a) Interpretation: A structure for the compound whose 1 HNMR spectrum given is to be proposed. Given: M.F = C 8 H 9 Br I.R: 820 cm -1 . 1 HNMR spectrum: 7.07δ (2H, doublet); 7.39δ (2H, doublet); 2.58δ (2H, quartet); 1.20δ (3H, triplet). Concept introduction: In 1 HNMR spectrum aromatic protons give a broad peak in the range 6.5δ-8.0δ, the primary alkyl protons around 0.7δ-1.3δ, the secondary alkyl protons around 1.2δ-1.6δ, and a tertiary alkyl protons in between 1.4δ-1.8δ. The multiplicity of a signal gives an idea about the protons present in the adjacent carbons. In I.R, the o-disubstituted benzenes absorb around 735-770 cm -1 , m-disubstituted benzenes absorb around 690-710 cm -1 and p-disubstituted benzenes absorb around 810-840 cm -1 . To propose: A structure for the compound whose 1 HNMR spectrum given is to be proposed. Given: M.F = C 8 H 9 Br I.R: 820 cm -1 . 1 HNMR spectrum: 7.07δ (2H, doublet); 7.39δ (2H, doublet); 2.58δ (2H, quartet); 1.20δ (3H, triplet).

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter 15.SE, Problem 47AP

Interpretation Introduction

a)

Interpretation:

A structure for the compound whose ¹HNMR spectrum given is to be proposed. Given: M.F = C₈H₉Br I.R: 820 cm^-1.

¹HNMR spectrum: 7.07δ (2H, doublet); 7.39δ (2H, doublet); 2.58δ (2H, quartet); 1.20δ (3H, triplet).

Concept introduction:

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

In I.R, the o-disubstituted benzenes absorb around 735-770 cm^-1, m-disubstituted benzenes absorb around 690-710 cm^-1 and p-disubstituted benzenes absorb around 810-840 cm^-1.

To propose:

A structure for the compound whose ¹HNMR spectrum given is to be proposed. Given: M.F = C₈H₉Br I.R: 820 cm^-1.

¹HNMR spectrum: 7.07δ (2H, doublet); 7.39δ (2H, doublet); 2.58δ (2H, quartet); 1.20δ (3H, triplet).

Expert Solution

Answer to Problem 47AP

A structure for the compound with M.F = C₈H₉Br with spectral characteristics: I.R: 820 cm^-1, ¹HNMR spectrum: 7.07δ (2H, doublet); 7.39δ (2H, doublet); 2.58δ (2H, quartet); 1.20δ (3H, triplet) is

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 2

Explanation of Solution

The molecular formula of the compound is C₈H₉Br.

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 3

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 4

Thus the compound has four unsaturation units like double bonds and/or rings. The two doublets at 7.07δ and 7.39δ each corresponding to two protons indicate the compound is aromatic and each proton responsible for the doublet has a proton on the adjacent carbon. The two proton quartet at 2.58δ (benzylic) and three proton triplet at 1.20δ (primary) could be accounted for only if an ethyl substituent is present on the ring along with bromine. The I.R absorption at 820 cm^-1 indicates that the ethyl group and bromine are arranged in para positions. Hence the compound is p-bromoethyl benzene.

Conclusion

A structure for the compound with M.F=C₈H₉Br with spectral characteristics: I.R: 820 cm^-1, ¹HNMR spectrum: 7.07δ (2H, doublet); 7.39δ (2H, doublet); 2.58δ (2H, quartet); 1.20δ (3H, triplet) is

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 5

Interpretation Introduction

b)

Interpretation:

A structure for the compound whose ¹HNMR spectrum given is to be proposed.

Given: M.F = C₉H₁₂ I.R: 750 cm^-1

¹HNMR spectrum: 7.13δ (4H, broad); 2.64δ (2H, quartet); 2.31δ (3H, singlet); 1.19δ (3H, triplet).

Concept introduction:

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

In I.R, the o-disubstituted benzenes absorb around 735-770 cm^-1, m-disubstituted benzenes absorb around 690-710 cm^-1and p-disubstituted benzenes absorb around 810-840 cm^-1.

To propose:

A structure for the compound whose ¹HNMR spectrum given is to be proposed. Given: M.F = C₉H₁₂ I.R: 750 cm^-1.

¹HNMR spectrum: 7.13δ (4H, broad); 2.64δ (2H, quartet); 2.31δ (3H, singlet); 1.19δ (3H, triplet).

Expert Solution

Answer to Problem 47AP

A structure for the compound whose 1HNMR spectrum given is to be proposed. Given: M.F = C₉H₁₂, I.R: 750 cm^-1, ¹HNMR spectrum: 7.13δ (4H, broad); 2.64δ (2H, quartet); 2.31δ (3H, singlet); 1.19δ (3H, triplet) is

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 7

Explanation of Solution

The molecular formula of the compound is C₉H₁₂.

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 8

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 9

Thus the compound has four unsaturation units like double bonds and/or rings. The four proton broad band at 7.13δ indicates that the compound is aromatic with two substituent groups attached to the benzene ring. The two proton quartet at 2.64δ (benzylic) and three proton triplet at 1.19δ (primary alkyl) could be accounted for only if an ethyl group is present on the ring. The three proton singlet at 2.31δ (benzylic) can be attributed to a methyl group attached to the ring. Thus the two substituent groups attached to the benzene ring are ethyl and methyl. The I.R absorption at 750 cm^-1 requires that the ethyl and methyl groups to be arranged in ortho positions. Hence the compound is o-ethyltoluene.

Conclusion

A structure for the compound whose ¹HNMR spectrum given is to be proposed. Given: M.F = C₉H₁₂, I.R: 750 cm^-1, ¹HNMR spectrum: 7.13δ (4H, broad); 2.64δ (2H, quartet); 2.31δ (3H, singlet); 1.19δ (3H, triplet) is

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 10

Interpretation Introduction

c)

Interpretation:

A structure for the compound whose ¹HNMR spectrum given is to be proposed. Given: M.F = C₁₁H₁₆ I.R: 820 cm^-1.

¹HNMR spectrum: 7.27δ (2H, doublet); 7.06δ (2H, doublet); 2.30δ (3H, singlet); 1.31δ (9H, singlet).

Concept introduction:

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

In I.R, the o-disubstituted benzenes absorb around 735-770 cm^-1, m-disubstituted benzenes absorb around 690-710 cm^-1 and p-disubstituted benzenes absorb around 810-840 cm^-1.

