UNEQUAL COUPLING TREE DIAGRAMS SPLITTING DIAGRAMS aka “TREE” DIAGRAMS

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. There are two requirements for the n+1 rule to work: 1) All 3J values must be the same all along the chain. 2) There must be free rotation or inversion (rings) to make all of the hydrogens on a single carbon be nearly equivalent.

C H 3Ja = 3Jb THE TYPICAL SITUATION WHERE THE n+1 RULE APPLIES Hydrogens can interchange their positions by rotations about the C-C bonds. C H 3Ja = 3Jb This makes all the hydrogens on each of the carbon atoms equivalent. All the couplings along the chain have the same J value.

WHAT HAPPENS WHEN THE J VALUES ARE NOT EQUAL ? 3Ja = 3Jb C C C H H H 3Ja 3Jb In this situation each coupling must be considered independently of the other. A “splitting tree” is constructed

USE THESE VALUES C H 7 Hz 3 Hz

CONSTRUCTING A TREE DIAGRAM SPLITTING FROM HYDROGENS TO THE LEFT H H H -CH2-CH2-CH2- LEVEL ONE C C C The largest J value is usually used first. H H H Two neighbors gives a triplet. 3Ja = 7 The next splittings will be added to each leg of the first splitting. Each level of the splitting uses the n+1 rule.

CONSTRUCTING A TREE DIAGRAM ADD SPLITTING FROM HYDROGENS TO THE RIGHT H H H -CH2-CH2-CH2- C C C C H H H H 3Ja = 7 3Jb = 3 FIRST LEVEL SECOND LEVEL LEVEL TWO triplet of triplets The smaller splitting is used second. EACH LEG OF LEVEL ONE IS SPLIT It is also a triplet.

WHEN BOTH 3J VALUES ARE THE SAME The n+1 rule is followed ….. LEVEL ONE Splitting from hydrogens on the left -CH2-CH2-CH2- INTENSITIES n+1 = (4 + 1) = 5 1:2:1 1:2:1 LEVEL TWO 1:2:1 Splitting from hydrogens on the right + 1:2:1 1:4:6:4:1 Splittings overlap ….. because of overlapping legs. You get the quintet predicted by the n+1 rule. WHEN THE n+1 RULE APPLIES WE CAN JUMP TO THE FINAL RESULT - NO TREE NEEDED

2-PHENYLPROPANAL A case where there are unequal J values.

Spectrum of 2-Phenylpropanal b d J = 7 Hz c TMS a c J = 2 Hz d b

3J1 = 7 Hz 3J2 = 2 Hz ANALYSIS OF METHINE HYDROGEN’S SPLITTING 7 Hz Rather than the expected quintet ….. the methine hydrogen is split by two different 3J values. quartet by -CH3 3J1 = 7 Hz 3J2 = 2 Hz doublet by -CHO ANALYSIS OF METHINE HYDROGEN’S SPLITTING quartet of doublets

PURE ETHANOL

ETHANOL HO-CH2-CH3 400 MHz Old sample Rapid exchange catalyzed by impurities hydrogen on OH is decoupled HO-CH2-CH3 triplet broad singlet quartet

ETHANOL 400 MHz expansion expansion doublet of triplet quartets Ultrapure sample (new) Slow or no exchange 400 MHz triplet

J = 7 J = 5 J = 7 J = 5 triplet quartet of doublets triplet

VINYL ACETATE ALKENE HYDROGENS

PROTONS ON C=C DOUBLE BONDS COUPLING CONSTANTS PROTONS ON C=C DOUBLE BONDS 3J-cis = 8-10 Hz 3J-trans = 16-18 Hz protons on the same carbon 2J-geminal = 0-2 Hz For protons on saturated aliphatic chains 3J ~ 8 Hz

NMR Spectrum of Vinyl Acetate 60 MHz

Analysis of Vinyl Acetate H 3 O HC HA HB 3J-trans > 3J-cis > 2J-gem HC HB HA 3JAC 3JBC 3JBC cis trans trans 3JAC 2JAB 2JAB cis gem gem

2,4-DINITROANISOLE BENZENE HYDROGENS

2,4-DINITROANISOLE 400 MHz 9.0 8.0 7.0

2,4-DINITROANISOLE 8.72 ppm 8.43 ppm 7.25 ppm

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