Applications of NMR Spectroscopy in Inorganic Chemistry

The Chemical Shift An isolated nucleus such as hydrogen nucleus has a precessional frequency of 100 MHz at field strength of 2.3487 T. But all nuclei are associated with electrons, which revolve around them. As a charged particle in circular motion, the electrons also produce a magnetic moment, and this is called as secondary or induced magnetic field. When placed in an applied magnetic field, the induced magnetic field of the electrons may oppose the net magnetic field experienced by the nucleus i.e., the electrons may circulate in such a way its magnetic field is opposing the applied field. This circulation is known as diamagnetic circulation. In diamagnetic circulation the electrons and the nucleus are spinning in opposite direction to each other. On the other hand if both are spinning in the same direction the circulation may be paramagnetic one.

Thus the nucleus can be shielded from the applied field by diamagnetic circulation of electrons. The extent of shielding will be constant for a given atom in an isolated condition, but will vary with the electron density about an atom at om in a molecule. The above equation may be generalized as Bi = B 0 (1- σi) where Bi is the net magnetic field felt by a particular nucleus ―i‖ its shielding constant is σi For example we know that oxygen atom is more electro negative than carbon and hence the electron density about the H atom in C-H bonds should be considerably higher than that in O-H bonds.  We can expect σCH > σOH and hence BCH = B0 (1- σCH) < BOH = B0 (1- σOH) i.e., the C-H protons are highly shielded and felt less applied field. But, the O-H protons are less shielded and felt more applied field. Due to this reason the C-H proton precess with a smaller Larmor frequency than that of O-H proton. Thus in order to come into resonance with a radiation of particular frequency (~100MHz at a field strength of 2.3487T), a C-H proton requires a greater applied field than the O-H proton. For example when we record the 1 H N.M.R. spectrum for methanol we can see that the O-H proton will give the signal first i.e., at comparatively lower field than the C-H proton. The C-H proton proton signal is obtained at higher field to that of O-H proton.

Dr. D. Ilangeswaran, M.Sc., M.Phil., Ph.D.,

In N.M.R. spectrum thus the identical nuclei can give rise to different absorption positions when present in different chemical surroundings as in the case of methyl and alcohol protons in methanol. The separation between absorption peaks is usually referred to as chemical shift. Internal Standard For the measurement of chemical shift we have to take tetramethylsilane, Si(CH 3)4 (TMS) as an internal standard. With reference to the signal of TMS we have to find out the shift in the position of other signals. Since TMS is immiscible with water for aqueous samples we may use (CH3)3SiCH2CH2CH2SO3 Na as an internal standard. For 13C spectrum also these two substances may be used as a reference. The advantages of TMS are 1. Its resonance is sharp and intense since all 12 H nuclei are equivalent and hence absorb at same position. 2. The resonance position of TMS is too high field of almost all other H absorptions in organic molecules and hence can be easily easil y recognized. o 3. It is a low boiling liquid (b.P.27 C) and so can be readily removed from the most samples after use. Conventionally N.M.R. spectra are displayed with the field increasing from the left, which places TMS signal to the extreme right. Two measurement scales are used,  scale and  scale. In  scale, the TMS signal is marked as 10p.p.m. and the chemical shift value is decreasing towards left. In  scale the TMS signal is marked as 0p.p.m. and the chemical shift value is increasing towards left. The relationship of these two scale is  =10 - . The Coupling Constant Generally the peaks of proton N.M.R. spectrum are splitting into two or more & the distance between the adjacent peaks in a multiplet is known as coupling constant. The reason for this is the spin interaction between the two protons when they are present very close proximity to each other. There are several kinds of spin interactions. When compared to gaseous molecules the spin-spin interaction will be more in solid substances and takes place to a smaller extent in liquid molecules. Factors Influencing Coupling Constant, J Values 1. Geminal Coupling The protons connected to the same atom are separated only by 2 chemical bonds are known as geminal protons. The geminal coupling constant values varies from +5Hz to -30Hz depending on the bond angles between these 2 protons. 2. Vicinal Coupling The protons present on the adjacent atoms are separated by 3 chemical bonds are known as vicinal protons. Usually the vicinal coupling constant values vary from 0 to 9 Hz depends on the dihedral angle. When the dihedral angle,  increases from 0 to 90 the J value decreases from 9 to 0 and again it increases if the  value rises from 90 to 180. Karplus equation given below helps to calculate the vicinal coupling constant, if  if  value is provided. J = 8.5 cos 2 - 0.28 (if  (if  = 0 – 0 – 90 90o) J = 9.5 cos 2 - 0.28 (if  (if  = 90 – 90 – 180 180o)


