Chapter 17.SE, Problem 55AP

A compound of unknown structure gave the following spectroscopic data:

Mass spectrum: M⁺=88.1

IR: 3600 cm^-1

¹Î—NMR: 1.4 δ (2 H, quartet, J=7 Hz); 1.2 δ (6H, singlet): 1.0 δ (1 H, singlet); 0.9 δ (3 H, triplet, J=7 Hz)

¹³CNMR: 74, 35, 27, 25 δ

(a) Assuming that the compound contains C and H but may or may not contain O, give three possible molecular formulas

(b) How many protons (H) does the compound contain?

(c) What functional groups(s) does the compound contain?

(d) How many carbons does the compound contain?

(e) What is the molecular formula of the compound?

(f) What is the structure of the compound?

(g) Assign peaks in the molecule’s ¹HNMR spectrum corresponding to specific protons.

Interpretation Introduction

a) Three molecular formulas possible for the compound are to be given assuming that the compound contains C and H but may or may not contain oxygen.

Interpretation:

A compound has the spectral data:

Mass spectrum: M+=88.1; IR=3600 cm^-1;

¹HNMR: 1.4 δ (2H, quartet, J=7Hz); 1.2 δ (6H, singlet); 1.0 δ (1H, singlet); 0.9 δ (3H, triplet, J=7Hz)

¹³CNMR: 74, 35, 27, 25 δ

Concept introduction:

Molecular ion peak in the mass spectrum gives an idea about the relative molecular mass of the compound. From the molecular mass and knowing the atomic masses of C, H and O the possible molecular formulas for the compound can be arrived.

In IR the O-H stretching of alcohols and phenols are observed in the range 3200-3550 cm^-1. In ¹HNMR spectrum, the alcoholic proton absorption occurs in the range 3.4 δ - 4.5 δ while that of phenolic proton occur in the range 3.0 to 8.0. The absorption due to 10 alkyl group (CH₃) is seen around 0.7 δ - 1.3 δ, that due to a 20 alkyl group (CH₂) is seen around 1.2 δ - 1.6 δ while that due to 30 alkyl group (CH) is seen around 1.4 δ - 1.8 δ The multiplicity of a signal gives an idea about the protons present in the adjacent carbons.

In ¹³CNMR spectrum, the carbon bonded to –OH absorb in the range 50-80 δ. The primary alkyl carbon absorb in the range 10-15 δ, a secondary alkyl radical in the range 16-25 δ while a tertiary alkyl in the range 25-35 δ.

To give:

a) Three molecular formulas possible for the compound assuming that the compound contains C and H but may or may not contain oxygen.

Answer to Problem 55AP

a) Three molecular formulas possible for the compound assuming that the compound contains C and H and O are C₅H₁₂O, C₄H₈O₂ and C₃H₄O₃.

Explanation of Solution

a) The IR absorption at 3600cm^-1 indicates the presence of an alcoholic group in the compound. Hence assuming the compound contains C, H and O, three possible formulas are possible for the compound with molecular mass 88. They are C₅H₁₂O, C₄H₈O₂ and C₃H₄O₃.

Conclusion

Three molecular formulas possible for the compound assuming that the compound contains C and H but may or may not contain oxygen are C₅H₁₂O, C₄H₈O₂ and C₃H₄O₃.

Interpretation Introduction

b) The number of protons in the compound is to be given.

Interpretation:

A compound has the spectral data:

Mass spectrum: M+=88.1; IR=3600 cm^-1;

¹HNMR: 1.4 δ (2H, quartet, J=7Hz); 1.2 δ (6H, singlet); 1.0 δ (1H, singlet); 0.9 δ (3H, triplet, J=7Hz)

¹³CNMR: 74, 35, 27, 25 δ

Concept introduction:

Molecular ion peak in the mass spectrum gives an idea about the relative molecular mass of the compound. From the molecular mass and knowing the atomic masses of C, H and O the possible molecular formulas for the compound can be arrived.

