NMR and MAss spectroscopy for engineering .pptx

NMR and MAss.pptx

1. Organic Chemistry 19HS121 – Unit 5 1

2. Unit 5: Structural Elucidation of Organic Compounds 19HS121 – Unit 5 2

3. Structural Elucidation of Organic Compounds UNIT - V - Structural Elucidation of Organic Compounds 19HS121 – Unit 5 3 IR (Principle, Identification of Functional Groups) NMR (Princ i ple , C h e mi c al Shift , 1 H - NM R - E t h yl Alcohol, Cis-Trans Isomers) Mass-Principle, Fragmentation (Nitrogen Rule) Introduction to XRD Reference Books Silverstein, Robert M Bassler, G. Clayton Morrill, Terence C, “Spectroscopic Identification of Organic Compounds”, Wiley publishers, 1981. COs Course Outcomes 5 Apply the principles of various instrumental techniques for structure determination of organic compounds.

4. 1 9 HS 121 – Unit 5 4 https:// www.slideshare.net/wkkok1957/ib-chemistry-on-infrared-spectroscopy-53759660?from_action=save

5. Unit 5: Spectroscopy 19HS121 – Unit 5 5 Spectroscopy The study of the interaction between radiations and matter as a function of wavelength λ. Interactions with particle radiation or a response of a material to an alternating field or varying frequency ν. Spectrum: A plot of the response as a function of wavelength or more commonly frequency is referred to as a spectrum. Method of “Seeing the unseeable” Atoms Mo l ecules Using electromagnetic radiation to obtain information about atoms and molecules that are too small to see.

6. 1 h 9 t H tp S1 s 2 :/ 1 / – w U w ni w t 5 .slideshare.net/wkkok1957/ib-chemistry-on-infrared-spectroscopy-53759660?from_action=save 6 Unit 5: Spectroscopy 1. UV-Visible radiations--------excitation of electrons---------UV-visible spectrum 2. IR- radiations---------------------vibration changes --------------------IR Spectrum 3. Radio frequency-------------spin rotational changes-------------N.M.R spectrum

7. 19HS121 – Unit 5 7 Unit 5: Spectroscopy Infrared (IR) spectroscopy measures the bond vibration frequencies in a molecule and is used to determine the functional group. Nuclear magnetic resonance (NMR) spectroscopy analyzes the environment of the hydrogens in a compound. This gives useful clues as to the alkyl and other functional groups present. Mass spectrometry (MS) fragments the molecule and measures their mass. MS can give the molecular weight of the compound and functional groups. X-ray diffraction (XRD) is a powerful nondestructive technique for characterizing crystalline materials. It provides information on structures, phases, preferred crystal orientations (texture), and other structural parameters, such as average grain size, crystallinity, strain, and crystal defects.

8. Unit 5: NMR Spectroscopy NMR spectroscopy is concerned with the study of absorption of EMR in the radio frequency region (4-900 MHz) , by nuclei of the atoms which causes nuclear spin transition. 19HS121 – Unit 5 8 NMR is used to study the wide variety of nuclei: 1 H NMR , 19 F NMR 13 C NMR , 29 Si NMR 15 N NMR, 31 P NMR 1 H NMR spectroscopy is one of most powerful tool for elucidating the number of hydrogen (protons) in the compound.

9. NMR Spectroscopy NMR spectroscopy is concerned with the study of absorption of radio frequency radiation, which causes nuclear spin transition in the nuclei of the atoms . NMR spectra mainly used in structure elucidation of organic molecules . Unit 5: NMR Spectroscopy Nuclear Magnetic Resonance Spectroscopy (NMR) Involve nucleus (proton + neutron) NOT electron Proton + neutron = Nucleons Nucleons like electron have spin and magnetic moment (acts like tiny magnet) 19HS121 – Unit 5 9

