The carbon-bound hydrogens of pyrrole (p. 236) appear at \(\delta 6.05\) and \(6.62 \mathrm{ppm}\) in the \({ }^{1} \mathrm{H} \mathrm{NMR}\) spectrum, upfield of the typical aromatic signal region. Does this observation mean that pyrrole is not aromatic?

Short Answer

Expert verified
No, pyrrole is aromatic. The shifts are affected by the electron density due to nitrogen.

Step by step solution

01

Understand the Context

Given the chemical shift values \( \delta 6.05 \mathrm{ppm} \) and \( \delta 6.62 \mathrm{ppm} \), the question asks to determine the aromaticity of pyrrole based on its \({ }^{1} \mathrm{H} \mathrm{NMR} \) spectrum.
02

Identify Typical Aromatic Range

Aromatic protons typically resonate in the \( \delta 6.00 - \delta 8.50 \mathrm{ppm} \) range in \({ }^{1} \mathrm{H} \mathrm{NMR} \). Compare these values with the provided \( \delta 6.05 \mathrm{ppm} \) and \( \delta 6.62 \mathrm{ppm} \).
03

Analyze the NMR Data

Notice that the hydrogen signals observed at \( \delta 6.05 \mathrm{ppm} \) and \( \delta 6.62 \mathrm{ppm} \) fall within the typical aromatic proton range, although they are on the lower end.
04

Consider Electronic Effects

Pyrrole is a heterocyclic aromatic compound with nitrogen, which can affect electron density and shift the hydrogen signals slightly upfield compared to benzene. This aligns with the observed values.
05

Determine Aromaticity

Given that the \({ }^{1} \mathrm{H} \mathrm{NMR} \) signals fall close to the aromatic range, and taking into account the electron-donating nature of nitrogen in pyrrole, conclude that pyrrole is aromatic.

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Key Concepts

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

Pyrrole NMR Data
To understand the aromaticity of pyrrole, let's first dive into its Nuclear Magnetic Resonance (NMR) data. NMR spectroscopy is a powerful tool to analyze and identify organic compounds because it tells us about the chemical environment around hydrogen atoms (or other nuclei) in a molecule. For pyrrole, the carbon-bound hydrogens show peaks at \(\delta 6.05\mathrm{ppm}\) and \(\delta 6.62\mathrm{ppm}\). These values are crucial because they help us determine if pyrrole is aromatic.

When these hydrogens experience the magnetic field in the NMR spectrometer, the chemical shift values they produce are due to their electronic environment. In this case, we observe that pyrrole's hydrogens resonate between \(\delta 6.05\mathrm{ppm}\) and \(\delta 6.62\mathrm{ppm}\), which will be important for our next steps in analyzing whether pyrrole is aromatic or not.
Aromatic Proton Chemical Shift
In NMR spectroscopy, the chemical shift of aromatic protons usually falls within a specific region, typically between \( \delta 6.00\mathrm{ppm} \) and \(\ \delta 8.50 \mathrm{ppm} \). This range helps chemists identify aromatic compounds because aromatic rings have delocalized electrons that create a unique magnetic environment.

After comparing, we see that pyrrole's \( \delta 6.05 \mathrm{ppm} \) and \( \delta 6.62 \mathrm{ppm} \) signals fall within this aromatic region, though they are on the lower end. This implies that pyrrole's hydrogens are experiencing an aromatic magnetic environment, but some electron density effects slightly shift the signals upfield.

Nitrogen, present in pyrrole, donates electron density. This donation affects the magnetic shielding of the carbon-bound hydrogens, explaining why their chemical shift is at the lower limit of the typical aromatic values.
Heterocyclic Aromatic Compounds
Now, let’s explore heterocyclic aromatic compounds like pyrrole. These compounds contain a ring structure with at least one atom that is not carbon, such as nitrogen in pyrrole. Despite the presence of a heteroatom, these rings can still be aromatic if they follow Hückel's rule of aromaticity.

