EBK EXPERIMENTAL ORGANIC CHEMISTRY: A M
EBK EXPERIMENTAL ORGANIC CHEMISTRY: A M
6th Edition
ISBN: 9781305687875
Author: Gilbert
Publisher: CENGAGE LEARNING - CONSIGNMENT
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Chapter 10.5, Problem 17E

(a)

Interpretation Introduction

Interpretation:Absorbance corresponding to carbon-carbon double bond and the vinylic hydrogen atoms in the IR and NMR spectra of 1-hexene should be specified.

Concept introduction: Infrared absorption frequencies values are given as follows:

  EBK EXPERIMENTAL ORGANIC CHEMISTRY: A M, Chapter 10.5, Problem 17E , additional homework tip  1

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.

The typical chemical shifts values are given as follows:

  EBK EXPERIMENTAL ORGANIC CHEMISTRY: A M, Chapter 10.5, Problem 17E , additional homework tip  2

(a)

Expert Solution
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Explanation of Solution

IR spectrum of 1-hexene is as follows:

  EBK EXPERIMENTAL ORGANIC CHEMISTRY: A M, Chapter 10.5, Problem 17E , additional homework tip  3

Three signals at frequencies 985 cm1 , 1650 cm1 and 3000 cm1 correspond to vinylic bending, C=C and C=CH absorption signals respectively.

(b)

Interpretation Introduction

Interpretation: The various resonances corresponding to 1H NMR of hydrogen nuclei found in 1-hexene 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 follows:

  EBK EXPERIMENTAL ORGANIC CHEMISTRY: A M, Chapter 10.5, Problem 17E , additional homework tip  4

The NMR spectrum of 1-hexene is given as follows:

  EBK EXPERIMENTAL ORGANIC CHEMISTRY: A M, Chapter 10.5, Problem 17E , additional homework tip  5

Signals for olefinic hydrogen and two proton doublet for terminal olefinic protons at δ 5.82 , 5.03 and 4.97 are evident from the spectrum.

The slight downfield signal of the methylene proton signal attached immediately next to olefin appears at 2.18.For each methylene protons set that is farther the δ value is close to 1.

Therefore various resonances corresponding to 1H NMR of hydrogen nuclei found in 1-hexene

are assigned as follows:

  EBK EXPERIMENTAL ORGANIC CHEMISTRY: A M, Chapter 10.5, Problem 17E , additional homework tip  6

(c)

Interpretation Introduction

Interpretation: The various resonances corresponding to 13C NMRofcarbon nuclei found in 1-hexene should be assigned.

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

Thus if a fragment is CH2CH3 then it would show two carbon signals for each of theses carbon.

(c)

Expert Solution
Check Mark

Explanation of Solution

The typical chemical shifts values are given as follows:

  EBK EXPERIMENTAL ORGANIC CHEMISTRY: A M, Chapter 10.5, Problem 17E , additional homework tip  7

  13C NMR data is given as follows:

  δ 14.0, 22.5, 31.6, 33.8, 114.3, 139.2

A total of 6 distinct signal peaks are given this indicates that there are six types of non-equivalent carbon in 1-hexene .

The signal near to 114.3 is for terminal olefinic carbon and near 139.2 is for other olefinic carbon. For the rest of the sp3 hybridized carbons signals appear near diminished δ

values assigned as follows:

  EBK EXPERIMENTAL ORGANIC CHEMISTRY: A M, Chapter 10.5, Problem 17E , additional homework tip  8

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Chapter 10 Solutions

EBK EXPERIMENTAL ORGANIC CHEMISTRY: A M

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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