To propose:

A structure for the compound whose ¹HNMR spectrum given is to be proposed. Given: M.F = C₁₁H₁₆ I.R: 820 cm^-1.

¹HNMR spectrum: 7.27δ (2H, doublet); 7.06δ (2H, doublet); 2.30δ (3H, singlet); 1.31δ (9H, singlet).

Expert Solution

Answer to Problem 47AP

A structure for the compound whose ¹HNMR spectrum given is to be proposed. Given: M.F = C₁₁H₁₆; I.R: 820 cm^-1; ¹HNMR spectrum: 7.27δ (2H, doublet); 7.06δ (2H, doublet); 2.30δ (3H, singlet); 1.31δ (9H, singlet).

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 12

Explanation of Solution

The molecular formula of the compound is C₁₁H₁₆.

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 13

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 14

Thus the compound has four unsaturation units like double bonds and/or rings. The two doublets at 7.27δ and 7.06δ each corresponding to two protons indicate the compound is aromatic and each proton responsible for the doublet has a proton on the adjacent carbon. The three proton singlet at 2.58δ (benzylic) could be accounted for a methyl group attached to the benzene ring. The remaining four carbons and nine hydrogens indicate the presence of a tert-butyl group attached to the benzene ring which is confirmed by the nine proton singlet at 1.31δ (primary alkyl). Thus methyl and tert-butyl are the two substituent groups on the benzene ring. The I.R absorption at 820 cm^-1 indicates that the two substituent groups are arranged in para positions. Hence the compound is p-t-butyltoluene.

Conclusion

A structure for the compound whose 1HNMR spectrum given is to be proposed. Given: M.F = C₁₁H₁₆; I.R: 820 cm^-1; ¹HNMR spectrum: 7.27δ (2H, doublet); 7.06δ (2H, doublet); 2.30δ (3H, singlet); 1.31δ (9H, singlet).

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 15

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

Part II. Identify whether the two protons in blue are homotopic, enantiopic, diasteriotopic, or heterotopic. a) HO b) Bri H HH c) d) H H H Br 0

None

Choose the option that is decreasing from biggest to smallest. Group of answer choices: 100 m, 10000 mm, 100 cm, 100000 um, 10000000 nm 10000000 nm, 100000 um, 100 cm, 10000 mm, 100 m 10000000 nm, 100000 um, 10000 mm, 100 cm, 100 m 100 m, 100 cm, 10000 mm, 100000 um, 10000000 nm

Answer 2

Question

Chapter 15.SE, Problem 47AP

Interpretation Introduction

a)

Interpretation:

A structure for the compound whose ¹HNMR spectrum given is to be proposed. Given: M.F = C₈H₉Br I.R: 820 cm^-1.

¹HNMR spectrum: 7.07δ (2H, doublet); 7.39δ (2H, doublet); 2.58δ (2H, quartet); 1.20δ (3H, triplet).

Concept introduction:

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

In I.R, the o-disubstituted benzenes absorb around 735-770 cm^-1, m-disubstituted benzenes absorb around 690-710 cm^-1 and p-disubstituted benzenes absorb around 810-840 cm^-1.

To propose:

A structure for the compound whose ¹HNMR spectrum given is to be proposed. Given: M.F = C₈H₉Br I.R: 820 cm^-1.

¹HNMR spectrum: 7.07δ (2H, doublet); 7.39δ (2H, doublet); 2.58δ (2H, quartet); 1.20δ (3H, triplet).

Expert Solution

Answer to Problem 47AP

A structure for the compound with M.F = C₈H₉Br with spectral characteristics: I.R: 820 cm^-1, ¹HNMR spectrum: 7.07δ (2H, doublet); 7.39δ (2H, doublet); 2.58δ (2H, quartet); 1.20δ (3H, triplet) is

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 2

Explanation of Solution

The molecular formula of the compound is C₈H₉Br.

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 3

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 4

Thus the compound has four unsaturation units like double bonds and/or rings. The two doublets at 7.07δ and 7.39δ each corresponding to two protons indicate the compound is aromatic and each proton responsible for the doublet has a proton on the adjacent carbon. The two proton quartet at 2.58δ (benzylic) and three proton triplet at 1.20δ (primary) could be accounted for only if an ethyl substituent is present on the ring along with bromine. The I.R absorption at 820 cm^-1 indicates that the ethyl group and bromine are arranged in para positions. Hence the compound is p-bromoethyl benzene.

Conclusion

A structure for the compound with M.F=C₈H₉Br with spectral characteristics: I.R: 820 cm^-1, ¹HNMR spectrum: 7.07δ (2H, doublet); 7.39δ (2H, doublet); 2.58δ (2H, quartet); 1.20δ (3H, triplet) is

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 5

Interpretation Introduction

b)

Interpretation:

A structure for the compound whose ¹HNMR spectrum given is to be proposed.

Given: M.F = C₉H₁₂ I.R: 750 cm^-1

¹HNMR spectrum: 7.13δ (4H, broad); 2.64δ (2H, quartet); 2.31δ (3H, singlet); 1.19δ (3H, triplet).

Concept introduction:

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

In I.R, the o-disubstituted benzenes absorb around 735-770 cm^-1, m-disubstituted benzenes absorb around 690-710 cm^-1and p-disubstituted benzenes absorb around 810-840 cm^-1.

To propose:

A structure for the compound whose ¹HNMR spectrum given is to be proposed. Given: M.F = C₉H₁₂ I.R: 750 cm^-1.

¹HNMR spectrum: 7.13δ (4H, broad); 2.64δ (2H, quartet); 2.31δ (3H, singlet); 1.19δ (3H, triplet).

Expert Solution

Answer to Problem 47AP

A structure for the compound whose 1HNMR spectrum given is to be proposed. Given: M.F = C₉H₁₂, I.R: 750 cm^-1, ¹HNMR spectrum: 7.13δ (4H, broad); 2.64δ (2H, quartet); 2.31δ (3H, singlet); 1.19δ (3H, triplet) is

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 7

Explanation of Solution

The molecular formula of the compound is C₉H₁₂.

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 8

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 9

Thus the compound has four unsaturation units like double bonds and/or rings. The four proton broad band at 7.13δ indicates that the compound is aromatic with two substituent groups attached to the benzene ring. The two proton quartet at 2.64δ (benzylic) and three proton triplet at 1.19δ (primary alkyl) could be accounted for only if an ethyl group is present on the ring. The three proton singlet at 2.31δ (benzylic) can be attributed to a methyl group attached to the ring. Thus the two substituent groups attached to the benzene ring are ethyl and methyl. The I.R absorption at 750 cm^-1 requires that the ethyl and methyl groups to be arranged in ortho positions. Hence the compound is o-ethyltoluene.

