Physical Chemistry CHEM 3510
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This 7 page Class Notes was uploaded by Mr. Clementine Gottlieb on Wednesday October 21, 2015. The Class Notes belongs to CHEM 3510 at Tennessee Tech University taught by Staff in Fall. Since its upload, it has received 24 views. For similar materials see /class/225695/chem-3510-tennessee-tech-university in Chemistry at Tennessee Tech University.
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Date Created: 10/21/15
CHEM 3510 Fall 2003 Nuclear Magnetic Resonance NMR Spectroscopy 7 We Will focus our attention to the smallest nucleus the proton 7 Similar to the electrons the same nuclei have anintrinsic spin angular momentum I The nuclear spin eigenvalues equations for proton are analogous to the electron ones A 1 1 A 1 1 12a 1h2a 12 1h2 22 1 B 2 2 B 1 011701 11137515 Z 2 Z 2 7 This intrinsic spin angular momentum confers the property of a magnetic dipole The dipole moment is related to the spin angular momentum by gNIgN NII Where gN nuclear g factor a constant close to 1 characteristic to each nucleus mN mass of nucleus BN nuclear magneton y magnetogyric ratio characteristic to each nucleus the higher it is the easier to observe the nucleus in NMR 7 Placed in a magnetic eld the nucleus or electron act like a small magnet 7 What happened With the energy levels in the presence of a magnetic eld 7 The presence of magnetic eld raises the degeneracy of the energy levels 7 The energy of the two levels that have m1 rl 2 degenerate in the absence of magnetic eld changes as E 7hymIBZ Where m1 rl2 for proton BZ strength of magnetic eld units are Tesla T and gauss G Where 1G 10394T 7Energy difference between m1 12 level and m1 712 level Held off Field on for the case of proton AE Em1 7127 Em1 12 71sz 5 AE depends on the strength of the magnetic eld CHEM 3510 Fall 2003 7 Example the energy difference between the levels for a magnetic eld of 21 100 gauss 211 T the energy of the radiation required for a transition is AE 1054 gtlt10734 J s rad391267522 gtlt107 rad T391 s 1211T AE 595 gtlt10726 J m 36 gtlt10396 kJmol m 30 gtlt1074 cm39l AE h V 3 V 90 MHZ radiofrequency region of electromagnetic radiation 7 There is a direct correlation between the magnetic field strength and the frequency of radiation called resonance frequency 732 27239 AEhvcgtv HzcgtayBZrad391 7 Example What magnetic field strength must be applied so a free proton will have a spin transition at 60 MHz 3 BZ 14100 G 7 7 NMR spectrometers can operate at fixed radiofrequency resonance frequency and variable magnetic field or at fixed magnetic field and variable frequency of radiation The spectra is not dependant on which setup is used 7 Early NMR spectrometers worked with magnetic fields of approximately 14000 T that sets the frequency of proton transitions of about 60 MHz These days the NMR instruments can work at 750 or even 900 MHz Highfrequency instruments are preferred because they give greater resolution 7 Not all protons resonate at the same frequency The external magnetic field is shielded by other magnetic fields generating by surrounding electrons The actual magnetic field acting on the nucleus is smaller than that of the external magnetic field 7 The electronically generated magnetic field B 8180 is proportional to the applied magnetic field Belec OBO 7 The sign is negative because the electronically magnetic field opposes applied magnetic field 7 The proportionality constant 039 is called shielding constant 10 5 and it depends on the electronic or chemical environment around the nucleus 27w a 333 O39Bl O39B DB 2 z 0 0 0 0 y1o ylo CHEMKSIEI FallZEIEIK e The resmance frequmcy ufa H emst r y i 17 a H I H e The hesdhenee heduehey depend dhme quot331811 eld quhexnslxummtBn sh dafme 351qu mu retard Mam heduehey rMasweLheresunance 39equency arext anmlanalrefa39enceurstandard tmmmhylsdan TMS SD n wm hm he msn39umenl em rTmmmmhylsLlane Sim13 has 12 equivalmthydrugmatumsxsrelauvely hummve an ugen 2mm absurh 2mm elds u erhydmgen mm m urgam cumpuunds5u TMS 53131 eppm enhe edge unhe 35113 e The eumpansuh wnh hewemu the sundast daneusmgthe charmml 5 5H m enhsgmmm Vipeme epphed Haydn eld 7 5 54 Mun erinw mms Vipeme 2 yrpemdmm rBd Vin b t v 176 u 2 H 2 5 752 5 750x10 rExamples if CHgil Hi OCH3 I CHEM 3510 Fall 2003 7 Which of the protons from above are