H Nuclear spin and Magnetic moment Nucleus of atoms is made up of protons and neutrons. Like clectrons, protons and neutrons also have property of spinning about their own axis. According to quantum mechanics value of angular momentum for each is Va h/2n). The net resultant of the angular momentum of all nuclear particles is called as nuclear spin and this may be represent as follows :~ h Nuclear spin = [ve 0] Here Is spin quantum number of nucleus, whose value may be 0,1/2 , 1, 3/2 ete. ‘There are three important principles for nuclear spin . - wmSpectroscopy 3 (i) If sum of the protons and neutrons is even then Tis. integral [I=1,2,3 etc]. eg. 7H, 'yNete (Gi) I'sum of proton and neutrons is odd them lis half integral, that is I= 12, 3/2, 5/2. eg 1H, PFC SP gto, (iii) If both protons and neutrons are even numbered, I=zero eg. °C, '60 and give no NMR signal. Number of proton Odd Even Odd Even Number of Neutron | Odd Odd Even Brea spin quantum number | 12,3. 72,372, |~_172,372 0 Example 7H, FN Be tn, GF Bc, fo iP.gBr | Rs Therefore it is evident from the above table that those nuclei which have zero spin quantum number. ie, those nuclei which do not spin can not be observed by NMR Spectroscopy. For example !2C, '60, 32S etc. Whereas those nuclei whose spin quantum number is 1/2 show absorption signal in NMR spectroscopy, for example 'H, MB,YC,'5N,!9F77Al;1P."Sn These nuclei have magnetic dipole. Protons and neutrons have magnetic dipole and when sum of protons and neutrons is even number then they counter balanced magnetic effect of each other, whereas when it is in odd number then some magnetic dipole exists in it. sik] Theory of NMR spectroscopy ‘Nucleus having magnetic moment behaves like a tiny magnet and spins on its own axis, When proton is placed in an external magnetic field then it interacts ata certain frequency in radio frequency region as a result of which it attains two types of configuration with respect to external magnetic field, (i) Alignment with the field i) Alignment against the field If proton is in alignment with the ‘external magnetic field then it may absorb energy and goes to higher energy level and again it can come back to lower energy level (more stable) by releasing some energy. This transfer of proton from one energy level to other energy {Alignment against the field level is called as flipping. The energy necessary for this is AE~hv which is obtained from the radio frequency region of electromagnetic radiations. Difference of energy between two nergy levels depends on the applied magnetic field.Organic Chemistry B.Sc. Part-II aE = wate Ho of isotope of each Here 1 = Nuclear magnetic moment which is a characteristic property nucleus Hy = Applied field, By = Constant which is called as nuclear magneton, We know that, AE = hv Hy 9, [y= HBsHo Therefore hy = a or Y= aT - " In NMR spectroscopy generally strength of applied magnetic field is nearly 10° gauss therefore frequency of absorbed radiation for proton is = y = 279505104 x10# SEOs NO 10" This frequency is of the te chien aor By the absorption of radiation of such frequency in this region by the proton, changes its alignment withi the magnetic field. There is only one proton in hydrogen nucleus therefore value of its spin quantum ‘number is either + ¥: or -Y4 which will depend on the direction of spin of nucleus, for lower energy level E,=+1/21 H and for higher energy level itis Ey =~ 1/2 Hl therefore difference of energy between Tegions is AE=E,-E,= 1/2 wH(-1/2uH,) =H Here 1 = magnetic moment of spinning nucleus. According to Bohr frequency relation AE=hy therefore hy = Hy =4x107Hz BHo y= ho or ik (9 4) repress ts15.4 Expermental method of N. R. spectroscopy When a sample is placed between two magnetic field in which one is Stationary and other isa variable at radio frequency then asa result of interaction between sample and the electromagnetic waves, sample absorbs energy of a specific field Absorption of energy is obtained in the form of NMR signal by radio frequency detector and amplifier. Nuclear Magnetic resonance spectrometer is of two type :- (i) Low resolution instrument (ii) High resolution instrument fr RF transmitter} RF receiver and detector generator coil [ Recorder Fig, 1.1 Line diagram of NMR spectrometer Description of main components of an NMR spectrometer is as follows :- 1. Sample tube :- Smm.O.D. glass tube is taken as sample tube in general 60 MH, instrument, and in this 5 to 20 mg of substance is taken in 0.5 cc solvetit. Solvent used should have following properties :- f {6 Organic Chemistry B- aa ) It should not has any proton. (ii) It should be inert. (ii) should be non polar. (iv) Its boiling point should be low ated (v) Itshould be comparatively cheap. sty Senet Hence carbon tetrachloride is a suitable solvent. In addition ©° oa ‘of CHC ent chloroform CDCI, is used in form of solvent although toma TP quite gives very litle trouble inthe spectra. Carbon disulphide may 8° ee mote OE of 2, Sample probe :- Sample probe is a device which holds e i a} axis at the position in the field In this, sample tube is rotated at its !one!