N2 reactions. Discuss why this result does not contradict the relative electronegativities of carbon and nitrogen.
04

Optimize the geometry of methyl iodide

Now, perform the geometry optimization for methyl iodide using the given HF/3-21G method. Analyze the resulting structure and focus on the bonding between the carbon and iodine atoms.
05

Examine the LUMO of methyl iodide

After optimizing the geometry, examine the LUMO of the methyl iodide molecule. Describe the shape of the LUMO and identify areas where the electron density is most likely to be accepted.
06

Relate the LUMO of methyl iodide to iodide's leaving group tendency

Based on your observation from Step 5, determine whether the LUMO of methyl iodide can anticipate the loss of iodide following the attack by cyanide. Explain how the LUMO's characteristics can provide an understanding of why iodide leaves after the nucleophilic attack by cyanide in S_N2 reactions.

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 魅影直播!

Key Concepts

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

Cyanide Ion
The cyanide ion, represented as CN鈦, is a simple but potent nucleophile. This means it has a high tendency to donate a pair of electrons during chemical reactions. Interestingly, in nucleophilic substitution, it's the carbon atom in cyanide that's the active player, even though nitrogen is more electronegative. This behavior might seem counterintuitive at first. Electronegativity refers to an atom's ability to attract and hold onto electrons. Since nitrogen is more electronegative than carbon, one might expect it to be the nucleophile. However, the cyanide ion's unique molecular structure plays a key role in its behavior.

The highest occupied molecular orbital (HOMO) of cyanide is largely concentrated on the carbon atom. This means that the electron density, which is crucial for making nucleophilic attacks, is more available near carbon. Thus, cyanide acts as a carbon nucleophile in many reactions, consistent with what we observe chemically.
Electronegativity
Electronegativity is a fundamental concept in chemistry that explains how different atoms attract electrons. It's typically measured on a scale, with fluorine being the most electronegative element. When considering electronegativity in cyanide, we see an intriguing contrast.

Despite nitrogen being more electronegative, the overall electron distribution within the cyanide ion is such that carbon becomes the reactive center. This happens because molecular orbital interactions can override simple electronegativity values. Essentially, it's not just about how strongly nitrogen can hold electrons, but also about which atom is more ready to form new bonds. The real-world implications mean that even though nitrogen has the natural pull for electrons, the structure of cyanide enables the carbon to act as the nucleophile.
Molecular Orbital Theory
Molecular Orbital Theory helps us understand the behavior of electrons in a molecule by considering them as part of a whole entity, rather than just individual atoms. In cyanide, applying this theory allows us to predict and explain its reactivity patterns.

By examining the molecular orbitals, specifically the HOMO, we can see where the cyanide's electrons are most "ready" to participate in reactions. The HOMO is primarily localized on the carbon atom, making it the site of nucleophilic attack despite nitrogen's higher electronegativity.
  • The HOMO indicates potential reactivity spots within a molecule, helping predict and rationalize chemical behavior.
  • Understanding which atom holds the HOMO can help predict which part of the molecule will react during a chemical process.
This theory provides deep insights into molecular behavior beyond simple electronegativity.
Geometry Optimization
Geometry optimization is a computational technique used to find the most stable structure of a molecule. In the context of cyanide, this involves using the Hartree-Fock method with the 3-21G basis set to refine the molecular geometry until it reaches a minimum energy configuration. Once optimized, this setup helps us view the electron distribution and bonding characteristics more clearly.

The optimal geometry reveals why carbon serves as the nucleophilic center. For example, in methyl iodide, optimizing the geometry allows us to examine the lowest unoccupied molecular orbital (LUMO), predicting its behavior during a nucleophilic substitution. The LUMO reveals areas where the molecule can accept electrons, thus understanding the bond cleavage during reactions like the leaving of iodide.
  • Optimization provides a snapshot of the molecular structure at its most energetically favorable state.
  • This data guides chemists in predicting which parts of a molecule will be reactive.
Overall, geometry optimization aids in visualizing and rationalizing molecular reactions.

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.)

singlet and triplet carbenes exhibit different properties and show markedly different chemistry. For example, a singlet carbene will add to a cis- disubstituted alkene to produce only \(c i s\) -disubstituted cyclopropane products (and to a trans-disubstituted alkene to produce only trans- disubstituted cyclopropane products), whereas a triplet carbene will add to produce a mixture of cis and trans products. The origin of the difference lies in the fact that triplet carbenes are biradicals (or diradicals) and exhibit chemistry similar to that exhibited by radicals, whereas singlet carbenes incorporate both a nucleophilic site (a low-energy unfilled molecular orbital) and an electrophilic site (a high- energy filled molecular orbital); for example, for singlet and triplet methylene: It should be possible to take advantage of what we know about stabilizing radical centers versus stabilizing empty orbitals and use that knowledge to design carbenes that will either be singlets or triplets. Additionally, it should be possible to say with confidence that a specific carbene of interest will either be a singlet or a triplet and, thus, to anticipate its chemistry. The first step is to pick a model and then to establish the error in the calculated singlet-triplet energy separation in methylene where the triplet is known experimentally to be approximately \(42 \mathrm{kJ} / \mathrm{mol}\) lower in energy than the singlet. This can then be applied as a correction for calculated singlet-triplet separations in other systems. a. Optimize the structures of both the singlet and triplet states of methylene using both Hartree-Fock and B3LYP density functional models with the \(6-31 G^{*}\) basis set. Which state (singlet or triplet) is found to be of lower energy according to the HF/6-31G* calculations? Is the singlet or the triplet unduly favored at this level of calculation? Rationalize your result. (Hint: Triplet methylene contains one fewer electron pair than singlet methylene.) What energy correction needs to be applied to calculated singlet-triplet energy separations? Which state (singlet or triplet) is found to be of lower energy according to the B3LYP/6-31G" calculations? What energy correction needs to be applied to calculated energy separations? b. Proceed with either the HF/6-31G* or B3LYP/6-31G* model, depending on which leads to better agreement for the singlet-triplet energy separation in methylene. Optimize singlet and triplet states for cyanomethylene, methoxymethylene, and cyclopentadienylidene: Apply the correction obtained in the previous step to estimate the singlet-triplet energy separation in each. For each of the three carbenes, assign the ground state as singlet or triplet. Relative to hydrogen (in methylene), has the cyano substituent in cyanomethylene and the methoxy substituent in methoxymethylene led to favoring of the singlet or the triplet? Rationalize your result by first characterizing cyano and methoxy substituents as \(\pi\) donors or \(\pi\) acceptors, and then speculating about how a donor or acceptor would stabilize or destabilize singlet and triplet methylene. Has incorporation into a cyclopentadienyl ring led to increased preference for a singlet or triplet ground state (relative to the preference in methylene)? Rationalize your result. (Hint: Count the number of \(\pi\) electrons associated with the rings in both singlet and triplet states.)

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.

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.