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									              IR /NMR Problem Set Notes

1. The problems in the problem set are designed to test
   your ability to utilize information provided by the
   various spectroscopy tools provided to you in both
   the first and second semesters of Organic Lab.
      Mass Spectrometry; Ultraviolet Spectrometry;
      Infrared Spectrometry (IR); Nuclear Magnetic
      Resonance Spectrometry (NMR); Partial
      Elemental Analysis.
2. The goal in each problem in sections B of the
   problem set is to deduce the structure and identity of
   the compound.
3. An answer sheet is provided for each problem.
4. The student is expected to provide a complete
   analysis of the information provided for each type of
   spectra provided and then utilize the combined
   information to identify the compound.
5. The answers to the problems are to be typed in the
   indicated boxes.
6. Students are encouraged to utilize literature
   resources, including the web, to aid in the
   identification process. The following web sites are
   quite useful.
   • http://riodb01.ibase.aist.go.jp/sdbs/cgi-
     bin/direct_frame_top.cgi
   • http://chemfinder.cambridgesoft.com/
             IR /NMR Problem Set Notes

7. The correct solution to the spectroscopy
   identifications must include support arguments from
   as many of the following sources that are provided
   with the problem:
    Significance of Mass Spectrometry information
     (Molecular Ion Peak, Molecular Weight,
     Nitrogen, Halogens, Significant Fragments).
    Elemental Analysis calculations to determine the
     Molecular Formula. The % Carbon and the %
     Hydrogen is usually given. From the molecular
     weight and amounts of Carbon & Hydrogen
     present, the student determines the other
     elements present in the compound along with
     their respective percentages.
    Significance of Ultraviolet Log Absorptivity
     Coefficient (Log ) values, if given.
    Evaluation of principal IR Absorptions, if given.
    Evaluation of principal NMR (1H1 & 13C6) signals
     & splitting patterns (proton number, types, and
     location.)
             IR /NMR Problem Set Notes

8. The following information may be available:
    a. IR Spectrum (Functional Groups)
    b. 1H1 NMR Spectrum (No., Type, Location of
       Protons)
    c. 13C6 NMR Spectrum (No. & Type Carbon atoms)
   d. UV-Vis Molar Absorptivity (Molar Extinction
      Coefficient) –  & log 
       Conjugate systems (alternating double bonds)
                  -C-C=C-C=C-
         Aliphatic Conjugation shows values of
          in the range:
            = 3000 – 100,000   (Log  = 3.5 – 5)
         Aromatic Conjugation shows values of
          in the range:
             = 1000 – 3000     (Log  = 3.0 – 3.5)
       Carbonyl (C=O) compounds show values of
         in the range:
             = 30 – 300        (Log  = 1.5 – 2.5)
       Nitro (NO2) compounds show values of
         in the range:
            = 10 – 30          (Log  = 1.0 – 1.5
         IR /NMR Problem Set Notes
e. Mass Spectrum with Molecular Ion Peak
    Molecular Ion Peak represents Molecular
     Weight.
    Molecular Ion peak values that are Odd
     indicate the presence of an Odd number of
     Nitrogen atoms in the compound.
    Two Molecular Ion peaks with a relative
     abundance ratio of 3:1 indicate presence of
     a single Chlorine atom.
    Two Molecular Ion peaks with a relative
     abundance ratio of 1:1 indicated presence
     of a single Bromine atom.
    The Mass difference between the Base
     Peak (100% abundance) and the Molecular
     Ion Peak sometimes represents a
     significant fragment that has been removed
     from the molecule.
           IR /NMR Problem Set Notes

f. A partial elemental analysis of the compound
     The percentage values given represent the
      percent of the compound’s Molecular
      Weight attributed to that element.
     Usually %Carbon & %Hydrogen are given.
     The Molecular Ion peak(s), Molar
      Absorptivity Coefficient, and the principal
      functional groups from the IR spectrum
      often give clues as to additional elements
      present in the compound.
     The remaining Molecular Weight after the
      Carbon and Hydrogen have been
      accounted for is divided among the
      remaining elements in the compound.
Molecular Weight      = 58.08
% Carbon              = 62.1 %
% Hydrogen            = 10.4%
   62.1/100 * 58.08 = 36.07/12 = ~ 3  3 Carbon
   10.4/100 * 58.08 =   6.04/1 = ~ 6  6 Hydrogen
58.08 - (36.07 + 6.04) = 15.97/16 = ~ 1  1 Oxygen
15.97 / 58.08 * 100   = 27.5% Oxygen


         Molecular Formula – C3H6O
                Problem Set Example
The following information is provided. See following
slides for details.

   The Molecular Ion Peak from a Mass Spectrometer
    indicates a value of 165.
   Partial Elemental Analysis
       C - 65.44%
       H – 6.71%
   Mass Spectrum
   IR Spectrum
   NMR 1H1 Spectrum
             Problem Set Example


 The Molecular Ion Peak from a Mass Spectrometer
  indicates a value of 165.

  Analysis – Compound contains an odd number of
  Nitrogen atoms
 Partial Elemental Analysis
   C - 65.44% / 100 = 0.6544
   H – 6.71% / 100 = 0.0671
   Carbon
        0.6544 x 165 = 108 / 12     = 9 Carbon
   Hydrogen
        0.0671 x 165 = 11.1 / 1.01 = 11 Hydrogen
   Remaining Mass
        165 - (108 + 11.1) = 165 – 119.1 = 45.9 ~ 46
   Additional Elements (32 + 14 = 46)
        Oxygen – 2 x 16 = 32 = 2 Oxygen Atoms
           32 / 165 * 100 = 19.5 % Oxygen
        Nitrogen – 1 x 14 = 14 = 1 Nitrogen Atom
           14 / 165 * 100 = 8.5 % Nitrogen

 Molecular Formula
                  C9H11NO2
                  Problem Set Example
Mass Spectrum




Molecular Ion Peak
   165 = Molecular Weight of Compound
   Value is Odd
   Therefore compound contains an odd number of
   Nitrogen atoms.
   Sometimes the Molecular Ion peak is missing, in
   which case the problem will supply the pertinent
   information about the Molecular Ion Peak.
                 Problem Set Example
IR Spectrum


      Unsat’d
                        Aromatic
                        P-Disubst
                Sat’d


                                    CH2
                                                 Aromatic
                                      CH3        P-Disubst
     NH2

                                    C=C
                             C=0            CO




Carbonyl Group
Carbon Oxygen Group
Primary Amine Group
Saturated Alkane
Unsaturated Alkene/Aromatic
Methyl Group
Methylene Group
Aromatic C=C
Aromatic p-Disubstitution
                  Problem Set Example
NMR Spectrum
                                                3 see 2
                                                forms
                                                          3
                                                Triplet




                                2 see 3
                Aromatic        forms
             P-Disubstitution   Quartet
                                 2
              2            2



                                      Singlet
                                      2




Note: The Methylene Group ( Quartet at 4.3 ppm) is
      downfield from its normal position (2 ppm)
      probably under the Influence of an Electronegative
      Group.
      Singlets (4.0 ppm) are often produced by Protons
      on elements other than Carbon – OH, NH, NH2
                Problem Set Example

The Parts of the Molecule




The Molecule




Benzocaine
Ethyl p-AminoBenzoate
4-Aminobenzoic Acid Ethyl Ester

								
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