Conclusion

A structure for the compound whose ¹HNMR spectrum given is to be proposed. Given: M.F = C₉H₁₂, I.R: 750 cm^-1, ¹HNMR spectrum: 7.13δ (4H, broad); 2.64δ (2H, quartet); 2.31δ (3H, singlet); 1.19δ (3H, triplet) is

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 10

Interpretation Introduction

c)

Interpretation:

A structure for the compound whose ¹HNMR spectrum given is to be proposed. Given: M.F = C₁₁H₁₆ I.R: 820 cm^-1.

¹HNMR spectrum: 7.27δ (2H, doublet); 7.06δ (2H, doublet); 2.30δ (3H, singlet); 1.31δ (9H, singlet).

Concept introduction:

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

In I.R, the o-disubstituted benzenes absorb around 735-770 cm^-1, m-disubstituted benzenes absorb around 690-710 cm^-1 and p-disubstituted benzenes absorb around 810-840 cm^-1.

To propose:

A structure for the compound whose ¹HNMR spectrum given is to be proposed. Given: M.F = C₁₁H₁₆ I.R: 820 cm^-1.

¹HNMR spectrum: 7.27δ (2H, doublet); 7.06δ (2H, doublet); 2.30δ (3H, singlet); 1.31δ (9H, singlet).

Expert Solution

Answer to Problem 47AP

A structure for the compound whose ¹HNMR spectrum given is to be proposed. Given: M.F = C₁₁H₁₆; I.R: 820 cm^-1; ¹HNMR spectrum: 7.27δ (2H, doublet); 7.06δ (2H, doublet); 2.30δ (3H, singlet); 1.31δ (9H, singlet).

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 12

Explanation of Solution

The molecular formula of the compound is C₁₁H₁₆.

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 13

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 14

Thus the compound has four unsaturation units like double bonds and/or rings. The two doublets at 7.27δ and 7.06δ each corresponding to two protons indicate the compound is aromatic and each proton responsible for the doublet has a proton on the adjacent carbon. The three proton singlet at 2.58δ (benzylic) could be accounted for a methyl group attached to the benzene ring. The remaining four carbons and nine hydrogens indicate the presence of a tert-butyl group attached to the benzene ring which is confirmed by the nine proton singlet at 1.31δ (primary alkyl). Thus methyl and tert-butyl are the two substituent groups on the benzene ring. The I.R absorption at 820 cm^-1 indicates that the two substituent groups are arranged in para positions. Hence the compound is p-t-butyltoluene.

Conclusion

A structure for the compound whose 1HNMR spectrum given is to be proposed. Given: M.F = C₁₁H₁₆; I.R: 820 cm^-1; ¹HNMR spectrum: 7.27δ (2H, doublet); 7.06δ (2H, doublet); 2.30δ (3H, singlet); 1.31δ (9H, singlet).

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 15

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Chapter 15.SE, Problem 47AP

Interpretation Introduction

a)

Interpretation:

A structure for the compound whose ¹HNMR spectrum given is to be proposed. Given: M.F = C₈H₉Br I.R: 820 cm^-1.

¹HNMR spectrum: 7.07δ (2H, doublet); 7.39δ (2H, doublet); 2.58δ (2H, quartet); 1.20δ (3H, triplet).

Concept introduction:

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

In I.R, the o-disubstituted benzenes absorb around 735-770 cm^-1, m-disubstituted benzenes absorb around 690-710 cm^-1 and p-disubstituted benzenes absorb around 810-840 cm^-1.

To propose:

A structure for the compound whose ¹HNMR spectrum given is to be proposed. Given: M.F = C₈H₉Br I.R: 820 cm^-1.

¹HNMR spectrum: 7.07δ (2H, doublet); 7.39δ (2H, doublet); 2.58δ (2H, quartet); 1.20δ (3H, triplet).

Expert Solution

Answer to Problem 47AP

A structure for the compound with M.F = C₈H₉Br with spectral characteristics: I.R: 820 cm^-1, ¹HNMR spectrum: 7.07δ (2H, doublet); 7.39δ (2H, doublet); 2.58δ (2H, quartet); 1.20δ (3H, triplet) is

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 2

Explanation of Solution

The molecular formula of the compound is C₈H₉Br.

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 3

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 4

Thus the compound has four unsaturation units like double bonds and/or rings. The two doublets at 7.07δ and 7.39δ each corresponding to two protons indicate the compound is aromatic and each proton responsible for the doublet has a proton on the adjacent carbon. The two proton quartet at 2.58δ (benzylic) and three proton triplet at 1.20δ (primary) could be accounted for only if an ethyl substituent is present on the ring along with bromine. The I.R absorption at 820 cm^-1 indicates that the ethyl group and bromine are arranged in para positions. Hence the compound is p-bromoethyl benzene.

Conclusion

A structure for the compound with M.F=C₈H₉Br with spectral characteristics: I.R: 820 cm^-1, ¹HNMR spectrum: 7.07δ (2H, doublet); 7.39δ (2H, doublet); 2.58δ (2H, quartet); 1.20δ (3H, triplet) is

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 5

Interpretation Introduction

b)

Interpretation:

A structure for the compound whose ¹HNMR spectrum given is to be proposed.

Given: M.F = C₉H₁₂ I.R: 750 cm^-1

¹HNMR spectrum: 7.13δ (4H, broad); 2.64δ (2H, quartet); 2.31δ (3H, singlet); 1.19δ (3H, triplet).

Concept introduction:

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

In I.R, the o-disubstituted benzenes absorb around 735-770 cm^-1, m-disubstituted benzenes absorb around 690-710 cm^-1and p-disubstituted benzenes absorb around 810-840 cm^-1.

To propose:

A structure for the compound whose ¹HNMR spectrum given is to be proposed. Given: M.F = C₉H₁₂ I.R: 750 cm^-1.

¹HNMR spectrum: 7.13δ (4H, broad); 2.64δ (2H, quartet); 2.31δ (3H, singlet); 1.19δ (3H, triplet).

Expert Solution

Answer to Problem 47AP

A structure for the compound whose 1HNMR spectrum given is to be proposed. Given: M.F = C₉H₁₂, I.R: 750 cm^-1, ¹HNMR spectrum: 7.13δ (4H, broad); 2.64δ (2H, quartet); 2.31δ (3H, singlet); 1.19δ (3H, triplet) is

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 7

Explanation of Solution

The molecular formula of the compound is C₉H₁₂.