3. Long Range Coupling If protons are separated by more than 3 chemical b onds the J value is very small. 13

C NMR Spectra For a 13C nucleus, the value of I = ½ and has a precessional frequency of 20 MHz in an external magnetic field strength of 1.9 tesla. But at the same magnetic field strength the precessional frequency of proton is 80 MHz. MHz . Hence the magnetic moment of 13C nucleus is 1/4th the magnetic moment of H nucleus. At the same time the natural abundance of 13C nucleus is 1.1% only. So, the NMR signal of 13C nucleus is very weak. The problem of weak signal in simple molecules can be overcome by synthesizing 13C enriched samples. But this is very difficult in complex molecules. In practice by using pulsed FT method 13C NMR spectra are recorded with natural abundance of 13C, with the sensitivity enhanced by summation of several spectra. Chemical shift values vary from 0 to 200 ppm. As both 13C and H nucleus have I = ½, we can expect 13C - 13C and 13C – 1H couplings. The probability of 2 13C atoms being together in the same molecule is ver y low and so 13C - 13C coupling is not usually observed. Due to 13C – 1H coupling, the 13C spectrum will be more complex. This can be avoided by proton decoupling. The 1H nucleus can be decoupled by double irradiation at its resonance frequency (80 MHz at 1.9 T). This is an example of hetero nuclear decoupling. The following table gives the chemical shift values of some carbon atoms. Type of C atoms

Chemical Shift ( ) ppm

sp carb carbon on 0 – 80 – 80 sp carb carbon on 80 – 80 – 150 150 sp carbon 70 – 70 – 90 90 sp (aroma (aromatic) tic) 110 – 110 – 140 140 C – X – X 10 – 10 – 80 80 Alcoholic, ether, C – C – O O 40 – 40 – 80 80 C – N – N 20 – 20 – 80 80 C=O 180 - 200 Here also internal standard is tetra methyl silane (TMS) 19

31

F and P NMR The abundance of 19F is 100 % and its value of I = ½. Its precessional frequency is

56.46 MHz at 1.4 tesla. Internal standard is CFCl 3 or CF3COOH and  value varies from 0 to 240 ppm. Geminal F – F – F coupling ranges from 43 – 370 – 370 Hz and vicinal coupling varies from 1 19 0 to 39 Hz. Coupling between H – F also strong with geminal coupling value between 42 – 80 Hz and vicinal coupling varies between 1.2 – 1.2 – 29 29 Hz. 31

P nucleus also has I = ½ with  = 24.3 MHz at 1.4 T. For example, (CH 3)2PCF2CH3 has 84 lines. First the t he signal of P split into 3 by 2 F atoms, then each line into 4 by single CH 3 protons. Finally each of 12 lines into 7 by 2 CH 3 protons.


Problems 1. HPF2: The 31Pnmr spectrum shows a sextet (three doublets). The two F atoms first split the t he phosphorous signal into a triplet and then the H atom split each peak into a doublet. Two triplets may result if it were J P.H > JP.F (When there are two interacting magnetic nuclei nearby a sample nucleus the nucleus with larger coupling constant value will first cause the splitting of the signal of the sample nucleus followed by the other nucleus). When J When J P-F > J P-H three doublets

When J When J P-H > J P-F Two triplets

Spli ttin g due to one H atom

spli ttin g due to two F atoms atoms

2. Protons attached to metal ions are very highly shielded, the peaks often occurring 5 to 1 5 ppm on the high field f ield side of TMS. It is due to non-polar nature of M-H bond. The H nmr of 3HRh(CN)5 is a doublet (IRh = 1/2) which occurs at 10.6 ppm on the high field side of TMS. 3. The 31Pnmr of Phosphorous acid , HPO(OH) 2 is a doublet and that of Hypophosphorous acid, H2PO(OH) is a triplet. Therefore, the structures are as given below. There is no coupling due to -OH proton as they are far removed. O O