best shielded Better shielded 5 lower 039 smaller electronically generated magnetic eld 5 smaller external eld required 5 smaller frequency of resonance frequency 5 smaller value for chemical shift Answer H in CH3I than H of CH3 in HCOOCH3 than H of HCO in HCOOCH3 7 The greater the electron density around a nucleus the greater the external magnetic eld required to produce resonance the smaller the chemical shi and the more up eld the resonance Will occur 7 If the hydrogen is bonded to a carbon also bond to a halogen then the resonance occurs down eld CH4 CH3C1 CHZClZ CHCl3 5 023 305 533 726 7 The more electronegative the halogen the further down eld the resonance occurs CH31 CH3Br CH3C1 CH3F 5 216 268 305 426 7 The further the halogen the more up eld the resonance CH3 7C1 CH3 7CH2C1 CH3 7CH2CH2C1 5 305 142 104 J 7 Similar behavior is found for other electron 5 Withdrawing groups like the oxygen containing derivatives like alcohols aldehydes ketones esters etc to 7 The relative area of the peaks is proportional to the number of protons H atoms This can be done automatically by the NMR spectrometer no chemical shm relative to OH 4 1 Brl C 2 25 3 5 Electmnegatlvity of halogen CHEM 3510 Fall 2003 I a H resonances GE M l Magnellc held v nmg current 5 5 no m m ou mo 0 7 Besides the effect ofextemal magnetic eld and 1 ex u a h the electronically generated magnetic eld one I 39 quot39 quot 1 i should also consider the effect of the spin of l neighboring H nuclei that leads to splitting of the L quot9 Ll l 7 The interaction between nuclear spins is called spinspin interaction The effect of signal in multiplets doublet 7 one neighbor triplet 7 two neighbors spinspin coupling is proportional to the spin angular momentum of both nuclei The proportionality constant is J12 called spinspin coupling constant J12 has units of Hz 7 The energy levels for 2 nonequivalent protons can be obtained using perturbation theory 1 2 597 mam 77 M MIWZ 597 f Brag 55 111 M 55 7111 M r own 5 7 7 f an 7a wgt WW mama 59an Zara 7J 9 qu 7 The resonance frequencies are given by J J vfvol70391i vfvol702i 7 Energy difference between the two doublets is proportional to V0 so higher V0 separate better the doublets CHEM 3510 Fall 2003 7 The results from above are obtained if vola1 7 all gtgt J12 Such system is called an AX system and the spectrum is called rstorder spectrum 7 For these types of spectra J m 5 Hz so multiplets separation should be much bigger than 100 Hz although typical values are about 100 Hz 7 Designation of spectra 7 Use A B C for each nonequivalent hydrogen atom Example A3 B2 etc depending on of atom 7 Use X Y Z ifthe chemical shi s are very different from the A B or C type hydrogen atoms 7 Example If there are two protons for Which 135 WTOZ EKZmquot J12 m vola1 7 all then one has an AB system Mr 7 The case of chemically equivalent protons 7 The spinspin coupling cannot be observed 7 Example CHZCl2 has only one signal a singlet This is an A2 system 7 Only 1 7gt 3 and 3 7gt 4 transitions are allowed and both have the same resonance condition 7 The singlet states are going up When spinspin interaction is included While the triplet is going down 7 For the rstorder spectra J12 ltlt V0 lo 1 7 all the number of peaks for each line is given by n 1 rule The number oflines in a signal is given by n 1 Where n is the number of neighboring protons neighboring protons protons separated by 3 bonds 7 Example CHCl2 7 CHZCl 7 Example CH3 7 CHZCl No of lines 1 2 3 4 Name singlet doublet triplet quartet Relative intensity 1 39 121 l33l Idealized pattern CHEM 3510 Fall 2003 ll 0 Vn1Ul Uzl secondrorder speetra eTlne n 1 rule does not apply The problem ls solyed uslng yananonsl metlnod e The case of 2 nonrequlvalent protons rThe Hamrltonran 11 423nm 5011 yBuGr mil a eTnere are 4 5pm Wave funenons for tlne unperturbed system 4a L1Lk2 4 2 M0499 4 3 1d2r 4 1 2 Mm A 2 Il 12 eUse allnear eombrnanon wc agt s24gt2 s34gt3 54 eMlmmrze energy STE 0 obtaln a secular determrnant rdrd2hJ4rE 0 0 0 0 rddzth475 hJ2 0 0 hJ2 drdrhwmg 0 0 0 0 1 421 Jew475 Where d hvn rq and 12 hvn rdz hvn1J JZ 312 3 2 4 4 t2 V02Ul 5221211 2 U 5 h 715 52 121 3954 mar 2 7 2 2 4 iThe allowedtransltlons are 1amp21amp3 2amp4 3amp4 for nonrequlvalents H 1amp3 3amp4 for equlvalent H eTlnere are 3 cases A2 Mth 4M nt IE
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