Hückel's rule states that for a molecule to be aromatic, it must have \(4n+2\) pi electrons, where \(n\) is a non-negative integer. In pyrrole, the ring includes 4 carbon atoms and 1 nitrogen atom, contributing to a total of 6 pi electrons (\(4 \times 1 + 2\)). As \(4n+2=6\) when \(n=1\), pyrrole satisfies Hückel's rule, confirming its aromaticity.

Furthermore, the aromatic nature of pyrrole influences its chemical shifts in NMR data. The nitrogen atom in the ring causes a slight modification in electron density around the hydrogens, affecting their magnetic environment. Therefore, the unique placement of the \( \delta 6.05 \mathrm{ppm} \) and \( \delta 6.62 \mathrm{ppm} \) shifts provides a telltale sign that even as a heterocyclic compound, pyrrole remains aromatic.

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Most popular questions from this chapter

The magnetic field strength is measured in units of tesla (T). Calculate the energy involved in an NMR transition at a field strength of \(4.7 \mathrm{~T}\). For \(\mathrm{H}, \gamma\) is \(2.7 \times 10^{8} \mathrm{~T}^{-1} \mathrm{~s}^{-1}\), and Planck's constant is \(9.5 \times 10^{-14}(\mathrm{kcal} / \mathrm{mol})(\mathrm{s})\).

Methoxyethene (methyl vinyl ether) has four signals in its \({ }^{1} \mathrm{H}\) NMR spectrum. The signals are at \(\delta=3.2 \mathrm{ppm}\) \((3 \mathrm{H}), \delta=3.9 \mathrm{ppm}(1 \mathrm{H}), \delta=4.0 \mathrm{ppm}(1 \mathrm{H})\), and \(\delta=6.4 \mathrm{ppm}\) (1H). Assign these signals as accurately as possible and explain your reasoning. Predict the coupling for each signal.

Compounds \(\mathbf{A}\) and \(\mathbf{B}\), containing only \(\mathrm{C}, \mathrm{H}\), and \(\mathrm{O}\), gave \(63.15 \% \mathrm{C}\) and \(5.30 \% \mathrm{H}\) upon elemental analysis. Their mass spectra do not show a molecular ion, but freezing point depression data indicate a molecular weight of \(150 \mathrm{~g} / \mathrm{mol}\). A reaction that converts alkenes to alkanes transforms both \(\mathbf{A}\) and \(\mathbf{B}\) into \(\mathbf{C}\). Spectral data for compounds \(\mathbf{A}, \mathbf{B}\), and \(\mathbf{C}\) are summarized below. Assign structures for compounds \(\mathbf{A}, \mathbf{B}\), and \(\mathbf{C}\) and explain your reasoning. Compound A IR (Nujol): \(1845(\mathrm{~m}), 1770(\mathrm{~s}) \mathrm{cm}^{-1}\) \({ }^{1} \mathrm{H} \mathrm{NMR}\left(\mathrm{CDCl}_{3}\right): \delta 2.03-2.90(\mathrm{~m}, 2 \mathrm{H})\) \(3.15-3.70(\mathrm{~m}, 1 \mathrm{H})\) \(5.65-6.25(\mathrm{~m}, 1 \mathrm{H})\) Compound B IR (Nujol): \(1825(\mathrm{~m}), 1755(\mathrm{~s}) \mathrm{cm}^{-1}\) \({ }^{1} \mathrm{H} \mathrm{NMR}\left(\mathrm{CDCl}_{3} / \mathrm{DMSO}-d_{6}\right): \delta 1.50-2.10(\mathrm{~m}, 1 \mathrm{H})\) 2.15-2.70 \((\mathrm{m}, 1 \mathrm{H})\) Compound C IR (neat): \(1852(\mathrm{~m}), 1786(\mathrm{~s}) \mathrm{cm}^{-1}\) \({ }^{1} \mathrm{H} \mathrm{NMR}\left(\mathrm{CDCl}_{3} / \mathrm{DMSO}-d_{6}\right): \delta 1.10-2.30(\mathrm{~m}, 4 \mathrm{H})\) \(3.05-3.55(\mathrm{~m}, 1 \mathrm{H})\)

