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I was provided with the following structure of an inorganic metal complex:

Pt metal complex

And asked to give information on its NMR spectrum shown below

NMR of Pt metal complex

The highfield signal is caused by a singlet with Platinum ($I = \frac{1}{2}$) due to the terminal $\ce{CH3}$ groups.

As $\ce{^{195}Pt}$ has an abuance of $33.8\%$ this meant that in $66.2\%$ of cases, the singlet will not split, whereas in around $33.8\%$ of cases a doublet would be observed due to the satellites, which would have intensities of $\frac{33.8}{2} = 16.9\%$ hence the "Triplet" was in the ratio of around $16:67:16$.

The low-field signal however for me was easy to deduce but more difficult to calculate ratios for. The Methyl's could couple to

  • No $\ce{^{195}Pt}$ resulting in a singlet
  • One $\ce{^{195}Pt}$ resulting in a doublet
  • Two $\ce{^{195}Pt}$ resulting in a singlet

The calculations given in the answer are shown:

  • No $\ce{^{195}Pt}$ resulting in a singlet: intensity $0.66 \times 0.66$
  • One $\ce{^{195}Pt}$ resulting in a doublet: intensity $2 \times 0.66 \times 0.33$
  • Two $\ce{^{195}Pt}$ resulting in a singlet: intensity $0.33 \times 0.33$ This results in a final ratio of 1:8:18:8:1

These calculations confused me. Why would for a singlet, the intensity be $0.66 \times 0.66$ if both $\ce{CH3}$ give the same signal and hence for the doublet would I need to multiply $0.66 \times 0.33$ by $2$?

These calculations also seemed erroneous. The calculation for the intensity of the doublet and singlet give the exact same answer and I could not reach the final ratio.

What is the logic behind these calculations?

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1 Answer 1

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Seems to me there is an error in the last line of this list:

  • No $\ce{^{195}Pt}$ resulting in a singlet: intensity $0.66 \times 0.66$
  • One $\ce{^{195}Pt}$ resulting in a doublet: intensity $2 \times 0.66 \times 0.33$
  • Two $\ce{^{195}Pt}$ resulting in a singlet: intensity $0.33 \times 0.33$ This results in a final ratio of 1:8:18:8:1

Simultaneous J-couplings to two $\ce{^{195}Pt}$ should result in a triplet, not a singlet, given that the J-couplings are identical in magnitude.

Therefore the expected relative intensities in the quintet are as follows:

$$\begin{align} &0.34^2\times \frac14 &= 0.029 \tag{1} \\&0.66\cdot0.34\times1 &=0.224 \tag{2} \\& 0.34^2\times \frac24 + 0.66^2 \times 1 &=0.493 \tag{3} \\& 0.66\times0.34\times1 &=0.224 \tag{4} \\& 0.34^2\times \frac14 &=0.029\ \tag{5} \end{align}$$

Note there is an implicit assumption that molecules with two $\ce{^{195}Pt}$ result in a 1:2:1 triplet contributing to peaks (1) (3) and (5) above, and that molecules with one $\ce{^{195}Pt}$ account entirely for peaks (2) and (4).

Normalized so that the central peak (3) has an integral of 18 the expected integrals become

$$\begin{align} & 1.0 \\&8.2 \\& 18.0 \\& 8.2 \\&1.0\end{align}$$

Separately I attempted digitizing and integrating the provided image of the spectrum. I obtained the following integrals:

$$\begin{align} & 0.25 \\&7.0 \\& 18.0 \\& 6.2 \\&0.7\end{align}$$

Given the resolution perhaps the disagreement is acceptable.

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  • $\begingroup$ My answer of course presumes that the nmr spectrum corresponds to the shown compound, and that its existence is even possible. After searching for a while I have not found anything remotely similar, but I cannot claim much expertise in platinum chemistry. $\endgroup$
    – Buck Thorn
    Dec 25, 2019 at 21:00

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