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Question

The ${}^{31}\text{P}\{^1\text{H}\}$ NMR spectrum of $2,2,6,6-\text{N}_4\text{P}_4\text{Cl}_4(\text{NMe}_2)_4$ is expected to show

The correct answer is
two triplets

NMR Spectrum Analysis: ${}^{31}\text{P}\{^1\text{H}\}$ Decoupled ${}^{31}\text{P}$ NMR

The ${}^{31}\text{P}\{^1\text{H}\}$ NMR experiment involves proton decoupling, meaning any coupling between ${}^{31}\text{P}$ and ${}^1\text{H}$ nuclei is removed. The observed splitting pattern in the ${}^{31}\text{P}$ spectrum arises from coupling between different ${}^{31}\text{P}$ nuclei (P-P coupling).

Molecular Structure and Symmetry

The compound $2,2,6,6-\text{N}_4\text{P}_4\text{Cl}_4(\text{NMe}_2)_4$ typically refers to a phosphazene cage structure containing a $\text{P}_4\text{N}_4$ framework. The formula indicates 4 phosphorus atoms, 4 nitrogen atoms, 4 chlorine substituents, and 4 dimethylamino ($\text{NMe}_2$) substituents distributed among the phosphorus atoms. The numbering ($2,2,6,6$) suggests a high degree of symmetry, leading to distinct sets of chemically equivalent phosphorus atoms.

In symmetrically substituted $\text{P}_4\text{N}_4$ cages, it is common to have two types of chemically equivalent phosphorus atoms. Let's denote these as Type A and Type B. Due to symmetry, there are likely 4 equivalent phosphorus atoms of Type A and 4 equivalent phosphorus atoms of Type B.

Predicting Signal Multiplicity from P-P Coupling

The multiplicity of a signal in NMR depends on the number of equivalent neighboring nuclei it couples to. The number of lines (multiplicity) is given by $2nI + 1$, where $n$ is the number of equivalent neighboring nuclei and $I$ is their nuclear spin ($I = 1/2$ for ${}^{31}\text{P}$).

  • Coupling to one equivalent neighboring ${}^{31}\text{P}$ nucleus ($n=1$) results in a doublet (2 lines).
  • Coupling to two equivalent neighboring ${}^{31}\text{P}$ nuclei ($n=1$) results in a triplet (3 lines).

Interpreting the Spectrum

For a symmetrical $\text{P}_4\text{N}_4$ cage structure with two distinct sets of phosphorus atoms (Type A and Type B):

  • If Type A phosphorus atoms primarily couple to two equivalent Type B phosphorus atoms through the cage framework, the signal for Type A atoms will appear as a triplet.
  • Similarly, if Type B phosphorus atoms couple to two equivalent Type A phosphorus atoms, their signal will also appear as a triplet.

Therefore, the ${}^{31}\text{P}\{^1\text{H}\}$ NMR spectrum is expected to show two distinct signals, each appearing as a triplet, corresponding to the two different types of phosphorus atoms in the molecule.

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Important Questions from Nmr

  1. According to Karplus equation, the vicinal proton-proton coupling constant is minimum when the value of dihedral angle is
  2. In $^1H$ NMR, the multiplicity pattern expected for the highlighted protons in the following compounds is

  3. The $^1H$ NMR of mixture of ethyl iodide and bromoform gives three signals at $\delta$ 6.80, 3.20 and 1.85 ppm with integration of 1, 3, 4.5, respectively. The molar ratio of ethyl iodide and bromoform is
  4. $^{13}C$ NMR spectrum of DMSO-$d_6$ gives a signal at $\delta$ 39.7 ppm as a
  5. The number of signals observed in the proton decoupled $^{13}\text{C}$ NMR spectrum of the following compound is

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