CHEM 781 Part 4 SpinSpin Coupling Introduction So

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CHEM 781 Part 4: Spin-Spin Coupling

CHEM 781 Part 4: Spin-Spin Coupling

Introduction • So far, spins were regarded spins isolated from each other. • However,

Introduction • So far, spins were regarded spins isolated from each other. • However, spins do have effect on neighboring spins – the magnetic moment of one spin will influence the effective field at its neighbors

Direct through space coupling (dipolar coupling) • The Energy of spin A will depend

Direct through space coupling (dipolar coupling) • The Energy of spin A will depend on the orientation of spin B relative to A. H-H ⊥ to B 0: parallel arrangement higher energy HH || B 0: parallel arrangement lower in energy

Magnitude of dipolar interaction • The interaction will depend on distance A-B and the

Magnitude of dipolar interaction • The interaction will depend on distance A-B and the angle of the vector A-B relative to the field: DAB is the dipolar coupling constant. • In solids, DAB will depend on the orientation of the internuclear vector relative to the field. Solid state proton spectra are therefore typically severely broadened as DAB can be several tens of k. Hz in magnitude. • In solution fast tumbling will average over all orientations and Dave = 0. Dipolar coupling is not directly observable in solution, but important for T 1, T 2 relaxation and NOE. Broad lines in 1 H NMR of larger proteins arise from short T 2 due to insufficient averaging.

Indirect (scalar) coupling • Interaction mediated by electron (hence also called through bond coupling

Indirect (scalar) coupling • Interaction mediated by electron (hence also called through bond coupling • Important is electron density at nucleus (Fermi-contact interaction). That means s-orbital participation is required. • This interaction does NOT depend on the orientation of the A-B vector to the field (hence the name scalar coupling).

Coupling through one bond: • • consider a P-H bond: (1 H-31 P): One

Coupling through one bond: • • consider a P-H bond: (1 H-31 P): One electron will be closer to H, the other closer to P, but always with anti-parallel orientation of spin (Pauli principle) The nucleus will prefer to be anti parallel to nearby electron spin (note that e- has negative charge and the magnetic moment will be aligned parallel); thus anti parallel orientation of 1 H and 31 P will be lower in energy than the parallel arrangement: Per definition, if anti parallel arrangement is lower in energy, J > 0 Sign of J is reversed if g values have opposite sign.

Coupling through two bonds • • • Electrons in neighboring orbitals tend to approach

Coupling through two bonds • • • Electrons in neighboring orbitals tend to approach each other with parallel spin (Hunds rule). This results in the parallel orientation of nuclear spins to be lower in energy J < 0

Dependence of spin-spin coupling on structure: • Understand splitting pattern, relate to number of

Dependence of spin-spin coupling on structure: • Understand splitting pattern, relate to number of neighbored spins • relate magnitude of coupling to chemical structure

Appearance of multiplets • First Order splitting: If the chemical shift difference (in Hz)

Appearance of multiplets • First Order splitting: If the chemical shift difference (in Hz) is large compared to the coupling constant J the splitting can be analyzed by first order rules: Each coupling partner will split the line into two lines of equal intensity • Equivalent spins: In the other extreme, coupling between equivalent spins does not give rise to a splitting. • Higher order spectra: If the difference in chemical shift is similar to magnitude of coupling, higher order effects are observed. One can intuitively understand that there has to be an intermediate regime between two equivalent and two very different spins. • As Δν 0 measured in Hz depends on the magnetic field B 0 the same ppm separation can give a higher order spectrum with Δν 0 ≈ JAB at a low field B 0 but a first order spectrum (Δν 0 ≫JAB) when using a stronger magnet with larger B 0.