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 8

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 9

Thus the compound has four unsaturation units like double bonds and/or rings. The four proton broad band at 7.13δ indicates that the compound is aromatic with two substituent groups attached to the benzene ring. The two proton quartet at 2.64δ (benzylic) and three proton triplet at 1.19δ (primary alkyl) could be accounted for only if an ethyl group is present on the ring. The three proton singlet at 2.31δ (benzylic) can be attributed to a methyl group attached to the ring. Thus the two substituent groups attached to the benzene ring are ethyl and methyl. The I.R absorption at 750 cm^-1 requires that the ethyl and methyl groups to be arranged in ortho positions. Hence the compound is o-ethyltoluene.

Conclusion

A structure for the compound whose ¹HNMR spectrum given is to be proposed. Given: M.F = C₉H₁₂, I.R: 750 cm^-1, ¹HNMR spectrum: 7.13δ (4H, broad); 2.64δ (2H, quartet); 2.31δ (3H, singlet); 1.19δ (3H, triplet) is

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 10

Interpretation Introduction

c)

Interpretation:

A structure for the compound whose ¹HNMR spectrum given is to be proposed. Given: M.F = C₁₁H₁₆ I.R: 820 cm^-1.

¹HNMR spectrum: 7.27δ (2H, doublet); 7.06δ (2H, doublet); 2.30δ (3H, singlet); 1.31δ (9H, singlet).

Concept introduction:

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

In I.R, the o-disubstituted benzenes absorb around 735-770 cm^-1, m-disubstituted benzenes absorb around 690-710 cm^-1 and p-disubstituted benzenes absorb around 810-840 cm^-1.

To propose:

A structure for the compound whose ¹HNMR spectrum given is to be proposed. Given: M.F = C₁₁H₁₆ I.R: 820 cm^-1.

¹HNMR spectrum: 7.27δ (2H, doublet); 7.06δ (2H, doublet); 2.30δ (3H, singlet); 1.31δ (9H, singlet).

Expert Solution

Answer to Problem 47AP

A structure for the compound whose ¹HNMR spectrum given is to be proposed. Given: M.F = C₁₁H₁₆; I.R: 820 cm^-1; ¹HNMR spectrum: 7.27δ (2H, doublet); 7.06δ (2H, doublet); 2.30δ (3H, singlet); 1.31δ (9H, singlet).

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 12

Explanation of Solution

The molecular formula of the compound is C₁₁H₁₆.

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 13

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 14

Thus the compound has four unsaturation units like double bonds and/or rings. The two doublets at 7.27δ and 7.06δ each corresponding to two protons indicate the compound is aromatic and each proton responsible for the doublet has a proton on the adjacent carbon. The three proton singlet at 2.58δ (benzylic) could be accounted for a methyl group attached to the benzene ring. The remaining four carbons and nine hydrogens indicate the presence of a tert-butyl group attached to the benzene ring which is confirmed by the nine proton singlet at 1.31δ (primary alkyl). Thus methyl and tert-butyl are the two substituent groups on the benzene ring. The I.R absorption at 820 cm^-1 indicates that the two substituent groups are arranged in para positions. Hence the compound is p-t-butyltoluene.

Conclusion

A structure for the compound whose 1HNMR spectrum given is to be proposed. Given: M.F = C₁₁H₁₆; I.R: 820 cm^-1; ¹HNMR spectrum: 7.27δ (2H, doublet); 7.06δ (2H, doublet); 2.30δ (3H, singlet); 1.31δ (9H, singlet).

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 15

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 15.SE, Problem 47AP

Interpretation Introduction

a)

Interpretation:

A structure for the compound whose ¹HNMR spectrum given is to be proposed. Given: M.F = C₈H₉Br I.R: 820 cm^-1.

¹HNMR spectrum: 7.07δ (2H, doublet); 7.39δ (2H, doublet); 2.58δ (2H, quartet); 1.20δ (3H, triplet).

Concept introduction:

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

In I.R, the o-disubstituted benzenes absorb around 735-770 cm^-1, m-disubstituted benzenes absorb around 690-710 cm^-1 and p-disubstituted benzenes absorb around 810-840 cm^-1.

To propose:

A structure for the compound whose ¹HNMR spectrum given is to be proposed. Given: M.F = C₈H₉Br I.R: 820 cm^-1.

¹HNMR spectrum: 7.07δ (2H, doublet); 7.39δ (2H, doublet); 2.58δ (2H, quartet); 1.20δ (3H, triplet).

Expert Solution

Answer to Problem 47AP

A structure for the compound with M.F = C₈H₉Br with spectral characteristics: I.R: 820 cm^-1, ¹HNMR spectrum: 7.07δ (2H, doublet); 7.39δ (2H, doublet); 2.58δ (2H, quartet); 1.20δ (3H, triplet) is

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 2

Explanation of Solution

The molecular formula of the compound is C₈H₉Br.

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 3

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 4

Thus the compound has four unsaturation units like double bonds and/or rings. The two doublets at 7.07δ and 7.39δ each corresponding to two protons indicate the compound is aromatic and each proton responsible for the doublet has a proton on the adjacent carbon. The two proton quartet at 2.58δ (benzylic) and three proton triplet at 1.20δ (primary) could be accounted for only if an ethyl substituent is present on the ring along with bromine. The I.R absorption at 820 cm^-1 indicates that the ethyl group and bromine are arranged in para positions. Hence the compound is p-bromoethyl benzene.

Conclusion

A structure for the compound with M.F=C₈H₉Br with spectral characteristics: I.R: 820 cm^-1, ¹HNMR spectrum: 7.07δ (2H, doublet); 7.39δ (2H, doublet); 2.58δ (2H, quartet); 1.20δ (3H, triplet) is

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 5

Interpretation Introduction

b)

Interpretation:

A structure for the compound whose ¹HNMR spectrum given is to be proposed.

Given: M.F = C₉H₁₂ I.R: 750 cm^-1

¹HNMR spectrum: 7.13δ (4H, broad); 2.64δ (2H, quartet); 2.31δ (3H, singlet); 1.19δ (3H, triplet).

Concept introduction:

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

In I.R, the o-disubstituted benzenes absorb around 735-770 cm^-1, m-disubstituted benzenes absorb around 690-710 cm^-1and p-disubstituted benzenes absorb around 810-840 cm^-1.

To propose:

A structure for the compound whose ¹HNMR spectrum given is to be proposed. Given: M.F = C₉H₁₂ I.R: 750 cm^-1.

