Question

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 1

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.

"BHA 10 BHA /2 1/2 - 72 3/2 32 lo-1-2-3 11 BNMR 10нме Гамме THOS 'Jw10=22.9H The to proton spectrum BH4 displays couplings "B. (quarter and B. Iseptet Teacher's Signature: