A quick look at the structure is usually enough to decide how many kinds of nonequivalent protons are present in a molecule by 19lned

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									Chapter 13
Introduction

• Mass Spectrometry (MS) – determines the size and formula

• Infrared (IR) Spectroscopy – determines the kinds of
                               functional groups present

• Nuclear Magnetic Resonance Spectroscopy (NMR) –
                          – determines the carbon-
                            hydrogen framework

• Ultraviolet Spectroscopy (UV) – determines if a conjugated
                                 p electron system is present
The Use of NMR Spectroscopy


• It is used to determine relative location of atoms
  within a molecule

• It is the most helpful spectroscopic technique in
  organic chemistry

• It is related to Magnetic Resonance Imaging (MRI)
  in medicine

• It maps carbon-hydrogen framework of molecules

• It depends on very strong magnetic fields
  1.       Nuclear Magnetic Resonance
           Spectroscopy (NMR)

• Nuclear Magnetic Resonance Spectroscopy (NMR) –
  is a technique used to map the carbon-hydrogen
  framework

       – It detects the energy absorption accompanying the transition
         between nuclear spin states that occurs when a molecule is
         placed in a strong magnetic field and irradiated with
         radiofrequency waves.
• The 1H and 13C nuclei:
   – have spins (i.e behave as if they were spinning about an
     axis)
   – are magnetic (i.e interact with an external magnetic field, Bo)
• Other magnetic nuclei include:
   – all nuclei with an odd number of protons (1H, 2H, 14N, 19F, 31P,...)
   – all nuclei with an odd number of neutrons (13C)

• Nonmagnetic nuclei include:
   – nuclei with even numbers of protons and neutrons (12C, 16O,…)
• In the absence of an external magnetic field, magnetic nuclear
  spins are oriented randomly

• In the presence of an external field, Bo, magnetic nuclear spins
  orient with (parallel to) or against (antiparallel to) the external
  field.

                                                       parallel



                                                antiparallel
• The parallel spin state (orientation) is slightly lower in
  energy and therefore favored (more populated).




                                                 parallel



                                          antiparallel
• When the oriented nuclei are irradiated with proper electromagnetic
  radiation frequency, energy is absorbed and the nuclei “spin-flip”
  from the lower-energy state (parallel) to the higher-energy state
  (antiparallel)

    Magnetic nuclei are said to be in resonance with the applied radiation
    (hence the name of nuclear magnetic resonance).

    The absorption of energy is detected, amplified and displayed as a
    nuclear magnetic resonance spectrum
• The exact frequency necessary for resonance depends on:

   – the strength of the external magnetic field
   – the identity of the nuclei
• The exact frequency necessary for resonance is proportional
  to field strength, Bo.

   – In the absence of an applied magnetic field, spin states have equal energies.
     In the presence of a magnetic field (Bo), spin states have unequal energies

   – If a very strong magnetic field is applied, the energy difference between the
     two spin states (DE) is larger, and higher-frequency (higher-energy) radiation
     is required for a spin-flip.
Practice Problem: The amount of energy required to spin-flip a
                  nucleus depends both on the strength of the
                  external magnetic field and on the nucleus. At a
                  field strength of 4.7 T, rf energy of 200 MHz is
                  required to bring a 1H nucleus into resonance , but
                  energy of only 187 MHz will bring a 19F nucleus
                  into resonance. Use the equation below to
                  calculate the amount of energy required to spin-flip
                  a 19F nucleus. Is this amount greater or less than
                  that required to spin-flip a 1H nucleus?




                    E=    1.20 X 10-4 kJ/mol .   and     n= c .
                                l (m)                       l
Practice Problem: Calculate the amount of energy required to
                  spin-flip a proton in a spectrometer operating
                  at 300 MHz. Does increasing the spectrometer
                  frequency from 200 to 300 MHz increase or
                  decrease the amount of energy necessary for
                  resonance?
   2.      The Nature of NMR Absorptions


• The absorption frequency is not the same for all 1H or
  13C nuclei in a molecule.