Upon direct photolysis or heating at \(220^{\circ} \mathrm{C}\), the dimer of 2,5 -dimethyl-3,4-diphenyl-2,4-cyclopentadien1-one (A) loses carbon monoxide and yields compound B, mp \(166-168^{\circ} \mathrm{C}\). Spectral data for compound \(\mathbf{B}\) are summarized below. Deduce the structure of compound \(\mathbf{B}\) and assign the peaks. Compound B Mass spectrum: \(\mathrm{m} / \mathrm{z}=492(\mathrm{M})\) IR \(\left(\mathrm{CCl}_{4}\right): 1704 \mathrm{~cm}^{-1}\) \({ }^{1} \mathrm{H}\) NMR \(\left(\mathrm{CDCl}_{3}\right): \delta 0.73(\mathrm{~s}, 3 \mathrm{H})\) $$ \begin{aligned} &0.73(s, 3 H) \\ &0.92(s, 3 H) \\ &1.51(s, 3 H) \\ &1.88(s, 3 H) \\ &6.6-7.5(\mathrm{~m}, 20 \mathrm{H}) \end{aligned} $$

Compound A was isolated from the bark of the sweet birch (Betula lenta). Chemical test (soluble in \(5 \%\) aqueous \(\mathrm{NaOH}\) solution but not in \(5 \%\) aqueous \(\mathrm{NaHCO}_{3}\) solution) for compound \(\mathbf{A}\) is positive for phenol (p. 233). The spectral data for compound \(\mathbf{A}\) are summarized below. Deduce the structure of compound A. (Be sure to assign the aromatic resonances in the \({ }^{1} \mathrm{H}\) NMR spectrum. Note that long- range coupling was not observed.) Hint: The \(\mathrm{p} K_{\mathrm{a}}\) of phenol is about 10, and that of benzoic acid is about \(4.2 .\) Compound A Mass spectrum: \(m / z=152(\mathrm{M}, 49 \%), 121(29 \%), 120\) (100\%), 92 (54\%) IR (neat): 3205 (br), 1675 (s), 1307 (s), 1253 (s), 1220 (s), and \(757(\mathrm{~s}) \mathrm{cm}^{-1}\) \({ }^{1} \mathrm{H} \mathrm{NMR}\left(\mathrm{CDCl}_{3}, 300 \mathrm{MHz}\right): \delta 3.92(\mathrm{~s}, 3 \mathrm{H})\) $$ \begin{aligned} &6.85(\mathrm{t}, J=8 \mathrm{~Hz}, 1 \mathrm{H}) \\ &7.00(\mathrm{~d}, J=8 \mathrm{~Hz}, 1 \mathrm{H}) \\ &7.44(\mathrm{t}, J=8 \mathrm{~Hz}, 1 \mathrm{H}) \\ &7.83(\mathrm{~d}, J=8 \mathrm{~Hz}, 1 \mathrm{H}) \\ &10.8(\mathrm{~s}, 1 \mathrm{H}) \end{aligned} $$ $$ { }^{13} \mathrm{C} \text { NMR }\left(\mathrm{CDCl}_{3}\right): \delta 52.1 \text { (g), } 112.7 \text { (s), } 117.7 \text { (d), } 119.2 $$ \({ }^{13} \mathrm{C}\) NMR \(\left(\mathrm{CDCl}_{3}\right): \delta 52.1(\mathrm{q}), 112.7(\mathrm{~s}), 117.7(\mathrm{~d}), 119.2\) (d), \(130.1\) (d), \(135.7\) (d), \(162.0\) (s), \(170.7\) (s) (hydrogen coupling indicated in parentheses) Use Organic Reaction Animations (ORA) to answer the following questions:

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