In IR the O-H stretching of alcohols and phenols are observed in the range 3200-3550 cm^-1. In ¹HNMR spectrum, the alcoholic proton absorption occurs in the range 3.4 δ - 4.5 δ while that of phenolic proton occur in the range 3.0 to 8.0. The absorption due to 10 alkyl group (CH₃) is seen around 0.7 δ - 1.3 δ, that due to a 20 alkyl group (CH₂) is seen around 1.2 δ - 1.6 δ while that due to 30 alkyl group (CH) is seen around 1.4 δ - 1.8 δ The multiplicity of a signal gives an idea about the protons present in the adjacent carbons.

In ¹³CNMR spectrum, the carbon bonded to –OH absorb in the range 50-80 δ. The primary alkyl carbon absorb in the range 10-15 δ, a secondary alkyl radical in the range 16-25 δ while a tertiary alkyl in the range 25-35 δ.

To give:

b) The number of protons in the compound.

Answer to Problem 55AP

b) The compound has 12 protons.

Explanation of Solution

b) IN ¹HNMR spectrum contain peaks accounting for the absorption for 12 protons in the molecule.

Conclusion

b) The compound has 12 protons.

Interpretation Introduction

c) The functional group(s) present in the compound is/are to be given.

Interpretation:

A compound has the spectral data:

Mass spectrum: M+=88.1; IR=3600cm^-1;

¹HNMR: 1.4 δ (2H, quartet, J=7Hz); 1.2 δ (6H, singlet); 1.0 δ (1H, singlet); 0.9 δ (3H, triplet, J=7Hz)

¹³CNMR: 74, 35, 27, 25 δ

Concept introduction:

Molecular ion peak in the mass spectrum gives an idea about the relative molecular mass of the compound. From the molecular mass and knowing the atomic masses of C, H and O the possible molecular formulas for the compound can be arrived.

In IR the O-H stretching of alcohols and phenols are observed in the range 3200-3550 cm^-1. In ¹HNMR spectrum, the alcoholic proton absorption occurs in the range 3.4 δ - 4.5 δ while that of phenolic proton occur in the range 3.0 to 8.0. The absorption due to 10 alkyl group (CH₃) is seen around 0.7 δ - 1.3 δ, that due to a 20 alkyl group (CH₂) is seen around 1.2 δ - 1.6 δ while that due to 30 alkyl group (CH) is seen around 1.4 δ - 1.8 δ The multiplicity of a signal gives an idea about the protons present in the adjacent carbons.

In ¹³CNMR spectrum, the carbon bonded to –OH absorb in the range 50-80 δ. The primary alkyl carbon absorb in the range 10-15 δ, a secondary alkyl radical in the range 16-25 δ while a tertiary alkyl in the range 25-35 δ.

To give:

c) The functional group(s) present in the compound.

Answer to Problem 55AP

c) The functional group in the compound is –OH (alcohol).

Explanation of Solution

c) In ¹³HNMR only four signals are observed. But 12 protons could not be accommodated on four carbons. At least five carbons are required. This justified by the ¹HNMR spectrum which an absorption integrating to six protons present on two equivalent carbons. Hence five carbons are present in the molecule.

Conclusion

c) The functional group in the compound is –OH (alcohol).

Interpretation Introduction

d) The number of carbons in the compound is to be given.

Interpretation:

A compound has the spectral data:

Mass spectrum: M+=88.1; IR=3600cm^-1;

¹HNMR: 1.4 δ (2H, quartet, J=7Hz); 1.2 δ (6H, singlet); 1.0 δ (1H, singlet); 0.9 δ (3H, triplet, J=7Hz)

¹³CNMR: 74, 35, 27, 25 δ

Concept introduction:

Molecular ion peak in the mass spectrum gives an idea about the relative molecular mass of the compound. From the molecular mass and knowing the atomic masses of C, H and O the possible molecular formulas for the compound can be arrived.