• All nuclei in molecules are surrounded by electrons
  with their own magnetic fields Blocal

• When an external field is applied to a molecule, the
  effective field (Beffective) felt by the nucleus is a bit
  smaller than the applied field (Bapplied) :

              Beffective = Bapplied - Blocal
• Electrons setting up local magnetic fields shield nearby
  nuclei from the full effect of the applied field
    Nuclei are shielded from the full effect of the applied field



• Since each specific nucleus in a molecule is shielded to
  a slightly different extent, the effective magnetic field is
  not the same for each nucleus:

    A distinct NMR signal for each chemically distinct   13C   or 1H
    nucleus in a molecule can be detected.

    Different signals appear for nuclei in different environments
Nuclear Magnetic Resonance (NMR) Spectrum


• NMR spectrum – plots the effective field strength felt by
  the nuclei (x-axis) versus the magnitude or intensity of
  NMR resonance signals (corresponding to the intensity of
  rf energy) (y-axis)

   – The intensity of NMR resonance signals is proportional to the
     molar concentration of the sample.

   – Each peak in the NMR spectrum corresponds to a chemically
     distinct 1H or 13C nucleus in the molecule

   – 1H or 13C spectra cannot be observed on the same spectrometer
     because of the difference in energy required to spin-flip the nuclei
1H NMR
Spectrum




13CNMR
Spectrum
• Chemically equivalent nuclei
   – are shielded to the same extent (i.e they have the same
     electronic environment)
   – always show a single absorption


• Example: Methyl Acetate


                      The 3H’s on –OCH3            The 3H’s on CH3C=O
                      are equivalent               are equivalent
NMR Spectrometer

• The Operation of a typical NMR spectrometer is illustrated:
The NMR Measurement

• The sample is dissolved in a solvent that does not have a
  signal itself and placed in a long thin tube

• The tube is placed within the gap of a magnet and spun

• Radiofrequency energy is transmitted and absorption is
  detected

• If two species interconverting faster than 103 times per
  second are present in a sample, NMR records only a
  single, averaged spectrum, rather than separate spectra of
  the two distinct species: “blurring effect”
• Species that interconvert give an averaged signal that can
  be analyzed to find the rate of conversion

• Example: Cyclohexane
Practice Problem: 2-Chloropropene shows signals for three kinds
                  of protons in its 1H NMR spectrum. Explain
   3.     Chemical Shifts


• Chemical shift - is the relative energy of resonance
                    of a particular nucleus resulting from
                    its local environment

                  - is the position on the NMR chart
                    where a nucleus absorbs
• NMR spectra show the applied field strength increasing
  from left to right:

   – Left part is downfield (or low-field)
   – Right part is upfield (or high-field)
• NMR spectra show the applied field strength increasing
  from left to right:

   – Nuclei that absorb on the downfield side are weakly shielded
   – Nuclei that absorb on the upfield side are strongly shielded
• NMR chart is calibrated versus a reference point, set
  as 0, tetramethylsilane [TMS]

   – TMS is used as a reference for both 1H and 13C
Measuring Chemical Shift

• Numeric value of chemical shift – is the difference between
  strength of magnetic field at which the observed nucleus
  resonates and field strength for resonance of a reference

   – Difference is very small but can be accurately measured
   – Taken as a ratio to the total field and multiplied by 106 so the
     shift is in parts per million (ppm)



• Absorptions normally occur downfield of TMS, to the left on
  the chart
Measuring Chemical Shift

• NMR charts are calibrated using an arbitrary scale called
  the delta scale (d), where 1d = 1 ppm (part per million) of
  spectrometer frequency

   – d is the number of parts per million (ppm) of the magnetic
     field expressed as the spectrometer’s operating frequency
     (it is a ratio and not a unit)


          Observed chemical shift (number of Hz away from TMS)
     d=
                 Spectrometer frequency in MHz
Measuring Chemical Shift

• The chemical shift of an NMR absorption in d units is
  constant, regardless of the operating frequency of the
  spectrometer

   – It is independent of instrument’s field strength

   – Example: A 1H nucleus at d = 2.0 ppm on a 60 MHz
     instrument also absorbs at d = 2.0 ppm on a 300 MHz.
Measuring Chemical Shift

• Interpretation of a spectrum is easier with higher operating
  spectrometer frequency (less likelihood of signal overlap)

