OWLv2 with LabSkills for Gilbert/Martin's Experimental Organic Chemistry: A Miniscale & Microscale Approach, 6th Edition, [Instant Access], 4 terms (24 months)
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Chapter 10.6, Problem 28E

(a)

Interpretation Introduction

Interpretation: Absorption associated with carboxylic acid and alkene functional groups and trans C=C in the IR spectrum of (E)-cinnamic acid should be identified.

Concept introduction:The IR spectrum of compound helps to detect the presence of the functional group in molecule. The functional groups that are present in compound absorb certain IR radiation at a characteristic wavelength or wavenumber. For example, the carbonyl group present in compound shows a characteristic peak around 1700 cm1 in the IR spectrum.

(a)

Expert Solution
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Explanation of Solution

The typical infrared absorption frequencies values are given as follows:

  OWLv2 with LabSkills for Gilbert/Martin's Experimental Organic Chemistry: A Miniscale & Microscale Approach, 6th Edition, [Instant Access], 4 terms (24 months), Chapter 10.6, Problem 28E , additional homework tip  1

IR spectrum of (E)-cinnamic acid is given as follows:

  OWLv2 with LabSkills for Gilbert/Martin's Experimental Organic Chemistry: A Miniscale & Microscale Approach, 6th Edition, [Instant Access], 4 terms (24 months), Chapter 10.6, Problem 28E , additional homework tip  2

The sharp absorption signal corresponding to carboxylic acid appears near 1700 cm1 . The signal corresponding to alkene appears at 1660 cm1 .

The fingerprint region that lies within 1250600 cm1 shows the evidence for trans- C=C bond. The signal near 965 cm1 is characteristic for C=CH bending.

(b)

Interpretation Introduction

Interpretation: The various resonances corresponding to 1H NMR of hydrogen nuclei found in (E)-cinnamic acid should be assigned.

Concept introduction:The scale used for the NMR spectrum is delta scale. δ denotes parts per million units. The lower values of δ denote upfield while higher values of δ denote downfield region. Each peak in the NMR spectrum corresponds to distinct hydrogen.

The area within each peak corresponds to the number of equivalent protons found at that chemical shift values.

Spin-spin splitting is observed as a result of the interaction amongst non-equivalent NMR active nuclei. This is independent of the strength of the external magnetic field. The formula to calculate the number of peaks in the 1H NMR spectra is as follows:

  Number of peaks=N+1

Where,

  • N is the number of equivalent protons on the adjacent carbon atoms.

Thus if a fragment is CH2CH3 then it would show a three proton triplet due to CH3 and a two proton quartet in accordance with N+1 the rule.

(b)

Expert Solution
Check Mark

Explanation of Solution

The typical chemical shifts values are given as follws:

  OWLv2 with LabSkills for Gilbert/Martin's Experimental Organic Chemistry: A Miniscale & Microscale Approach, 6th Edition, [Instant Access], 4 terms (24 months), Chapter 10.6, Problem 28E , additional homework tip  3

The 1H NMR spectra corresponding to (E)-cinnamic acid is given as follows:

  OWLv2 with LabSkills for Gilbert/Martin's Experimental Organic Chemistry: A Miniscale & Microscale Approach, 6th Edition, [Instant Access], 4 terms (24 months), Chapter 10.6, Problem 28E , additional homework tip  4

The signals are magnified and hence various resonances in 1H

NMR are assigned as follows:

  OWLv2 with LabSkills for Gilbert/Martin's Experimental Organic Chemistry: A Miniscale & Microscale Approach, 6th Edition, [Instant Access], 4 terms (24 months), Chapter 10.6, Problem 28E , additional homework tip  5

(c)

Interpretation Introduction

Interpretation: The manner the resonances from the vinylic hydrogens can be used to distinguish E and Z on the basis of 1H the NMR spectrum of (E)-cinnamic acid should be determined.