10. 10 Unit 5: NMR Spectroscopy Theory of NMR Spectroscopy Spin quantum number (I) is related to he atomic and mass number of the nucleus. Element with either odd mass or odd atomic number have the property of nuclear spin. Nuclei with even number of nucleon ( 12 C and 16 O) Even number of proton and neutron – NO net spin Nucleon spin cancel out each other –Nucleus have NO overall magnetic moment – NOT absorb radio wave Nuclei with odd number of nucleon ( 1 H, 13 C, 19 F, 31 P) -Nucleon have net spin – Nucleus have NET magnetic moment – Absorb radio wave -Nuclei with net spin – magnetic moment will interact with radio waves Nuclei have a “spin” associated with them (i.e., they act as if they were spinning about an axis) due to the spin associated with their proton and neutron. Nuclei are positively charged, their spin induces a magnetic field. I Atomic Mass Atomic Number Examples Half-integer Odd Odd 1 H, 19 F, 31 P (1/2) Half- integer Odd Even 13 C (1/2) Integer Even Odd 2 H, 14 N (1) Zero Even Even 12 C, 16 O (0) NMR active NMR inactive Net spin No net spin Spin cancel each other

11. 19HS121 – Unit 5 11 Unit 5: NMR Spectroscopy Principles of NMR Spectroscopy The theory behind NMR comes from the spin of the nucleus and it generated a magnetic filed. Without external applied magnetic filed, the nuclear spins are random in directions. But when an external magnetic field (B o ), is present in nuclei align themselves either with or against the filed of the external magnet. No external magnetic field With external magnetic field external magnetic field

12. 12 Unit 5: NMR Spectroscopy The emitted radio frequency is directly proportional to the strength of the applied filed. ν α B o ν = γB o / 2π Bo = External magnetic filed by proton, γ = Magnetogyric ratio (The ratio between the nuclear magnetic moment and angular moment). Principles of NMR Spectroscopy If an external magnetic filed is applied, an energy transfer (∆E) is possible between ground state and excited state. When the spin returns to its ground state level, the absorbed radio frequency energy is emitted at the same frequency level. The emi t t ed r a dio f r equen c y signal that gi v e t h e NMR sp e c t r u m of the concerned nucleus. β - spin state α - spin state Applied magnetic field strength increases (B o )

13. Unit 5: NMR Spectroscopy NMR Instrumentation 19HS121 – Unit 5 13

14. Unit 5: 1 HNMR Spectroscopy Proton NMR is much more sensitive than 13 C and the active nucleus ( 1 H) is nearly 100 % of the natural abundance. Shows how many kinds of nonequivalent hydrogens are in a compound. Theoretical equivalence can be predicted by seeing if replacing each H with “X” gives the same or different outcome. Equivalent H’s have the same signal while nonequivalent are “d i f fer e nt ” a nd as s u ch m a y c a u s e ad d it ion a l s p l itti ng (diastereoto pi c effect). 19HS121 – Unit 5 14

15. Unit 5: NMR Spectr os c opy Main features of 1 HNMR Spectra 1. Number of signals (diff absorption peak) – Number of diff. proton/chemical environment 2. Area under peak (Integration) Number of hydrogen in a particular proton/chemical environment (Integration trace) Ratio of number of hydrogen in each environment 3. Chemical shift Chemical environment where proton is in Spinning electron create own magnetic field, creating a shielding effect Proton which are shielded appear upfield. (Lower frequency for resonance to occur) Proton which are deshielded appear downfield. (Higher frequency for resonance to occur) Measured in ppm ( δ ) 4. Splitting pattern Due to spin-spin coupling Number of peak split is equal to number of hydrogen on neighbouring carbon+1 (n+1) peak 19HS121 – Unit 5 15

16. Information from 1 H-NMR spectra: Number of signals : How many different types of hydrogens in the molecule. Position of signals (chemical shift) : What types of hydrogens. Relative areas under signals (integration) : How many hydrogens of each type. Splitting pattern : How many neighboring hydrogens. Unit 5: NMR Spectroscopy NMR spectrum of CH 3 CH 2 Br 19HS121 – Unit 5 16

17. Number of signals: How many different types of hydrogens in the molecule. Protons in different environments give different NMR signals. Equivalent protons give the same NMR signal. Number of signals? Unit 5: NMR Spectroscopy All equivalent H’s 1 NMR signal C H 2 H CC H 2 2 H 2 C H H 2 CC H 2 H CC H H C C 3 3 H CC H , H C C H , H 3 CC H 3 C H 3 C H 3 C H 3 Br CC , H 3 CCC H 3 , H 3 CCC C H 4 , C H 3 C H 3 , H 2 C H 2 C H 3 C C H 3 , H 3 C B r , C H 3 C H 3 H C C H H C C H , Br C H H 3 C CH 3 , H 2 H 2 H 2 C C C l C C l H 3 C 2 H C C H 2 B r H 3 C C H C H 3 C H 2 B r H 3 C 2 H C O H H C , H C C H C H C H 3 B r , , 2 Types of H’s 2 NMR signals 3 Types of H’s 3 NMR signals H C H C C H C H O H C H 3 4 Types of H’s 4 NMR signals 19HS121 – Unit 5 17

