PowerPoint Presentation: UNEQUAL COUPLING TREE DIAGRAMS SPLITTING DIAGRAMS aka “TREE” DIAGRAMS
PowerPoint Presentation: WHERE DOES THE N+1 RULE WORK ? The n+1 rule works only for protons in aliphatic chains and rings, and then under special conditions. 1) All 3 J values must be the same all along the chain. There are two requirements for the n+1 rule to work: 2) There must be free rotation or inversion (rings) to make all of the hydrogens on a single carbon be nearly equivalent.
PowerPoint Presentation: C H H C H H C H H 3 J a = 3 J b Hydrogens can interchange their positions by rotations about the C-C bonds. All the couplings along the chain have the same J value. THE TYPICAL SITUATION WHERE THE n+1 RULE APPLIES This makes all the hydrogens on each of the carbon atoms equivalent.
PowerPoint Presentation: WHAT HAPPENS WHEN THE J VALUES ARE NOT EQUAL ? C H H C H H C H H 3 J a 3 J b 3 J a = 3 J b In this situation each coupling must be considered independently of the other. A “splitting tree” is constructed
PowerPoint Presentation: C H H C H H C H H 7 Hz 3 Hz USE THESE VALUES
PowerPoint Presentation: C H H C H H C H H 3 J a = 7 -CH 2 -CH 2 -CH 2 - CONSTRUCTING A TREE DIAGRAM The largest J value is usually used first. SPLITTING FROM HYDROGENS TO THE LEFT The next splittings will be added to each leg of the first splitting. LEVEL ONE Two neighbors gives a triplet. Each level of the splitting uses the n+1 rule.
PowerPoint Presentation: C H H C H H C H H 3 J a = 7 -CH 2 -CH 2 -CH 2 - CONSTRUCTING A TREE DIAGRAM C H H C H H C H H 3 J b = 3 triplet of triplets ADD SPLITTING FROM HYDROGENS TO THE RIGHT The smaller splitting is used second. FIRST LEVEL SECOND LEVEL LEVEL TWO It is also a triplet. EACH LEG OF LEVEL ONE IS SPLIT
PowerPoint Presentation: WHEN BOTH 3 J VALUES ARE THE SAME -CH 2 -CH 2 -CH 2 - ….. because of overlapping legs. You get the quintet predicted by the n+1 rule. The n+1 rule is followed ….. n+1 = (4 + 1) = 5 Splitting from hydrogens on the left Splitting from hydrogens on the right Splittings overlap 1:2:1 1:2:1 1:2:1 1:2:1 1:4:6:4:1 INTENSITIES + LEVEL ONE LEVEL TWO WHEN THE n+1 RULE APPLIES WE CAN JUMP TO THE FINAL RESULT - NO TREE NEEDED
PowerPoint Presentation: 2-PHENYLPROPANAL A case where there are unequal J values.
PowerPoint Presentation: Spectrum of 2-Phenylpropanal J = 2 Hz J = 7 Hz a b c d a b d c TMS
PowerPoint Presentation: 3 J 1 = 7 Hz 7 Hz 2 Hz 3 J 2 = 2 Hz the methine hydrogen is split by two different 3 J values. Rather than the expected quintet ….. ANALYSIS OF METHINE HYDROGEN’S SPLITTING quartet by -CH 3 doublet by -CHO quartet of doublets
PowerPoint Presentation: PURE ETHANOL
PowerPoint Presentation: ETHANOL Old sample Rapid exchange catalyzed by impurities quartet triplet broad singlet HO-CH 2 -CH 3 hydrogen on OH is decoupled 400 MHz
PowerPoint Presentation: ETHANOL Ultrapure sample (new) Slow or no exchange triplet doublet of quartets triplet 400 MHz expansion expansion
PowerPoint Presentation: quartet of doublets J = 7 J = 5 triplet triplet J = 7 J = 5
PowerPoint Presentation: VINYL ACETATE ALKENE HYDROGENS
PowerPoint Presentation: 3 J- cis = 8-10 Hz 3 J- trans = 16-18 Hz protons on the same carbon 2 J- geminal = 0-2 Hz PROTONS ON C=C DOUBLE BONDS COUPLING CONSTANTS For protons on saturated aliphatic chains 3 J ~ 8 Hz
NMR Spectrum of Vinyl Acetate: NMR Spectrum of Vinyl Acetate 60 MHz
PowerPoint Presentation: Analysis of Vinyl Acetate H C H B H A C H 3 C O O C H C C H A H B 3 J BC 3 J AC 3 J AC 3 J BC 2 J AB 2 J AB trans trans cis cis gem gem 3 J- trans > 3 J- cis > 2 J- gem
PowerPoint Presentation: 2,4-DINITROANISOLE BENZENE HYDROGENS
PowerPoint Presentation: 2,4-DINITROANISOLE 400 MHz 9.0 8.0 7.0
PowerPoint Presentation: 2,4-DINITROANISOLE 8.72 ppm 8.43 ppm 7.25 ppm
PowerPoint Presentation: END