# NMR of a bridging Methyl complex

I was provided with the following structure of an inorganic metal complex: And asked to give information on its NMR spectrum shown below 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?

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.

• 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. Dec 25 '19 at 21:00