Dissociation rates: trans-effect rule in square planar complexes

In lecture, the trans-effect was described.

A ligand $L^t$ with a higher trans-effect as $L$ (cis to $L^t$) leads to a faster dissociation of ligand $L^d$ (trans to $L^t$). I would expect that the attacking ligand $B$ is needed to make a stronger bond as the leaving ligand $L^d$ did.

In order to see if these assumptions are always valid in planar quadratic systems I've set up following problem according to the given trans-effect order:

$\ce{CO}$ has a stronger trans-effect as $\ce{CH3-}$ and therefore the ligand trans to $\ce{CO}$ will be exchanged faster as trans to the $\ce{CH3-}$. But the leaving group $\ce{I-}$ would also have a higher trans-effect as $\ce{Cl-}$ so I wouldn't expect a substitution in b). Are these statements correct?

First off, there are two nice articles on the trans effect on Wikipedia and LibreTexts that are worth reading.

You need to remember that there are two factors leading to the trans effect:

• Weakening the $\ce{Pt-X}$ bond trans to another group (i.e., a structural effect)
• Stabilizing the 5-coordinate transition state, since substitution reactions in square planar complexes are generally associative. (i.e., a kinetic effect)

(a) I don't think the $\ce{I-}$ will substitute for the $\ce{NH3}$ group. Instead, I think the $\ce{CO}$ "wins" the trans effect "competition" and so the trans $\ce{CH3-}$ will be replaced by the $\ce{I-}$.

Next, substitution with the $\ce{Cl-}$ should replace the $\ce{I-}$.

Now your second question seems to be if you'll actually get a substitution with $\ce{Cl-}$ since $\ce{I-}$ has a greater trans effect. Yes, you will. All of these are in equilibrium, so if you generate a lot of $\ce{I-}$ that can substitute back (and get starting material). But you'll definitely get the Cl-substituted product too if you separate the reaction mixture.

Indeed, the prototypical example for the trans effect is synthesis of trans-platin or cis-platin, starting from either $\ce{[PtCl4]^{-2}}$ or $\ce{[Pt(NH3)4]^{+2}}$. Even though $\ce{NH3}$ has a smaller trans effect, you can synthesize cis-platin from $\ce{[PtCl4]^{-2}}$ and substituting two $\ce{Cl-}$ with $\ce{NH3}$.

• The trans effect on Wikipedia is quite easy to understand because there are only two different groups involved. Did I correctly understood; for step a) you've only chosen the ligand which has the highest trans effect? I'd argue that in the starting material $\ce{CO}$ trans to $\ce{CH3-}$ is more stable as $\ce{CO}$ trans to $\ce{I-}$ because of the better donor properties of $\ce{CH3-}$ there is a better push-pull-stabilization. But I think my argument is wrong because it is a thermodynamic argument and not a kinetic argument. – laminin Oct 13 '14 at 19:19
• The unfortunate problem with trans effect is that there is a thermodynamic structural effect and a kinetic effect. Yes, you pick the leaving group trans to the strongest substituent. I'm not sure I follow the rest of your argument here. – Geoff Hutchison Oct 13 '14 at 20:16
• I'd argue that in the starting material $\ce{CO}$ trans to $\ce{CH3-}$ is more stable as $\ce{CO}$ trans to $\ce{I−}$ because of the better donor properties of $\ce{CH3-}$ there is a better push-pull-stabilization that means $\ce{CO}$ favors electrons from a strong trans placed donor for a good $\Pi$ Backbonding. The methyl group acts already as the better donor as iodine does. – laminin Oct 14 '14 at 8:25
• That's true, but besides the point. The $\ce{CO}$ group won't substitute in either case. – Geoff Hutchison Oct 14 '14 at 16:25
• @laminin That's the point. Unfortunately, people call "trans effect" for two different things, but the key in your question is the kinetic effect on the 5-coordinate transition state. So there's no "exception" because these are one and the same thing. The substituent trans to the highest "kinetic trans" effect will leave. – Geoff Hutchison Oct 21 '14 at 1:48