18. Unit 5: NMR Spectroscopy 2. Chemical Shift A chemical shift is defined as the difference in parts per million (ppm) between resonance frequency of the observed proton and tetramethylsilane (TMS) hydrogens (as an internal reference). (Or) Numeric value of chemical shift: difference between strength of magnetic field at which the observed nucleus resonates and field strength for resonance of a reference. TMS is the most common reference compound in NMR, it is set at δ = 0 ppm 19HS121 – Unit 5 18 Chemical Shift, δ = Frequency of signal – Frequency of reference Spectrometer frequency Difference is very small but can be accurately measured. X 10 6 Taken as a ratio to the total field and multiplied by 10 6 so the shift is in parts per million (ppm)

19. Unit 5: NMR Spectroscopy Chemical Shift The NMR spectrum is a plot intensity of NMR signals VS magnetic field (frequency) in reference to TMS NMR spectra show applied field strength increasing from left to right. Left part is downfield, the right is upfield. Nuclei that absorb on upfield side are strongly shielded where nuclei that absorb on downfield side is weakly shielded . The el e c t r o ns s ur ro u nding a n u cle u s a f f e ct t h e e f f e cti v e m agnet i c f i eld sensed by the nucleus. 19HS121 – Unit 5 19

20. Unit 5: NMR Spectroscopy Chemical Shift Shielding Protons: High electron density around a nucleus shields the nucleus from the external magnetic field and the signals are upfield in the NMR spectrum. Deshielding Protons: Lower electron density around a nucleus deshields the nucleus from the external magnetic field and the signals are downfield in the NMR spectrum. 19HS121 – Unit 5 20

21. Unit 5: NMR Spectroscopy Primary H’s Secondary H’s Tertiary H H’s attached with ester H’s attached with carbonyl H attached with aromatic H atom attached to alkyne H’s attached with halogen H’s attached with ether 2 o H’s attached with ester 1 o H’s attached with ester Alcohol H Vinylic and alkenyl H’s Phenolic H Aryl H’s Aldehyde H Acidic H 19HS121 – Unit 5 21

22. Unit 5: NMR Spectroscopy Chemical Shift 19HS121 – Unit 5 22

23. Unit 5: NMR Spectroscopy 3. Integration (relative areas under each signal): How many hydrogens of each type. 1. C H C H 3 3 a - 6H 2. C H 3 C H 2 C H 2 Br a - 3H b - 2H c - 2H a b a 3. C H 3 C H C H 3 Cl a - 6H b - 1H a : b : c = 3 : 2 : 2 a b c a a a : b = 6 : 1 3 B r a C H C H b C H c b H C c , H C 4. 5. a - 3H b - 2H c - 2H a : b : c = 3 : 2 : 2 C H 3 d O H 19HS121 – Unit 5 23 C H b C H c b H C c H C a a - 3H b - 2H c - 2H d - 1H a : b : c : d = 3 : 2 : 2 : 1

24. Unit 5: NMR Spectroscopy 4. Splitting pattern: Splitting pattern is calculated using n+1 rule. This is considered by the nearby hydrogen nuclei. 19HS121 – Unit 5 24 n – Number of protons in nearby nuclei n n+1 Pascal pattern (Ratio) Multiplicity 1 1 Singlet 1 2 1 : 1 Doublet 2 3 1 : 2 : 1 Triplet 3 4 1 : 3 : 3 : 1 Quartet 4 5 1 : 4 : 6 : 4 : 1 Quintet 5 6 1 : 5 : 10 : 10 : 5 : 1 Hextet 6 7 1 : 6 : 15 : 20 : 15 : 6 : 1 Septet a a 1. C H 3 C H 3 a - 6H - Singlet a b c C H 3 C H 2 C H 2 Br a - 3H - Triplet b - 2H - Hextet c - 2H - Triplet a b a C H 3 C H C H 3 Cl a - 6H - Doublet b - 1H - Septet * n must be equivalent neighboring hydrogens to give rise to a Pascal splitting pattern. If the neighbors are not equivalent, then you will see a complex pattern * the alcohol hydrogen –OH usually does not split neighboring hydrogen signals nor is it split. Normally a singlet of integration 1 between 1 – 5.5 ppm (variable).