Concept introduction:The scale used for the NMR spectrum is delta scale. δ denotes parts per million units. The lower values of δ denote upfield while higher values of δ denote downfield region. Each peak in the NMR spectrum corresponds to distinct hydrogen.

The area within each peak corresponds to the number of equivalent protons found at that chemical shift values.

Spin-spin splitting is observed as a result of the interaction amongst non-equivalent NMR active nuclei. This is independent of the strength of the external magnetic field. The formula to calculate the number of peaks in the 1H NMR spectra is as follows:

  Number of peaks=N+1

Where,

  • N is the number of equivalent protons on the adjacent carbon atoms.

Thus if a fragment is CH2CH3 then it would show a three proton triplet due to CH3 and a two proton quartet in accordance with N+1 the rule.

(c)

Expert Solution
Check Mark

Explanation of Solution

The typical chemical shifts values are given as follows:

  OWLv2 with LabSkills for Gilbert/Martin's Experimental Organic Chemistry: A Miniscale & Microscale Approach, 6th Edition, [Instant Access], 4 terms (24 months), Chapter 10.6, Problem 28E , additional homework tip  6

The signals are magnified and hence various resonances in 1H

NMR are assigned as follows:

  OWLv2 with LabSkills for Gilbert/Martin's Experimental Organic Chemistry: A Miniscale & Microscale Approach, 6th Edition, [Instant Access], 4 terms (24 months), Chapter 10.6, Problem 28E , additional homework tip  7

The coupling constant values indicate whether the alkene is E or Z. If E alkene is present the coupling constant should come tobe much higher than in the case of Z alkene that has a lower value of coupling constant.

This coupling constant is essentially the difference between two adjacent peaks in the NMR spectrum. Since the space between protons on olefin is more, therefore, it can be concluded that more coupling constant is indicative of E stereochemistryin (E)-cinnamic acid .

(d)

Interpretation Introduction

Interpretation: The resonances of 13C NMR should be assigned.

Concept introduction: The scale used for the 13C NMR spectrum is the delta scale with the only difference is larger chemical shift. The range lies from δ 0250 denotes parts per million unit. The lower values of δ denote upfield while higher values of δ denote downfield region. Each peak in the NMR spectrum corresponds to a distinct carbon.

(d)

Expert Solution
Check Mark

Explanation of Solution

The typical chemical shifts values are given as follows:

  OWLv2 with LabSkills for Gilbert/Martin's Experimental Organic Chemistry: A Miniscale & Microscale Approach, 6th Edition, [Instant Access], 4 terms (24 months), Chapter 10.6, Problem 28E , additional homework tip  8

The symmetry in the structure suggests there are four types of aromatic carbon that appear δ at 128.5, 128.6, 127.9 and 135.2.

The ethylene benzylic carbon appears at 116.6 while others olefinic and 144.2 respectively. The carbonyl carbon of carboxylic acid appears at 171.5.

Therefore these seven distinct carbon 13C

NMRresonances are assigned as follows:

  OWLv2 with LabSkills for Gilbert/Martin's Experimental Organic Chemistry: A Miniscale & Microscale Approach, 6th Edition, [Instant Access], 4 terms (24 months), Chapter 10.6, Problem 28E , additional homework tip  9

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Chapter 10 Solutions

OWLv2 with LabSkills for Gilbert/Martin's Experimental Organic Chemistry: A Miniscale & Microscale Approach, 6th Edition, [Instant Access], 4 terms (24 months)