          For 60 MHz instrument, 1d = 60 Hz
          For 300 MHz instrument, 1d = 300 Hz



• Absorptions normally occur downfield of TMS:
   – Almost all 1H NMR absorptions occur 0-10 ppm from TMS
   – Almost all 13C NMR absorptions occur 0-220 ppm from TMS.
Practice Problem: When the 1H NMR spectrum of acetone,
                  CH3COCH3, is recorded on an instrument
                  operating at 200 MHz, a single sharp
                  resonance at 2.1 d is seen.


     a) How many hertz downfield from TMS does the acetone
        resonance correspond to?

     b) If the 1H NMR spectrum of acetone were recorded at
        500 MHz, what would be the position of the absorption
        in d units?

     c) How many hertz downfield from TMS does this 500 MHz
        resonance correspond to?
Practice Problem: The following 1H NMR peaks were recorded on
                  a spectrometer operating at 200 MHz. Convert
                  each into d units.


              a) CHCl3; 1454 Hz

              b) CH3Cl; 610 Hz

              c) CH3OH; 693 Hz

              d) CH2Cl2; 1060 Hz]
   4.     13CNMR Spectroscopy: Signal
          Averaging and FT-NMR

• Carbon-13, 13C – is the only naturally occurring
                  carbon isotope with a nuclear spin
   – Natural abundance is only 1.1% of C’s in molecules
   – Sample is thus very dilute in this isotope



• Two techniques have been developed to detect the
  13C isotope in an organic sample by NMR:

   – Signal averaging (increases instrument sensitivity)
   – Fourier-transform NMR (increases instrument speed)
• Signal averaging: increases instrument sensitivity

   – Any individual 13C NMR spectrum is extremely “noisy”, but when
     hundreds of individual runs are added together by computer and
     then averaged, a greatly improved spectrum results.



• Fourier-transform NMR: increases instrument speed

   – In the FT-NMR technique, all signals are recorded simultaneously

   – The sample placed in a magnetic field of constant strength is
     irradiated with a short burst of rf energy.

   – All 1H and 13C in the sample resonate at once, and the complex
     composite signal is manipulated using so-called Fourier transforms
     before it can be displayed.
              Large amount of noise
 A single
    run




A average
of 200 runs
   5.     Characteristics of 13C NMR
          Spectroscopy

• The carbon NMR spectrum of a compound provides the
  number of different types of electronic environments of
  carbon atoms in a molecule

   – It displays a single sharp signal for each chemically
     distinct 13C nucleus in the molecule.

   – Most 13C resonances are between 0 to 220 ppm
     downfield from TMS
–   13CNMR spectrum displays a single sharp signal for each
    chemically distinct 13C nucleus in the molecule.



• Example: Methyl Acetate


                            The C on –OCH3     The C on –CH3C=O

    The C of C=O
Correlation of Environment with Chemical Shift

• Most 13C resonances are between 0 to 220 ppm downfield
  from TMS

• The exact chemical shift of each 13C depends on its electronic
  environment:
       – Electronegativity of nearby atoms
       – Hybridization of 13C atom
– Electronegativity of nearby atoms: C bonded to O, N, or
  halogen absorb downfield because O, N, or halogen pull
  electrons away from nearby 13C atoms, decreasing their
  electron density and “deshielding” them.

– Hybridization of 13C atom:
   – sp3 C signal is in the range 0-90 d
   – sp2 C signal is in the range 110-220 d
   – C=O signal is at the low-field end, in the range 160-220 d
   C=O signal is at the low-field end, in the range 160-220 d



 2-butanone




 para-bromo
acetophenone
Example:




   Only six C absorptions with different peak sizes
Practice Problem: Identify the different 13C NMR absorptions




      Five absorptions: 14.1, 60.5, 128.5, 130.3, and 166.0 d
Practice Problem: Predict the number of carbon resonance lines
                  you would expect in the 13C NMR spectra of
                  the following compounds:


       a) Methyl cyclopentane        b) 1-Methylcyclohexene

       c) 1,2-Dimethylbenzene        d) 2-Methyl-2-butene
Practice Problem: Propose structures for compounds that fit the
                  following descriptions:


        a) A hydrocarbon with seven lines in its 13C NMR
           spectrum

        b) A six-carbon compound with only five lines in its 13C
           NMR spectrum

        c) A four-carbon compound with three lines in its 13C
           NMR spectrum
Practice Problem: Assign the resonances in 13C NMR spectrum
                  of methyl propanoate, CH3CH2CO2CH3