In IR the O-H stretching of alcohols and phenols are observed in the range 3200-3550 cm^-1. In ¹HNMR spectrum, the alcoholic proton absorption occurs in the range 3.4 δ - 4.5 δ while that of phenolic proton occur in the range 3.0 to 8.0. The absorption due to 10 alkyl group (CH₃) is seen around 0.7 δ - 1.3 δ, that due to a 20 alkyl group (CH₂) is seen around 1.2 δ - 1.6 δ while that due to 30 alkyl group (CH) is seen around 1.4 δ - 1.8 δ The multiplicity of a signal gives an idea about the protons present in the adjacent carbons.

In ¹³CNMR spectrum, the carbon bonded to –OH absorb in the range 50-80 δ. The primary alkyl carbon absorb in the range 10-15 δ, a secondary alkyl radical in the range 16-25 δ while a tertiary alkyl in the range 25-35 δ.

To give:

d) The number of carbons in the compound.

Answer to Problem 55AP

d) The compound has 5 carbons.

Explanation of Solution

d) Its molecular formula has to be C₅H₁₂O as it is a five carbon alcohol with 12 hydrogen atoms.

Conclusion

d) The compound has 5 carbons.

Interpretation Introduction

e) Molecular formula of the compound is to be given.

Interpretation:

A compound has the spectral data:

Mass spectrum: M+=88.1; IR=3600cm-1;

¹HNMR: 1.4 δ (2H, quartet, J=7Hz); 1.2 δ (6H, singlet); 1.0 δ (1H, singlet); 0.9 δ (3H, triplet, J=7Hz)

¹³CNMR: 74, 35, 27, 25 δ

Concept introduction:

Molecular ion peak in the mass spectrum gives an idea about the relative molecular mass of the compound. From the molecular mass and knowing the atomic masses of C, H and O the possible molecular formulas for the compound can be arrived.

In IR the O-H stretching of alcohols and phenols are observed in the range 3200-3550 cm^-1. In ¹HNMR spectrum, the alcoholic proton absorption occurs in the range 3.4 δ - 4.5 δ while that of phenolic proton occur in the range 3.0 to 8.0. The absorption due to 10 alkyl group (CH₃) is seen around 0.7 δ - 1.3 δ, that due to a 20 alkyl group (CH₂) is seen around 1.2 δ - 1.6 δ while that due to 30 alkyl group (CH) is seen around 1.4 δ - 1.8 δ The multiplicity of a signal gives an idea about the protons present in the adjacent carbons.

In ¹³CNMR spectrum, the carbon bonded to –OH absorb in the range 50-80 δ. The primary alkyl carbon absorb in the range 10-15 δ, a secondary alkyl radical in the range 16-25 δ while a tertiary alkyl in the range 25-35 δ.

To give:

e) Molecular formula of the compound.

Answer to Problem 55AP

e) Molecular formula of the compound is C₅H₁₂O.

Explanation of Solution

e) The six proton singlet at 1.2 δ can be attributed to two methyl groups attached to a carbon without any proton. The two proton triplet at 1.4 δ can be assigned to CH₂ attached to a methyl which gives a triplet at 0.9 δ. The one proton singlet at 1.0 δ is due to the hydroxyl proton.

Conclusion

e) Molecular formula of the compound is C₅H₁₂O.

Interpretation Introduction

f) The structure of the compound is to be given.

Interpretation:

A compound has the spectral data:

Mass spectrum: M+=88.1; IR=3600cm^-1;

¹HNMR: 1.4 δ (2H, quartet, J=7Hz); 1.2 δ (6H, singlet); 1.0 δ (1H, singlet); 0.9 δ (3H, triplet, J=7Hz)

¹³CNMR: 74, 35, 27, 25 δ

Concept introduction:

Molecular ion peak in the mass spectrum gives an idea about the relative molecular mass of the compound. From the molecular mass and knowing the atomic masses of C, H and O the possible molecular formulas for the compound can be arrived.

