UNEQUAL COUPLING UNEQUAL COUPLING TREE DIAGRAMS TREE DIAGRAMS SPLITTING DIAGRAMS aka “TREE” DIAGRAMS
WHERE DOES THE N+1 RULE WORK ?WHERE DOES THE N+1 RULE WORK ?
The n+1 rule works only for protons in aliphatic chainsand rings, and then under special conditions.
1) All 3J 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.
CH
HCH
HCH
H3Ja = 3Jb
Hydrogens can interchange theirpositions byrotations aboutthe 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 thehydrogens on eachof the carbon atomsequivalent.
WHAT HAPPENS WHEN THE J VALUES ARE NOT EQUAL ?
CH
HCH
HCH
H3Ja
3Jb
3Ja = 3Jb
In this situation each coupling must be consideredindependently of the other.
A “splitting tree” is constructed
CH
HCH
HCH
H3Ja = 7
-CH2-CH2-CH2-
CONSTRUCTING A TREE DIAGRAMCONSTRUCTING 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 givesa triplet.
Each level of thesplitting uses then+1 rule.
CH
HCH
HCH
H3Ja = 7
-CH2-CH2-CH2-
CONSTRUCTING A TREE DIAGRAMCONSTRUCTING A TREE DIAGRAM
CH
HCH
HCH
H3Jb = 3
triplet of triplets
ADD SPLITTING FROM HYDROGENS TO THE RIGHT
The smaller splittingis used second.
FIRST LEVEL
SECOND LEVEL LEVEL TWO
It is also a triplet.EACH LEG OF LEVEL ONE IS SPLIT
WHEN BOTH 3J VALUES ARE THE SAME
-CH2-CH2-CH2-
….. because of overlapping legs.You get the quintet predicted bythe n+1 rule.
The n+1 rule is followed …..
n+1 = (4 + 1) = 5
Splitting fromhydrogens onthe left
Splitting fromhydrogens onthe right
Splittingsoverlap
1:2:11:2:1
1:2:11:2:1
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
CHCH3 CHO
3J1 = 7 Hz
7 Hz 2 Hz
3J2 = 2 Hz
the methine hydrogen is split by two different3J values.
Rather than the expectedquintet …..
ANALYSISANALYSISOF METHINEOF METHINEHYDROGEN’SHYDROGEN’SSPLITTINGSPLITTING
quartet by -CH3
doubletby -CHO
quartet of doublets
ETHANOLETHANOLOld sampleRapid exchange catalyzedby impurities
quartet
triplet
broadsinglet
HO-CH2-CH3
hydrogen on OHis decoupled
400 MHz
ETHANOLETHANOLUltrapure sample (new)Slow or no exchange
tripletdoublet ofquartets
triplet400 MHz
expansion expansion
• 3J-cis = 8-10 Hz • 3J-trans = 16-18 Hz
• protons on the same carbon 2J-geminal = 0-2 Hz H
H
HH
H
H
PROTONS ON C=C DOUBLE BONDSPROTONS ON C=C DOUBLE BONDSCOUPLING CONSTANTS
For protons on saturated aliphatic chains 3J 8 Hz
Analysis of Vinyl AcetateAnalysis of Vinyl Acetate
HC HB HA
CCHH33 CC
OO
OOCC
HHCC
CCHHAA
HHBB
3JBC
3JAC
3JAC3JBC
2JAB2JAB
trans trans
cis
cis
gem gem
3J-trans > 3J-cis > 2J-gem