E which is in the radio frequency range by yrp72563


									Chem 263                                                                  Sept. 8, 2009

NMR Continued (focusing on 1H NMR)

If you apply a magnetic field (BO) to the nuclei there are two possible energy states, one
with the spin aligned with the field and the other opposed.

                                     ΔE = hν (this is in the radio frequency range)
                                    !E which is in the radio frequency range


Absorption from the lower energy state to the higher one becomes possible, which allows
a spectrum to be observed:

Chemical Shift

Chemical shift is the change in the frequency of absorption. The chemical shift is
typically shown in units of parts per million (ppm) and is compared to a reference. For
hydrogens this reference is usually tetramethylsilane (TMS).

The hydrogen atoms in TMS are highly shielded and the signal for TMS is at 0 ppm.

Deshielding occurs when electrons are pulled away from a hydrogen atom, which can be
caused by attaching an electron withdrawing group, such as a carbonyl. When the
electrons are pulled away, the nuclei is more exposed to the magnetic field resulting in a
downfield chemical shift.
                                                                                     H3C Si CH3           = TMS
                                                              H                          CH3

                                                        R     C N
     O                        H
                                        H        H
R        H                           C C                                            CH3
                                               R C O

10           9            8   7         6      5       4          3       2          1         0      (ppm)

Downfield                                                                                          Upfield
De-Shielded                                                                                        Shielded

                                  Intensity:       1 (2H’s)                   3 (2CH3’s)

H                    CH3

 1   C           C   2

H                    CH3

                                                   5                          1.5                   ppm
The intensity depends on the number of hydrogen atoms, in this case 2:6 which is the
same as 1:3

HC                       Cl
                                               HA                 HC HB
         C       C

HB                       HA
1-chloroethene aka
                                               8                  6   5                  ppm
  vinyl chloride

                                   HB and HC are different due to the distance from the Cl;
                                   HB is trans to the Cl while
                                   HC is cis to the Cl
Diastereotopic Protons

 H3C                       HA
                                           C-2 is a stereogenic centre, there is no plane of symmetry in the molecule
             (S)                           HA and HB are in different chemical environments and have different
          C            C                   chemical shifts; they are diastereotopic
              2    3
     HO                     CH3

The methyl group is only one signal because the C-C bond can freely rotate and NMR
cannot distinguish between the H’s. The barrier to bond rotation is only a few kilocalories
per mole and about 15 to 20 kilocalories per mole of energy are available at room
temperature. Hence rotation is very rapid.

If you substitute Ha or Hb with deuterium (labeled as D), which is the isotope of
hydrogen with one proton and one neutron, you get diastereomers; therefore, these two
hydrogens are diastereotopic

1                                                            1
H 3C                       HA                                H3C                    D
                                D                                                       HB
         C             C
                                                                    C           C
                   3                                                        3
 H                                                            H
  HO                       CH3                                 HO                   CH3
                                4                                                       4
     SS isomer                                                     SR isomer

Newman Projection
Round circle in structure below is back carbon (carbon 3 of 2-butanol) – carbon 2 is
directly attached to it and is represented as the dot (point) where the bonds meet in front.

              HA (R)                                                HB (S)
     H                 CH3                  rotate            H              CH3

H3C                    HB (S)                             HA (R)             CH3
              OH                                                    OH

Anti Staggered conformation                              Gauche conformation
(most stable)

When there is stereogenic center present, the two hydrogens, HA and HB, of the methylene
(CH2) group cannot be brought into the same chemical environment. Bond rotation can
bring HB to the position where HA was before, as represented by the above Newman
projections of the two conformations, but the CH3’s are occupying different positions and
therefore the two methylene H’s are not in the same chemical environment.

Spin-Spin Splitting (Coupling)

      Ha     Hb

X                     Y        X≠Y

      X      Y

HA is not a singlet due to the influence of HB on its spin and vice versa. This diagram
shows the energy for HA, the spins for HB are shown in red only to show the effect on HA
             HA           HA   HB spin


JAB = coupling constant between atoms A and B, measured in Hertz (Hz)

The HA coupling constant corresponds to the energy difference caused by the HB proton
in alignment with the field and against the field.
This energy difference is equal to the effect that the HA proton has on HB, so both of these
coupling constants are equal.

The coupled signal is centered around the same chemical shift value that the signal would
have if there were no coupling.

Limitations of Coupling
   1.      2 to 3 bonds separating nuclei (typically not seen further apart than 3 bonds)
   2.      usually no coupling across O, N, S, C=O
                 HA            HB

           eg. C          O    C    HA and HB would not couple due to the oxygen

Note that there are exceptions to these rules



       Cl Hb
  Ha           Cl
       Cl Hb                                                    Jab         Jab

achiral (Not chiral)
                                    doublet                      triplet

       triplet                      doublet

    Jab        Jab                    Jab

          Ha                           Hb                0    (ppm)

Pascal’s Triangle

                                               1 1
                                              1 2 1
                                             1 3 31
                                            1 4 6 4 1

above represents intensities of singlet, doublet (1 to 1), triplet (1 to 2 to 1), quartet ( 1 to 3
to 3 to 1), quintet (1 to 4 to 6 to 4 to 1)

The degree of coupling (also referred to as multiplicity) can be described by the rule:
2nI + 1
where I = spin and n = the number of equivalent hydrogens that are near (2 or 3 bonds
away) from the hydrogen that is being examined
Since I = ½ for 1H, this rule can be simplified to the n+1 rule
Consider examples below and the types hydrogens (a and b)

                    (a)                                                 (b)
                                                                  (a)  CH3
        (b)          O
                                                                 H3C O    CH3
              diethyl ether                                            CH3
                                                               methyl tert-butyl ether
                                                            methyl 1,1-dimethylethyl ether

              (a)         (b)
                                                                        (b)           TMS

                          1     0     ppm                                             0      ppm
        quartet triplet                                 singlet     singlet
        2CH2’s    2 CH3’s                            1CH3          3CH3’s
Intensity would be 4:6 or 2:3                        Intensity would be 3:9 or 1:3

No stereogenic centers present. No coupling across oxygen.
Both molecules should have 2 signals. Both have ether functional groups.

The intensity (of the entire coupled signal) is determined by the number of hydrogens that
make the particular signal, for example, the quartet above is for the 2 CH2’s which is a
total of 4 hydrogens, while the triplet is for the 2 CH3’s which is a total of 6 hydrogens.

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