The 1H NMR spectrum of the borohydride ion BH4 is given below' The element boron has two isotopes, b

1
The 1H NMR spectrum of the borohydride ion (BH4-)
is given below. The element boron has two
isotopes, both of which are NMR active 10B
(I3, 20) and 11B (I3/2, 80).
                                                 
         
Rationalize the appearance of the observed 1H
spectrum.  
2
Note that if you could integrate the 7 lines from
the spectrum arising from the molecules
containing 10B and then the 4 lines the spectra
arising from the molecules containing 11B, the
relative intensities should be 14 (reflecting
the ratio of the natural abundances of the 10B
and 11B).If you measure the coupling constants
from the spectra, you would find that 1JH-11B is
approximately 80 Hz and that 1JH-10B is
approximately 27 Hz. The ratio between these is
80/27 3 reflecting the fact that the
magnetogyric ratio of 11B (g11B) is approximately
3 times larger than the magnetogyric ratio and
10B (g10B) and that coupling is proportional to
the product of the magnetogyric ratios of the
nuclei which are coupled.                        
                                   
3
CH2D2 molecule is tetrahedral and the two protons
are equivalent. They have the same chemical shift
and will be split by coupling to the two
deuterium nuclei. D has a spin of 1. You would
expect the proton signal to be split into 5 lines
(2nI 1 where n 2 and I1) and 5 lines can be
clearly seen in the spectrum with a splitting
2JH-D 2.0 Hz.
The 1H NMR spectrum of methane (CH4) consists of
a single resonance and because the four protons
have exactly the same chemical shift, no H-H
couplings appear in the spectrum so it is not
possible to measure 2JH-H from the spectrum. The
1H NMR spectrum of dideuteromethane (CD2H2) is
given below. Rationalize the splitting pattern
and explain how you could estimate 2JH-H using
this spectrum.
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• (CH3O)3PO
• i)Rationalise the multiplicity of signals in each
  spectrum.
• ii)What would happen to the appearance of the 13C
  spectrum if the 1H spectrum was irradiated with
  strong broad-band irradiation?
• iii)What would happen to the appearance of the 1H
  spectrum if the 31P spectrum was irradiated with
  strong broad-band irradiation?
• iv)What would happen to the appearance of the 13C
  spectrum if the 31P spectrum was irradiated with
  strong broad-band irradiation?
• v)What would happen to the appearance of the 1H
  spectrum if the 13C spectrum was irradiated with
  strong broad-band irradiation?

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The geometry of trimethyl phosphite is
tetrahedral at phosphorus. All three of the CH3O
groups are equivalent. There would be 1 resonance
in the 13C NMR spectrum, 1 resonance in the 1H
NMR spectrum and 1 resonance in the 31P NMR
spectrum. Note that 31P is spin ½ and 100
abundant, 1H is spin ½ and effectively 100
abundant and 13C is spin ½ but only about 1
abundant.

• The 13C NMR spectrum is a doublet of quartets.
  The quartet splitting arises from the 3 protons
  directly attached to the carbon. The doublet
  splitting must arise from the 2-bond coupling
  between C and P. The 1H NMR spectrum is a doublet
  and the splitting must arise from the 3-bond
  coupling between the protons and phosphorus.
• If the 1H NMR spectrum was irradiated with strong
  Rf radiation (ie decoupled), the quartet
  splitting would disappear in the 13C spectrum so
  the 13C spectrum would appear as a doublet.
• If the 31P NMR spectrum was irradiated, the
  doublet splitting would disappear in the 1H
  spectrum so the 1H resonance would appear as a
  singlet.
• If the 31P NMR spectrum was irradiated, the
  doublet splitting would disappear in the 13C
  spectrum so the 13C spectrum would appear as a
  quartet.
• If the 13C NMR spectrum was irradiated, there
  would be almost no effect on the 1H NMR spectrum.
  Almost all of the protons in the sample would be
  attached to the NMR-silent nucleus 12C (not 13C)
  because the natural abundance of 13C is only
  about 1. If you could see the (small)
  13C-satellites in the spectrum, the 13C splitting
  in these would be removed.

6
Lead (Pb) has several naturally occurring
isotopes including 204Pb (1.5), 206Pb (24),
207Pb (22) and 208Pb (52). Of these isotopes
only 207Pb has a non-zero nuclear spin (I1/2).
What would you expect to observe for the 1H NMR
spectrum of tetramethyllead given that 2J207Pb-H
60Hz ?

• Me4Pb is a tetrahedral molecule so all 4 of the
  ethyl groups are equivalent. There will be only
  one resonance in the 1H NMR spectrum and this
  will be split by coupling to the Pb nucleus. The
  1H resonance will be split into a doublet by
  coupling to 207Pb. The observed spectrum will be
  the superposition of the contributing sub-spectra
  - one singlet (about 76) and one doublet
  (splitting 60 Hz and about 24 of the signal
  intensity).

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Information from Coupling Constants

• Direct dipole-dipole coupling between nuclear
  spins is averaged to zero by molecular tumblings
  in isotropic fluids so it does not concern us
• It is indirect coupling, transmitted by the
  valence electrons of the molecule, that is
  observed.
• As with chemical shifts, the theoretical approach
  does not usually lead to accurate predictions of
  the coupling constants. one does use empirical
  rules and correlations.

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Factors that Contribute to Coupling Constants3
components arising from nucleus-electron
interactions

• Firstly, the magnetic moment of one nucleus
  interacts with the field produced by orbital
  motion of electrons, which in turn interact with
  the second nuclear moment.
• Secondly, there is a dipole interaction involving
  the electron spin magnetic moments
• The final contribution is derived from spins of
  electrons in orbitals which have non-zero
  probability of being at the nucleus (e.g. derived
  from s-atomic orbitals)
• The final factor, known as Fermi Contact Term, is
  by far most important for 1H-1H couplings, but
  for other nuclei the situation is not so simple.

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One Bond Coupling

• In situations where orbital and dipolar terms are
  not of great importance, one bond couplings
  depend on the amount of s-character of the
  internuclear bond (i.e. hybridization)
• This gives rise to empirical relationships which
  are very useful.
• For example values of 1JCH in hydrocarbons are
  125, 160 and 250 Hz for nominal sp3, sp2 and sp
  hybridization (giving simple proportionality to
  s-character).
• But electronegative substituents increase
  coupling greatly (e.g. CHCl3 1JCH 209 Hz)
• In other cases, empirical relationships have been
  worked out sNsC -80 1JNC
• Other, less simply quantified, relationships
  exist for other couplings such as 1JPC, 1JNH, or
  1JPF.

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One Bond Coupling

• Other, less simply quantified, relationships
  exist for other couplings such as 1JPC, 1JNH, or
  1JPF.
• Coordination numbers from coupling constants
  (1JPH 180 and 400 Hz for 3- and 4-coordinate P,
  respectively).
• 1JPF provides information about the
  axial-equatorial disposition of F in a tbp-P
• Single and double bonds between two coupling
  nuclei can be identified using 1J (home work
  1JP-Se and 1JPSe).
• P(III) and P(V) can be readily distinguished from
  1J.
• Coupling between M and L in metal complexes
  depend on the coordination number of the metal
  (s-contribution to each M-L bond) e.g. 1JPtP for
  Pt(PEt3)2Cl2 and Pt(PEt3)2Cl4