                                          1     3    4

          2
    6.     DEPT 13C NMR Spectroscopy


• DEPT-NMR (distortionless enhancement by polarization
  transfer)

      – is one of the improved pulsing and computational
        methods that give additional information

      – can help distinguish among signals due to CH3, CH2,
        CH, and C

      – determines the number of H’s bonded to each C
A DEPT experiment is done in three stages:
   • First stage: is to run an ordinary spectrum (called a
     broadband-decoupled spectrum):
         – It locates the chemical shifts of all C’s

   • Second stage: is called a DEPT- 90 run:
         – only signals due to CH carbons appear

   • Third stage: is called DEPT- 135 or INEPT (insensitive
     nuclear enhancement by polarization transfer) run:
         – positive signals for CH3 and CH carbons
         – negative signals for CH2 carbons.
         – zero signal for carbons having no hydrogen (C)
• Information from all three spectra can be used to determine the
  number of protons attached to each carbon.
Example: DEPT-NMR spectrum for 6-methyl-5-hepten-2-ol
                                                 CH3 CH3 CH3
Broadband-decoupled
spectrum
                              C

                              6   5     2    4     8371




                                  CH    CH
DEPT- 90
spectrum




DEPT- 135
spectrum
                                             CH2   CH2
Practice Problem: Assign a chemical shift to each carbon in 6-
                  methyl-5-hepten-2-ol
Practice Problem: Estimate the chemical shift of each carbon in
                  the following molecule. Predict which carbons
                  will appear in the DEPT-90 spectrum, which
                  will give positive peaks in the DEPT-135
                  spectrum, and which will give negative peaks
                  in the DEPT-135 spectrum.
Practice Problem: Propose a structure for an aromatic
                  hydrocarbon, C11H16, that has the following 13C
                  NMR spectral data.


       Broadband decoupled 13C NMR: 29.5, 31.8, 50.2, 125.5,
                                   127.5, 130.3, 139.8 d

       DEPT-90: 125.5, 127.5, 130.3 d

       DEPT-135: positive peaks at 29.5, 125.5, 127.5, 130.3 d
                 negative peak at 50.2 d
      7.     Uses of 13C NMR Spectroscopy

•   13C  NMR spectroscopy - provides useful information for
    structure determination which cannot be provided by
    infrared spectroscopy or mass spectroscopy
• Example: Does the E2 elimination of alkyl halide, 1-chloro-
  methyl cyclohexane, with strong base give the more highly
  substituted alkene, 1-methylcyclohexene (Zaitsev’s product)
  or methylenecyclohexane?


                        Five sp3 C peaks        Three sp3 C peaks
                        Two sp2 C peaks         Two sp2 C peaks
• Example: Difference in symmetry of products is directly
  observed in the spectrum. The spectrum of the reaction
  product clearly identifies 1-methylcyclohexene as the product
  formed in the E2 reaction.




                                        Five sp3 C peaks (d 20-50)




         Two sp2 C peaks (d 100-150)
Practice Problem: Addition of HBr to a terminal alkyne leads to
                  the Markovnikov addition product, with the Br
                  bonding to the more highly substituted carbon.
                  How could you use 13C NMR to identify the
                  product of the addition of 1 equivalent of HBr to
                  1-hexyne?
     8.      1HNMR Spectroscopy and Proton
             Equivalence

•   1H   NMR spectroscopy
    – is much more sensitive than 13C because the active
      nucleus (1H) is nearly 100 % of the natural abundance
    – shows how many kinds of nonequivalent hydrogens are
      present in a compound
• A quick look at the structure is usually enough to decide
  how many kinds of nonequivalent protons are present in a
  molecule.