25. 4. Splitting pattern: The interaction between the spins of neighbouring nuclei in a molecule may cause the splitting of NMR spectrum. This is known as spin-spin coupling or splitting. The Splitting pattern is related to the number of equivalent H-atom at the nearby nuclei. Unit 5: NMR Spectroscopy 19HS121 – Unit 5 25

26. Unit 5: NMR Spectroscopy 4. Rules for splitting pattern: Chemically equivalent protons (H) do not show splitting (spin-spin coupling). Only nonequivalent protons couple (splitting) All equivalent H in same chemical environment will produce a same peak Spin spin coupling – occur when proton have different chemical shift. Protons on adjacent carbons normally will couple. 19HS121 – Unit 5 26

27. Unit 5: NMR Spectroscopy Unit 5: NMR Spectroscopy Splitting pattern: CH 3 spilt to doublet by 1 adjacent H CH 3 e x pe r i en c e 2 s li ghtl y d i f f e ren t magnetic filed (MF) due to adjacent H Spilt with relative intensity 1:1 Doublet CH 3 spilt to Triplet by 2 adjacent H CH 3 experience 3 different magnetic filed (MF) due to 2 adjacent H Triplet Spilt with relative intensity 1:2:1 CH 3 spilt to quartet by 3 adjacent H CH 3 experience 4 different magnetic filed (MF) due to 3 adjacent H 19HS121 – Unit 5 27 Quartet Spilt with relative intensity 1:3:3:1

28. Unit 5: NMR Spectroscopy Coupling constant: The distance between the peaks in a given multiplet is a measure of the splitting effect known as coupling constant. It is denoted by symbol J, expressed in Hz. Coupling constants are a measure of effectiveness of spin-spin coupling (splitting pattern) and very useful in 1H NMR of complex structures. 19HS121 – Unit 5 28

29. Unit 5: NMR Spectroscopy 1 HNMR spectrum of Ethanol 19HS121 – Unit 5 29

30. Unit 5: NMR Spectroscopy 19HS121 – Unit 5 30

31. Unit 5: NMR Spectroscopy 19HS121 – Unit 5 31

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33. Unit 5: NMR Spectroscopy a 19HS121 – Unit 5 33 b H 3 C a b H b H CH 3 a a = δ 1.54 ppm b = δ 5.37 ppm Cis-2-butene Coupling constant J = 7-12 Hz

34. Unit 5: NMR Spectroscopy a a b H 3 C H H CH 3 b a b Trans-2-butene a = δ 1.58 ppm b = δ 5.55 ppm Coupling constant J = 12-18 Hz 19HS121 – Unit 5 34

35. Unit 5: NMR Spectroscopy CIS/TRANS Isomers - NMR Unsymmetrically substituted double bond compound leads to non-equivalent alkenyl hydrogens which leads to spin -spin coupling. Within a set of cis/trans isomers, the coupling constant for the trans isomer, J, is always larger than cis isomer. 19HS121 – Unit 5 35

36. Unit 5: NMR Spectroscopy https:/ /w w w.chegg.com/homework-help/coupling-c-protons-b-c-protons-cis-trans-alkenes-shown-figur-chapter-14-problem-30p-solution-9780134066585-exc 19HS121 – Unit 5 36

37. 19HS121 – Unit 5 37 Assignment: Why OH signal gives a singlet instead of triplet

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39. 1 9 HS 121 – Unit 5 39 http s://w ww.y outube.com/watch?v=2oPUyIbPxLo http s://w ww.y outube.com/watch?v=VUIPYnWLSDE

40. Unit 5: Mass Spectroscopy 19HS121 – Unit 5 40

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47. 19HS121 – Unit 5 47 When something is moving in a circle, it is having Centripetal force. Q) What is causing the centripetal force? Ans: Magnetic Force by the magnet F m = F C Where, F C = Centripetal force F M = Magnetic force qvB = mv 2 /r qB = mv/r r= mv/qB Where, r = Radius of curvature v = velocity q = Charge B = Magnetic field If we know the radius of curvature , then we can find the mass.

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