Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: AussieKenDoll on July 08, 2020, 10:26:23 AM
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P-31 NMR of PF3 , F nuclei equally couple to P, I=1/2 and gives a quartet of 1:3:3:1
But the F-19 NMR of BF3 coupling to the B-11; I=3/2 gives quartet of 1:1:1:1
How do we know if a quartet gives 1:3:3:1 or 1:1:1:1?
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Hi Ken,
When a -CH2 is in-between an oxygen atom and a CH3 group, the methylene group is usually observed to b a 1:3:3:1 quartet. Do you know about Pascal's triangle and how this relates to coupling?
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yes I know about the pascal triangle and its relation to coupling but how does this 1:1:1:1 comes?
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You stated correctly that a single B-11 nucleus is I = 3/2 (it is one type of quadrupolal nucleus). What does this actually mean in terms of what spin states the nucleus can be found in? How is this different from spin-1/2 nuclei?
What can you deduce about the approximate relative populations of the nuclear spin states of B-11?
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it can take -3/2,-1/2,1/2,3/2 spin states
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Yes, and the populations of those four states are roughly equal to one another. For a single spin-1/2 nucleus the populations of the -1/2 and +1/2 are also roughly equal.
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yes they are nearly equal! but how it relates to if a quartet gives 1:3:3:1 or 1:1:1:1 in NMR??
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A single F-19 nucleus "feels" whether the B-11 is in the -3/2, -1/2, +1/2, or +3/2 state. By feeling, I mean that the F-19 experiences a slightly different effective magnetic field. Because the four states are equally populated, the four F-19 signals are equally intense. See if you can apply this to the other situation about which you enquired.
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What does the P-31 nucleus experience?