Second Order Spectra BCMB/CHEM 8190

Transcription

1 Second Order Spectra BCMB/CHEM 8190

2 First Order and Second Order NMR Spectra The "weak coupling" or "first order" approximation assumes that, for simple coupled systems, the difference between the Larmor frequencies of the coupled nuclei is large compared to the coupling constant between them: Δυ >> J When the frequency difference approaches the coupling constant, the spectra are said to be "second order" Δυ ~ J The simple rules presented for first order spectra (multiplicity, number of peaks, peak intensities, chemical shifts of multiplets, values and measurement of coupling constants) do not necessarily apply to second order spectra Importantly, as B 0 increases, Larmor frequencies decrease and Δυ decreases, but J is B 0 independent, so Δυ J and spectra become second order Spin systems for second order spectra will use letters close in the alphabet (AB, ABC), to indicate similarity in frequencies

3 Second Order Spectrum Example: Acrylonitrile Shown is an "ideal" first order spectrum (AMX) for acrylonitrile Are 1 peaks, equal intensities (predicted for first order) 1 H 1 C H A B C A N C 1 H B For δ A =6.3 ppm, δ B =6.1 ppm, δ C =5.6 ppm: AMX first order spectrum 800 MHz 60 MHz J AB = - Hz, Δυ AB = 160 Hz Δυ AB = 1 Hz J AC = 11 Hz, Δυ AC = 560 Hz Δυ AC = 4 Hz J BC = 16 Hz, Δυ BC = 400 Hz Δυ BC = 30 Hz At lower B 0 (as Δυ J ) the spectrum becomes second order In first order spectra, can't discern the sign of the coupling (by looking at the spectra). Not always the case for second order

4 Field dependence of second order spectra As Δυ J, a number of noticeable effects - peak intensities deviate from what is expected for first order - number of peaks changes (can get peaks for "forbidden" transitions) - signals become unrecognizable with respect to first order expectations 60 MHz 90 MHz 50 MHz 490 MHz

5 Field dependence of second order spectra simulated spectra

6 Peak Intensities and Transition Probabilities Recall for pairs of spin ½ nuclei, the product wavefunctions (αα, βα, αβ, and ββ), in the limit of first order spectra (AX), are good solutions to Schrödinger's equation - they form a complete orthonormal set Can write wavefunctions for more complex Hamiltonians by making linear combinations of all members of the set (this is how second order spectra are treated) ψ = c 1 αα + c αβ + c 3 βα + c 4 ββ = Recall, the probability of being in some state, ψ, is ψ ψ Thus, the probability (ρ) of (mostly) being in one (j) of our welldefined basis states is ρ jj c j c j The transition probability, the probability of starting in one state (k) and ending in another (l), is ρ k l ( c k c ) ( k c l c ) l j c j ϕ j

7 Peak Intensities and Transition Probabilities The transition probability, the probability of starting in one state (k) and ending in another (l), is ρ k l ( c k c ) ( k c l c ) l If we know we start in state k (c* k c k = 1), then we only need to solve for the probability that we are going to end up in state l (peak intensities) This requires consideration of the time dependence (for transitions to occur), hence Schrödinger's time dependent equation (time derivative of the wavefunction) Ĥψ(t) = i! d(ψ(t)) dt ψ = c j ϕ j j So, to see how c l changes with time (in unit time, what is the probability of going from state k to l) d(c l ) dt = φ l Ĥ' φ k integrated from t = 0 to 1 Last thing: have to square the result to get intensities ρ k l φ l Ĥ' φ k

8 Understanding second order spectra Pure first order (AX) spectrum: HF - even at low field strength, Larmor frequencies of 1 H and 19 F are MHz apart (at.35 T, 100 MHz 1 H and 94 MHz 19 F, Δυ ~ 6,000,000 Hz), and the one-bond HF coupling constant is ~500 Hz, spin 1 so, Δυ >>> J spin ββ αβ βα αα J AX =0 J AX >0 spin 1 spin spin 1 spin E 1 H spectrum 19 F spectrum Transition energies when J = 0 (i.e. no coupling) - for spin 1, αα βα and αβ ββ energies are equal - for spin, αα αβ and βα ββ energies are equal Transition energies when J > 0 (i.e. positive coupling) - when J > 0, ΔE for transitions to/from the highest energy state (ββ) increases compared to J = 0, ΔE for transitions to from the lowest energy state (αα) decrease compared to J = 0