™ 7 " several hundred rotations per minute by an air driven turbine assumed ie 3: Magnet: Its main fetion sto produce magnets He pointto conse ‘magnet of length 30 om. ormore provides a field of about 4000 Le opencous AIS? for is that, the magnetic field produced by magnet should be ho! ae resolution work, magnet should be stringent. dareceiver coil art 4, Magnetic field sweep :- There is a transmitter CO aied sample tube thas been found that effective magnetic fick! OY + veging 2 rect ure rill gauss without any loss in the homogeneity of the field by © through these coils. ‘Although change in strength of magnetic field is itself ronisation is done with the recorder ations of a ae frequency transmitter :- Radio frequency calor rodnoes with 60 definite frequency. It works as a source coil. Generally radio frequency wok, Its joined MH, capacity is used. Frequency should be constant for high resolution Wor witha radio frequency receiver coil (RF receiver coil). witha detest aitisal é. Detector - Radio frequency receiver col is perpendicular 1° source ol andi around of sample tube. It detects the radio frequency ‘ga produced by ae in sample tube. Its joined with amplifier and recorder. fee ae rector rt fequeey signals re passed to recorder rough amplifier. Here a graph is obtained between applied frequency of radio frequency source and intensity of absorbance. Generally cathod ray oscillograph recorder is used. Procedure :- To obtain NMR spectrum of any compound, its 5-20 mg, quantity is dissolved in appropriate solvent (in which there is no proton) and is taken in asample tube. Sample ube is placed between the poles of an electromagnet. On al sides of sample tubes transmitter and receiver coils are present. Intensity of magnetic field isc jn such way by magnetic field sweap device that emitted frequency of spinning nuclei present in sample becomes equal to the radio frequency of oscillator and in this position proton absorbs energy. Rei cil gives ial tothe detector and NMR spevrumis obtained in records and linear with= Spectrum of ethyl bromide. ifr Chemical shift In NMR spectrum the number of signals give us information about different kind of protons present in the molecule. Similarly the position of the signal tells us what kind of Protons they are i.e. whether they are aliphatic or aromatic, primary, secondary tertiary or adjacent to halogen or oxygen or other atom or group etc. These different kinds of protons have different electronic environments and itis this electronic environment that determines where a proton absorbs in the NMR spectrum. When an atom is placed in outer magnetic field, electrons present in molecule come in spin state and produce their magnetic field. It is called induced magnetic field. This induced magnetic field may reinforce the applied magnetic field or oppose it. If induced magnetic field reinforces the applied magnetic field, a small external field will be needed for flipping the spin of the proton. Such a proton is said to be deshielded and the absorption is known as down field. If thé induced magnetic field opposes the external magnetic field, stronger magnetic field will be required to flip the proton. The proton is then said to be shielded and the -absorption is said to be up field, foe Hence by shielding or deshielding of protons by electron makes a change in the position of NMR signal in the spectrum, it is called chemical shift. Although the chemical shift is measured as a field or frequency, in reality it is a ratio of necessary’ change in the field to the applied field or of the necessary change in frequency to the standard frequency, hence is a dimensionless constant usually denoted by 5 and expressed in parts per million (ppm). Chemical shift is affected by induced magnetic field by the following two ways. * ()Ifinduced magnetic field vpposes applied majmetic field then effective magnetic field will be less then applied magnetic field. It is called positive shielding and NNR signal is shifted towards up’field. &R Organic Chemistry B.Sc. induced magnetic field Effective magnetic field Applied magnetic field metic field, effective magnetiC (Gi) If induced magnetic field reinforces the se ve shielding or deshieldin& field will be more than applied magnetic field. It's © and NMR signals are shifted towards down field. Induced magnetic field rtecive magnetic field Applied magnetic Held No displacement Down field shit Up field shit Chemical shift — Deshielding ‘No Shielding. Shielding To determine the chemical shift of different protons in a molecule a compound is used as reference. Itis generally tetra-methyl silane ie. TMS. f cHs-$i—cH, CH, Itis used as reference since itis - (i) chemically inert (ii) magnetically isotropic (iii) volatile in nature (b.p. 300K) (iv) soluble in common organic solvent (v) shows a strong ‘absorption signal which is at upfield in comparison to all types of protons since maximum shielding is found in its all equivalent protons, Hence TMS is used as an extemal reference. Displacement between the signals of targeted proton and that of TMS is called chemical shift. TMS may be used as external reference when D;0 or H,0 is taken as solvent. In the case of aqueous solution, sodium 22-day -2-llepentone5-sulpboane (DSS) is used as internal reference, Protons Which i ot l cI C1 C905 NE fa Spectroscopy is have same value of chemical shift show one type of absorption signal ie these protons are called equivalent protons. Protons which show different chemical shifts are called non- equivalent protons. NMR scale :- Unit which is used to denote chemical shift, is in parts per mullion of applied magnetic field. We know that chemical shift of any proton may be determine with the help of TMS. Chemical shift is generally expressed in terms of 8 scale. 