P H

P OH

H

OH

H HO 4. The 31Pnmr of P 4S3 shows two peaks with 3:1 ratio. The more intense peak is a doublet and the less intense peak is a quartet. S does not couple as I S = 0. This spectrum indicates that there are three equivalent and one unique P atoms. P S

S S

P

P P


5. Solutions of 1:1 TiF 62- and TiF4 in ethanol give 19Fnmr consisting of two peaks with intensity ratio 4:1 (ITi48 = 0). The low intensity peak is a pentet and the high intensity peak is a doublet. Therefore the structure is [TiF5(OHC2H5)]. 6. Ammonia dissolved in water undergoes very fast proton exchange with water. The NMR spectrum consists of only one peak, which is the average resonance frequency of all the protons on nitrogen and oxygen. oxygen. l9 7. The Fnmr spectrum of TiF 4 in donor solvents at — at — 30 30 oC consists of two triplets of equal intensity. This corresponds to the cis-structure. However, the spectrum at 0 oC shows a single fluorine peak. This may correspond to the trans structure. But actually the change in the spectrum at higher temperature is not due to cis - trans conversion but it is due to the dissociation reaction of type: TiF 4.2B → TiF4.B + B

At high temperature, this dissociation is fast enough as to make all the four fluorine atoms equivalent. 8. Nuclear quadrupole relaxation changes the splitting pattern. When the relaxation is rapid the spin state of the nucleus is rapidly changed and as a result splittings do not occur. Slow relaxation rates cause normal splittings. Intermediate exchange rates often result in a broadening of the peaks. Relaxation effects are often encountered for nuclei, which have quadrupole moment because these nuclei are very efficiently relaxed by the fluctuating electric field gradients, which arise from the thermal motion of the polar solvent and solute molecules. For N14H3 (I=1), three broad HNMR signals are obtained and forN 15 H3 (I=1/2; no quadrupole relaxation), a sharp doublet is obtained. 9. Spin-spin coupling depends also on the number of intervening bonds. In saturated molecules of light elements JH _ H falls off rapidly as the number of bonds between the two interacting nuclei increases and usually is negligible for coupling of nuclei separated by more than three bonds. 10. Long-range couplings are observed in unsaturated compounds. Here the spin-spin coupling is transmitted through -bonds, which are delocalized over the entire molecule. 11. Long-range coupling occurs through space instead of  or  bonds in cases where coupling involves an atom other than hydrogen. For the compound SF 5.CF2CF3, JF-F (of the CF2 group and trans- F) is -5 cps while J F _ F (of CF2 group and the cis-F) is -16 cps. In the latter case there is through-space coupling.


12. F19 nmr of CH2=CF2 shows four lines of equal intensity. The two F and two H atoms are non - equivalent. The two fluorines have the same chemical shift. This peak is split by the TWO H – ATOMS one by one into four lines of equal intensity (it is not 1:2:1 as If the 2F or 2H were equivalent). trans H-F coupling

cis H-F coupling

Ha

Fa C

Hb

C Fb

13. C1F3: Two fluorines are of one type and one F is of other type. The F 19nmr shows a (1F, triplet) and a (2F, doublet). F

Cl (14) P4O136-: (15) P4S3:

F F Two peaks viz., (2P, triplet) and (2P, triplet). Two peaks viz., (3P, doublet) and (1P, quartet)

16. If the molecules being studied are undergoing very rapid exchange reactions, the nmr spectrum drastically affected. A mixture of CH3COOH and H 2O does not show two separate O-H resonances from water and acid but instead shows only one. The -OH protons are exchanging positions very rapidly between the acid and the water. 17. In N.N-dimethylacetamide, the methyl groups a,b,c are non-equivalent at room temperature as the double bond character of C-N bond restricts the rotation about C-N axis. As a result, the methyl groups b and a are cis and trans to O-atom and give rise to two peaks. At high temperatures, rotation becomes free and as a result there is no distinction between cis and trans methyl groups. Hence, b and c give single peak.