Two spins: AX, AB and A 2 spin systems AX: Dn. AB >> JAB

Two spins: AX, AB and A 2 spin systems AX: Dn. AB >> JAB First order splitting AB: Dn. AB ≈ JAB Higher order multiplet A 2: Dn. AB = 0, JAB ≠ 0 No splitting

Spin-spin coupling – energy levels Without field, coupling will split the energy levels into

Spin-spin coupling – energy levels Without field, coupling will split the energy levels into a singlet and a degenerate triplet state A Field will leave singlet state unchanged, but split triplet state according to Iz. Only transitions within the triplet state will be observed If the two spins are non equivalent, the separation of singlet and triplet state no longer holds. S -> T transitions allowed

Something to think about: • For dihydrogen (H 2), the interconversion of Singlet and

Something to think about: • For dihydrogen (H 2), the interconversion of Singlet and Triplet state is so slow, that the two forms are considered as two separate isomers, para- and ortho-hydrogen. • At room temperature, the ratio is 25/75 para/ortho hydrogen, consistent with k T >> h JHH (DE = h Dn = h. J) • but at 77 K the ratio para/ortho it is 52/48, and at 20 K it is 99. 8/0. 2. – What is the magnitude of the spin-spin coupling required to explain the results ? – What else could be going on ? – How could this be useful for NMR ?

Multiple coupling partners • • If a nucleus is coupled to several different nuclei,

Multiple coupling partners • • If a nucleus is coupled to several different nuclei, coupling patterns will superimpose: CHB-CHA-CHC: HA coupled to HB and HC, lets assume JAB > JAC in general up to 2 n lines are possible if all coupling different

special case JAB = JAC • This is the same as a CH 2

special case JAB = JAC • This is the same as a CH 2 group • The two inner lines overlap, resulting in a 1 : 2 : 1 triplet.

Rules for coupling to a group of equivalent spins Suppose a spin with a

Rules for coupling to a group of equivalent spins Suppose a spin with a group of n neighbored spins I: • Number of lines given by 2 n I + 1 • for I = ½ intensities are given by Pascals Triangle • For I ≠ ½ the intensities will be different (see multiplets of deuterated solvents in 1 H and 13 C spectra) Note that this is just a special case of multiple coupling partners. All first order couplings patterns can be regarded as a superposition of individual doublets. •

Example CH-CH 2 group • One proton coupled to two equivalent nuclei: The two

Example CH-CH 2 group • One proton coupled to two equivalent nuclei: The two nuclei allow four combinations of their spin states, two of which are equivalent: spin states (+ ½ + ½) (+ ½ ; - ½ ) (-½ ; + ½ ) (-½ ; - ½ ) total spin of CH 2 group +1 0 -1 Intensity of line ∼ number of equivalent combinations of spin states 1 2 1

Example CH-CH 3 group • proton coupled to three equivalent nuclei spin states (+½

Example CH-CH 3 group • proton coupled to three equivalent nuclei spin states (+½ ≡ + -½ ≡ -) total spin group (+ + +) (+ - -)(- + +)(- - +) (+ - -)(- + -)(- - +) (- - -) + 3/2 + ½ - 3/2 Intensity of line ∼ number of equivalent combinations of spin states 1 3 3 1

Exercise: Sketch the 31 P spectrum of PF 6

Exercise: Sketch the 31 P spectrum of PF 6

Coupling to nuclei with I > ½ important to understand spectra of deuterated solvents

Coupling to nuclei with I > ½ important to understand spectra of deuterated solvents Example: 13 C spectrum of CDCl 3 (one coupling partner of I=1): Example: 13 C spectrum of DMSO-d 6 (CD 3 -SO-CD 3) (three coupling partners with I=1) m. I Itot Intensity (+1 +1 +1) +3 1 (+1 +1 0)(+1 0 +1)(0 +1 +1) +2 3 (+1 +1 -1)(+1 -1 +1)(-1 +1 +1)(0 0 +1)(0 +1 0)(+1 0 0) +1 6 (0 0 0)(+1 -1 0)(+1 0 -1)(0 +1 -1)(-1 +1 0)(-1 0 +1)(0 -1 +1) 0 7 (-1 -+1+1)(-1 +1 -1)(+1 -1 -1)(0 0 -1)(0 -1 0)(-1 0 0) -1 6 (-1 -1 0)(-1 0 -1)(0 -1 -1) -2 3 (-1 -1 -1) -3 1

 • Note that the proton signals of deuterated solvents arise from partially protonated