6 value for any proton is equal at either 60 MHz or 100 MHz. Itis a dimension less unit. It does not depend upon frequency of applied radiation. By dividing v with the applied frequency and multiplied by 10°, Thus a peak at 60 Hz (v=60) from TMS at an applied frequency of 60 MHz ‘would beat 1,00 (5 scale). Hz /esc'600 640 480 420 360 500 240 180 120 60 1109 8 7 6 5 4 3 2 1 0\&opm) Fig. 1.4 NMR scale on 60 MHzand 100 MHz 108 =, 60x10 8 1.00ppm ‘The same peak at applied frequency of 100 MHz would be at v 100, but would still be a 51.00. 6 To0x108 <1" “6 100ppm Absorption signal of TMS is assumed as zero in this scale, ‘tscale:- In this scale, absorption signal of TMS is assumed at 10 and value of chemical shift of rest of the proton is in decreasing order to TMS. = 1000-5“4 Organic Chemistry B.Sc. Partin theilding, up field a 0 9 8 7 6 5 4 3 2 1° 9 Although chemical shift is measured in terms of either magnetic field or frequency but actually it isa ratio of applied field and desirable change in magnetic field or is the ratio of standard frequency and desirable change in the frequency. It is expressed in 6 ppm. 5 ppm= SHR x 108 HR Here Hy and H,.are magnetic field of sample and reference respectively. Spm Vs=YR 5106 MS Where vg and vp refer to the sample and resonance frequency. : Absorption signals of almost all important protons lie in between 0-10 in & scale, ‘Some important NMR absorption signals are as follows. Table 1.1 S.N. | Types of profone 8 | t 1 Cyclopropane A 98 2 Primary R-CH; 09 oa ‘Secondary R-CHy-R 87 ‘Tertiary R-CH-R 85 Allyl C=C-CHs 7 83 Benzylic Ar-CH 2,-3.0 78-10 ‘Amino R-NH, ese 9-5 Hydroxylic R-OH 1-55 9-45 9 Phenolic Ar-OH 4-12 6-(-2) 10. | Fluoride uc-F 4-45 6-55 Bromide uC-Br 25-4 15-6 | Acetylinic cec-H AlkenicSpectroscopy 15 a To] Aleohor AC= On 14 35-6 1s. | Ester HC-COOR 2-22 8-78 16. | Ether HC-OR 33-40 67-60 17 | Acid HiC- coon 2-26 80-74 18 | Chloride CH,-Cl 30 10 R-CH,-Cl 34 66 R,CH-Cl 40 60 19. | Aldehyde RCHO 9-10 1-0 Itis clear from the above table that as deshielding of proton is more, absorption signal obtained at down field i.e. value of 6 will be more. Factors affecting chemical shift Chemical shift ofa proton is affected by the following factors. (1) Electronegativity and Inductive effect: A proton is covalently bonded to carbon, nitrogen, oxygen or other atoms by a sigma bond in an organic compound. When compound is placed ina strong magnetic field, the electrons of sigma bond circulate and hence generate a small induced magnetic field which opposes the applied magnetic field. The shield experienced by a given nucleus is directly related to the electron density surrounding it. A nearby electronegative atoms thus withdraws electron density from the neighbourhood of the proton, so that, a small applied field is required to cause the spin state of the proton to flip. The higher is the electronegativity ofthe atom, greater isthe deshielding caused to proton (i) This effect is illustrated by the 6 values for the protons in methyl halide. CH,-F CH,-Cl CH,-Br CH,-1 5426 30 2.82 216 The order of electronegativity of halogens is as follows. F>CI>Br>1 Since F is most electronegative, hence it will greatly deshielded the methyl protons hence absorption signals observed at downfield i.e. at high 5 value. The order of electronegativity of carbon, nitrogen and oxygen is as follows. Cc=C-H << Cyn «< nett Vinylic | Aromatic Aldehydic Sppm 46-59 6.0-8.5 9.0-10.0 halogen atoms. As aresult ofSpectroscopy re (2) vander Ws 's deshielding - Some Protons come in stearically hindered state in molecules having more substituted ‘sroups, Electron cloud of bulky group repell the electron cloud of proton and proton get deshielded, thus proton gives absorption signal at down field or having high 8 value Itis called vander Waal’s deshielding Applied magnetic field ()Effect of magnetic anisotropy or space effect - Circulation of electron, especially the welectrons, about nearby nuclei generates an induced field which can either oppose (causes shielding) or reinforce (causes. ‘deshielding) the applied field at the proton, depending upon the location of proton or the space occupied by the protons. The occurrence of shielding or deshielding can be determined by the location of proton in the space and so this effect is known as space effect. It is of two types. (2) Diamagnetic anisotropic :- When applied magnetic fieldis in opposite direction to applied field thats it opposes it then proton gets shielded and NMR absorption signal will be observed at up field that is at low 6 value, For example: Alkyne (b) Paramagnetic anisotropic :- When induced magnetic field reinforces the applied magnetic field then proton gets deshielded and NMR absorption signal are obtained at down field that is at high 6 values, For example: alkene, benzene, aldehyde ete. It can be explained by the following example - (D Alkyne :- It's typical example is acetylene, Molecule of acetylene is linear and it’s triple bond is symmetrical with the axis. If applied magnetic field and axis of acetylene molecule are in similar direction then electrons of triple bond, spin perpendicular to applied ‘magnetic field as a result of which induced magnetic field opposes applied magnetic field that is shielding of proton takes place