Cis and trans coupling appear differently on 1H NMR spectrum
Here are a couple of terms to know:
- Vicinal - Coupling between hydrogens on adjacent carbons.
- Geminal - Coupling between nonequivalent hydrogens on the same carbon atom.
Type of coupling | Name | Range (Hz) | Typical (Hz) |
Vicinal, cis | 6-14 | 10 | |
Vicinal, trans | 11-18 | 16 | |
Geminal | 0-3 | 2 | |
None | 4-10 | 6 | |
Allylic | 0.5-3.0 | 2 |
When alkynel hydrogen atoms are not symmetrically substituted on a double bonded carbon, the hydrogens of a cis and trans isomer will yield a different shift on the NMR spectrum. Because the coupling constant is smaller in a cis isomer than in a trans isomer, the NMR spectrums of the two isomers are different conveying the hydrogens in a cis isomer to be slightly more upfield to-- the right of the spectrum-- and trans hydrogens to be more downfield to the left.
Sometimes coupling will lead to very complicated patterns as a result of the J values that vary widely due to the relationship between the hydrogens involved. When this occurs, information can still be derived to determine the structure of a molecule by looking at the number of signals, the chemical shift of each one, integration, and splitting patterns similarly to identifying alkane NMR.
Alkenyl carbons are deshielded in 13C NMR
For background information on 13C NMR, please refer to 13C Nuclear Magnetic Resonance from the previous chapter. Compared to alkane carbons with one bond, alkene carbons show a relatively low field shift on the 13C NMR spectrum and absorb about 100 ppm lower field. Also, in broad-band decoupled 13C NMR, sp2 carbons absorb as sharp single lines so with these two methods, it is easy to determine the presence of a double bond in 13C NMR spectrum.
Here are the common 13C Chemical Shift Ranges:
Note that the carbon-carbon double bonds are found in the range between 100-170 ppm. Carbon atoms on alkenes that are attatched to another carbon group are found more downfield than carbon alkenes attatched to hydrogens.
Let's try a 1H NMR practice problem with C4H7Cl:
Remember from previous sections that to solve an NMR spectrum with double bonds, we must know the Degrees of Unsaturation. From this, we get degrees of unsaturation= (9-7)/2=1 so there is one pi bond or ring in our molecule. Next we must look at the integration of the NMR spectrums.
- ppm= 1.8 with 3H reveals a CH3attached to an unsaturated functional group
- ppm= 4.0 with integration of 2H is a CH2 most likely attached to the Cl
- ppm = 4.9 and 5.1 are singlets of 1H and must be our two alkene hydrogens
There are three ways to attach the discovery we made, but only one of them are the correct answer. Since the coupling of the two alkene hydrogens are small whereas vicinal hydrogens tend to be large, we conclude that the hydrogens are geminal and appear on the same carbon. This leaves the two other groups to be located on the other alkene carbon.
Stereo Chemical Information from coupling constant
Distinction between cis and trans isomers
Trans coupling is greater than for cis. From the value the isomers can be identified as in the following example.
- For coupling the two hydrogens must have different chemical shift.
- Trans coupling (range :11 to 19 Hz) is greater than cis coupling (range: 5 to 14 Hz).
- 3JHH means coupling between two hydrogens separated by a distance of three bonds (vicinal coupling). If it is coupling between hydrogens at the same carbon then it is 2JHH (geminal coupling)
In some molecules the cis-trans, distinction is made even from the position of the absorption, as in cis and trans stilbenes. In trans case the hydrogens are deshielded by two aromatic rings while in the cis case it is by one ring.
Variation of dihedral angle with coupling constant
Dihedral angle: Consider the following Newman and Sawhorse projections for ethane.
- Looking from the front Hb is eclipsed by Ha the angle between the plane containing Ha and Hb is zero (they are in the same plane).
- The second conformation has Hb at a different spatial position, they are not in the same plane. The plane containing Hb is 600 to the right of the plane containing Ha. This is the dihedral angle, in the first case the dihedral angle is zero.
- In the third case though the two atoms are in the same plane Hb has undergone a rotation of 1800 and that is the dihedral angle.
- The dihedral angle is used when two are separated by three bonds.
- Martin Karplus was the first to study the variation of coupling constant 3JHH with dihedral angle.
- The magnitude of the coupling constant between hydrogens of adjacent carbon atoms depend directly on the dihedral angle.
- This magnitude is greatest when this is zero or 1800 and smallest when it is 900.
- When it is zero side-side overlap of the two C-H bond orbitals will be maximum, when it is 1800 back lobes overlap is maximum leading to higher “J”. When it is 600 C-H bond orbitals overlap is at a minimum since they are perpendicular with each other leading to lower “J”.
Stereochemistry of cyclohexane derivatives
The information from the above can be used to assign the configuration of the anomeric carbon in cyclic sugars. In the case of α- glucose at the anomeric carbon the equatorial proton is has a chemical shift of δ 5.2 which is down field by 0.5 for the corresponding axial proton in β-glucose.
Coupling Constants : Cis-Trans Isomers of Olefins
The magnitude of the 3J(H,H) coupling constant in olefins follows the general rule Jtrans > Jcis and this is a convenient way to distinguish between cis and trans isomers:
3J is also influenced by the olefin's substituents, the distinction is meaningful only for one by direct comparison of the isomers or for olefins with very similar substitution pattern.
( E ) - hex - 2 - en - 1 - yl acetate
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BHUSAWAL, MAHARASHTRA , INDIA
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The first 2/3 of this are blatantly copied from libretexts (https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkenes/Properties_of_Alkenes/Nuclear_Magnetic_Resonance_(NMR)_of_Alkenes). You should delete this blog...
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