Why is proton coupling through bonds and how does this affect identical protons?

I had the opportunity to work with a NMR spectrometer and decided to learn some of the theory behind it. After finishing [Clayden's](http://www.amazon.co.uk/Organic-Chemistry-Jonathan-Clayden/dp/0198503466) chapter on $^{1}\textrm{H}$ NMR (Ch. 11) I'm still left with some questions, mostly concerning coupling.


Proton coupling is described as the interaction between neighboring (2-4 bonds away) protons magnetic field. This can split the signals by a constant amount (the coupling constant), which depend on the environment of the protons but not on the strength of the magnetic field. Ideally creating a binomial distribution when the neighboring protons have the same shift.


* How does a proton create a magnetic field that influences the neighboring protons (mainly) through bonding and not through space?
* Why do identical protons not influence each other? Or more generally: why is it that the more identical the influencing protons are with the influenced proton (i.e. $\Delta\delta$ getting equal to/lower than $J$), the more their "outer" signals vanish? Especially since $J$ is a constant, I would expect that for example two doublets would just merge into one doublet and not into only a singlet.





> 
> How does a proton create a magnetic field that influences the neighboring protons (mainly) through bonding and not through space?
> 
> 
> 


Through space coupling is known as [dipolar coupling](http://en.wikipedia.org/wiki/Magnetic_dipole%E2%80%93dipole_interaction#Dipolar_coupling_and_NMR_spectroscopy) and is actually much greater than through bond coupling ([J-coupling](http://www.chem.wisc.edu/areas/reich/nmr/05-hmr-03-jcoupl.htm)). However, in low viscosity liquids the chemical shift effects of dipolar coupling are averaged to essentially a single value by molecular rotation. Relaxation effects of dipolar coupling are still observed. 



> 
> Why do identical protons not influence each other?
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> 
> 


For two protons there will be four states $\alpha \alpha , \alpha \beta , \beta \alpha$, and $\beta \beta$. In the general case, each state may have a different energy and there can be four possible one quantum transitions ($\alpha \alpha$ to $\alpha \beta$, $\alpha \alpha$ to $\beta \alpha$, $\alpha \beta$ to $\beta \beta$, and $\beta \alpha$ to $\beta \beta$), each transition having a unique energy and hence 4 peaks (two doublets). 


However, when the two protons are equivalent, $\beta \alpha$ and $\alpha \beta$ will be the same energy, $E\_2$ let's say. So, letting the energy of $\alpha \alpha$ be $E\_1$ and $\beta \beta$ be $E\_3$, and given a one quantum transition, the possible energies for transtions are only $E\_3$ - $E\_2$ and $E\_2$- $E\_1$. But $E\_3$ and $E\_1$ are equally spaced above and below $E\_2$, so there is only one energy of transition. 


For a more-rigorous explanation see: <http://www.chemie.uni-hamburg.de/nmr/insensitive/tutorial/en.lproj/energy_levels.html>


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> 
> How does a proton create a magnetic field that influences the neighboring protons (mainly) trough bonding and not trough space?
> 
> 
> 


A proton doesn't interact with another proton by producing a magnetic field *per se*, but it interacts via the bonding electrons *via* the [Fermi contact interaction](http://en.wikipedia.org/wiki/Fermi_contact_interaction). And because the Wikipedia article is kind of dry on this matter, here a figure that very well illustrates what is going on:


![Fermi Contact Interaction](https://i.stack.imgur.com/R3j3T.png)


This interaction is not the only source for indirect coupling of spins. If we only consider protons, then we're fine. But if you'd like to consider fluorine-fluorine couplings, you also have to take into account the dipolar interaction between the nuclear and electron spins. This dipolar interaction also provides a source for anisotropic $J$ interactions.




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> 
> Why do identical protons not influence each other? Or more general: why is it that how more identical the influencing protons are with the influenced proton (i.e. $\Delta\delta$ getting equal to/lower than $J$), the more their "outer" signals vanish? Especially since $J$ is a constant, I would expect that for example two doublets would just merge into one doublet and not into only a singlet.
> 
> 
> 


Also here, a picture is worth more than a thousand words (I'll spare you the calculation of this spectrum):


![Spectrum for an interacting two-spin system](https://i.stack.imgur.com/K8JYI.png)


The $k$ is defined as follows:
$$ k = \left| \frac{\Omega\_2 - \Omega\_1}{2\pi J} \right|$$


As you can clearly see, the outer peaks start getting smaller the smaller the difference in chemical shift ($\Omega$) of the two spins tends towards zero. This is sometimes called ["roof effect"](http://chemwiki.ucdavis.edu/Physical_Chemistry/Spectroscopy/Magnetic_Resonance_Spectroscopies/Nuclear_Magnetic_Resonance/NMR%3A_Experimental/NMR%3A_Interpretation#High-order_splitting_pattern).


The intensity of the left- and rightmost lines can be calculated (assuming $J>0$) with the following formula:
$$ I\_{13} = I\_{34} = \frac{1 - \sin(2\alpha)}{4} $$
where
$$ 2\alpha = \tan^{-1}\left(\frac{2\pi J}{\Omega\_1 - \Omega\_2} \right) $$


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