Chemical Forums

Please login or register.

Login with username, password and session length

Sponsored links

Pages: [1]   Go Down

Author Topic: This 19FNMR Coupling Pattern  (Read 2197 times)

0 Members and 1 Guest are viewing this topic.

workingundergrad1896

  • New Member
  • **
  • Mole Snacks: +0/-0
  • Offline Offline
  • Posts: 3
This 19FNMR Coupling Pattern
« on: January 03, 2018, 11:13:46 AM »

Hi, can someone please help me understand this coupling pattern in this 19F spectra, I've spent hours reading through what textbooks I have and searching literature but I must be missing something as I cannot seem to come to an answer.



I understand that the flourine atom will be coupling to the protons 3 and 4 bonds away, and that it does couple to carbon but carbon coupling is not seen in the flourine spectra so is not relevant here. Does the flourine atom have quadrupolar coupling to the boron atom? I understand that coupling to the boron nucleus would create a quartet because of I = 3/2, but how does this splitting combine with the proton splitting to give what appears to be this 1:3:3:2:4:2:3:3:1 pattern?

Or is the pattern caused by the fact that the protons 3 bonds away are equivalent, but the protons 4 bonds away are not equivalent because of conformations of the boronate ester ring, and thus cause different coupling? If this is so can you explain or provide any resources as to why this specific pattern is caused?

If anyone has any relevant references or resources that could help thanks so much
Logged

mjc123

  • Chemist
  • Sr. Member
  • *
  • Mole Snacks: +195/-8
  • Offline Offline
  • Posts: 1364
Re: This 19FNMR Coupling Pattern
« Reply #1 on: January 03, 2018, 11:12:31 PM »

It looks to me, to a first approximation anyway, like a triplet of triplets, such as you would expect from coupling with two different sets of two equivalent Hs. If you number the peaks 1,2,3.. from left to right, the triplets are {1,2,4}, {3,5,7} and {6,8,9} and the peaks should have relative intensities 1:2:2:1:4:1:2:2:1. Doesn't look too far off to me. Try to measure the peak splittings accurately - evidently 0.01 ppm isn't precise enough, as the peaks are all shown as 0.01 ppm apart, and it's evident to the eye that the splittings are not all the same. Why not try simulating it, if you have software (you could even try it in excel) - estimate the coupling constants, determine peak positions and intensities, clothe with a lineshape - and see if it looks like your spectrum.
Logged

Babcock_Hall

  • Chemist
  • Sr. Member
  • *
  • Mole Snacks: +221/-16
  • Online Online
  • Posts: 3349
Re: This 19FNMR Coupling Pattern
« Reply #2 on: January 04, 2018, 03:51:18 AM »

The two ortho-hydrogen atoms are isochronous (meaning that they have the same chemical shift), but they are not magnetically equivalent.  The same is true of the two meta-hydrogen atoms.  The concept of magnetically equivalent versus non-equivalent nuclei is often ignored at the undergraduate level.  But I will give a poor man's explanation.  To be magnetically equivalent, two nuclei must first be isochronous, and they must also couple to other nuclei with the same coupling constant.  If they couple to a given nucleus with different coupling constants, then they are not magnetically equivalent, and first-order coupling rules do not apply.
« Last Edit: January 04, 2018, 04:31:29 AM by Babcock_Hall »
Logged

wildfyr

  • Global Moderator
  • Full Member
  • ***
  • Mole Snacks: +92/-7
  • Offline Offline
  • Posts: 811
Logged

Flatbutterfly

  • Regular Member
  • ***
  • Mole Snacks: +4/-0
  • Offline Offline
  • Posts: 45
Re: This 19FNMR Coupling Pattern
« Reply #4 on: January 10, 2018, 08:53:35 AM »

Second-order effects in NMR patterns are indeed best taught at advanced undergraduate or graduate level and so I won’t go into details here.  So how do you report your F-19 NMR spectrum?  Experimental: list the range of chemical shifts (δ) and put (m) after values (m = multiplet).  In your discussion: the F-19 NMR spectrum of the cmpd is the X part of an X NMR pattern and as such is a second-order spectrum that is not readily interpreted like a first-order spectrum.

Second-order effects occur when the chemical shift difference between two resonances  is comparable to their coupling (Δδ(AB) ≈  J(AB)) and is indicated by letters close together in the alphabet. [1]  It also occurs when the nuclei are chemically equivalent but not magnetically equivalent which is denoted by primes. (see earlier ref).
[1] https://www.chem.wisc.edu/areas/reich/nmr/05-hmr-15-AABB.htm

Boron has two isotopes both of which are NMR active: B-10 (~20%, spin 3) B-11 (~80%, spin 3/2).  Because there is an electric field gradient at B possible coupling (i.e., 2nI+ 1) will almost certainly not occur due rapid self-decoupling by the quadrupolar nuclei.
Logged

Irlanur

  • Chemist
  • Full Member
  • *
  • Mole Snacks: +31/-4
  • Offline Offline
  • Posts: 419
Re: This 19FNMR Coupling Pattern
« Reply #5 on: May 23, 2018, 08:19:52 AM »

I just stumbeled upon this, sorry that this is a bit late, but:

Why would this be a second order spectrum if one only observes the 19F ? Without thinking about it too much I would say that the chemical shift terms of the protons do not influence the 19F Resonance frequencies.
Logged

Babcock_Hall

  • Chemist
  • Sr. Member
  • *
  • Mole Snacks: +221/-16
  • Online Online
  • Posts: 3349
Re: This 19FNMR Coupling Pattern
« Reply #6 on: May 24, 2018, 02:50:25 AM »

Irlanur,

Good point.  I think it would be a different story if there were two fluorine atoms and two hydrogens.
Logged
Pages: [1]   Go Up
 

Mitch Andre Garcia's Chemical Forums 2003-Present.

Page created in 0.046 seconds with 20 queries.