Monday, April 11, 2016

INEPT

The sensitivity of a low γ, spin I = ½ nucleus is determined by the difference in populations between the low energy and high energy states, governed by the Boltzmann distribution. If the low γ, spin I = ½ nucleus is coupled to a proton the energy level diagram is more complicated than simply two levels and is shown in the figure below where a 13C-1H spin pair is used as an example.
The populations of the states involved in the 13C transitions and hence the sensitivity of the 13C signal can be altered by inverting the H1 or H2 1H transitions with 180° pulses. This is illustrated in the figure below.
In the left panel, the H1 transition of a 13C-1H spin pair is inverted (i.e. the populations of the two energy levels of the H1 transition are swapped). This also affects the populations of the energy levels involved in the C1 and C2 13C transitions. After inversion of the H1 1H transition, the intensities of the C1 and C2 13C transitions have changed from their equilibrium value of 1:1 to an enhanced value of 5:-3. If the H2 transition is inverted (right-hand panel), the C1:C2 intensity ratio is -3:5. In both cases the sensitivity of the 13C doublet has been enhanced compared to its equilibrium value. This enhancement is called INEPT (Insensitive Nuclei Enhanced by Polarization Transfer) and is one of the most common sensitivity enhancement techniques used in NMR pulse sequences. The simplest implementation of INEPT is shown in the figure below along with the vector diagrams.
Phase cycling can be employed to obtain a -4:4 anti-symmetric doublet, rather than doublets with components of unequal magnitude. This is represented in the figure below.
A refocusing element can be added to the end of the sequence to refocus the anti-symmetric doublets and data can be collected with proton decoupling.
The result is a singlet with 4 times  (i.e. γHC) the intensity of the singlet one would expect under equilibrium conditions without an NOE.  For 15N, one obtains a sensitivity gain of ~10. The results of these implementations of INEPT are compared to the equilibrium situation in the figure below.
INEPT has the additional advantage that its repetition rate is determined by the 1H T1 rather than the 13C T1.  This is a tremendous additional sensitivity improvement when multiple scans are collected because the 1H T1 is often shorter than the 13C T1 by an order of magnitude.  One can collect approximately ten times as many scans per unit time.  This advantage is even more significant for 15N.  Reverse INEPT is used in the collection of 1H data for  carbon-proton pairs to suppress the protons bound to 12C. 

9 comments:

Unknown said...

Very well explained

Unknown said...

In the first figure- After the application of a 180 pulse to the carbon channel that inverts the magnetization of carbon to negative z axis does this carbon magnetization evolve during the delay of 1/4J that has been applied. If it does, how can we think about it?

Glenn Facey said...

Bhuwan,
The only evolution of the 13C magnetization during the 1/4J period after it has been inverted by the pi pulse is T1 relaxation which is neglected in the explanation and the figure.
Glenn

Imran Ali said...

What if τ=1/2J

Glenn Facey said...

Imran,
If doing 13C NMR, J is the one bond carbon - proton coupling constant. If doing 15N NMR, J is the one bond 15N - proton coupling constant.
Glenn

Anonymous said...

Dear glenn
What if hydrogen of NH is replased by Deuterium?
How INEPT spectra will look like?

Glenn Facey said...

Anonymous,
If one replaces the proton on an N-H with deuterium. 1H-15N INEPT would not be possible.
Glenn

Rostyslav said...

Dear Glenn, what would the Equilibrium Energy Level Diagram look like in the case of a 31P-13C pair? What will be the gain factors of 13C signals in this case?

Glenn Facey said...

Rostyslav,
The energy level diagram would look similar to that for 1H and 13C except the populations would be based on the gyromagnetic ratios for 31P and 13C. The overall gain for the refocused 31P-13C INEPT with 31P decoupling would be 1.6.

Glenn