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The $\ce{^{119}Sn}$ chemical shift of the dimeric tin hydride shown below (R=terphenyl) changes on addition of $\ce{D2}$ from a 1:2:1 triplet at $\pu{657.9 ppm}$ (due to coupling between $\ce{Sn}$ and bridging $\ce{H}$) to a doublet at $\pu{650.6 ppm}$. I assume that's because of reversible dissociation of one of the bridging $\ce{H}$, which allows formation of a mixed species where both $\ce{H}$ and $\ce{D}$ are present (the doublet being due to coupling with the remaining $\ce{H}$). However, I'm not quite certain why that alters the $\ce{^{119}Sn}$ chemical shift (I know the $\ce{Sn-D}$ bond will be shorter/stronger due to the lower ZPE). I would be interested in any thoughts/comments.

Tin hydride

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    $\begingroup$ You can probably find some relevant literature if you search for "nmr isotope shift". Not sure what level of theory you're looking for: some of the older papers can get really theoretical! $\endgroup$
    – orthocresol
    Mar 31 at 14:17
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OP's main question: Why would substitution of $\ce{H}$ with $\ce{D}$ alter chemical shift?

OP has correctly indicated that the $\ce{Sn-D}$ bond will be shorter/stronger than $\ce{Sn-H}$ due to the lower ZPE, which may cause the chemical shift difference. That is the major reason for isotope effect in chemical kinetics mainly due to bond strength (shorter the bond distance, stronger the bond). However, here, how change of isotope effected the chemical shift is caused by increasing the electron density of the adjacent atom ($\ce{Sn}$). Since $\ce{Sn-D}$ distance is shorter than $\ce{Sn-H}$, the sharing electron pair in $\ce{Sn-D}$ bond is closer to $\ce{Sn}$ tan that in $\ce{Sn-H}$ bond. Thus, it is safe to say that $\ce{Sn}$ of $\ce{Sn-D}$ is shielded more than that of $\ce{Sn-H}$. Therefore, we can expect up-field shift (red-shift) in relevant $\ce{Sn}$ chemical shift. That's what you have experienced $(\pu{650.6 ppm} - \pu{657.9 ppm} = -\pu{7.3 ppm})$ when you have changed the $\ce{H}$ with $\ce{D}$ of the dimer. This phenomena is common among deuterated organic compounds (Ref.1):

$$ \begin{array}{llrrr} \hline \text{Hydrated cpd} & \text{deuterated cpd} & \delta \ \left(\ce{^{13}C-H}\right)/\pu{ppm} & \delta \ \left(\ce{^{13}C-D}\right)/\pu{ppm} & \Delta \delta/\pu{ppm}\\ \hline \ce{CHCl3} & \ce{CDCl3} & 77.36 & 77.16 \pm 0.06 & -0.20 \\ \ce{CH3C#N} & \ce{CD3C#N} & 1.79 & 1.32 \pm 0.02 & -0.47 \\ \ce{(CH3)2C=O} & \ce{(CD3)2C=O} & 30.60 & 29.84 \pm 0.01 & -0.76 \\ \ce{C6H6} & \ce{C6D6} & 128.62 & 128.06 \pm 0.02 & -0.56 \\ \ce{CH3-OH} & \ce{CD3-OH} & 49.86 & 49.00 \pm 0.01 & -0.86 \\ \hline \end{array} $$

These chemical shifts are also depend on the other factors such as temperature and solvent. For comparison purposes, all chemical shifts of hydrated compounds have been recorded in their deuterated versions (e.g., $\ce{CHCl3}$ in $\ce{CDCl3}$) at a constant temperature (Ref.1).

Note: I also realized that $^1\!J_{\ce{Sn-D}}$ constant must be zero or negligible according to your observation. I tried to find the value of $^1\!J_{\ce{Sn-D}}$, but failed. Note that in deuterated organic solvents, $^1\!J_{\ce{C-D}}$ varies between $\pu{20 Hz}$ and $\pu{30 Hz}$ (Ref.1).

References:

  1. Hugo E. Gottlieb, Vadim Kotlyar, Abraham Nudelman, "NMR Chemical Shifts of Common Laboratory Solvents as Trace Impurities," J. Org. Chem. 1997, 62(21), 7512-7515 (DOI: https://doi.org/10.1021/jo971176v).
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