Chemistry 125: Lecture 60 March 23, 2011 NMR Spectroscopy Chemical Shift and Diamagnetic Anisotropy, Spin-Spin Coupling This For copyright notice see final page of this file

Components of Effective Magnetic Field. Applied Field Molecular Field: Net electron orbiting - “Chemical Shift” (Range ~12 ppm for 1 H, ~ 200 ppm for 13 C) Nearby magnetic nuclei - “Spin-Spin Splitting” (In solution J HH 0-30 Hz ; J CH Hz) B effective B molecular (diamagnetic) B applied

The Chemical Shift: Electron Orbiting and Diamagnetic Anisotropy

Chemical Shift and Shielding high electron density shielded upfield high e - density low chemical shift low frequency deshielded downfield low e - density high chemical shift high frequency CH 3 C C-H ! ??? TMS B effective B molecular (diamagnetic) B applied Note: Electron orbiting to give B is driven by B; so B  B.  (ppm) Alkyl R-H H C H CHCH X X = O, Hal, N RC CHCH O H O OHOH O R-OH (depends on conc, T) ++ --

ZERO! Suppose molecule in fluid undergoes rotational averaging. net from average over sphere net from average around circle  1/r 3 Electrons Orbiting Other Nuclei Diamagnetism from Orbiting Electrons Ignore electrons on other atoms! B applied PPM Suppose the studied nucleus is fixed relative to the other nucleus by bond(s).

ZERO! net from average over sphere Electrons Orbiting Other Nuclei Unless orbiting depends on molecular orientation B applied Diamagnetic “Anisotropy” (depends on orientation) NOT suppose less orbiting for this molecular orientation reinforces B applied

B0B0 Diamagnetic Anisotropy Benzene “Ring Current” B 0 can only drive circulation about a path to which it is perpendicular. If the ring rotates so that it is no longer perpendicular to B 0, the ring current stops. Net deshielding of aromatic protons; shifted downfield

Aromaticity: PMR Chemical Shift Criterion HCCl 3 TMS   electrons (4  3) + 2 DIAMAGNETIC ANISOTROPY! ? DIAMAGNETIC ANISOTROPY 8 H2 H TMS 10  electrons (distorted – less overlap & ring current)  (ppm) Boekelheide (1969)

HCCl TMS Aromaticity: PMR Chemical Shift Criterion   electrons (4  3) + 2 DIAMAGNETIC ANISOTROPY! DIAMAGNETIC ANISOTROPY Metallic K adds 2  electrons to give 16 (4n) -2 CH 3 signals shift downfield by 26 ppm despite addition of “shielding” electrons. “Anti-Aromatic” Dianion  (ppm) Shrink Scale Boekelheide (1969) THF solvent

Diamagnetic Anisotropy Acetylene “Ring Current” H H H H The H nuclei of benzene lie beside the orbital path when there is ring current. (B 0 at H reinforced; signal shifts downfield). The H nuclei of acetylene lie above the orbiting path when there is ring current. (B 0 at H diminshed; signal shifts upfield). H H Warning! This handy picture of diamagnetic anisotropy due to ring current may well be nonsense! (Prof. Wiberg showed it / /to be nonsense for 13 C.)

Spin-Spin Splitting

 (ppm) CH 3 C OCH 2 CH 3 O Triplet (1:2:1) C. C. H H Four (2 2 ) sets of molecules that differ in spins of adjacent H nuclei “Spin Isomers” so similar in energy that equilibrium keeps them equally abundant Chem 220 NMR Problem 1 (of 40)

CH 3 C OCH 2 CH 3 O C. C. H H H  (ppm) Quartet (1:3:3:1) 7.3 Triplet (1:2:1) Eight (2 3 ) sets of molecules that differ in spins of adjacent H nuclei 7.3 Influence of CH 2 on CH 3 must be the same as that of CH 3 on CH 2 and independent of B o J in Hz vs. Chemical Shift in (Orbiting driven by B o ) Chem 220 NMR Problem 1 (of 40) binomial coefficients : : 4: 1:

DMSO-d 5 CD 3 SCD 2 H O HO-CH 2 -CH Hz5.1 Hz Doublet of Quartets 1.8 Hz ? Hz 13 CH 3 1:4:6:4:1 Quintet? ?  ppm × 400 MHz J = 7.2 Hz 1.1% of C D is a weaker magnet than H. ? H2OH2O 1:2:3:2:1 Quintet Subtle Asymmetry  (ppm)  D can be oriented 3 ways in B o.

What determines the Strength of Spin-Spin Splitting?

Isotropic J H-H is mediated by bonding electrons (the anisotropic through-space part is averaged to zero by tumbling)

Not spatial proximity! Might overlap be greater for anti C-H bonds ?? HOMO-3 When the “up” electron of this MO is on Nucleus A only its “down” electron is available to be on Nucleus B In tumbling molecules, nuclear spins communicate not through space, but through paired electrons on the nuclei. Through-space interaction of dipoles averages to zero on tumbling. J = 0-3 Hz J = Hz J = 6-12 Hz J = 6-8 Hz J = 1-3 HzJ = 0-1 Hz 3.07 Å 1.85 Å 2.38 Å J depends on the s-orbital content of molecular orbitals.

good p  -p  good s-s bad p  -p  bad s-s 2 bad s-p  good s-p  ; good p  -s ++  ++ Examine the overlap of the components. Which gives better overlap? s-p  > s-s or p  -p  (See Lecture 12) Backside overlap is counterintuitive. Better Overlap!

C Overlap Overlap Integral Å s-p  s-s C C C C C C p-p 

H 2-13 Hz, depends on conformation (overlap) 13 Hz2 Hz H gauche ~7 Hz 11 Hz (approximate way to measure a rigid torsional angle!) No “handle” for rf if same chem shift (see Frame 26 below) invisible

End of Lecture 60 March 23, 2011 Copyright © J. M. McBride Some rights reserved. Except for cited third-party materials, and those used by visiting speakers, all content is licensed under a Creative Commons License (Attribution-NonCommercial-ShareAlike 3.0).Creative Commons License (Attribution-NonCommercial-ShareAlike 3.0) Use of this content constitutes your acceptance of the noted license and the terms and conditions of use. Materials from Wikimedia Commons are denoted by the symbol. Third party materials may be subject to additional intellectual property notices, information, or restrictions. The following attribution may be used when reusing material that is not identified as third-party content: J. M. McBride, Chem 125. License: Creative Commons BY-NC-SA 3.0