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Chapter 23: Problem 15

Offer plausible explanations of the following facts: a. Aza-2,4-cyclopentadiene (pyrrole) is unstable in acid solution and polymerizes. (Consider the effect of adding a proton to this molecule at the nitrogen and at carbon.) b. 1,3-Diaza-2,4-cyclopentadiene (imidazole) is a much stronger base than 1,3-diazabenzene (pyrimidine). c. The triaminomethyl cation, \(\left(\mathrm{NH}_{2}\right)_{3} \mathrm{C}\), is an exceptionally weak acid.

Short Answer

Expert verified
Pyrrole polymerizes in acid due to loss of stability; imidazole is a stronger base than pyrimidine due to better resonance; the triaminomethyl cation is a weak acid because of strong electron donation.

Step by step solution

01

Understanding pyrrole's behavior in acidic solutions

Pyrrole (aza-2,4-cyclopentadiene) is unstable in acidic solutions due to its susceptibility to protonation. When a proton is added, it can attach to the nitrogen atom or a carbon atom in the ring. Protonation at the nitrogen decreases aromatic stabilization, destabilizing the molecule and leading to polymerization as the structure seeks to regain stability.
02

Comparing Base Strength of Imidazole and Pyrimidine

Imidazole (1,3-diaza-2,4-cyclopentadiene) is a stronger base than pyrimidine (1,3-diazabenzene) because its nitrogen atoms are in a more planar five-membered ring, allowing for a better overlap of p-orbitals and thus more effective conjugation and resonance stabilization of the conjugate acid. In pyrimidine, the nitrogen atoms are in a six-membered aromatic ring structure, where resonance stabilization upon protonation is less significant, making it a weaker base.
03

Evaluating Acidity of Triaminomethyl Cation

The triaminomethyl cation \((\mathrm{NH}_2)_3\mathrm{C}^+\) is an exceptionally weak acid due to the strong electron-donating effect of the three amino groups. These groups increase the electron density on the central carbon atom, placing a positive charge stabilizing effect that opposes deprotonation, thus making it more challenging for the structure to lose a proton.

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Key Concepts

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

Protonation
Protonation is a fundamental concept in organic chemistry, referring to the addition of a proton (H鈦�) to a molecule. When considering pyrrole (aza-2,4-cyclopentadiene), the process of protonation can dramatically influence its stability.
When a proton approaches pyrrole in an acidic solution, it can attach either to the nitrogen atom or to a carbon atom in the ring. Each of these interactions has different consequences for the molecule's stability.
  • Protonation at Nitrogen: While nitrogen can happily accept a proton due to its lone pair, this act disrupts the aromatic stability of pyrrole. Aromatic compounds are characterized by their "ring of stabilization," and adding a proton interferes with this, leading to instability.
  • Protonation at Carbon: This too weakens the aromatic structure, although not as severely as when nitrogen is protonated. Yet, both cases reduce the aromatic stabilization enough to make the molecule unstable.
As a result of this instability, pyrrole can undergo polymerization where it reacts with itself, forming long chains in a bid to regain aromatic stabilization.
Acid-Base Strength
Acid-base strength is pivotal in understanding the behavior of compounds during chemical reactions, particularly when assessing structures like imidazole and pyrimidine. Here, we're diving into why imidazole is a stronger base than pyrimidine.
The strength of a base is determined by its ability to accept protons. For imidazole (1,3-diaza-2,4-cyclopentadiene), its structure is the key factor.
  • Imidazole's 5-Membered Ring: Imidazole has nitrogen atoms incorporated in a five-membered ring, making it planar. This configuration allows for excellent overlap of p-orbitals, enhancing resonance and conjugation when it forms a conjugate acid. The greater the resonance, the more stable the conjugate acid, and subsequently, the stronger the base.
  • Pyrimidine's 6-Membered Ring: In contrast, pyrimidine (1,3-diazabenzene) has nitrogen in a six-membered ring. This limits the resonance stabilization once it accepts a proton, resulting in a weaker base compared to imidazole.
So, resonance and the ring structures make imidazole better at accepting protons, highlighting its stronger basic properties.
Aromatic Stabilization
Aromatic stabilization plays a crucial role in determining the acidity of certain cations, such as the triaminomethyl cation \(\left(\mathrm{NH}_2\right)_3 \mathrm{C}^+\). This cation is atypically a weak acid, and aromatic stabilization concepts can explain why.
Acidity is often judged by how easily a compound can lose a proton. If a structure is stabilized by donating electrons, it resists losing a proton, thus appearing as a weak acid.
  • Electron-Donating Amino Groups: In the triaminomethyl cation, three amino groups donate electrons to the central carbon atom. This electron-donating capacity creates an electron-rich environment that stabilizes the positive charge.
  • Charge Distribution: Because electron donation stabilizes the positive charge, the tendency to lose a proton is minimized. As a result, the cation is a weak acid, less likely to donate a proton compared to typical acids.
Thus, the interplay between electron donation and charge distribution highlights how aromatic structures can stabilize and influence acidity.

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Most popular questions from this chapter

Predict the products expected from the reactions of the following amines with nitrous acid (prepared from \(\mathrm{NaNO}_{2}+\mathrm{HCl}\) in aqueous solution): a. 2 -methylpropanamine b. azacyclopentane c. 2-butenamine d. 3-amino-2,3-dimethyl-2-butanol

a. Explain why 1,3-diazacyclopentadiene (imidazole) is a much stronger acid than azacyclopentadiene (pyrrole). b. Would you expect benzenamine to be a stronger or weaker acid than cyclohexanamine? Give your reasoning.

Assess the possibility of O-alkylation in the reaction of \(\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{2} \mathrm{Br}\) with \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{SO}_{2} \mathrm{NH}^{\ominus} \mathrm{Na}^{\oplus} .\) Give your reasoning.

Prominent peaks in the mass spectrum of a basic nitrogen compound have \(m / e\) values of \(87,72,57\), and 30 . The NMR spectrum shows only three proton resonances, having intensity ratios of \(9: 2: 2\) at \(0.9,1.3\), and \(2.3 \mathrm{ppm}\). Assign a structure to the compound and account for the fragment ions \(m / e 72,57\), and 30 .

Compound A is chiral and is a liquid with the formula \(\mathrm{C}_{5} \mathrm{H}_{11} \mathrm{O}_{2} \mathrm{~N}\). A is insoluble in water and dilute acid but dissolves in sodium hydroxide solution. Acidification of a sodium hydroxide solution of chiral A gives racemic A. Reduction of chiral A with hydrogen over nickel produces chiral compound B of formula \(\mathrm{C}_{5} \mathrm{H}_{13} \mathrm{~N}\). Treatment of chiral B with nitrous acid gives a mixture containing some chiral alcohol \(\mathrm{C}\) and some 2 -methyl-2-butanol. Write structures for compounds \(\mathrm{A}, \mathrm{B}\), and C that agree with all the given facts. Write balanced equations for all the reactions involved. Show your reasoning. In this type of problem, one should work backward from the structures of the final products, analyzing each reaction for the structural information it gives. The key questions to be inferred in the preceding problem are (a) What kind of chiral compound or compounds could give 2-methyl-2-butanol and a chiral alcohol with nitrous acid? (b) What kinds of compounds could give \(\mathrm{B}\) on reduction? (c) What does the solubility behavior of A indicate about the type of compound that it is? (d) Why does chiral A racemize when dissolved in alkali?
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