• Example: Methyl Acetate


                     The 3H’s on –OCH3       The 3H’s on CH3C=O
                     are equivalent          are equivalent
• The nonequivalence of two H’s can be determined by
  seeing if replacing each H with “X” gives the same or
  different structure

   – Equivalent H’s have the same signal while
     nonequivalent H’s have different signals

   – There are degrees of nonequivalence
Nonequivalent H’s

• Replacement of each H with “X” gives a different
  constitutional isomer
   – The H’s are in constitutionally heterotopic environments
   – They have different NMR absorption (d)
   – They are nonequivalent under all circumstances
Equivalent H’s

• Replacement of either H with “X” gives the same
  structure
   – The H’s are in homotopic (identical) environments
   – They have same NMR absorption (d)
   – They are equivalent
Enantiotopic Distinctions

• Replacement of the pro-R or pro-S H with “X” gives a
  different enantiomer
   – The H’s are in enantiotopic (mirror images of each other)
     environments
   – They have same NMR absorption (d)
   – They are electronically equivalent
Diastereotopic Distinctions

• In chiral molecules, replacement of the pro-R or pro-S
  H with “X” gives a different diastereomer
   – The H’s are in diastereotopic environments
   – They will have different NMR absorption (d)
   – They are chemically and electronically nonequivalent
Practice Problem: Identify the indicated sets of 1H’s as unrelated,
                  homotopic, enantiopic, or diasterotopic




(a) Enantiotopic       (b) Diastereotopic     (c) Diastereotopic
(d) Diastereotopic     (e) Diastereotopic     (f) Homotopic
Practice Problem: How many kinds of electronically
                  nonequivalent protons are present in each of
                  the following compounds, and thus how many
                  NMR absorptions might you expect in each?


               a) CH3CH2Br

               b) CH3OCH2CH(CH3)2

               c) CH3CH2CH2NO2

               d) Methylbenzene

               e) 2-Methyl-1-butene

               f) cis-3-Hexene
Practice Problem: How many absorptions would you expect the
                  following compound to have in its 1H NMR
                  spectrum?
 9. Chemical Shifts in 1H NMR Spectroscopy

• Chemical shifts (d) are due to Blocal :

   – Nuclei that are strongly shielded require higher Bapplied
     and thus absorb on the upfield side
   – Nuclei that are weakly shielded require lower Bapplied
     and thus absorb on the downfield side

• Most 1H chemical shifts fall within the range 0-10 d.
• Most 1H chemical shifts fall within the range 0-10 d.
   – H’s attached to sp3 C have higher field signals
   – H’s attached to sp2 C have lower field signals
   – H’s on C bonded to electronegative atoms (O, N, X)
     absorb at the low-field end

   Saturated d 0-1.5    Allylic d 1.5-2.5   Y = O, N, X d 2.5-4.5

                                                          Cl




   Vinylic d 4.5-6.5   Aromatic d 6.5-8.0
Practice Problem: Each of the following compounds has a single
                  1H NMR peak. Approximately where would you

                  expect each compound to absorb?


              a) Cyclohexane

              b) CH3COCH3

              c) Benzene       O O
                               II II
              d) Glyoxal     H—C—C—H

              e) CH2Cl2

              f) (CH3)3N
Practice Problem: Identify the different kinds of nonequivalent
                  protons in the following molecule, and tell
                  where you would expect each to absorb:
    10. Integration of 1H NMR Absorptions:
            Proton Counting

• In the 1H NMR spectrum, the relative intensity of a
  signal (integrated area) is proportional to the number
  of protons causing the signal

   – This information is used to deduce the structure
   – Integrated areas are presented as a “stair-step” line
   – For narrow peaks, the heights are the same as the
     areas and can be measured with a ruler
• Example: In ethanol (CH3CH2OH), the signals have the
  integrated ratio 3:2:1


• Example: In methyl 2,2-dimethylpropanoate, the signals
  have the integrated ratio 1:3
Practice Problem: How many peaks would you expect in the 1H
                  NMR spectrum of p-xylene? What ratio of
                  peak areas would you expect on integration of
                  the spectrum? Refer to Table 13.3 for
                  approximate chemical shifts, and sketch what
                  the spectrum would look like. Note: Aromatic
                  rings have two resonance forms.
Practice Problem: Identify the number of peaks and the ratio of
                  peak areas on integration of the spectrum



                                        2.4-2.7 d (6H’s)




                               6.5-8.0 d (4H’s)




                     Two peaks 3:2
    11. Spin-Spin Splitting in 1H NMR
        Spectra

• Spin-spin splitting is when peaks often split into
  multiple peaks due to interactions between
  nonequivalent protons on adjacent carbons

      – The set of peaks is called a multiplet
        (2 = doublet, 3 = triplet, 4 = quartet)

      – Coupling is the interaction of the spins
        of nearby nuclei.
• Example: Bromoethane (CH3CH2Br)




  The -CH2Br protons have           The -CH3 protons have
  four peaks (quartet) at 3.42 d    three peaks (triplet) at 1.68 d
• Coupling is the interaction of the spins of nearby
  nuclei.