and NMR absorption signal is observed at up field that is at low 8 value. Hy REN, Applied — Sys! magnetic ——> field ‘ Inc magnetic lines Spinning x of forces electron Fig, 1.5: Shielding of acetylinic proton i.e, diamagnetic anisotropic effect18 Organic Chemistry B.Sc, p,, rey (AI) Alkene - Geometry of alkene molecule is such that double bond is onthe agg plane and hydrogen atoms are perpendicular to the plane. If applied magnetic fold ic perpendicular direction to double bond then x electrons of double bond spin at right any and due to this induced magnetic field, @ opposes the applied magnetic field Hy with respect to carbon atoms, i, diamagnetic anisotropic effect. @ reinforce the applied magnetic field Hy with respect to hydrogen atom (proton) there is a paramagnetic anisotropic effect. With the result of this proton get deshielded and NMR absorption signal is obtained at dovm field ie. at igh 8 value Applied magnetic field Fig. 1.6: Paramagnetic anisotropic effectin alkene (III) Aldehyde :-Like alkene in aldehydes also, induced magnetic field, Applied magnetic field Fig, 1.7; Paramagnetic anisotropic effect in aldehyde (i) opposes the applied magnetic fieldwith respect to carbon atom | that is there will be diamagnetic anisotropic effectSpectroscopy 19 ec (1) with respect to hydrogen atom (proton) it reinforces the applied magnetic field ie there will be paramagnetic anisotropic effect due to which proton will be deshielded and NMR absorption signal will be observed at low field i.e at high 8 value (IV) Benzene -In benzene x electrons that is aromatic sextet is delocalised on the ring. In the presence of applied magnetic field, ring electrons produce ring current as a result of spin. This induced ring current produces — (i) diamagnetic anisotropic effect at the center of aromatic ring and (it) paramagnetic anisotropic effect at the surface of aromatic ring. Therefore aromatic protons get deshielded and NMR absorption signal is observed down field whereas protons situated above and below the plane of aromatic ring gets shielded as a result of which NMR signal is observed at up field. Applied magnetic field Fig. 8 : Ring current effectin benzene (V) Acetophenone :- In acetophenone due to ring current effectin aromatic rng, protons of ring resonate at low field (high 6 value). Ortho protons of ring show partially down field shift because additional deshielding takes place at ‘ortho position due to carbonyl group. Therefore ortho, meta and para protons show absorption signals at 8, 7.85, 7.40 and 7.30 ppm respectively. According to the figure Fig. 1.9 Shielding (+) and deshielding (-) 1.9, carbonyl group and benzene ring region of acetophenone are coplaner.20 Organic Chemistry B.Sc. Part-yy ee ee main magnetic field (H,) is perpendicular tothe plane of molecule then x electron o¢ >C=0 group — (i) shields conical zone above and below and (ii) deshields the lateral zone due to which both ortho proton gets deshielded equally. (VI) Annulene :- At low temperature proton situated at the outside of annulene ring are highly 7 deshielded, whereas proton situated inside the ring are strongly shielded. Asaresult of which 8 value obtained at 9.3 and 0.3 ppm, respectively, Fig. 1.10 shielded and deshielded protons of annulene [14] (3) Hydrogen bonding :- If hydrogen bond is present in any compound then NMR absorption signal is observed at down field in these compound. In hydrogen bond proton is attached to more electronegative atom due to which electron density at proton reduces as a result of which proton gets deshielded and absorption signal is observed at down field that is at high 5 value. Chemical displacement towards low field depends on the strength of hydrogen bond. Since no displacement in the NMR signal is observed in the molecules having intramolecular hydrogen bonding hence on the basis of it, intramolecular hydrogen bonding may be differentiated from intermolecular hydrogen bonding. For example, chemical displacement of N-H proton in amine and O-H proton in alcohol depends on the nature of the solvent and concentration of solvent. For example, hydroxylic proton of pure alcohol gives an absorption signal at 5 5.35ppm whereas in the presence of nonpolar solvents like CCI, at 5-20% concentration hydroxylic proton gives absorption signal at 5 2-4ppm. Inthe ‘same way at infinite dilution or in vapour state, absorption signal is obtained at 8 0.Sppm. Chemical displacement of hydroxylic proton depends upon temperature and concentration, this can be explained with the help of hydrogen bonding. In concentrated solution proton is deshielded by hydrogen bonding and shows down field resonance but when dilution of amine and alcohol is done with non-polar solvent then hydrogen bond becomes weak and shows resonance at high field i.e. at § value (2-4 ppm). In pheaol, absorption signal is observed at 8 4-12 ppm but on increasing dilution with non-polar solvent absorption signal of -OH proton shifts at upfield. Inenol-form of acetyl acetone, due to intramolecular hydrogen bonding a six member! ring is formed as a result of which absorption signal of hydrogen bonded proton is obsene! at high down field.Spectroscopy 21 (b) (a) Hyper bo, at = to, (c) (d) 4.9, =55, d= 1.98 Inthe same way salicylic acid gives absorption signal at 6 10-12 ppm showing high Aeshielding effect due to intramolecular hydrogen bonding, as & In concentrated