(n+1) Rule
- Page ID
- 14970
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)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, \(H_a\), \(H_b\), and \(H_c\), are equivalent. Thus, 1H NMR spectrum of 1 \(H_a\)s only one peak. \(H_a\), \(H_b\), and \(H_c\) are coupled to no hydrogen nuclei. Thus, for \(H_a\), \(H_b\), and \(H_c\), n=0; (n+1) = (0+1) = 1. The multiplicity of the peak of \(H_a\), \(H_b\), and \(H_c\) is one. The peak \(H_a\)s one line; it is a singlet. eg. 2:
There are two sets of equivalent hydrogen nuclei in 2:
- Set 1: \(H_a\)
- Set 2: \(H_b\), \(H_c\)
Thus, the 1H NMR spectrum of 2 \(H_a\)s two peaks, one due to \(H_a\) and the other to \(H_b\) and \(H_c\).
The peak of \(H_a\): There are two vicinal hydrogens to \(H_a\): \(H_b\) and \(H_c\). \(H_b\) and \(H_c\) are equivalent to each other but not to \(H_a\). Thus, for \(H_a\), n=2; (n+1) = (2+1) = 3. The multiplicity of the peak of \(H_a\) is three. The peak \(H_a\)s three lines; from the Pascal’s triangle, it is a triplet.
The peak of \(H_b\) and \(H_c\): There is only one vicinal hydrogen to \(H_b\) and \(H_c\): \(H_a\). \(H_a\) is not equivalent to \(H_b\) and \(H_c\). Thus, for \(H_b\) and \(H_c\), n=1; (n+1) = (1+1) = 2. The multiplicity of the peak of \(H_b\) and \(H_c\) is two. The peak \(H_a\)s 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: \(H_a\)
- Set 2: \(H_b\)
- Set 3: \(H_c\)
peak of \(H_a\):
multiplicity of the peak of \(H_a = 2 \times 2 = 4\). To determine the splitting pattern of the peak of \(H_a\), use the Pascal’s triangle, based on the observation that, for alkenyl hydrogens, \(J_{cis} > J_{gem}\).
The peak of \(H_a\) is a doublet of a doublet.
peak of \(H_b\):
multiplicity of the peak of \(H_b = 2 \times 2 = 4\). To determine the splitting pattern of the peak of \(H_b\), use the Pascal’s triangle, based on the observation that, for alkenyl hydrogens, \(J_{trans} > J_{gem}\).
The peak of \(H_b\) is a doublet of a doublet.
peak of \(H_c\):
multiplicity of the peak of \(H_c = 2 \times 2 = 4\). To determine the splitting pattern of the peak of \(H_c\), use the Pascal’s triangle based on the observation that, for alkenyl hydrogens, \(J_{trans} > J_{cis}\).
The peak of \(H_c\) is a doublet of a doublet.