# (n+1) Rule

The (n+1) Rule, an empirical rule used to predict the multiplicity and, in conjunction with Pascal’s triangle, splitting pattern of peaks in 1H and 13C NMR spectra, states that if a given nucleus is coupled (see spin coupling) to n number of nuclei that are equivalent (see equivalent ligands), the multiplicity of the peak is n+1.

eg. 1: The three hydrogen nuclei in 1, Ha, Hb, and Hc, are equivalent.  Thus, 1H NMR spectrum of 1 has only one peak.  Ha, Hb, and Hc are coupled to no hydrogen nuclei.  Thus, for Ha, Hb, and Hc, n=0; (n+1) = (0+1) = 1.  The multiplicity of the peak of Ha, Hb, and Hc is one.  The peak has one line; it is a singlet.

eg. 2: There are two sets of equivalent hydrogen nuclei in 2:

Set 1: Ha
Set 2: Hb, Hc

Thus, the 1H NMR spectrum of 2 has two peaks, one due to Ha and the other to Hb and Hc.

The peak of Ha:  There are two vicinal hydrogens to Ha: Hb and Hc.  Hb and Hc are equivalent to each other but not to Ha.  Thus, for Ha, n=2; (n+1) = (2+1) = 3.  The multiplicity of the peak of Ha is three.  The peak has three lines; from the Pascal’s triangle, it is a triplet.

The peak of Hb and Hc:  There is only one vicinal hydrogen to Hb and Hc: Ha.  Ha is not equivalent to Hb and Hc.  Thus, for Hb and Hc, n=1; (n+1) = (1+1) = 2.  The multiplicity of the peak of Hb and Hc is two.  The peak has two lines, from the Pascal’s triangle, it is a doublet.

To determine the multiplicity of a peak of a nucleus coupled to more than one set of equivalent nuclei, apply the (n+1) Rule independently to each other.

eg: There are three set of equivalent hydrogen nuclei in 3:

Set 1: Ha
Set 2: Hb
Set 3: Hc peak of Ha: multiplicity of the peak of Ha = 2×2 = 4.  To determine the splitting pattern of the peak of Ha, use the Pascal’s triangle, based on the observation that, for alkenyl hydrogens, Jcis > Jgem. The peak of Ha is a doublet of a doublet.

peak of Hb: multiplicity of the peak of Hb = 2×2 = 4.  To determine the splitting pattern of the peak of Hb, use the Pascal’s triangle, based on the observation that, for alkenyl hydrogens, Jtrans > Jgem. The peak of Hb is a doublet of a doublet.

peak of Hc: multiplicity of the peak of Hc = 2×2 = 4.  To determine the splitting pattern of the peak of Hc, use the Pascal’s triangle based on the observation that, for alkenyl hydrogens, Jtrans > Jcis. The peak of Hc is a doublet of a doublet.