   – The tiny magnetic field produced by one nucleus
     (Bnucleus 1) affects the magnetic field felt by neighboring
     nuclei (Beffective 2)

   – Nuclear spins can have two possible alignments with
     respect to the magnetic field (can align either with or
     against the applied field)
Example: The origin of spin-spin splitting in bromoethane


• The -CH3 protons in bromoethane are
  neighbored by two magnetic -CH2Br
  protons

• Each -CH2Br proton has its own nuclear
  spin which can align either with or
  against the applied field, affecting the
  -CH3 protons

• These spin states are represented as
  1:2:1 triplet
Example: The origin of spin-spin splitting in bromoethane


•   The three different spin alignments of
    the two -CH2Br protons are:

    a. Both proton spins align with Bapplied,
       Beffective is larger; Bapplied is reduced        a   b   c
       (downfield)

    b. One proton spin aligns with Bapplied,
       and one proton spin aligns against
       Bapplied, no effect

    c.   Both proton spins align against
         Bapplied, Beffective is smaller; Bapplied is
         increased (upfield)
Example: The origin of spin-spin splitting in bromoethane


• The -CH2Br absorption is split into a
  1:3:3:1 quartet

• There are four possible combinations of
  the three spins of -CH3 protons
• The “n + 1 rule” states that:

   – Protons that have n equivalent neighboring protons
     show n + 1 peaks in their NMR spectrum

   – The splitting is into one more peak than the number of
     H’s on the adjacent carbon
• Example: 2-bromopropane


  The -CHBr proton is split by 6      The 6 equivalent -CH3 protons
  equivalent -CH3 protons (n = 6):    are split by 1 -CHBr proton (n = 1):
  a septet                            a doublet

  n = 6 leads to 6+1 = 7 peaks        n = 1 leads to 1+1 = 2 peaks




                                     Integration 6:1
• The distance between peaks in a multiplet is called the
  coupling constant, denoted J
      – It is measured in Hz and is in the 0-18 Hz range
      – It is dependent on the molecular geometry but
        independent of the spectrometer field strength
• Some coupling patterns and relative intensities of lines:
      – If two multiplets have the same coupling constant (J), the
        H’s causing those multiplets are adjacent in the molecule
      – The relative intensities are in proportion of a binomial
        distribution
Rules for Spin-Spin Splitting1

1. Chemically equivalent protons do not show spin-spin
   splitting. The equivalent protons may be on the same
   carbon or on different carbons
Rules for Spin-Spin Splitting2

2.   The signal of a proton that has n equivalent neighboring
     protons is split into a multiplet of n+1 peaks with coupling
     constant J. Protons that are farther than two carbon atoms
     apart don’t usually couple, although they sometimes show
     small coupling when they are separated by a p bond.
Rules for Spin-Spin Splitting3

3.   Two groups of protons coupled to each other have the
     same coupling constant J.
Rules for Spin-Spin Splitting


1.   Chemically equivalent protons do not show spin-
     spin splitting.

2.   The signal of a proton that has n equivalent
     neighboring protons is split into a multiplet of n+1
     peaks with coupling constant J.

3.   Two groups of protons coupled to each other have
     the same coupling constant J.
• Example: para-methoxypropiophenone
                              The -OCH3 protons are
                              unsplit (n = 0): a singlet

Each of two types of equivalent                  The 2 equivalent O=CCH2 protons
Ar-H protons is split by its                     are split by 3 –CH3 protons (n = 3):
neighbor (n = 1): a doublet                      a quartet and vice-versa, a triplet
Why is there no splitting of carbon signals into
mutiplets in 13C NMR?