solution of amine, protons show absorption signal at low field, this is duc tothe presence ofinter-molecular hydrogen bonding because intermolecular hydrogen bonding depends on the concentration therefore on increasing dilution, bond becomes ‘weak and proton gives absorption signal at high field ie. atlow 8 value. When concentration reduces, intermolecular hydrogen bond becomes weak andparamagnetic effect diminishes. (4) Temperature :- Effect of temperature on chemical shiftis very less, Proton of -OH, ~NH and -SH which take part in hydrogen bonding shows a shift of absorption signal towards up field on increase in temperature because on rise in temperature hydrogen bond ‘becomes weak and diminishes finaly, ppm a=1.98, Spin-Spin Spliting or Spin-Spin Coupling Interaction between spin of one proton to its neighbouring proton or protons is called as Spin-Spin Coupling, Effective area around a nucleus decreases or increases by ‘neighbouring protons. For example : Inthe spectrum of ethanol, absorption signal obtained by proton of methyl group, is effected by the magnetic field produced by proton of methylene ‘group. In the same way proton of methylene group is effected by the magnetic field produced by proton of methyl group. As a result of this absorption signal of methyl and methylene Sroup undergoes splitting, Signal of methyl protons is obtained in the form of triplet and ‘signal of methylene protons is in the form of quartet in 1:3 3: | ratio. To express spin-spin splitting, milligauss or hertz scale is used. From the high resolution spectrum of ethyl alcohol itis clear that NMR absorption Signal shown by methyl and methylene proton, splitinto extrapeaks therefore that secondary environmental effect which is imposed on chemical shiftis called as spin-spin splitting. In the effect of applied magnetic field chemical shift is produced as a result of spin of electron around the nucleus. Due to spin of electron a small magnetic field is produced which generally opposes the applied field due to which for the resonance of shielded nucleus a partial change occurs in frequency or field. Im the effect of secondary environment, one set of nuclei affects the resonating behaviour of other set of nuclei that is interaction or coupling takes place between two22 Afferent sets of protons or Now we wil study he eet of protons of metal 7 7 ng potesule can snow Proton in ethyl alcohol, Two protons of methylene grouP prese! following four types of arrangements due to theit sp!" \ —H, — Applied magnetic field , Inthe first type of combination both methylene proton are pal oa the applied magnetic field. In second type of combination paired Poca other. If spin i where as in between both types of combination spins are opposite 0 ig field on mety paired and is in opposite direction to external field then applied mae ation absorPtion ‘proton reduces hence high field is required for resonance and in this signal will shift to high field. ‘There are three types of protons in ethyl alcohol = —., ,d and alige ® © 2 CH;— CH, — 0H psonion “Thus inthis type of spectrum three absorption signal should be obtained. eae signal of methyl proton due to spin-spin coupling with methylene proton SPI its triplet where as methylene proton due to spin-spin coupling gives quartet. Similarly a doublet and a septate is observed in 2-chloropropane. cH,-CHCI-CH, @ ® @ Methine proton on spin-spin coupling splits into (6+1) =7 peaks and gives septate whereas methyl protons splitting (1+1=2) gives doublet. FK:) Spin-Spin Coupling Constant Coupling constant is a measure of effectiveness of spin-spin coupling Distance between the centre of two nearby peaks of a multiplet is constant and this is called as spin-spin coupling constant. (@ Coupling constant is shown by J. Its unit is cycles/sec. or Hz. (i) Value of coupling constant does not depends on the applied magnetic field (iv) Value of coupling constant is generally between 1-20Hz. (¥) Value of coupling constant depends on the structural interrelations between the ccouppled proton. (vi) Value of coupling constant of cis proton is less in comparison to coupling constant trans-proton in geometrical isomers. (vi) If chemical shift of two interacting group of proton is more then analysis of NMR spectrum become easy. In other words if then its analysis becomes easy This is called as first order NMR. Forexample in ethyl alcohol, absorption signa of ‘methyl, methylene and hydroxyl protons are observed in the terms of triplet, quar ,.ll‘Spectroscopy 23 and singlet, respectively, Here the value of coupling constant of peaks of methyl and ‘methylene proton is more than 7, where the difference between both the peaks is 140 ited [£57] then analysis of NMR spectrum becomes complex and this iscalled as complex order NMR. For example in complex order NMR of ethyl alcohol absorption signal of methyl, methylene and hydroxyl proton are observed in the form of triplet, Ooctate and triplet, respectively (oat) With the help of coupling constant (J) we can be differentiate - (a) between a doublet and two singlet (b) between two doublet and a quartet. ‘With the change in radio frequency ifno chang is observed between two peaks them it will be doublet. On the other hand if change between two peaks get observed with the change in radio frequency then they may be two singlet, Similar differentiation may be made between two doublet