¹HNMR spectrum: 7.13δ (4H, broad); 2.64δ (2H, quartet); 2.31δ (3H, singlet); 1.19δ (3H, triplet).

Expert Solution

Answer to Problem 47AP

A structure for the compound whose 1HNMR spectrum given is to be proposed. Given: M.F = C₉H₁₂, I.R: 750 cm^-1, ¹HNMR spectrum: 7.13δ (4H, broad); 2.64δ (2H, quartet); 2.31δ (3H, singlet); 1.19δ (3H, triplet) is

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 7

Explanation of Solution

The molecular formula of the compound is C₉H₁₂.

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 8

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 9

Thus the compound has four unsaturation units like double bonds and/or rings. The four proton broad band at 7.13δ indicates that the compound is aromatic with two substituent groups attached to the benzene ring. The two proton quartet at 2.64δ (benzylic) and three proton triplet at 1.19δ (primary alkyl) could be accounted for only if an ethyl group is present on the ring. The three proton singlet at 2.31δ (benzylic) can be attributed to a methyl group attached to the ring. Thus the two substituent groups attached to the benzene ring are ethyl and methyl. The I.R absorption at 750 cm^-1 requires that the ethyl and methyl groups to be arranged in ortho positions. Hence the compound is o-ethyltoluene.

Conclusion

A structure for the compound whose ¹HNMR spectrum given is to be proposed. Given: M.F = C₉H₁₂, I.R: 750 cm^-1, ¹HNMR spectrum: 7.13δ (4H, broad); 2.64δ (2H, quartet); 2.31δ (3H, singlet); 1.19δ (3H, triplet) is

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 10

Interpretation Introduction

c)

Interpretation:

A structure for the compound whose ¹HNMR spectrum given is to be proposed. Given: M.F = C₁₁H₁₆ I.R: 820 cm^-1.

¹HNMR spectrum: 7.27δ (2H, doublet); 7.06δ (2H, doublet); 2.30δ (3H, singlet); 1.31δ (9H, singlet).

Concept introduction:

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

In I.R, the o-disubstituted benzenes absorb around 735-770 cm^-1, m-disubstituted benzenes absorb around 690-710 cm^-1 and p-disubstituted benzenes absorb around 810-840 cm^-1.

To propose:

A structure for the compound whose ¹HNMR spectrum given is to be proposed. Given: M.F = C₁₁H₁₆ I.R: 820 cm^-1.

¹HNMR spectrum: 7.27δ (2H, doublet); 7.06δ (2H, doublet); 2.30δ (3H, singlet); 1.31δ (9H, singlet).

Expert Solution

Answer to Problem 47AP

A structure for the compound whose ¹HNMR spectrum given is to be proposed. Given: M.F = C₁₁H₁₆; I.R: 820 cm^-1; ¹HNMR spectrum: 7.27δ (2H, doublet); 7.06δ (2H, doublet); 2.30δ (3H, singlet); 1.31δ (9H, singlet).

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 12

Explanation of Solution

The molecular formula of the compound is C₁₁H₁₆.

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 13

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 14

Thus the compound has four unsaturation units like double bonds and/or rings. The two doublets at 7.27δ and 7.06δ each corresponding to two protons indicate the compound is aromatic and each proton responsible for the doublet has a proton on the adjacent carbon. The three proton singlet at 2.58δ (benzylic) could be accounted for a methyl group attached to the benzene ring. The remaining four carbons and nine hydrogens indicate the presence of a tert-butyl group attached to the benzene ring which is confirmed by the nine proton singlet at 1.31δ (primary alkyl). Thus methyl and tert-butyl are the two substituent groups on the benzene ring. The I.R absorption at 820 cm^-1 indicates that the two substituent groups are arranged in para positions. Hence the compound is p-t-butyltoluene.

Conclusion

A structure for the compound whose 1HNMR spectrum given is to be proposed. Given: M.F = C₁₁H₁₆; I.R: 820 cm^-1; ¹HNMR spectrum: 7.27δ (2H, doublet); 7.06δ (2H, doublet); 2.30δ (3H, singlet); 1.31δ (9H, singlet).

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 15

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 15.SE, Problem 47AP

Interpretation Introduction

a)

Interpretation:

A structure for the compound whose ¹HNMR spectrum given is to be proposed. Given: M.F = C₈H₉Br I.R: 820 cm^-1.

¹HNMR spectrum: 7.07δ (2H, doublet); 7.39δ (2H, doublet); 2.58δ (2H, quartet); 1.20δ (3H, triplet).

Concept introduction:

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

In I.R, the o-disubstituted benzenes absorb around 735-770 cm^-1, m-disubstituted benzenes absorb around 690-710 cm^-1 and p-disubstituted benzenes absorb around 810-840 cm^-1.

To propose:

A structure for the compound whose ¹HNMR spectrum given is to be proposed. Given: M.F = C₈H₉Br I.R: 820 cm^-1.

¹HNMR spectrum: 7.07δ (2H, doublet); 7.39δ (2H, doublet); 2.58δ (2H, quartet); 1.20δ (3H, triplet).

Expert Solution

Answer to Problem 47AP

A structure for the compound with M.F = C₈H₉Br with spectral characteristics: I.R: 820 cm^-1, ¹HNMR spectrum: 7.07δ (2H, doublet); 7.39δ (2H, doublet); 2.58δ (2H, quartet); 1.20δ (3H, triplet) is

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 2

Explanation of Solution

The molecular formula of the compound is C₈H₉Br.

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 3

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 4

Thus the compound has four unsaturation units like double bonds and/or rings. The two doublets at 7.07δ and 7.39δ each corresponding to two protons indicate the compound is aromatic and each proton responsible for the doublet has a proton on the adjacent carbon. The two proton quartet at 2.58δ (benzylic) and three proton triplet at 1.20δ (primary) could be accounted for only if an ethyl substituent is present on the ring along with bromine. The I.R absorption at 820 cm^-1 indicates that the ethyl group and bromine are arranged in para positions. Hence the compound is p-bromoethyl benzene.

Conclusion

A structure for the compound with M.F=C₈H₉Br with spectral characteristics: I.R: 820 cm^-1, ¹HNMR spectrum: 7.07δ (2H, doublet); 7.39δ (2H, doublet); 2.58δ (2H, quartet); 1.20δ (3H, triplet) is

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 5

Interpretation Introduction

b)

Interpretation:

A structure for the compound whose ¹HNMR spectrum given is to be proposed.

