344 Organic Chemistry Laboratory Fall 2013 Lecture 2 More 1 H-NMR Spectroscopy June 18 2013
Finished Lecture 1 here 1.18 6.63 7.02 2.79 3.51
1.18 CH 3 Septet & doublet NH 2 Typical i Pr group pattern 6 CH 3.51 2 2.79 1
Coupling constant J (Hz) – indicates strength of coupling J ~ 7 Hz for alkyl (sp 3 ) systems 20 Septet 15 1:6:15:20:15:6:1 15 6 6 1 1
Derivation of splitting diagrams 1) Two different coupling constants in a simple alkyl chain (J ab much larger than J bc ) H a H b H c Y C C C X apply strongest coupling first! H b H a H b H c J ab > J bc J ab J bc triplet of triplets
Derivation of splitting diagrams 2) Two identical coupling constants in a simple alkyl chain (J ab equal or almost equal to J bc ) H a H b H c H b Y C C C X H a H b H c Overlap of peaks in each signal J ab = J bc due to similarity of coupling J ab constants J ab and J bc Pentet observed! This is the splitting pattern J bc predicted by the n+1 rule for H b when n = 4
Derivation of splitting diagrams 2) Two identical coupling constants in a simple alkyl chain (J ab equal or almost equal to J bc ) H a H b H c H b Y C C C X H a H b H c J ab = J bc J ab For H b : n = 4 n + 1 = 5 1:4:6:4:1 pentent J bc When coupling constants are equal, splitting results in n + 1 peaks (i.e. a “normal” splitting pattern) THIS IS THE MOST COMMON CASE FOR A SIMPLE ALKYL GROUP
Signals from benzene ring protons
Coupling constants in aromatic systems NH 2 H a H a J ab J ab H b H b Me Me J ab = J ortho = 6 – 12 Hz
Why is H a more shielded than H b ? NH 2 Ha Ha H a J ab H b 6.63 Hb Hb 7.02 Consider the substituents Me Me J ab Consider resonance structures (when applicable) 2 2
H a shielded relative to H b which can be rationalized by resonance effects. (Shielding/deshielding by alkyl groups is usually minor.) -0.28 -0.28 -0.21 -0.21 NBO charges B3LYP/6-31G(d) -NH 2 , -NR 2 , -OMe, -OH, etc. are common electron-donating groups. Electron-donating groups increase e - density at the o and p C-atoms. This shields the H-atoms at those positions relative to benzene H-atoms.
OMe Ha Ha Hb Hb NO 2 3.91 OMe 3
OMe H b H a Ha Ha 8.18 6.95 Hb Hb NO 2 Why is H a more shielded than H b ? or…..why is H b so deshielded? 2 2 Consider the resonance structures What is this?
H b deshielded relative to H a which can be rationalized by resonance effects. Recall, -OCH 3 is electron-donating which helps shield the H a atoms. -0.32 -0.27 -0.19 -0.20 NBO charges B3LYP/6-31G(d) -NO 2 , -NR 3 + , -CF 3 , -CO 2 R etc. are common electron-withdrawing groups. Electron-withdrawing groups reduce e - density at the o and p C-atoms. This deshields the H-atoms at those positions relative to benzene H-atoms.
1.93 Me 3 2 2 1 1 1
Coupling constants in alkene systems J ab = J cis = 6 – 12 Hz H a H b J bc = J gem = 1 – 3 Hz R H c J ac = J trans = 12 – 18 Hz J trans > J cis > J gem
H a is trans to H c H a is cis to H b Aromatic ring protons (by chem. shift and integration) 7.23 Alkene protons (by chem. shift and integration) 7.11 5.70 5.17 But which is which? 6.61 2 2 1 1 1 Consider J values
Derivation of splitting diagrams 3) Splitting in alkene systems (J ac > J ab ) apply strongest coupling first! H a H b H a X H c J ac = J trans Same concept as example 1) J ab = J cis J ac J trans > J cis Doublet of doublets J ab Hb-Hc geminal coupling not shown
Hb H d Ha Hc 7.23 H e Hd Hd 7.11 He He 5.70 Me 5.17 H a 6.61 Doublet of doublets 2 with J trans & J cis 2 1 1 1
H c Hb H b Ha Hc J trans > J cis Hd Hd 5.70 ppm 5.17 ppm He He H a J trans J cis Me 17.2 Hz 10.5 Hz J trans Scale now in Hz
H c Hb J bc H b J bc Ha Hc Hd Hd He He Me J ac trans + gem coupling J ab cis + gem coupling
3.89 OMe 3
Coupling constants in aromatic systems OMe H d H a J ab ≈ J ad ≈ J bd = J meta = 1-3 Hz J ac = J para = 0 - 1 Hz H c NO 2 H b J ortho > J meta > J para = 6 – 12 Hz
Write down the relationships between the protons! OMe Hd Ha 6.8 ppm Hc NO 2 7.5 ppm Hb H c is ortho to H b 7.48 ortho to H d 7.2 ppm (para to H a ) 7.71 7.7 ppm 7.79 7.23 1 1 1 1
Derivation of splitting diagrams 4) Splitting in aromatic systems (J cb = J cd ) X Can also apply to H a , just draw diagram H c using meta coupling constants H d H a H c Y J cb H b J cb = J ortho Overlap of peaks in each signal J cd = J ortho due to similarity of coupling constants J cb and J cd H c is ortho to H b Same concept as example 2) J cd J cd ortho to H d Doublet of doublets but…….. (para to H a ) central peaks overlap to give appearance of 1:2:1 triplet (ITS NOT A REAL TRIPLET!!!) Ha-Hc para coupling not shown
OMe Hd Ha Hc NO 2 Hb As practice, draw the splitting diagrams for H b and H d - list relationships of each proton - apply n+1 rule to these relationships - factor in the coupling constants
H b is most deshielded (proximity to NO 2 group) H d is most shielded (proximity to OMe group) NMR Chemical Shifts B3LYP/6-31G(d) OMe 6.8 ppm 7.5 ppm Hd Ha 7.48 H a H c 7.71 Hc NO 2 7.2 ppm Hb 7.7 ppm H b H d 7.79 7.23 1 1 1 1
O Me Experiment NH Information OH O N -Acetylanthranillic acid Inset Spectrum 7.0 - 8.9 ppm TMS HCCl 3 impurity “missing” impurity OH peak Integration values
Strategy and Tactics for solving NMR spectra How many different types of H-atoms? Indicated by how many groups of signals What types of H-atoms? Indicated by the chemical shift of each signal How many H-atoms of each type are there? Indicated by the integration of the signal for each group What is the connectivity of the molecule? Indicated by the splitting pattern and coupling constant of each signal What other evidence do you have? Use GC-MS, 13 C-NMR, IR, melting point etc. as complimentary information Practice and Ask Questions!
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