In IR the O-H stretching of alcohols and phenols are observed in the range 3200-3550 cm^-1. In ¹HNMR spectrum, the alcoholic proton absorption occurs in the range 3.4 δ - 4.5 δ while that of phenolic proton occur in the range 3.0 to 8.0. The absorption due to 10 alkyl group (CH₃) is seen around 0.7 δ - 1.3 δ, that due to a 20 alkyl group (CH₂) is seen around 1.2 δ - 1.6 δ while that due to 30 alkyl group (CH) is seen around 1.4 δ - 1.8 δ The multiplicity of a signal gives an idea about the protons present in the adjacent carbons.

In ¹³CNMR spectrum, the carbon bonded to –OH absorb in the range 50-80 δ. The primary alkyl carbon absorb in the range 10-15 δ, a secondary alkyl radical in the range 16-25 δ while a tertiary alkyl in the range 25-35 δ.

To give:

f) The structure of the compound.

Answer to Problem 55AP

f) The structure of the compound is

Explanation of Solution

f) Thus the structure of the alcohol is

Conclusion

The structure of the compound is

Interpretation Introduction

g) The peaks in the ¹HNMR of the molecule are to be assigned to specific protons.

Interpretation:

A compound has the spectral data:

Mass spectrum: M+=88.1; IR=3600cm^-1;

¹HNMR: 1.4 δ (2H, quartet, J=7Hz); 1.2 δ (6H, singlet); 1.0 δ (1H, singlet); 0.9 δ (3H, triplet, J=7Hz)

¹³CNMR: 74, 35, 27, 25 δ

Concept introduction:

Molecular ion peak in the mass spectrum gives an idea about the relative molecular mass of the compound. From the molecular mass and knowing the atomic masses of C, H and O the possible molecular formulas for the compound can be arrived.

In IR the O-H stretching of alcohols and phenols are observed in the range 3200-3550 cm^-1. In ¹HNMR spectrum, the alcoholic proton absorption occurs in the range 3.4 δ - 4.5 δ while that of phenolic proton occur in the range 3.0 to 8.0. The absorption due to 10 alkyl group (CH₃) is seen around 0.7 δ - 1.3 δ, that due to a 20 alkyl group (CH₂) is seen around 1.2 δ - 1.6 δ while that due to 30 alkyl group (CH) is seen around 1.4 δ - 1.8 δ The multiplicity of a signal gives an idea about the protons present in the adjacent carbons.

In ¹³CNMR spectrum, the carbon bonded to –OH absorb in the range 50-80 δ. The primary alkyl carbon absorb in the range 10-15 δ, a secondary alkyl radical in the range 16-25 δ while a tertiary alkyl in the range 25-35 δ.

To assign:

g) The peaks in the ¹HNMR of the molecule to specific protons.

Answer to Problem 55AP

g) 1.2 δ (6H, singlet) given by protons marked as ‘a’.

1.4 δ (2H, quartet, J=7Hz) given by protons marked as ‘b’.

0.9 δ (3H, triplet, J=7Hz) given by protons marked as ‘c’.

1.0 δ (1H, singlet) given by proton marked as‘d’.

Explanation of Solution

g) 1.2 δ (6H, singlet) given by protons marked as ‘a’.

1.4 δ (2H, quartet, J=7Hz) given by protons marked as ‘b’.

0.9 δ (3H, triplet, J=7Hz) given by protons marked as ‘c’.

1.0 δ (1H, singlet) given by proton marked as‘d’.

Conclusion

g) 1.2 δ (6H, singlet) given by protons marked as ‘a’.

1.4 δ (2H, quartet, J=7Hz) given by protons marked as ‘b’.

0.9 δ (3H, triplet, J=7Hz) given by protons marked as ‘c’.

1.0 δ (1H, singlet) given by proton marked as‘d’.