How can I calculate the percent abundance of boron 10 and boron 11 from an NMR spectrum? I understand how to do it with a GC-MS spectrum, and I know the answer should be about 20% for boron 10 and 80% for boron 11, but I do not understand how to analyze the NMR spectrum or what numbers I should use for the calculations.
ANSWER
General Concept spin and magnetic field spin :-
When S1/2
a nuclear quadrupole moment results leading to line broadening due
to additional transition possible. When a nucleus with spin S place
in Magnetic Field, it has 2S+1
orientation, namely the ms value. The ms value are
S, S-1,.... -S-1, -S. Boron
10 has S= 3/2 and the possible ms
value are 3/2, 1/2, -1/2, -3/2.As per selection
rule ∆ms =+-1 and the allowed transition 3/2 to 1/2 to -1/2
to -3/2. This can explain taking an example BH4^-
Boron has two isotopes boron 10 and boron 11
for both isotopes I is not equal to Zero. Boron 10 has
S=3 with an abundance of 20%. And has Boron 11 has
S=3/2 with an abundance 80% . In BH4- has
tetrahedral structure and four proton attached equivalent. Hence
there will be one signal in 1H NMR spectrum. However this signal
will be split by boron 11 into 2×3/2 + 1= 4 lines. All these line
are in equal intensity since different orientation have equal
probability, Similarly this proton couple with Boron 10 will be
split into 2×3 + 1= 7 line. Again all line are in equal intensity
because all the orientation have equal probability. Since Boron 10
has 20% abundance, each line intensity will be 20/7 nearly equal to
3 % intensity. Abundance of this isotopes both 80 and 20% equal
share five lines.
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