You have bottles containing three isomeric compounds, \(\mathbf{A}, \mathbf{B}\), and \(\mathbf{C}\), each of the formula \(\mathrm{C}_{10} \mathrm{H}_{14}\). The only significant bands in the \(\mathrm{IR}\) spectra of all three compounds are at about 3050,2950 , and \(1610 \mathrm{~cm}^{-1}\). The \({ }^{1} \mathrm{H}\) NMR spectrum of A shows signals at \(\delta 2.13(\mathrm{~s}, 6 \mathrm{H})\) and \(6.88(\mathrm{~s}, 1 \mathrm{H}) ;\) B shows signals at \(\delta 2.18(\mathrm{~s}, 3 \mathrm{H}), 2.25(\mathrm{~s}, 3 \mathrm{H})\), and \(6.89(\mathrm{~s}, 1 \mathrm{H})\); and \(\mathrm{C}\) shows signals at \(\delta 2.11(\mathrm{~s}, 3 \mathrm{H})\), \(2.19(\mathrm{~s}, 3 \mathrm{H}), 2.22(\mathrm{~s}, 6 \mathrm{H})\), and \(6.80(\mathrm{~s}, 2 \mathrm{H})\). The \({ }^{13} \mathrm{C}\) NMR spectrum of \(\mathrm{A}\) shows lines at \(\delta 131.02,133.52\), and \(19.04 \mathrm{ppm}\). Compound \(\mathbf{B}\) shows signals at \(\delta 133.76\), \(133.79,126.88,20.65\), and \(15.78\) ppm. Compound C shows signals at \(\delta 136.15\), \(134.43,131.69,128.29,20.74,20.20\), and \(14.86 \mathrm{ppm}\). What are the structures?

Step by step solution

01

- Analyze IR Spectra

The IR spectra for all three compounds show bands at 3050, 2950, and 1610 cm⁻¹. These absorptions are consistent with aromatic compounds, indicating that all three have benzene rings.

02

- Analyze 1H NMR Spectrum of A

The 1H NMR spectrum of A shows signals at δ 2.13 (s, 6H) and δ 6.88 (s, 1H). The signal at δ 2.13 indicates two equivalent methyl groups (each with 3 hydrogens), and the signal at δ 6.88 suggests a hydrogen bond on a benzene ring.

03

- Analyze 13C NMR Spectrum of A

The 13C NMR spectrum of A shows lines at δ 131.02, 133.52, and 19.04 ppm. The signals at δ 131.02 and 133.52 indicate aromatic carbons, and the signal at δ 19.04 indicates methyl carbons. This suggests A is 1,4-dimethylbenzene.

04

- Analyze 1H NMR Spectrum of B

The 1H NMR spectrum of B shows signals at δ 2.18 (s, 3H), δ 2.25 (s, 3H), and δ 6.89 (s, 1H). The presence of two singlets for the methyl groups and a singlet for the benzene hydrogen suggests a different symmetrical configuration.

05

- Analyze 13C NMR Spectrum of B

The 13C NMR spectrum of B shows lines at δ 133.76, 133.79, 126.88, 20.65, and 15.78 ppm. This supports two slightly different methyl groups and their respective aromatic carbons. B is 1,3-dimethylbenzene.

06

- Analyze 1H NMR Spectrum of C

The 1H NMR spectrum of C shows signals at δ 2.11 (s, 3H), δ 2.19 (s, 3H), δ 2.22 (s, 6H), and δ 6.80 (s, 2H). Four signals total with the two hydrogens at δ 6.80 suggesting an aromatic structure with multiple substituents.

07

- Analyze 13C NMR Spectrum of C

The 13C NMR of C shows signals at δ 136.15, 134.43,131.69, 128.29, 20.74, 20.20, and 14.86 ppm. The distinct chemical shifts for multiple methyl groups indicate a 1,2,4,5-substituted benzene ring. C is 1,2,3,4-tetramethylbenzene.

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Key Concepts

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

IR spectroscopy

IR spectroscopy is an essential tool for identifying functional groups within a molecule. In the provided exercise, the IR spectra of all three compounds show bands at 3050, 2950, and 1610 cm⁻¹. These characteristic absorptions are consistent with the presence of aromatic compounds, suggesting that all three structures contain benzene rings.
The band at approximately 3050 cm⁻¹ is due to C-H stretching vibrations in aromatic rings, while the bands at 2950 and 1610 cm⁻¹ correspond to C-H stretching in alkyl groups and C=C stretching in the aromatic ring, respectively. By comparing these IR signals, we can confirm that the given compounds share a common aromatic framework.

1H NMR spectroscopy

1H NMR spectroscopy provides information about the hydrogen atoms in a molecule. In the exercise, each compound shows unique 1H NMR signals, enabling us to differentiate between isomers. First, compound A shows signals at δ 2.13 (singlet for 6 hydrogens) and δ 6.88 (singlet for 1 hydrogen). This indicates the presence of two equivalent methyl groups and a lone aromatic hydrogen, suggesting a symmetrical substitution pattern.
Compound B shows signals at δ 2.18 and δ 2.25 (each singlet for 3 hydrogens) and δ 6.89 (singlet for 1 hydrogen). The presence of two distinct methyl signals implies a pair of non-equivalent methyl groups attached to the benzene ring.
Lastly, compound C has signals at δ 2.11, δ 2.19, δ 2.22 (singlets for 3, 3, and 6 hydrogens, respectively), and δ 6.80 (singlet for 2 hydrogens). The multiple methyl singlets suggest three distinct environments for the methyl groups, consistent with a multi-substituted aromatic ring.

13C NMR spectroscopy

13C NMR spectroscopy reveals detailed information about the carbon atoms in a molecule. In the exercise, each compound exhibits different 13C NMR spectra that help identify their unique structures. Compound A shows three signals at δ 131.02, 133.52, and 19.04 ppm, indicating aromatic carbons and methyl carbons in a symmetrical arrangement, consistent with 1,4-dimethylbenzene.
Compound B presents five signals at δ 133.76, 133.79, 126.88, 20.65, and 15.78 ppm, supporting the presence of two non-equivalent methyl groups in a less symmetrical structure, implying a 1,3- substitution pattern.
Compound C displays seven signals at δ 136.15, 134.43, 131.69, 128.29, 20.74, 20.20, and 14.86 ppm, indicating a complex aromatic environment with multiple methyl groups, matching with a structure like 1,2,4,5-tetramethylbenzene.

Aromatic compounds

Aromatic compounds are characterized by their stability and ring-like structures with conjugated pi-electrons. The exercise confirms that all three compounds are aromatic due to their IR and NMR spectral features pointing to benzene ring structures. Aromaticity is further supported by the specific chemical shifts observed in both 1H NMR and 13C NMR spectra.
Aromatic rings, such as benzene, display unique reactivity and spectral characteristics due to the delocalized electrons within the ring. These properties help in identifying and confirming the presence of such rings in complex molecules.

Chemical structure analysis

Chemical structure analysis involves integrating data from various spectroscopic techniques to elucidate the exact structure of a compound. In this exercise, we combined information from IR spectroscopy, 1H NMR, and 13C NMR to identify the isomeric compounds. By examining IR spectra, we determined the aromatic nature of the molecules.
1H NMR provided insights into the hydrogen environments, such as methyl groups and aromatic hydrogens. Finally, 13C NMR gave us detailed information about the distinct carbon environments in each isomer, helping to pinpoint the specific substitution patterns on the benzene rings.
This comprehensive analysis approach illustrates how different spectroscopic techniques complement each other to reveal the full chemical structure of a compound. Always consider all available data collectively for accurate identification.