• Note that the proton signals of deuterated solvents arise from partially protonated isotopomers. • The proton spectrum of deuterated DMSO will be that of CD 3 -SO-CHD 2, so there will be two coupling partners with I = 1. Exercise: Sketch the 19 F spectrum of BF 4 - and Sb. F 6 -

Magnitude of coupling constants • Coupling constant J does NOT depend on field B

Magnitude of coupling constants • Coupling constant J does NOT depend on field B 0 and is thus measured in units of Hz • J is proportional to g values of involved nuclei Therefore, C-D couplings in deuterated solvents are scaled compared to the ones in the equivalent protonated compound • The magnitude of J depends on the electronic and steric environment

Couplings through one bond 1 J HH : dihydrogen H-H 276 Hz ~> How

Couplings through one bond 1 J HH : dihydrogen H-H 276 Hz ~> How would one measure this ? 1 J CH : always positive, typically between 80 – 250 Hz. • depends on s-orbital contribution to the bond. In hydrocarbons one finds J = 500 Hz • (%-s) where (%-s) = 0. 25 for sp 3, 0. 33 for sp 2 and 0. 5 for sp-orbitals • Substituents have an inductive effect (EN substituents cause an increase in JCH): CH 3 -Li CH 3 -H CH 3 -Cl CH 2 Cl 2 CHCl 3 CHCF J/Hz 98 125 150 178 205 239 H 2 C=CHX X = J/Hz H 156. 4 -CHO 162. 3 CN 178 Cl 195 F 200

Stereo chemical effects on 1 JCH • Effect of nitrogen lone pair in imines

Stereo chemical effects on 1 JCH • Effect of nitrogen lone pair in imines and similar compounds: 1 JCH (trans) < 1 JCH (cis) with respect to the lone pair

Couplings through two bonds • • Called geminal couplings Can be positive or negative

Couplings through two bonds • • Called geminal couplings Can be positive or negative 2 J HH depends on hybridisation of connecting C-atom: Similar for 2 JCH: Two-bond H-C correlations often weak or absent in aromatic systems

Coupling through three bonds • • 3 J Typically positive Vicinal coupling HH The

Coupling through three bonds • • 3 J Typically positive Vicinal coupling HH The most important one in structure determination • Magnitude depends on dihedral angle φ: 3 J = A + B cos(φ) + C cos 2(φ) A B C calculated for HC-CH fragment -0. 28 -0. 5 9. 8 empirical to fit experimental data 2 -1 10

Vicinal coupling constants in selected compounds Always: • • 3 J(φ=0 ) < 3

Vicinal coupling constants in selected compounds Always: • • 3 J(φ=0 ) < 3 J(φ=180 ) in olefins, 3 Jcis < 3 Jtrans In cyclohexane (chair conformation): coupling between two axial protons larger than between equatorial or axial with equatorial protons: 3 Ja, a > 3 Ja, e ≈ 3 Je, e 2 J 1, 2 = -14. 5 Hz 3 J 1, 3 = 2. 2 Hz 3 J 1, 4 = 10. 6 Hz 3 J 2, 3 = 5. 6 Hz • Cyclopropanes: φtrans ≈ 0 , φcis ≈ 135 =>3 Jcis > 3 Jtrans

Multiple conformations lead to averaging of couplin constant • In flexible aliphatic chains fast

Multiple conformations lead to averaging of couplin constant • In flexible aliphatic chains fast conformational interconversion occurs ( rotation about C-C single bond) => average coupling of 7 - 8 Hz observed • Similar considerations also apply to other couplings over 3 bonds (3 JCH, 3 JPH, 3 Jcc etc)

Long range couplings presence of double bonds and/or favored conformations in rings required: •

Long range couplings presence of double bonds and/or favored conformations in rings required: • allylic, naphthalene or condensed aromatics, selected ring systems

Extended Vector picture to describe couplings • The two multiplet lines can be described

Extended Vector picture to describe couplings • The two multiplet lines can be described by two vectors rotating in opposite directions (assuming Ω = 0). • Separation of these vectors into its x- and y-components yields the corresponding rotation equation for coupling assuming M = -M 0 AIy. C at t = 0: