Proton NMR spectrum of a-xylene. There are three types of protons in a-xylene, but only two absorptions are
seen in the spectrum. The aromatic protons Hb and He are accidentally equivalent, producing a broadened
peak at 07.1
The area under the blue line is corresponding to the number of protons.
A proton in the NMR spectrometer is subjected to both the external magnetic field and
the induced field of the shielding electrons. This splitting of signals into multiplets,
called spin-spin splitting, results when two different types of protons are close enough
that their magnetic fields influence each other. Such protons are said to be magnetically
Splitting of the - CH2Br group in 1 , 1 ,2-tribromoethane. When the nearby Ha proton is aligned with the
external magnetic field, it deshields Hb; when Ha is aligned against the field, it sb.elds H Proton NMR spectrum of ethylbenzene. The ethyl group appears as a triplet at81.( - CH3) and a quartet at82.6
(-CH2 - ) . The aromatic protons appear as a multiplet near 87 .2.
Bonded to adjacent carbons: three bonds between protons (vicinal hydrogens)
spin-spin splitting is normally observed
Bonded to the same carbon: two bonds between protons (germinal hydrogens)
spin-spin splitting is normally observed (if nonequivalent)
Leaning of a multiplet. A multiplet often "leans" upward toward the protons that
are causing the splitting. The ethyl multiplets in ethylbenzene lean toward each other
PROBLEM 13- 10
An unknown compound (C3H2NCl) shows moderately strong IR absorptions around 1 650 cm-1
and 2200 cm -1 . Its NMR spectrum consists of two doublets (J = 14 Hz) at 155.9 and 157.1.
Propose a structure consistent with these data.
A: 13- 10 The formula C3H2NCI has three elements of unsaturation. The IR peak at 1 650 cm- I
indicates an alkene, while the absorption at 2200 cm- I must be from a nitrile (not enough
carbons left for an alkyne). These two groups account for the three elements of unsaturation.
The NMR gives the coupling constant for the two protons as 1 4 Hz. This large J value shows the
two protons as trans (cis, J = 10 Hz; geminal , J = 2 Hz). The structure must be the one in the box
. N triple bond C—CH=CH—Cl. The atom must have an odd number of nutrons for NMR signal to be visible.
NMR for C13 is very hard because only 1 % of the carbon atoms in a sample
are the magnetic 13C isotope, the sensitivity of I3C NMR is decreased by a factor of 100.
Also, the gyromagnetic ratio of l3C is only one-fourth that of the proton decreasing sensetivity.