The structure for a hydrocarbon with the following spectral characteristics is to be identified. Mass spectrum: M + at 120. 1 HNMR spectrum: 7.25Î´ (5H, broad singlet); 2.90Î´ (1H, septet, J=7 Hz); 1.22Î´ (6H, doublet, J=7 Hz). Concept introduction: The molecular ion peak gives the approximate molecular mass of the compound. In 1 HNMR spectrum aromatic protons give a broad peak in the range 6.5Î´-8.0Î´, the primary alkyl protons around 0.7Î´-1.3Î´, the secondary alkyl protons around 1.2Î´-1.6Î´, and a tertiary alkyl protons in between 1.4Î´-1.8Î´. The multiplicity of a signal gives an idea about the protons present in the adjacent carbons. To identify: The structure for a hydrocarbon with the following spectral characteristics. Mass spectrum: M + at 120. 1HNMR spectrum: 7.25Î´ (5H, broad singlet); 2.90Î´ (1H, septet, J=7 Hz); 1.22Î´ (6H, doublet, J=7 Hz).

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Answer 1

Question

Chapter 15.SE, Problem 40AP

Interpretation Introduction

Interpretation:

The structure for a hydrocarbon with the following spectral characteristics is to be identified.

Mass spectrum: M⁺ at 120.

¹HNMR spectrum: 7.25Î´ (5H, broad singlet); 2.90Î´ (1H, septet, J=7 Hz); 1.22Î´ (6H, doublet, J=7 Hz).

Concept introduction:

The molecular ion peak gives the approximate molecular mass of the compound. In ¹HNMR spectrum aromatic protons give a broad peak in the range 6.5Î´-8.0Î´, the primary alkyl protons around 0.7Î´-1.3Î´, the secondary alkyl protons around 1.2Î´-1.6Î´, and a tertiary alkyl protons in between 1.4Î´-1.8Î´. The multiplicity of a signal gives an idea about the protons present in the adjacent carbons.

To identify:

The structure for a hydrocarbon with the following spectral characteristics.

Mass spectrum: M⁺ at 120. 1HNMR spectrum: 7.25Î´ (5H, broad singlet); 2.90Î´ (1H, septet, J=7 Hz); 1.22Î´ (6H, doublet, J=7 Hz).

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Answer 2

Question

Chapter 15.SE, Problem 40AP

Interpretation Introduction

Interpretation:

The structure for a hydrocarbon with the following spectral characteristics is to be identified.

Mass spectrum: M⁺ at 120.

¹HNMR spectrum: 7.25Î´ (5H, broad singlet); 2.90Î´ (1H, septet, J=7 Hz); 1.22Î´ (6H, doublet, J=7 Hz).

Concept introduction:

The molecular ion peak gives the approximate molecular mass of the compound. In ¹HNMR spectrum aromatic protons give a broad peak in the range 6.5Î´-8.0Î´, the primary alkyl protons around 0.7Î´-1.3Î´, the secondary alkyl protons around 1.2Î´-1.6Î´, and a tertiary alkyl protons in between 1.4Î´-1.8Î´. The multiplicity of a signal gives an idea about the protons present in the adjacent carbons.

To identify:

The structure for a hydrocarbon with the following spectral characteristics.

Mass spectrum: M⁺ at 120. 1HNMR spectrum: 7.25Î´ (5H, broad singlet); 2.90Î´ (1H, septet, J=7 Hz); 1.22Î´ (6H, doublet, J=7 Hz).

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Answer 3

Question

Chapter 15.SE, Problem 40AP

Interpretation Introduction

Interpretation:

The structure for a hydrocarbon with the following spectral characteristics is to be identified.

Mass spectrum: M⁺ at 120.

¹HNMR spectrum: 7.25Î´ (5H, broad singlet); 2.90Î´ (1H, septet, J=7 Hz); 1.22Î´ (6H, doublet, J=7 Hz).

Concept introduction:

The molecular ion peak gives the approximate molecular mass of the compound. In ¹HNMR spectrum aromatic protons give a broad peak in the range 6.5Î´-8.0Î´, the primary alkyl protons around 0.7Î´-1.3Î´, the secondary alkyl protons around 1.2Î´-1.6Î´, and a tertiary alkyl protons in between 1.4Î´-1.8Î´. The multiplicity of a signal gives an idea about the protons present in the adjacent carbons.

To identify:

The structure for a hydrocarbon with the following spectral characteristics.

Mass spectrum: M⁺ at 120. 1HNMR spectrum: 7.25Î´ (5H, broad singlet); 2.90Î´ (1H, septet, J=7 Hz); 1.22Î´ (6H, doublet, J=7 Hz).

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Answer 4

Question

Chapter 15.SE, Problem 40AP

Interpretation Introduction

Interpretation:

The structure for a hydrocarbon with the following spectral characteristics is to be identified.

Mass spectrum: M⁺ at 120.

¹HNMR spectrum: 7.25Î´ (5H, broad singlet); 2.90Î´ (1H, septet, J=7 Hz); 1.22Î´ (6H, doublet, J=7 Hz).