and one quartet Coupling constant may be explained withthe help of following examples ) GG Inthe NOR spectrum of this compound two absorption signals will be obtained. Due tothe effect of wo equivalent Protons (a), proton (b) will give a triplet. Distance between two neighbouring peaks in this triplet is same. This distance between two adjacent peaks is called as coupling constant relative ratio ofits three peaks is 1:2:1, (2) DCC, - In the NMR spectrum of this compound two absorption signal will obtained. Due to the effect of three equivalent Protons (a), proton (b) will give quartet and in this relative rato of four peaks I He 6) at i Cl ori. 1,1,2trichloroethane -In the NMR spectrum of this compound ‘two absorption signals will be obtained. Value of coupling constant in each will be constant. ‘Due to the effect of two equivalent proton (a), proton (b) will give triplet whereas due to the effect of proton (b), protons (a) will give doublet,: ry BSC. Part-I ist (4) Gig Cit, Chi, Three absorption sigmals a6 observ’ ngives propyl iodide. Due to the effect of proton (a) and (©) (5 protons). Pr Jyy76:8H2 Type T3HE 2.1.86 and 3.17 PP Absorption signals of (a) (b) and (c) are obtained a ingretive se 12:1 respectively. Due to effect of proton (b), proton (a) will ve tiple! ing constant are Jp The same type of splitting is observed for proton (c), value of their coup! fe taherefF© s 68 Hzand J, =7.3 Hz, respectively because the value of nang proton which Fe the methylene proton (b) is effected by the effect of such two groups magnetically non-equivalent. Thus number of absorPu9 be (341) (241)=12. Inhigh resolution spectrum this ‘yPe off low resoluti served for methyl Equivalent proton doesnot undergo interaction with each other is observed. For example in ethane, benzene, etc al rotons ET only one singlet is obtained but in 2-butanone i-h,-co-Shs tree pss oF are there and these are non-equivalent. There is mo Pro jacent carbon ae (a) types of proton hence there will be no splitting in absorption signal but OF carbon atom adjacent to (b) type of proton’ there are three protons, therefore acc ding to (ot) raf it will give a quartet whereas atom adjacent 10 (©) type of protons, two protons are present and therefore it wil give triplet. ooo ‘ {@) Incl Cy CH, methylene proton are affected by two equivalent methyl protons therefore septet will be obtained but if in a compound: proton is affected by non- equivalent protons, for example: if proton of atom (B) are effected by rotons oftwo sane Aand C which are non-equivalent then absorption signal of atom (B) will split into (, + 1) (ng + 1) peaks. Here ny and ne are the number of protons of A and C respectively. (xi) Relativeratio of peak area of a multiplets as follows (a) doublet 1:1 (b) triplet 1 = (©) quartet 1:3:3:1 (d) quintet 1:4:6:4:1 (@) sextet 1:5:10:10:5:1 (O)heptet, 1:6: 15:20:15 :6:1 Multiplicity of ion si i ‘ho oe iplicity of absorption signal and relative peak area ratio can be shown by PascalSpectroscopy 25 a ——SSSS—SSSC“‘“‘$$ OS Singlet 1 Doublet tw Triplet bt Quartet 13:31 Quintet 1:46:41 Senet 1:5:10:10:5:1 Hepret 1:6:15:20:15:6:1 Maltiplet Quintet Quartet Triplet Doublet Singlet Fig.1.11 ‘The technique in which proton or group of equivalent protons is irradiated with radio frequency in such a way that coupling effect Produced by adjacent proton is completely finished is called as spin decoupling technique. Determination of Peak Areas Area under NMR absorption peaks is proportional to the number of nucleus responsible for that peak. There are many such factors which broadens the absorption signal due to which its height reduces. Some important factors are purity of sample, chemical exchange reaction and spinning rate of sample etc. In modem NMR recorder there is a electronic integrator which directly shows relative peak area of the absoption signals. For example a curducn @ & oo @ CH; CHy-Cl (CH CH, CH-CI CH CCH a:b=3:2 a:b:c=3:2:2 a:b=6:1 2NMR signals 3NMR signals 2NMR signals With the help electronic integrator in NMR spectrum a ledder like curve is obtained the length of this ledder like curve is noted and with its help the number of different types of Proton can be calculated, for example, NMR spectrum of isopropyl chlorideis as followsaa Fig. 1.12 n comes out s type of ledder like curve ratio of relative peak area of a and b provon 7 to be 7.2 : 1.2 which is equal to 6 : 1. On the basis of it, number of proton present oe 4 mea can be calculated. In the same way ratio of relative peak area of p-tert buty lene is 8.8: 2.9: 3.8 which is equal to the ratio of number of protons i.e: 9: ee © CH; © © HC CH; a:b:c= wre © (iieeemieene certo enSpectroscopy 27 [Bz poiications of nuclear magnetic resonance spectroscopy With the help of NMR spectroscopy, (1) Functional group present in any compound may be identified (2) Types of nucleus present in any compound may be calculated. (3) Relative position of different groups present in compound may be known. (4) Mechanism of the reaction may be studied. For example, in the aqueous solution of sodium hydroxide, exchange of OH proton of H,O and NaOH takes place in less than 107 seconds as a result of which only one absorption signal is obtained. If proton exchange takes place in more than 10-* seconds ie. exchange is relatively slow then in this condition two absorption signals are obtained. NaOH +HOH* === > NaOH? + HOH (5) On the basis of resonance of protons, unknown compound can be identified. The attachment of proton with oxygen or nitrogen or sulphur atom in the compound can be identified as follows = (When protonis attached with oxygen: (a) Alcohol for example, ethanol Cli, Cit, OH In pure alcohol proton attached with oxygen atom shows absorption signal at 3 ppm while inthe presence of carbon-tetra chloride OH absorption signal observed at 4 ppm. On changing the solvent change in the value of OH absorption signal can be explained on the basis of hydrogen bonding. In pure ethanol due to hydrogen bonding, clectron density around the proton decreases because shared electrons are attached by the more electronegative oxygen atom and due to decrease in electron density proton gets, deshielded and -OH absorption signal shift towards low field that is at high 8 value. In presence of non-polar solvent such as CCl, inter-molecular hydrogen bonding reduces and atmore dilution it diminishes, As a result of which electron density around proton becomes ‘more i. due to shielding effect absorption signal shifts towards high field i. at low 8 value. (b) Water :- In most ofthe solvents or compound water is present as an impurity. If quantity of water is more i.e. if there isa possibility of hydrogen bonding than absorption signal is observed at 8= 4.7 ppm. Ifittle quantity of water is present then it gives absorption signals at 5 1.5 ppm. (©) Phenol :- Like alcoholic proton, phenolic protons give a sharp singlet due to fast exchange. Range ofthis peak depends on the solvent concentration and temperature and its absorption signal is observed at 8= 4.0-7.Sppm. (4) Carboxylic acid :-In the presence of nonpolar solvent carboxylic acids are obtained in he form of dimer. Due to this reason absorption signal is obtained at 8 = 10-13ppm but in presence of polar solvents dimerisation of carboxylic acid does not takes places, Carboxylic(Pee __LL— 28 oo emistey BSS Perey nn let position, whi ae fst exchange with the protons of water or leah! °° ioe . on the concentration of acid. xchange of hydrogen (1D When proton is attached with nitrogen = Ifthe =¥° es and neighbouring Present on nitrogen is fast then NH proton willdecouple with HOE! orodion signal carbon atom and will give a sharp singlet and this also does 2°* split ee ene of proton present on carbon atom. Irate of exchange ofhySO8=™ 807 PS sbeorpion toms moderate then NH proton wil decouple partially and Wi EE 8 very Low Peak but this does nt spit the neighbouring CH proton but if xn ea on then NH absorption peak will be broad as well sit will couple with P neighbouring carbon atom for example :- In CH;-N-C-0 CH, CH; (®) NH proton gives a broad absorption peak at 5= 5.16ppm = (i) On coupling wth NH proton CH, prota, wl gvea doublet at 5 =? hue (iii) Whereas ethony protons will give a quartet and triplet TesPOSt (1 Amine in aliphatic amine NH proton gives an absorption peak at = 3. re of the Position of peak and its splitting depends on concentration, temperature and natu a 0-8.0ppm, (I) Amide :-In amide absorption signals are obtained at aa TRE] erreaion oft spocrum of some important compounds | and (1) Ethyl bromide :- In ethyl bromide two types of protons 1. methyl protons tained. methylene protons are present. Therefore two types of NMR absorption signal are o Fie eo fofe Be HOH: 100 0 oe eee CH;CH,Br @ @) 3H Gh + L 1 ri 1 L wb 6 5 4 3 1 Fig. 1.13: NMR spectrum of ethyl bromideSpectroscopy » Absorption signal of three equivalent protons of methy! group due 1 the efecto methylene proton will spit into n+ Lie. 2+ |» speaks and wll give triplet at 1 Tope whereas absorption signal oftwo equivalent protons of methylene group will split due tothe effect of three protons of methyl group into 3 1 = 4 peaks ¢ a quartet will be oes “ano of relative intensity of methyl group is | 2. Land for methylene soup is With the help of electronic integrator it can be observed that it has only one methyl group and one methylene group because ratio of ledder like curve is 3.2, respectively (2) Ethanol :- There are three types of protons in ethanol therefore inthis three types of absorption signal should be obtained. Due to the deshielding effect of oxyyen atom OH proton absorption signal will be observed at low field. at high 8 value, after this signal of methylene proton and at high field with respect to it absorption signal of methyl proton will be observed. With the help of electronic integrator ratio of intensity of absorption signals is calculated. Due to effect of methylene group, three equivalent protons of methyl yroup gives a (n+ 1)=3 triplet. In the presence of the impurity of acid or moisture in alcohol singlet is observed due to -OH absorption signal whereas in pure alcohol -OH absorption signals is observed in the form of a triplet |: 2: | because -OH proton coupled with the protons of adjacent methylene protons and will spit into (2 + 1) = 3 absorption peaks Methylene proton gives a multiplet having eight peaks, 500 400 300 200 100 ° 1 L L 1 He CH;CH,OH @ &) © ae 7 6 5 4 3 2 4 oO Fig. 1.14 NMR spectrum of pure ethyl alcohol Methylene proton is affected by the protons of two different chemical environment therefore multiplicity of methylene proton willbe affected by (CH, and -OH) protons therefore multiplet of methylene proton will be (n+ 1)(m +1) ie. (3 +1)(1+1)=(4)(2)=8. Le. eight peaks will be obtained in multiplet. (3) 1,1,2-Tribromo ethane :- There are two types of protons in this compound, iB: -CHoBr one bromine atom i attached to proton (b) while two bromine stoms are attached to proton (a) to the high electronegativity of bromine deshielding of (a) typeistry BS. Party, eres bsrtey 30 Proton will more c therefore signa i observed at ow field (i.e 8 Signal of (b) types of proton is obtained at comparatively hil 200 a 500 400 300 00 Hy CHBry- ChB @ = 8 7 6 5 4 Fig. 1.18: NMR spectrum of | tribromoethane ; Absorption signal of (a) types of protons will split into (2 * 1) = 3 peaks and will gives triplet at §= 5. 