9 Understanding second order spectra Peak intensities for "pure" first order (AX) 1 H spectrum 19 F spectrum spectrum: HF - the intensities of the signals in each multiplet should be identical - we can calculate the probabilities for single quantum (Δm = 1) interconversions, which should, therefore, all be the same For the first order system, we'll write the Hamiltonian as Ĥ' =γb 1 (Îx1 + Îx) - we can leave out the time dependence (rotating frame) - use B 1 (rf field) which interacts with the magnetic moment (γi x, assuming B 1 acts along x-axis) Example: evaluate for ββ αβ transition ( ρ k l φ l Ĥ' φ k ) ρ ββ αβ ββ γb 1 (I x1 + I x ) αβ (recall Îx α =1 β Î x β =1 α) = γb 1 ββ (I x1 ) αβ + ββ (I x ) αβ = γb 1 1 ββ ββ +1 ββ αα = γb 1 [ 1 + 0] Evaluate all four transitions: ρ αα αβ = ρ αα βα = ρ αβ ββ = ρ βα ββ What about αα ββ???

10 Understanding second order spectra Consider a "pure" second order (A ) spectrum (H, H O, etc.) - expect single peak (equivalent nuclei have identical chemical shifts) - coupling does not result in peak splitting...why?. Problem: αα, αβ, βα, ββ are not individually good solutions to Schrödinger's equation for equivalent nuclei - recall, when not first order, must include I x I x and I y I y in the scalar coupling term of the Hamiltonian (first order approximation no longer valid) - also, αβ and βα imply they are distinguishable (can't be for identical nuclei) - αα and ββ are OK Solution: use linear combinations of αβ and βα (αβ + βα) (αβ βα) So, we can then solve for the 4 possible transition probabilities ρ k l φ l Ĥ' φ k ρ ββ (αβ+βα ) ββ γb 1 (I x1 + I x ) ( αβ + βα) / ρ ββ (αβ βα ) ββ γb 1 (I x1 + I x ) ( αβ βα) / ρ ββ (αα+βα ) αα γb 1 (I x1 + I x ) ( αβ + βα) / ρ ββ (αα βα ) αα γb 1 (I x1 + I x ) ( αβ βα) /

11 Understanding second order spectra Solve for one of the transitions ( ) / ρ ββ (αβ+βα ) ββ γb 1 (I x1 + I x ) αβ + βα = γb 1 ββ I x1 ( αβ + βα) / +γb 1 ββ I x ( αβ + βα) / = γb 1 ββ I x1 αβ ( ) / + ββ I x1 ( βα) / + ββ I x ( αβ) / + ββ I x ( βα) / = γb 1 (1 ) ββ ββ + (1 ) ββ αα + (1 ) ββ αα + (1 ) ββ ββ = γb 1 (1 ) (1 ) = γb 1 1 =1 (γb 1 ) All solutions Result - two transitions of equal energy and equal probability (i.e. one peak) - two transitions of with different energies, but zero probability (not allowed) ρ ββ (αβ+βα ) 1 (γb 1 ) ρ ββ (αβ βα ) = 0 ρ αα (αβ+βα ) 1 (γb 1 ) ββ (αβ+βα)/ (αβ-βα)/ E ρ αα (αβ βα ) = 0 αα

12 "Intermediate" behavior of nd order spectra AB systems are "intermediate" between AX and A - as Δυ J an AX system becomes AB - inner peaks of doublets more intense than outer peaks ("roof effect") - J can still be measured as distance between peak centers - chemical shifts are not center of doublets, but weighted averages of peak intensities ("center of gravity"), and can be determined knowing outer lines are not observed, but couplings still exist H O δ A δ B = (δ 1 δ 4 )(δ δ 3 ) AB HF

13 ABX systems ABX systems raise additional complications - for 'X', can't necessarily deduce coupling constant from peak separation - example: coupling of H5 (A), H5' (B) and H4 (X) in ribose rings Base O φ H 5 C H 4 H 5 ' OH J 45? HO OH - may consider using Karplus relationship to deduce torsion angle from J 45 and J 45' - splitting in ABX spectrum in X signal is equal to (J 45 +J 45' )/ - can't conclude J 45 = J 45' Helpful to simulate spectra X - software: S.A. Smith, J. Magn. Res., 166, 75 (1994) M.Veshtort, R.G.Griffin, J Magn. Res., 178, 48-8 (006)