18. NMR spectrum of methylammonium chloride in water at pH = I shows (i) a quartet methyl peak (split by three NH3 protons); (ii) a sharp water peak; (iii) three broad peaks from NH 3 protons (I N=1). Broadening is due to quadrupole nature of Nitrogen. No fine structure due to – CH CH3 protons is observed because of quadrupole broadening by nitrogen. nitrogen. When pH is increased, all bands begin to broaden. At pH = 5, the spectrum shows: (i) one sharp peak due to – to – CH CH3; and (ii) a broad peak due to all other protons. 11 19. B3H8 : B nmr spectrum is a nonet which results from a splitting of three equivalent borons by eight equivalent protons. Because of the fast intramolecular intramole cular hydrogen exchange all B and all H become equivalent.

20. P3 N3Cl4F2: The 19Fnmr spectrum shows it is a AB2X2 system. The two F atoms are X 2, the P atom to which the fluorines are attached is A and the other two P atoms are B 2; 1.1difluoride-3.3.5,5-tetrachloride.

Cl

N

Cl P N

P

F

Cl P

Cl

N F

21. P4 N4Cl6(NHC6H5)2: 31Pnmr shows two triplets of equal intensity. The structure is given below.

Cl

Cl N P P Cl

H5C6HN N

N

Cl P N P Cl Cl NHC 6H5 Dr. D. Ilangeswaran, M.Sc., M.Phil., Ph.D.,

22. NF3: The F l9nmr shows a sharp singlet at -205 oC. As the temperature is raised the peak broadens and at 20 °C, a sharp triplet (I N14 = 1) is obtained. This change is opposite to that normally obtained for exchange processes. At lower temperatures, the slow molecular motions are most effective for quadrupole relaxation of N 14. At higher temperature, the relaxation is not as effective and the life-time of a given state for N 14 nucleus is sufficient to cause spin-spin splitting. 23. The N 14nmr of azoxybenzene exhibits only a singlet. The field gradient at the N-O nitrogen is so large as to make this resonance unobservable. Only the other nitrogen shows the signal. O

-

N

24.

+

N

31

Pnmr spectrum of PF 2H(I5 NH2)2 shows 90 lines as explained below. F

NH2 P

H

F NH2

One P31 signal gets split into a doublet by one H Each line in the doublet is split into a triple by two F's (2 X 3 = 6 lines) li nes) Each of these six lines is split into triplet by two 15N’ 15N ’s (I N15 = ½) (6 X 3 = 18 lines) Each of these 18 lines is split into a quintet by the four NH 2 protons (18 X 5 = 90 lines) 25. The proton noise decoupled P 31 spectrum of a mixture of trans - [PtCl4(PEt3) and trans [PtBr 4(PEt3) shows six lines (each with l95Pt satellites), showing the existence of six complexes viz, [PtBr 4Cl(PEt)3], [PtBr 3Cl(PEt)3], [PtBr 2Cl2(PEt)3] two isomers, [PtBrCl3(PEt)3] and [PtCl 4(PEt)3]. Effect of Quadrupole Moments in NMR Nuclei with quadrupole moments undergo spin-lattice relaxation rapidly. Therefore the signal of a nucleus, which is attached to a quadrupole nucleus, is extensively broadened. The nmr signals of a quadrupole nucleus are broadened so extensively that no spectrum is obtained. The quadrupole relaxation process is due to the interaction of electric field gradient with the quadrupole moment. The field gradient is caused by the asymmetry of the electronic cloud. (1) In halide ion, the charge distribution is spherical and gives rise ris e to only small field gradient

at the nucleus (longer  (longer ). Hence, halide ions give sharp peaks.