Given: M.F = C₉H₁₂ I.R: 750 cm^-1

¹HNMR spectrum: 7.13δ (4H, broad); 2.64δ (2H, quartet); 2.31δ (3H, singlet); 1.19δ (3H, triplet).

Concept introduction:

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

In I.R, the o-disubstituted benzenes absorb around 735-770 cm^-1, m-disubstituted benzenes absorb around 690-710 cm^-1and p-disubstituted benzenes absorb around 810-840 cm^-1.

To propose:

A structure for the compound whose ¹HNMR spectrum given is to be proposed. Given: M.F = C₉H₁₂ I.R: 750 cm^-1.

¹HNMR spectrum: 7.13δ (4H, broad); 2.64δ (2H, quartet); 2.31δ (3H, singlet); 1.19δ (3H, triplet).

Expert Solution

Answer to Problem 47AP

A structure for the compound whose 1HNMR spectrum given is to be proposed. Given: M.F = C₉H₁₂, I.R: 750 cm^-1, ¹HNMR spectrum: 7.13δ (4H, broad); 2.64δ (2H, quartet); 2.31δ (3H, singlet); 1.19δ (3H, triplet) is

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 7

Explanation of Solution

The molecular formula of the compound is C₉H₁₂.

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 8

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 9

Thus the compound has four unsaturation units like double bonds and/or rings. The four proton broad band at 7.13δ indicates that the compound is aromatic with two substituent groups attached to the benzene ring. The two proton quartet at 2.64δ (benzylic) and three proton triplet at 1.19δ (primary alkyl) could be accounted for only if an ethyl group is present on the ring. The three proton singlet at 2.31δ (benzylic) can be attributed to a methyl group attached to the ring. Thus the two substituent groups attached to the benzene ring are ethyl and methyl. The I.R absorption at 750 cm^-1 requires that the ethyl and methyl groups to be arranged in ortho positions. Hence the compound is o-ethyltoluene.

Conclusion

A structure for the compound whose ¹HNMR spectrum given is to be proposed. Given: M.F = C₉H₁₂, I.R: 750 cm^-1, ¹HNMR spectrum: 7.13δ (4H, broad); 2.64δ (2H, quartet); 2.31δ (3H, singlet); 1.19δ (3H, triplet) is

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 10

Interpretation Introduction

c)

Interpretation:

A structure for the compound whose ¹HNMR spectrum given is to be proposed. Given: M.F = C₁₁H₁₆ I.R: 820 cm^-1.

¹HNMR spectrum: 7.27δ (2H, doublet); 7.06δ (2H, doublet); 2.30δ (3H, singlet); 1.31δ (9H, singlet).

Concept introduction:

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

In I.R, the o-disubstituted benzenes absorb around 735-770 cm^-1, m-disubstituted benzenes absorb around 690-710 cm^-1 and p-disubstituted benzenes absorb around 810-840 cm^-1.

To propose:

A structure for the compound whose ¹HNMR spectrum given is to be proposed. Given: M.F = C₁₁H₁₆ I.R: 820 cm^-1.

¹HNMR spectrum: 7.27δ (2H, doublet); 7.06δ (2H, doublet); 2.30δ (3H, singlet); 1.31δ (9H, singlet).

Expert Solution

Answer to Problem 47AP

A structure for the compound whose ¹HNMR spectrum given is to be proposed. Given: M.F = C₁₁H₁₆; I.R: 820 cm^-1; ¹HNMR spectrum: 7.27δ (2H, doublet); 7.06δ (2H, doublet); 2.30δ (3H, singlet); 1.31δ (9H, singlet).

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 12

Explanation of Solution

The molecular formula of the compound is C₁₁H₁₆.

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 13

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 14

Thus the compound has four unsaturation units like double bonds and/or rings. The two doublets at 7.27δ and 7.06δ each corresponding to two protons indicate the compound is aromatic and each proton responsible for the doublet has a proton on the adjacent carbon. The three proton singlet at 2.58δ (benzylic) could be accounted for a methyl group attached to the benzene ring. The remaining four carbons and nine hydrogens indicate the presence of a tert-butyl group attached to the benzene ring which is confirmed by the nine proton singlet at 1.31δ (primary alkyl). Thus methyl and tert-butyl are the two substituent groups on the benzene ring. The I.R absorption at 820 cm^-1 indicates that the two substituent groups are arranged in para positions. Hence the compound is p-t-butyltoluene.

Conclusion

A structure for the compound whose 1HNMR spectrum given is to be proposed. Given: M.F = C₁₁H₁₆; I.R: 820 cm^-1; ¹HNMR spectrum: 7.27δ (2H, doublet); 7.06δ (2H, doublet); 2.30δ (3H, singlet); 1.31δ (9H, singlet).

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 15

Answer 6

Chapter 15.SE, Problem 47AP

Interpretation Introduction

a)

Interpretation:

A structure for the compound whose ¹HNMR spectrum given is to be proposed. Given: M.F = C₈H₉Br I.R: 820 cm^-1.

¹HNMR spectrum: 7.07δ (2H, doublet); 7.39δ (2H, doublet); 2.58δ (2H, quartet); 1.20δ (3H, triplet).

Concept introduction:

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

In I.R, the o-disubstituted benzenes absorb around 735-770 cm^-1, m-disubstituted benzenes absorb around 690-710 cm^-1 and p-disubstituted benzenes absorb around 810-840 cm^-1.

To propose:

A structure for the compound whose ¹HNMR spectrum given is to be proposed. Given: M.F = C₈H₉Br I.R: 820 cm^-1.

¹HNMR spectrum: 7.07δ (2H, doublet); 7.39δ (2H, doublet); 2.58δ (2H, quartet); 1.20δ (3H, triplet).

Expert Solution

Answer to Problem 47AP

A structure for the compound with M.F = C₈H₉Br with spectral characteristics: I.R: 820 cm^-1, ¹HNMR spectrum: 7.07δ (2H, doublet); 7.39δ (2H, doublet); 2.58δ (2H, quartet); 1.20δ (3H, triplet) is

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 2

Explanation of Solution

The molecular formula of the compound is C₈H₉Br.

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 3

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 4

Thus the compound has four unsaturation units like double bonds and/or rings. The two doublets at 7.07δ and 7.39δ each corresponding to two protons indicate the compound is aromatic and each proton responsible for the doublet has a proton on the adjacent carbon. The two proton quartet at 2.58δ (benzylic) and three proton triplet at 1.20δ (primary) could be accounted for only if an ethyl substituent is present on the ring along with bromine. The I.R absorption at 820 cm^-1 indicates that the ethyl group and bromine are arranged in para positions. Hence the compound is p-bromoethyl benzene.

