Organic Chemistry
Organic Chemistry
8th Edition
ISBN: 9781305580350
Author: William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. Foote
Publisher: Cengage Learning
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Chapter 14, Problem 14.38P
Interpretation Introduction

Interpretation:

The protein (P) gives a three type of cluster and what mass-charge values do these clusters appear in the mass spectrum, the correct reason has to be explained with the given molecular mass values.

Concept Introduction:

Electrospray ionization (ESI):

  • • This technique used in mass spectrometry to produced different ions.
  • • It is especially useful in producing ions from macromolecules because it overcomes the propensity of these molecules to fragment when ionized.
  • • The primary ion source used in liquid chromatography-mass spectrometry, because it is liquid-gas interface that is capable of coupling liquid chromatography with mass spectrometry. 

Mass spectrum: It generates multiple ions from the sample under investigation, it than separates them according to their specific mass to charge ration (m/z) and records the relative abundance (RA) of each ion types.

      MelectronbeamM+.+e-[Molecule]Molecular ionelectron[Radicalcation]

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Relative Intensity Part VI. consider the multi-step reaction below for compounds A, B, and C. These compounds were subjected to mass spectrometric analysis and the following spectra for A, B, and C was obtained. Draw the structure of B and C and match all three compounds to the correct spectra. Relative Intensity Relative Intensity 100 HS-NJ-0547 80 60 31 20 S1 84 M+ absent 10 30 40 50 60 70 80 90 100 100- MS2016-05353CM 80- 60 40 20 135 137 S2 164 166 0-m 25 50 75 100 125 150 m/z 60 100 MS-NJ-09-43 40 20 20 80 45 S3 25 50 75 100 125 150 175 m/z
Part II. Given two isomers: 2-methylpentane (A) and 2,2-dimethyl butane (B) answer the following: (a) match structures of isomers given their mass spectra below (spectra A and spectra B) (b) Draw the fragments given the following prominent peaks from each spectrum: Spectra A m/2 =43 and 1/2-57 spectra B m/2 = 43 (c) why is 1/2=57 peak in spectrum A more intense compared to the same peak in spectrum B. Relative abundance Relative abundance 100 A 50 29 29 0 10 -0 -0 100 B 50 720 30 41 43 57 71 4-0 40 50 60 70 m/z 43 57 8-0 m/z = 86 M 90 100 71 m/z = 86 M -O 0 10 20 30 40 50 60 70 80 -88 m/z 90 100

Chapter 14 Solutions

Organic Chemistry

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