(2) Solutions of I - (I127= 5/2) show a NMR signal. When iodine is added the triiodide, I 3-, is formed destroying cubic symmetry of the iodide ion so that quadrupole broadening becomes effective and the signal disappears. Broadening depends on the amount of iodine added and hence the rate constant for the reaction I- + I2 → I3- can be calculated. Contact Shifts (NMR of Paramagnetic Species) The unpaired electron present in the paramagnetic compounds broadens the spectrum by both dipolar and electron spin-nuclear spin coupling mechanisms. The magnetic ma gnetic moment of an electron is ~10000 times larger than the nuclear magnetic moment. Therefore, paramagnetic ions produce large magnetic fields which make dipolar-spin lattice relaxation very rapid. If the electron relaxation is very slow, then splittings due to s = ±1/2 will be observed. If the relaxation is fast, the magnetic nucleus senses only the time-averaged magnetic field of the two spin orientations and a single peak is observed. Intermediate relaxation rates will broaden the spectrum. spectr um. The presence of unpaired electrons makes the electron electr on relaxation ver y rapid; this causes a very large chemical shift (~3000-5000 cps) of the resonance in the NMR spectrum. This is referred to as contac contactt nmr shi ft . The relationship between contact shift and magnetic field for proton magnetic resonance is given as:

Where, e gyromagnetic ratio for the electron,  N is that for the magnetic nucleus, S is the electron spin multiplicity, H is (Hcomplex - Hligand) and v = (vcomplex - vligand),  is Bohr megneton and g is the ratio of magnetic moment to the total angular momentum of the electron. The condition requisite to observing a contact shift in nmr is

e is electronic relaxation time, s is chemical exchange time and A N is the contact interaction constant (electron spin - nuclear spin coupling constant). Example: In nickel(ll)aminotroponeiminate, one electron is transferred to Ni(II) from the ligand. As a result, the unpaired electrons present at  and  positions are aligned with magnetic field while that at  is anti-aligned with the field.  and  carbons have positive spin density and  carbon negative spin density, i.e.,  and  protons are shielded and  proton is deshielded.


Fluxional Behaviour These molecules possess more than a single conformation representing an energy minimum. Several such minima may be present and accessible with ordinary thermal energies. NMR techniques are very useful in studying fluxional molecules. The fluxional behaviour is studied by taking NMR spectrum at lower and higher temperatures. At lower temperature, the rate of fluxional behaviour will be slow and we can obtain individual lines for each site. At higher temperature, due to inter – inter – conversion, conversion, a time – time – averaged spectrum will result. For example:

1) A 1) A complex formed between tetramethylallene and iron carbonyl Below -60 °C the H nmr shows three peaks in the ratio 1:1:2 representing the three cis – proton, – proton, three trans-protons and six protons in a plane perpendicular to C – Fe – Fe bond. As the temperature is raised, the spectrum collapses to a single resonance for the average environment of the 12 protons as the iron migrates around the allene -system. CH3

H3C C H3C

CH3

H3C

Fe(CO) 4 C Fe(CO) 4

CH3

H3C

CH3

2)  3 -allyl complex The HNMR spectrum at low temperature shows two doublets, representing the cisand trans- protons and one multiplet, corresponding to the non-terminal hydrogen. Upon warming, the spectrum changes with collapse of the two doublets into one, establishing that there are four terminal hydrogen atoms and one non-terminal hydrogen atom. This change is due to the rapid inter-conversion, which makes H s and Ha indistinguishable.

H H a

H b

C C

M

H C

H a

H b

H a

H b


C C

M C

H H a

H b

H a

H b

C C

C M

H a

H b

3) Mercury 3) Mercury cydopentadienyl compound, cp2 Hg This complex shows two conformers (a) and (b). In (a) the divalent Hg is  bonded to two  3-C5H8 rings. In (b) it is an ionic compound, a  5-complex. Structure (a) shows 3 protons resonances with intensity ratio 1:2:2, whereas all the protons are equivalent in (b). The Hnmr spectrum at -70 oC shows a single peak.

H Hg H

2+

Hg

(a)

(b)

4) Ferrocenophane 4) Ferrocenophane A sharp singlet for the methylenic protons and two narrow multiplets for the α and  protons.

5) P31nmr spectrum of PF4 NMe2 shows triplets of triples (two equatorial and two axial F's interact separately with P31 signal) at low temperature and on warming, it gives a regular quintet (all the four F's become equivalent).

F

F P

F


F

CH3 N CH3

Applications of NMR Spectroscopy in Inorganic Chemistry

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