# 6.1 Determination of energetics of fluxional molecules by nmr  (Page 3/4)

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$\frac{1}{t}=\frac{1}{{t}_{a}}+\frac{1}{{t}_{b}}$

For reference, the exact lineshape function (assuming two equivalent groups being exchanged) is given by the Bloch Equation, [link] , where g is the intensity at frequency v , and where K is a normalization constant

$g\left(v\right)=\frac{\text{Kt}\left({v}_{a}+{v}_{b}{\right)}^{2}}{\left[0\text{.}5\left({v}_{a}+{v}_{b}\right)-v{\right]}^{2}+{4\pi }^{2}{t}^{2}\left({v}_{a}-v{\right)}^{2}\left({v}_{b}-v{\right)}^{2}}$

## Low temperatures to coalescence temperature

At low temperature (slow exchange), the spectrum has two peaks and Δ v >>t. As a result, [link] reduces to [link] , where T 2a’ is the spin-spin relaxation time. The linewidth of the peak for species a is defined by [link] .

$g\left(v{\right)}_{a}=g\left(v{\right)}_{b}=\frac{{\text{KT}}_{2a}}{1+{T}_{2a}^{2}\left({v}_{a}-v{\right)}^{2}}$
$\left({\mathrm{\Delta v}}_{a}{\right)}_{1/2}=\frac{1}{\pi }\left(\frac{1}{{T}_{2a}}+\frac{1}{{t}_{a}}\right)$

Because the spin-spin relaxation time is difficult to determine, especially in inhomogeneous environments, rate constants at higher temperatures but before coalescence are preferable and more reliable.

The rate constant k can then be determined by comparing the linewidth of a peak with no exchange (low temp) with the linewidth of the peak with little exchange using [link] , where subscript e refers to the peak in the slightly higher temperature spectrum and subscript 0 refers to the peak in the no exchange spectrum.

$k=\frac{\pi }{\sqrt{2}}\left[\left({\mathrm{\Delta v}}_{e}{\right)}_{1/2}-\left({\mathrm{\Delta v}}_{0}{\right)}_{1/2}\right]$

Additionally, k can be determined from the difference in frequency (chemical shift) using [link] , where Δ v 0 is the chemical shift difference in Hz at the no exchange temperature and Δ v e is the chemical shift difference at the exchange temperature.

$k=\frac{\pi }{\sqrt{2}}\left({\mathrm{\Delta v}}_{0}^{2}-{\mathrm{\Delta v}}_{e}^{2}\right)$

The intensity ratio method, [link] , can be used to determine the rate constant for spectra whose peaks have begun to merge, where r is the ratio between the maximum intensity and the minimum intensity, of the merging peaks, I max /I min

$k=\frac{\pi }{\sqrt{2}}\left(r+\left({r}^{2}-r{\right)}^{1/2}{\right)}^{-1/2}$

As mentioned earlier, the coalescence temperature, T c is the temperature at which the two peaks corresponding to the interchanging groups merge into one broad peak and [link] may be used to calculate the rate at coalescence.

$k=\frac{{\mathrm{\pi \Delta v}}_{0}}{\sqrt{2}}$

## Higher temperatures

Beyond the coalescence temperature, interchange is so rapid (k>>t) that the spectrometer registers the two groups as equivalent and as one peak. At temperatures greater than that of coalescence, the lineshape equation reduces to [link] .

$g\left(v\right)=\frac{{\text{KT}}_{2}}{\left[1+{\mathrm{\pi T}}_{2}\left({v}_{a}+{v}_{b}-2v{\right)}^{2}\right]}$

As mentioned earlier, determination of T 2 is very time consuming and often unreliable due to inhomogeneity of the sample and of the magnetic field. The following approximation ( [link] ) applies to spectra whose signal has not completely fallen (in their coalescence).

$k=\frac{0\text{.}\text{5π}{\Delta v}^{2}}{\left({\mathrm{\Delta v}}_{e}{\right)}_{1/2}-\left({\mathrm{\Delta v}}_{0}{\right)}_{1/2}}$

Now that the rate constants have been extracted from the spectra, energetic parameters may now be calculated. For a rough measure of the activation parameters, only the spectra at no exchange and coalescence are needed. The coalescence temperature is determined from the NMR experiment, and the rate of exchange at coalescence is given by [link] . The activation parameters can then be determined from the Eyring equation ( [link] ), where k B is the Boltzmann constant, and where ΔH - TΔS = ΔG .

$\text{ln}\left(\frac{k}{T}\right)=\frac{{\mathrm{\Delta H}}^{‡}}{\text{RT}}-\frac{{\mathrm{\Delta S}}^{‡}}{R}+\text{ln}\left(\frac{{k}_{B}}{h}\right)$

For more accurate calculations of the energetics, the rates at different temperatures need to be obtained. A plot of ln(k/T) versus 1/T (where T is the temperature at which the spectrum was taken) will yield ΔH , ΔS , and ΔG . For a pictorial representation of these concepts, see [link] .

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what are the products of Nano chemistry?
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
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da
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Bhagvanji
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Application of nanotechnology in medicine
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Damian
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Professor
I think
Professor
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Alexandre
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Rafiq
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Damian
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scanning tunneling microscope
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Rafiq
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Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
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Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
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write examples of Nano molecule?
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The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
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Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
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Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
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biomolecules are e building blocks of every organics and inorganic materials.
Joe
how did you get the value of 2000N.What calculations are needed to arrive at it
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how to find Rutherford scattering parameters angles
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