Conclusion

A structure for the compound with M.F=C₈H₉Br with spectral characteristics: I.R: 820 cm^-1, ¹HNMR spectrum: 7.07δ (2H, doublet); 7.39δ (2H, doublet); 2.58δ (2H, quartet); 1.20δ (3H, triplet) is

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 5

Answer 7

Chapter 15.SE, Problem 47AP

Interpretation Introduction

a)

Interpretation:

A structure for the compound whose ¹HNMR spectrum given is to be proposed. Given: M.F = C₈H₉Br I.R: 820 cm^-1.

¹HNMR spectrum: 7.07δ (2H, doublet); 7.39δ (2H, doublet); 2.58δ (2H, quartet); 1.20δ (3H, triplet).

Concept introduction:

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

In I.R, the o-disubstituted benzenes absorb around 735-770 cm^-1, m-disubstituted benzenes absorb around 690-710 cm^-1 and p-disubstituted benzenes absorb around 810-840 cm^-1.

To propose:

A structure for the compound whose ¹HNMR spectrum given is to be proposed. Given: M.F = C₈H₉Br I.R: 820 cm^-1.

¹HNMR spectrum: 7.07δ (2H, doublet); 7.39δ (2H, doublet); 2.58δ (2H, quartet); 1.20δ (3H, triplet).

Answer 8

Expert Solution

Answer to Problem 47AP

A structure for the compound with M.F = C₈H₉Br with spectral characteristics: I.R: 820 cm^-1, ¹HNMR spectrum: 7.07δ (2H, doublet); 7.39δ (2H, doublet); 2.58δ (2H, quartet); 1.20δ (3H, triplet) is

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 2

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

The molecular formula of the compound is C₈H₉Br.

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 3

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 4

Thus the compound has four unsaturation units like double bonds and/or rings. The two doublets at 7.07δ and 7.39δ each corresponding to two protons indicate the compound is aromatic and each proton responsible for the doublet has a proton on the adjacent carbon. The two proton quartet at 2.58δ (benzylic) and three proton triplet at 1.20δ (primary) could be accounted for only if an ethyl substituent is present on the ring along with bromine. The I.R absorption at 820 cm^-1 indicates that the ethyl group and bromine are arranged in para positions. Hence the compound is p-bromoethyl benzene.

Answer 13

Conclusion

A structure for the compound with M.F=C₈H₉Br with spectral characteristics: I.R: 820 cm^-1, ¹HNMR spectrum: 7.07δ (2H, doublet); 7.39δ (2H, doublet); 2.58δ (2H, quartet); 1.20δ (3H, triplet) is

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 5

Answer 14

Interpretation Introduction

b)

Interpretation:

A structure for the compound whose ¹HNMR spectrum given is to be proposed.

Given: M.F = C₉H₁₂ I.R: 750 cm^-1

¹HNMR spectrum: 7.13δ (4H, broad); 2.64δ (2H, quartet); 2.31δ (3H, singlet); 1.19δ (3H, triplet).

Concept introduction:

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

In I.R, the o-disubstituted benzenes absorb around 735-770 cm^-1, m-disubstituted benzenes absorb around 690-710 cm^-1and p-disubstituted benzenes absorb around 810-840 cm^-1.

To propose:

A structure for the compound whose ¹HNMR spectrum given is to be proposed. Given: M.F = C₉H₁₂ I.R: 750 cm^-1.

¹HNMR spectrum: 7.13δ (4H, broad); 2.64δ (2H, quartet); 2.31δ (3H, singlet); 1.19δ (3H, triplet).

Expert Solution

Answer to Problem 47AP

A structure for the compound whose 1HNMR spectrum given is to be proposed. Given: M.F = C₉H₁₂, I.R: 750 cm^-1, ¹HNMR spectrum: 7.13δ (4H, broad); 2.64δ (2H, quartet); 2.31δ (3H, singlet); 1.19δ (3H, triplet) is

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 7

Explanation of Solution

The molecular formula of the compound is C₉H₁₂.

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 8

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 9

Thus the compound has four unsaturation units like double bonds and/or rings. The four proton broad band at 7.13δ indicates that the compound is aromatic with two substituent groups attached to the benzene ring. The two proton quartet at 2.64δ (benzylic) and three proton triplet at 1.19δ (primary alkyl) could be accounted for only if an ethyl group is present on the ring. The three proton singlet at 2.31δ (benzylic) can be attributed to a methyl group attached to the ring. Thus the two substituent groups attached to the benzene ring are ethyl and methyl. The I.R absorption at 750 cm^-1 requires that the ethyl and methyl groups to be arranged in ortho positions. Hence the compound is o-ethyltoluene.

Conclusion

A structure for the compound whose ¹HNMR spectrum given is to be proposed. Given: M.F = C₉H₁₂, I.R: 750 cm^-1, ¹HNMR spectrum: 7.13δ (4H, broad); 2.64δ (2H, quartet); 2.31δ (3H, singlet); 1.19δ (3H, triplet) is

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 10

Answer 15

Interpretation Introduction

b)

Interpretation:

A structure for the compound whose ¹HNMR spectrum given is to be proposed.

Given: M.F = C₉H₁₂ I.R: 750 cm^-1

¹HNMR spectrum: 7.13δ (4H, broad); 2.64δ (2H, quartet); 2.31δ (3H, singlet); 1.19δ (3H, triplet).

Concept introduction:

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

In I.R, the o-disubstituted benzenes absorb around 735-770 cm^-1, m-disubstituted benzenes absorb around 690-710 cm^-1and p-disubstituted benzenes absorb around 810-840 cm^-1.

To propose:

A structure for the compound whose ¹HNMR spectrum given is to be proposed. Given: M.F = C₉H₁₂ I.R: 750 cm^-1.

¹HNMR spectrum: 7.13δ (4H, broad); 2.64δ (2H, quartet); 2.31δ (3H, singlet); 1.19δ (3H, triplet).

Answer 16

Expert Solution

Answer to Problem 47AP

A structure for the compound whose 1HNMR spectrum given is to be proposed. Given: M.F = C₉H₁₂, I.R: 750 cm^-1, ¹HNMR spectrum: 7.13δ (4H, broad); 2.64δ (2H, quartet); 2.31δ (3H, singlet); 1.19δ (3H, triplet) is

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 7

Answer 17

Expert Solution

Answer 18

Expert Solution

Answer 19

Expert Solution

Answer 20

Explanation of Solution

The molecular formula of the compound is C₉H₁₂.