Ch. 10.2 - Prob. 11ECh. 10.2 - Prob. 12ECh. 10.2 - Prob. 13ECh. 10.2 - Prob. 14ECh. 10.2 - Prob. 15ECh. 10.2 - Prob. 16ECh. 10.2 - Prob. 17ECh. 10.3 - Prob. 1ECh. 10.3 - Prob. 2ECh. 10.3 - Prob. 3ECh. 10.3 - Prob. 4ECh. 10.3 - Prob. 5ECh. 10.3 - Prob. 6ECh. 10.3 - Prob. 7ECh. 10.3 - Prob. 8ECh. 10.3 - Prob. 9ECh. 10.3 - Prob. 10ECh. 10.3 - Prob. 11ECh. 10.3 - Prob. 12ECh. 10.3 - Prob. 13ECh. 10.3 - Prob. 14ECh. 10.3 - Prob. 15ECh. 10.3 - Prob. 16ECh. 10.3 - Prob. 17ECh. 10.3 - Prob. 18ECh. 10.3 - Prob. 19ECh. 10.3 - Prob. 20ECh. 10.3 - Prob. 21ECh. 10.3 - Prob. 22ECh. 10.3 - Prob. 23ECh. 10.3 - Prob. 24ECh. 10.3 - Prob. 25ECh. 10.3 - Prob. 26ECh. 10.3 - Prob. 27ECh. 10.3 - Prob. 28ECh. 10.3 - Prob. 29ECh. 10.3 - Prob. 30ECh. 10.3 - Prob. 31ECh. 10.3 - Prob. 32ECh. 10.5 - Prob. 1ECh. 10.5 - Prob. 2ECh. 10.5 - Prob. 3ECh. 10.5 - Prob. 4ECh. 10.5 - Prob. 5ECh. 10.5 - Prob. 6ECh. 10.5 - Prob. 7ECh. 10.5 - Prob. 8ECh. 10.5 - Prob. 9ECh. 10.5 - Prob. 10ECh. 10.5 - Prob. 11ECh. 10.5 - Prob. 12ECh. 10.5 - Prob. 14ECh. 10.5 - Prob. 15ECh. 10.5 - Prob. 16ECh. 10.5 - Prob. 17ECh. 10.5 - Prob. 18ECh. 10.5 - Prob. 19ECh. 10.5 - Prob. 20ECh. 10.5 - Prob. 23ECh. 10.6 - Prob. 1ECh. 10.6 - Prob. 2ECh. 10.6 - Prob. 3ECh. 10.6 - Prob. 4ECh. 10.6 - Prob. 5ECh. 10.6 - Prob. 6ECh. 10.6 - Prob. 7ECh. 10.6 - Prob. 8ECh. 10.6 - Prob. 9ECh. 10.6 - Prob. 10ECh. 10.6 - Prob. 11ECh. 10.6 - Prob. 12ECh. 10.6 - Prob. 13ECh. 10.6 - Prob. 14ECh. 10.6 - Prob. 15ECh. 10.6 - Prob. 16ECh. 10.6 - Prob. 17ECh. 10.6 - Prob. 18ECh. 10.6 - Prob. 20ECh. 10.6 - Prob. 21ECh. 10.6 - Prob. 22ECh. 10.6 - Prob. 23ECh. 10.6 - Prob. 24ECh. 10.6 - Prob. 25ECh. 10.6 - Prob. 26ECh. 10.6 - Prob. 28ECh. 10.6 - Prob. 29ECh. 10.6 - Prob. 30ECh. 10.7 - Prob. 1ECh. 10.7 - Prob. 2ECh. 10.7 - Prob. 3ECh. 10.7 - Prob. 4ECh. 10.7 - Prob. 5ECh. 10.7 - Prob. 6ECh. 10.7 - Prob. 7ECh. 10.7 - Prob. 8ECh. 10.7 - Prob. 9ECh. 10.7 - Prob. 10ECh. 10.7 - Prob. 11ECh. 10.7 - Prob. 12ECh. 10.8 - Prob. 1ECh. 10.8 - Prob. 2ECh. 10.8 - Prob. 4ECh. 10.8 - Prob. 5ECh. 10.8 - Prob. 6ECh. 10.8 - Prob. 7ECh. 10.8 - Prob. 8ECh. 10.8 - Prob. 9ECh. 10.8 - Prob. 10ECh. 10.8 - Prob. 11ECh. 10.8 - Prob. 12ECh. 10.8 - Prob. 13ECh. 10.8 - Prob. 14ECh. 10.8 - Prob. 15E
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