1. Below is shown the 1H NMR spectrum of (KBH4). Explain the appearance of the spectrum. (spectra attached)
The answer given is:
Both 10B (I = 3, nat. ab. 20 %) & 11B (I = 3/2, nat. ab. 80 %) NMR active and couple of 1H. Hence 80 % of the sample contains 11BH4 which gives rise to a 1:1:1:1 1H NMR signal, whilst the remaining sample contains 10BH4 which gives rise to a 1:1:1:1:1:1:1 septet signal. Overall integration of the two overlapping signals should be 4:1
Can someone help me understand how the answer is obtained, I know in this forum you don't answer questions, rather you try directing the enquirer towards the answer but here I have the answer but don't understand it!
So I know its a 1H NMR spectrum, and I also know both 10B and 11B are NMR active and I also know they abundance being 20% and 80%, where does l = 3 and 3/2 come from? I also know that since 11B is more abundant the sample contains more of it than it does 10B. But why does this equate to 11B giving rise to a 1:1:1:1 signal and the other gives rise to septet signal? Also why is the integration of the two 4:1?