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 8

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 9

Thus the compound has four unsaturation units like double bonds and/or rings. The four proton broad band at 7.13δ indicates that the compound is aromatic with two substituent groups attached to the benzene ring. The two proton quartet at 2.64δ (benzylic) and three proton triplet at 1.19δ (primary alkyl) could be accounted for only if an ethyl group is present on the ring. The three proton singlet at 2.31δ (benzylic) can be attributed to a methyl group attached to the ring. Thus the two substituent groups attached to the benzene ring are ethyl and methyl. The I.R absorption at 750 cm^-1 requires that the ethyl and methyl groups to be arranged in ortho positions. Hence the compound is o-ethyltoluene.

Answer 21

Conclusion

A structure for the compound whose ¹HNMR spectrum given is to be proposed. Given: M.F = C₉H₁₂, I.R: 750 cm^-1, ¹HNMR spectrum: 7.13δ (4H, broad); 2.64δ (2H, quartet); 2.31δ (3H, singlet); 1.19δ (3H, triplet) is

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 10

Answer 22

Interpretation Introduction

c)

Interpretation:

A structure for the compound whose ¹HNMR spectrum given is to be proposed. Given: M.F = C₁₁H₁₆ I.R: 820 cm^-1.

¹HNMR spectrum: 7.27δ (2H, doublet); 7.06δ (2H, doublet); 2.30δ (3H, singlet); 1.31δ (9H, singlet).

Concept introduction:

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

In I.R, the o-disubstituted benzenes absorb around 735-770 cm^-1, m-disubstituted benzenes absorb around 690-710 cm^-1 and p-disubstituted benzenes absorb around 810-840 cm^-1.

To propose:

A structure for the compound whose ¹HNMR spectrum given is to be proposed. Given: M.F = C₁₁H₁₆ I.R: 820 cm^-1.

¹HNMR spectrum: 7.27δ (2H, doublet); 7.06δ (2H, doublet); 2.30δ (3H, singlet); 1.31δ (9H, singlet).

Expert Solution

Answer to Problem 47AP

A structure for the compound whose ¹HNMR spectrum given is to be proposed. Given: M.F = C₁₁H₁₆; I.R: 820 cm^-1; ¹HNMR spectrum: 7.27δ (2H, doublet); 7.06δ (2H, doublet); 2.30δ (3H, singlet); 1.31δ (9H, singlet).

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 12

Explanation of Solution

The molecular formula of the compound is C₁₁H₁₆.

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 13

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 14

Thus the compound has four unsaturation units like double bonds and/or rings. The two doublets at 7.27δ and 7.06δ each corresponding to two protons indicate the compound is aromatic and each proton responsible for the doublet has a proton on the adjacent carbon. The three proton singlet at 2.58δ (benzylic) could be accounted for a methyl group attached to the benzene ring. The remaining four carbons and nine hydrogens indicate the presence of a tert-butyl group attached to the benzene ring which is confirmed by the nine proton singlet at 1.31δ (primary alkyl). Thus methyl and tert-butyl are the two substituent groups on the benzene ring. The I.R absorption at 820 cm^-1 indicates that the two substituent groups are arranged in para positions. Hence the compound is p-t-butyltoluene.

Conclusion

A structure for the compound whose 1HNMR spectrum given is to be proposed. Given: M.F = C₁₁H₁₆; I.R: 820 cm^-1; ¹HNMR spectrum: 7.27δ (2H, doublet); 7.06δ (2H, doublet); 2.30δ (3H, singlet); 1.31δ (9H, singlet).

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 15

Answer 23

Interpretation Introduction

c)

Interpretation:

A structure for the compound whose ¹HNMR spectrum given is to be proposed. Given: M.F = C₁₁H₁₆ I.R: 820 cm^-1.

¹HNMR spectrum: 7.27δ (2H, doublet); 7.06δ (2H, doublet); 2.30δ (3H, singlet); 1.31δ (9H, singlet).

Concept introduction:

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

In I.R, the o-disubstituted benzenes absorb around 735-770 cm^-1, m-disubstituted benzenes absorb around 690-710 cm^-1 and p-disubstituted benzenes absorb around 810-840 cm^-1.

To propose:

A structure for the compound whose ¹HNMR spectrum given is to be proposed. Given: M.F = C₁₁H₁₆ I.R: 820 cm^-1.

¹HNMR spectrum: 7.27δ (2H, doublet); 7.06δ (2H, doublet); 2.30δ (3H, singlet); 1.31δ (9H, singlet).

Answer 24

Expert Solution

Answer to Problem 47AP

A structure for the compound whose ¹HNMR spectrum given is to be proposed. Given: M.F = C₁₁H₁₆; I.R: 820 cm^-1; ¹HNMR spectrum: 7.27δ (2H, doublet); 7.06δ (2H, doublet); 2.30δ (3H, singlet); 1.31δ (9H, singlet).

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 12

Answer 25

Expert Solution

Answer 26

Expert Solution

Answer 27

Expert Solution

Answer 28

Explanation of Solution

The molecular formula of the compound is C₁₁H₁₆.

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 13

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 14

Thus the compound has four unsaturation units like double bonds and/or rings. The two doublets at 7.27δ and 7.06δ each corresponding to two protons indicate the compound is aromatic and each proton responsible for the doublet has a proton on the adjacent carbon. The three proton singlet at 2.58δ (benzylic) could be accounted for a methyl group attached to the benzene ring. The remaining four carbons and nine hydrogens indicate the presence of a tert-butyl group attached to the benzene ring which is confirmed by the nine proton singlet at 1.31δ (primary alkyl). Thus methyl and tert-butyl are the two substituent groups on the benzene ring. The I.R absorption at 820 cm^-1 indicates that the two substituent groups are arranged in para positions. Hence the compound is p-t-butyltoluene.

Answer 29

Conclusion

A structure for the compound whose 1HNMR spectrum given is to be proposed. Given: M.F = C₁₁H₁₆; I.R: 820 cm^-1; ¹HNMR spectrum: 7.27δ (2H, doublet); 7.06δ (2H, doublet); 2.30δ (3H, singlet); 1.31δ (9H, singlet).

EP ORGANIC CHEMISTRY,24 MONTH-OWLV2, Chapter 15.SE, Problem 47AP , additional homework tip 15

Answer 30

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

Answer to Problem 47AP

Explanation of Solution

Answer to Problem 47AP

Explanation of Solution

Answer to Problem 47AP

Explanation of Solution

Want to see more full solutions like this?

Chapter 15 Solutions