•   No coupling of a 13C nucleus with nearby carbons
    and/or hydrogens is seen because:

    –   The low natural abundance makes it unlikely that two
        13C nuclei will be adjacent



    –   The method used for 13C spectrum, broadband
        decoupling, cancels the 1H’s magnetic fields
• Example: Propose a structure for C5H12O with the following
  1H NMR data:

             –   0.92 d (3H, triplet, J = 7Hz)
             –   1.20 d (6H, singlet)
             –   1.50 d (2H, quartet, J = 7Hz)
             –   1.64 d (1H, broad singlet)


     Degree of unsaturation = ((2x5+2)-12)/2 = 0
     The compound is saturated, either an alcohol or an ether
Practice Problem: Predict the splitting patterns you would expect
                  for each proton in the following molecules:
Practice Problem: Draw structures for compounds that meet the
                  following descriptions:



           a) C2H6O; one singlet

           b) C3H7Cl; one doublet and one septet

           c) C4H8Cl2O; two triplets

           d) C4H8O2; one singlet, one triplet, and one quartet
Practice Problem: Propose a structure for a compound with a
                  formula C4H10O consistent with the following
                  1H NMR spectrum




    Degree of unsaturation = ((2x4+2)-10)/2 = 0
    The compound is saturated, either an alcohol or an ether

    CH3CH2OCH2CH3

                                  One quartet       One triplet
                                   3.5d              1.2d
  12. More Complex Spin-Spin Splitting
      Patterns

• Spectra can be more complex due to:
      – overlapping signals
      – multiple nonequivalence


• The n + 1 rule correctly predicts splitting caused by
  equivalent protons but not by nonequivalent protons
   – Splittings caused by nonequivalent protons are more complex.
• Example: Toluene


    The 5 Ar-H protons are not equivalent
    but give a complex overlapping pattern
• Example: Trans-cinnamaldehyde
                 The 5 Ar-H proton signals                The =CH proton is split
                 overlap into a complex pattern           by two nonequivalent
                                                          protons: a doublet of
The O=CH proton is                The =CH proton is       doublets
split by its neighbor             split by its neighbor
(n = 1): a doublet                (n = 1): a doublet
• A tree diagram for the C2 proton of trans-cinnamaldehyde shows how
  it is coupled to the C1 and C3 protons with different coupling constants




  The peaks nearer the signal of the
  coupled partner are always larger;
  those farther from the signal are
  always smaller
Practice Problem: 3-Bromo-1-phenyl-1-propene shows a complex
                  NMR spectrum in which the vinylic proton at C2
                  is coupled with both the C1 vinylic proton (J =
                  16 Hz) and the C3 methylene protons (J = 8 Hz)
                  Draw a tree diagram for the C2 proton signal.
                  Explain the five-line multiplet observed.




    J 1-2 = 16 Hz
    J 2-3 = 8 Hz
    13. Uses of 1H NMR Spectroscopy


•   1H  NMR spectroscopy- is a technique used to identify
    likely products in the laboratory quickly and easily
• Example: Does the hydroboration/oxidation of methylene-
  cyclohexane yield cyclohexylmethanol or 1-methylcyclohexanol?
• Example: The 1H NMR spectrum clearly identifies the
  reaction product.

              2-proton triplet (d 3.40)




                          3-proton singlet (d 1)
Practice Problem: How could you use 1H NMR to determine the
                  regiochemistry of electrophilic addition to
                  alkenes? For example, does addition of HCl to
                  1-methylcyclohexene yield 1-chloro-1-
                  methylcyclohexane or 1-chloro-2-
                  methylcyclohexane?
Chapter 13
Practice Problem: The compound whose 1H NMR spectrum is
                  shown has the molecular formula C3H6Br2.
                  Propose a structure.
Practice Problem: The compound whose 1H NMR spectrum is
                  shown has the molecular formula C4H7O2Cl
                  has an I.R absorption peak at 1740 cm-1.
                  Propose a structure.
Practice Problem: Assign as many of the resonances as you can
                  to specific carbon atoms in the 13C NMR
                  spectrum of ethyl benzoate.
Practice Problem: The 1H and 13C NMR spectra of compound A,
                  C8H9Br, are shown. Propose a structure for A,
                  and assign peaks in the spectra to your structure

								
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