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 10.2, Problem 14E

(a)

Interpretation Introduction

Interpretation: To specify the absorptions which are associated with the carbon-carbon double bond and the hydrogen linked with the first carbon atom.

Concept Introduction: The spectroscopy which includes the interaction of infrared radiation with matter is known as IR spectroscopy. It comprises various techniques, related to absorption spectroscopy. IR spectroscopy is also used to study and recognizing chemical substances.

(a)

Expert Solution
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Answer to Problem 14E

Absorptions associated with the carbon-carbon double bond and the hydrogen linked with C1 carbon atom of 2-methyl -1-butene is as follows:

    Wave numberType of bond
    1500cm1C=C stretching
    3000cm1C=CH stretching

Explanation of Solution

The structure of 2-methyl -1-butene is 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 10.2, Problem 14E , additional homework tip  1

Important IR stretching frequencies are as follows:

    Wave number

      (cm)

    Type of bond
    C=C16801600
    C=N16501550
    C=O17801650
    CO12501050
    CH33002700
    OH (alcohol) 36503200
    OH (carboxylic acid) 33002500
    OWLv2 with LabSkills for Gilbert/Martin's Experimental Organic Chemistry: A Miniscale & Microscale Approach, 6th Edition, [Instant Access], 4 terms (24 months), Chapter 10.2, Problem 14E , additional homework tip  222602220
    OWLv2 with LabSkills for Gilbert/Martin's Experimental Organic Chemistry: A Miniscale & Microscale Approach, 6th Edition, [Instant Access], 4 terms (24 months), Chapter 10.2, Problem 14E , additional homework tip  322602100
    NH35003300
    OWLv2 with LabSkills for Gilbert/Martin's Experimental Organic Chemistry: A Miniscale & Microscale Approach, 6th Edition, [Instant Access], 4 terms (24 months), Chapter 10.2, Problem 14E , additional homework tip  41600and 15001430

From the given spectral data of 2-methyl -1-butene, absorptions associated with the C=C double bond and the hydrogen attached to the C1 carbon atom is as follows:

    Wave numberType of bond
    1500cm1C=C stretching
    3000cm1C=CH stretching

(b)

Interpretation Introduction

Interpretation: To assign the various resonances to the hydrogen nuclei of 2-methyl -1-butene in 1H NMR spectrum.

Concept Introduction: Nuclear magnetic resonance (NMR) is a spectroscopy technique that works on the basis that the nucleus of the atoms absorbs electromagnetic radiation in the radio frequency region.

  1HNMR or proton nuclear magnetic resonance is a tool for explaining the number of protons or hydrogen in the compound.

(b)

Expert Solution
Check Mark

Answer to Problem 14E

Assignment of 1H NMR of 2-methyl -1-butene is 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 10.2, Problem 14E , additional homework tip  5

Explanation of Solution

From the given NMR data of 2-methyl -1-butene, assignment of 1H NMR of 2-bromo-2methylbutane is 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 10.2, Problem 14E , additional homework tip  6

Main features of 1H NMR of 2-methyl -1-butene are as follows:

  1. The methyl protons ( CH3 ) give rise to a signal at δ 1.75 as a 3H, singlet as protons are allylic and no adjacent protons are present.
  2. The vinylic proton =CH2 gives signal of 2H at δ 4.80 .
  3. The allylic protons ( CH2 ) gives a quartet signal at δ 2.00 .
  4. The terminal protons in ( CH3 ) gives peak around at δ 1.05

(c)

Interpretation Introduction

Interpretation: To assign the various resonances to the hydrogen nuclei of 2-methyl -1-butene in 13C NMR spectrum.

Concept Introduction: Carbon-13 NMR has several merits over proton NMR in terms of its power to explain biochemical and organic structures. Carbon-13 NMR provides information related to the backbone of molecules instead of the periphery. In addition, the chemical shift range for Carbon-13 for most organic compounds is around 200 ppm related with 10 to 15 ppm for the proton.

(c)

Expert Solution
Check Mark

Answer to Problem 14E

The assignment of 13C NMR of 2-methyl -1-butene is 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 10.2, Problem 14E , additional homework tip  7

Explanation of Solution

The range of chemical shift in 13C NMR spectra are-

    StructureChemical shift (ppm)
    RCH2OH5065
    RCH2R1555
    RCH31040
    R3CH2060
    C=C120-140

From the given NMR data of 2-methyl -1-butene, the assignment of 13C NMR of 2-methyl -1-butene is 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 10.2, Problem 14E , additional homework tip  8

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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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