Concept introduction:

The molecular ion peak gives the approximate molecular mass of the compound. In ¹HNMR spectrum aromatic protons give a broad peak in the range 6.5Î´-8.0Î´, the primary alkyl protons around 0.7Î´-1.3Î´, the secondary alkyl protons around 1.2Î´-1.6Î´, and a tertiary alkyl protons in between 1.4Î´-1.8Î´. The multiplicity of a signal gives an idea about the protons present in the adjacent carbons.

To identify:

The structure for a hydrocarbon with the following spectral characteristics.

Mass spectrum: M⁺ at 120. 1HNMR spectrum: 7.25Î´ (5H, broad singlet); 2.90Î´ (1H, septet, J=7 Hz); 1.22Î´ (6H, doublet, J=7 Hz).

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Answer 5

Chapter 15.SE, Problem 40AP

Interpretation Introduction

Interpretation:

The structure for a hydrocarbon with the following spectral characteristics is to be identified.

Mass spectrum: M⁺ at 120.

¹HNMR spectrum: 7.25Î´ (5H, broad singlet); 2.90Î´ (1H, septet, J=7 Hz); 1.22Î´ (6H, doublet, J=7 Hz).

Concept introduction:

The molecular ion peak gives the approximate molecular mass of the compound. In ¹HNMR spectrum aromatic protons give a broad peak in the range 6.5Î´-8.0Î´, the primary alkyl protons around 0.7Î´-1.3Î´, the secondary alkyl protons around 1.2Î´-1.6Î´, and a tertiary alkyl protons in between 1.4Î´-1.8Î´. The multiplicity of a signal gives an idea about the protons present in the adjacent carbons.

To identify:

The structure for a hydrocarbon with the following spectral characteristics.

Mass spectrum: M⁺ at 120. 1HNMR spectrum: 7.25Î´ (5H, broad singlet); 2.90Î´ (1H, septet, J=7 Hz); 1.22Î´ (6H, doublet, J=7 Hz).

Answer 6

Chapter 15.SE, Problem 40AP

Interpretation Introduction

Interpretation:

The structure for a hydrocarbon with the following spectral characteristics is to be identified.

Mass spectrum: M⁺ at 120.

¹HNMR spectrum: 7.25Î´ (5H, broad singlet); 2.90Î´ (1H, septet, J=7 Hz); 1.22Î´ (6H, doublet, J=7 Hz).

Concept introduction:

The molecular ion peak gives the approximate molecular mass of the compound. In ¹HNMR spectrum aromatic protons give a broad peak in the range 6.5Î´-8.0Î´, the primary alkyl protons around 0.7Î´-1.3Î´, the secondary alkyl protons around 1.2Î´-1.6Î´, and a tertiary alkyl protons in between 1.4Î´-1.8Î´. The multiplicity of a signal gives an idea about the protons present in the adjacent carbons.

To identify:

The structure for a hydrocarbon with the following spectral characteristics.

Mass spectrum: M⁺ at 120. 1HNMR spectrum: 7.25Î´ (5H, broad singlet); 2.90Î´ (1H, septet, J=7 Hz); 1.22Î´ (6H, doublet, J=7 Hz).

Answer 7

Chapter 15.SE, Problem 40AP

Interpretation Introduction

Interpretation:

The structure for a hydrocarbon with the following spectral characteristics is to be identified.

Mass spectrum: M⁺ at 120.

¹HNMR spectrum: 7.25Î´ (5H, broad singlet); 2.90Î´ (1H, septet, J=7 Hz); 1.22Î´ (6H, doublet, J=7 Hz).

Concept introduction:

The molecular ion peak gives the approximate molecular mass of the compound. In ¹HNMR spectrum aromatic protons give a broad peak in the range 6.5Î´-8.0Î´, the primary alkyl protons around 0.7Î´-1.3Î´, the secondary alkyl protons around 1.2Î´-1.6Î´, and a tertiary alkyl protons in between 1.4Î´-1.8Î´. The multiplicity of a signal gives an idea about the protons present in the adjacent carbons.

To identify:

The structure for a hydrocarbon with the following spectral characteristics.

Mass spectrum: M⁺ at 120. 1HNMR spectrum: 7.25Î´ (5H, broad singlet); 2.90Î´ (1H, septet, J=7 Hz); 1.22Î´ (6H, doublet, J=7 Hz).

Answer 8

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Answer 9

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Answer 11

Ch. 15.1 - Prob. 1P Ch. 15.1 - Give IUPAC names for the following compounds:Ch. 15.1 - Prob. 3P Ch. 15.2 - Pyridine is a flat, hexagonal molecule with bond...Ch. 15.3 - Prob. 5P Ch. 15.4 - Draw the five resonance structures of the...Ch. 15.4 - Prob. 7P Ch. 15.4 - Prob. 8P Ch. 15.5 - Prob. 9P Ch. 15.5 - Prob. 10P

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