This is the reaction that took place:


This is the information from the $\ce{^1H}$ NMR spectrum:

$$\begin{array}{|c|c|c|} \hline \delta \textbf{ / ppm} & \textbf{Relative intensity} & \textbf{Multiplicity} \\ \hline 1.65 & 3 & \text{d, } J = 6.2\ \mathrm{Hz} \\ \hline 1.73 & 15 & \text{s} \\ \hline 2.88 & 1 & \text{dd, } J = 11.0,1.5\ \mathrm{Hz} \\ \hline 3.28 & 1 & \text{dd, } J = 6.6,1.5\ \mathrm{Hz} \\ \hline 3.58 & 1 & \text{dq, } J = 7.0,6.2\ \mathrm{Hz} \\ \hline 3.96 & 1 & \text{ddd, } J = 11.0,7.0,6.6\ \mathrm{Hz} \\ \hline \end{array}$$

What is the product formed?

I know the product only contains carbon, hydrogen, rhodium, and chlorine, and I think its molecular formula is $\ce{C14H21ClRh}$ (perhaps $\ce{H22}$ instead of $\ce{H21}$). My initial thought was that the diene could displace one chloride ligand to form $\ce{[RhCp^*Cl(C4H6)]}$.

  • $\begingroup$ There are two approaches to solving this - one coming from NMR analysis, and the other coming from understanding the chemistry. Ideally they lead to the same solution! The NMR data can only give one possible solution here. Do you know what [RhCp*Cl2]2 is used for? This will help you along the right path. $\endgroup$ – long Oct 10 '16 at 21:58

The precatalyst system [RhCp*Cl2]2 was developed by Peter Maitlis and his group at the University of Sheffield in the 1970's and 80's. It has been used as a hydrogenation catalyst under a number of conditions, and is readily commercially available. Under the right conditions, ethanol can act as a suitable transfer hydrogenation source (not quite as good as 2-propanol, but better than methanol), but in this case, in the presence of a suitable base (sodium carbonate), the rhodium dimer will react to give a stable product.

$$\begin{array}{|c|c|c|} \hline \textbf{label} & \delta \textbf{ / ppm} & \textbf{Relative intensity} & \textbf{Multiplicity} \\ \hline \textbf{A} & 1.65 & 3 & \text{d, } J = 6.2\ \mathrm{Hz} \\ \hline \textbf{B} & 1.73 & 15 & \text{s} \\ \hline \textbf{C} & 2.88 & 1 & \text{dd, } J = 11.0,1.5\ \mathrm{Hz} \\ \hline \textbf{D} & 3.28 & 1 & \text{dd, } J = 6.6,1.5\ \mathrm{Hz} \\ \hline \textbf{E} & 3.58 & 1 & \text{dq, } J = 7.0,6.2\ \mathrm{Hz} \\ \hline \textbf{F} & 3.96 & 1 & \text{ddd, } J = 11.0,7.0,6.6\ \mathrm{Hz} \\ \hline \end{array}$$

From the NMR, it is possible to deduce what structure is likely, if we construct a valid spin system from the available data.

  • signal B: δ1.73 integral of 15 : we still have a Cp*
  • signal A: δ1.65 integral of 3 : looks a lot like we have a methyl group. It has a doublet coupling of 6.2Hz, which corresponds to signal E at δ3.58.
  • signal E: δ3.58 is single 1H which appears as a dq. We can account for the q coupling; it is next to a methyl group, which leaves the doublet coupling of 7.0Hz, which corresponds to signal F at δ3.96
  • signal F: δ3.96 is a single 1H with additional coupling to 2 other single 1H nuclei; one of 11.0Hz (signal C), which is pseudo-trans, and the other at 6.6Hz (pseudo-cis, signal D). This two signals have a mutual coupling of 1.5Hz, typical of a pseudo-gem couping on a double bond.

So that gives a spin system skeleton that looks like:

enter image description here

which, is able to form a stable π complex. Homework spoiler alert:

enter image description here

and a useful reference from many years ago....

K. Moseley, J. W. Kang and P. M. Maitlis J. Chem. Soc. A, 1970, 2875-2883. DOI: 10.1039/J19700002875


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