Lecture 7 NMR cont Chem 237 March 3 2015 Preparing NMR Samples What solvent do you use to dissolve sample Proper Solvent Level Diagram of NMR Spectrometer 300 MHz spectrometer 750 MHz spectrometer Examples of Splitting for Common Fragments 1 3 3 1 1 2 1 1 3 3 1 1 1 1 6 15 20 15 6 1 1 1 ppm 2 1ppm 2 02 1 1 92 0 1 81 9 1 71 8 1 61 7 1 51 6 1 41 5 ppm 2 1ppm 2 02 1 1 92 0 1 81 9 1 71 8 1 61 7 1 51 6 1 41 5 1 31 4 1 21 3 1 11 2 1 01 1 0 91 0 0 80 9 21 3 1 11 2 1 01 1 0 91 0 0 80 9 0 8 1 51 6 1 81 9 1 41 5 1 31 4 1 01 1 0 91 0 0 80 91 21 3 1 11 2 1 01 1 0 91 0 0 80 9 0 8 1 92 0 1 71 8 1 61 71 11 2 1 51 6 1 41 5 1 31 4 quartet triplet 1 21 3 quartet doublet septet doublet ppm 2 1ppm 2 02 1 0 8 area 2 area 3 CH2 CH3 area 1 CH area 3 CH3 area 1 CH 1 92 area 6 2 x CH3 Putting It All Together C6H12O When given a molecular formula calculate the DOU first This tells you the number of multiple bonds or rings in a structure 2C 2 H N X DOU 2 2 6 2 12 DOU 1 2 There is one ring or double bond in the structure ppm 1 02 8ppm 2 62 8 2 42 6 2 22 4 2 02 2 1 82 0 1 61 8 1 41 6 1 21 4 1 01 2 NMR LAB A Prelab 1 1 Give the number of degrees of unsaturation DOU for each of the following compounds N a C9H12O d N b C6HCl5O NH2 N c e N H N N What is a DOU Degree of Unsaturation How many rings and double bonds Nitrogen and halogens Structure easy to determine Molecular formula need to calculate NMR LAB A Prelab 21 For each of the following compounds give the number of signals resonances that you would expect to find in the 1H NMR spectrum Br S a c O b OH d H Expand structure to show all of the hydrogens Look for distinct chemical positions chemical shifts NMR LAB A Prelab 31 For each of the following compounds give the multiplicity s d t etc and the number of protons relative integrations for each type of proton indicated Be sure to give the integration value for the entire set of protons that are of the type indicated b d O Cl O a c e Expand structure to show all of the hydrogens For each H count carefully the number of equivalent neighbors Add 1 to the count to get the multiplicity For relative integrations consult the Pascal s Triangle NMR LAB A Prelab 1 4 Assign the given 1H NMR chemical shifts for each of the following compounds to the appropriate sets of protons as indicated i a b O Br O Cl Cl Cl 4 0 5 8 ppm cdef 1 0 1 7 1 8 4 1 ppm Use skills enumerated previously g h 1 0 2 3 3 6 ppm NMR LAB A Prelab 51 Many molecular fragments have easily identifiable coupling patterns by 1H NMR Match the following molecular fragments with the correct spectrum shown below HC a b H2 C C H2 CH3 H2 C c d C H2 CH3 HC CH3 1 2 3 4 Splitting Diagrams What splitting pattern results when coupling constants differ Jab Jbc Jab Jbc Spin Spin Splitting Coupling Constants typically 1 Hz Number of Intervening Bonds 2J 3J 4J H C H 2J 30 Hz Aromatic Systems H 3J 6 10 Hz H H C C H 3J 18 Hz H 4J 1 3 Hz H H H C C C 4J 5 Hz H 5J 1 Hz H Bond Length H H 1 40 1 34 H 3J 11 5 Hz H H 3J 8 0 Hz 1 42 H 3J 6 4 Hz Spin Spin Splitting Coupling Constants Alkene Stereochemistry H 3Jcis 7 11 Hz H Dihedral Angle dihedral angle H H H 3Jtrans 12 18 Hz H H H 2Jgem 0 3 Hz Hax Karplus Correlation Heq Heq Hax Hax Heq Heq Hax 3Jax ax 180 8 12 Hz 23 3Jax eq 60 2 5 Hz 3Jeq eq 60 2 5 Hz Spin Spin Coupling Splitting or Multiplicity a O e H d b O c a Which hydrogens are responsible for which peaks isopentyl formate 277 205 283 703 458 632 464 790 c 471 531 1254 370 541 524 534 871 528 253 521 679 514 124 507 537 501 009 494 588 488 322 477 552 d 1260 316 1266 288 C6H12O2 a a b 1270 2508 614 12601270 12501260 12401250 540 2508 614 520540500520 e 290 2508 614 280290 270280 d 480 2508 614 470480460470 Hz 1240 02400Hz27022002400 20002200 18002000 16001800 14001600 12001400 10001200 460 500 c 8001000 600800 b 400600 200400 0200 1 3 3 1 quartet 2 45 ppm J 7 5 Hz area 2 CH2 1 6 15 20 15 6 1 septet 2 63 ppm J 6 5 Hz area 1 CH 1 1 doublet 1 1 ppm J 6 5 Hz area 6 2 x CH3 3 1 991 99 11 ppm 2 8ppm 2 0542 62 8 1 0 1 025 1 000 0 975 1 2 1 triplet 1 0 ppm J 7 5 Hz area 3 CH3 2 488 2 463 2 438 2 413 2 695 2 674 2 652 2 630 2 609 2 587 2 565 C6H12O DOU 300 MHz 1 111 1 089 Putting It All Together 3 5 98 5 98 2 42 6 2 22 4 2 02 2 1 82 0 1 61 8 1 41 6 1 21 4 1 01 2 Putting It All Together C6H12O DOU 1 300 MHz 1 6 15 20 15 6 1 septet 2 63 ppm J 6 5 Hz area 1 CH 1 2 1 triplet 1 0 ppm J 7 5 Hz area 3 CH3 1 3 3 1 quartet 2 45 ppm J 7 5 Hz area 2 CH2 1 1 doublet 1 1 ppm J 6 5 Hz area 6 2 x CH3 Putting It All Together 1 111 1 089 Fragment analysis 1 025 1 000 0 975 That takes care of 5 of the 6 carbons This leaves only 1 carbon 1 oxygen and DOU The carbon and the oxygen must be 2 488 2 463 2 438 2 413 2 695 2 674 2 652 2 630 2 609 2 587 2 565 C6H12O DOU 300 MHz There is only one way that these fragments can be put together 3 1 991 99 11 2 8ppm …