7ppm whereas absorption signal of two equivalent proton of (b) will be split into (1 + 1) =2 by the spin-spin coupling of proton of (a) type ie. a doublet is observed at 34.1ppm. (4) Ethyl acetate - There are thre types of protons 2 ethyl acetate. & ®» : hy C-OCH-CH qo pe ite © ® @ Cis €-OCHCHs oO Fig. 1.16: NMR spectrum of ethyl acetate ‘Absorption signal of methyl proton (a) splits due tos spin- it proton into (2+ 1) = 34. triplets observed at 8 ee erat group is effected by the chemical environment of methyl group and pee €Spectroscopy 31 coupling to give quartet whereas methyl proton (c) gives singlet. In this triplet, singlet and quartet are obtained at 8 1.2, 2.0 and 4 Ippm, respectively. (5) Toluene :- In the NMR absorption spectrum of toluene two absorption signals are | obtained. CH; so Phenyl protons due to deshielding effect shows a singlet at 6 7.0ppm whereas methyl proton gives a singlet at § 2.2ppm. 600 500 400 300 200 100 : G Sr ) 7 6 5 4 3 2 : 7 Fig, 1.17: NMR spectrum of tolueneic Chemistry Bge pp 32 _Organis Panny ee idation of simple PEP] Problems pertaining to the structure elUc pic techniq £0} ues Compounds using UV, IR and NMR spectros: We have already studied IR and UV spectroscopic techniques in Previous classe, q brief summary related to this is given here, Structure of some simple oe compounds can be determined by IR, UV and PMR spectroscopy techniques by the following manner, () Alcohol: (A) Indentification from infrared spectrum - Thereare four types of bands are observed HH in saturated alcohol ni Coul |, C-C,C-H,C-0 and 0-H. Out of these two absorption HH bands are important. () O-H stretching vibrations :- Position and intensity of this stretching vibration ‘depends upon the hydrogen bonding. Possibility of hydrogen bonding in gaseous or dilute solution state is negligible. Hence-OH group is free from hydrogen bonding. In this state a sharp absorption band is observed at 3650-3584 em, In liquid or concentrate solution state of alcohol due to hydrogen bonding association takes place as a result of which absorption band is observed at lower wave number region 3584-3200 cm". Band width and shifting towards lower wavelength region depends on the extent of hydrogen bonding. (i) C-O stretching vibrations: A sharp absorption band is observed in the region 1260-1000 en! due to C-O stretching vibrations in alcohol and phenol, C-O stretching vibrations couple with neighbouring C-C stretching vibrations as a result of which nature ‘of alcohol affects the absorption bond, For primary alcohol itis observed at 1075-1025 cm! for secondary at! 150-1075 em and for tertiary alechol absorption bands appears at 1210- 1100 em" Wayolengh, ym 5 78 9 1012 45 101 3300" 30602600 2000 1800 1600 1400 1200 1060 aoo"“ag ‘Wave number, om! Fig 1.19 Infrared spectrum of -butanotSpectroscopy _ on Sey to it absorption of C-C stretching vibration is found in the region 1500-1600 ; Value of C-H stretching vibrations depends on the hybridization of carbon atom. In C(sp° ')-H bonds absorption bands due to stretching vibration is observed at 2800-3000 em’. In C(sp)-H it is observed at 3000-3100 cm-! and in C(sp)-H bands it is observed at 3300 cm”. Bending vibrations of C-H bandsare observed in comparatively at lower wave number region at about 1430-1470 cm-! Due to the presence of unsaturation and aromatic ring, absorption band due to C=C bond is observed at 1650 cm~! Infrared spectrum of 2-butanol is shown in fig. 1.19 In this due to hydrogen bond O-H stretching vibration absorption band is observed at 3300 cm". C-O stretching vibration absorption band is at 1100 cm“! which suggests the Presence of secondary alcohol. All C-atoms are in sp? hybridized state due to which (sp*-H) stretching vibration absorption band is observed in between 2800-2950 cm-! (B) Identification with UV spectrum :- Saturated alcohols show n —>0* transitions. In alcohol, in addition to presence of 6 electrons atom lone pair of electron is also present on oxygen. In comparison to o —> o* transition low energy is required for n — o* transition as aresult of which absorption bands are observed at higher wave length region in comparison to o —> o* transition. Generally saturated alcohol shows absorption band at about 200nm wave length region. For example, methanol shows absorption band at 184nm with Emax = 200 Absorption band in ethanol is observed at 200nm. Due to this reason only, for the structural determination of compounds, alcohol is usedas a solvent in near ultraviolet region. In phenol, ethelenic band (E-band) is observed at 210.5nm, € max 6200 and benzenoid band (B-band) is observed at 270nm, Emax 1450. : Catichol (OH-C,H,-OH) gives ethelenic band at 214nm, band at 276nm € ax = 2300. cm = 6300 and benzenoid Emax