TBR Chemistry 9.62 - transition metal complex substitution w/ phosphine ligands

[whistle47]

This seemed like a straightforward problem, but I guess not.

TBR Chemistry Ch 9 Passage IX Q62

The passage describes an experiment to study the chelating effect on transition metal complex substitutions. The reactant is an octahedral complex with three sites bound to a certain polymer and the other three sites occupied by three identical phosphines. C4H13N3 is added, and the passage states that the three amine functional groups will replace the three phosphine ligands, creating product polymer-M-C4H13N3.

62. If in one reaction the R group of the phosphine is methyl, what is observed when the methyl is replaced by ethyl?
A. Increased reaction rate with ethyl because steric hindrance of the leaving group has increased.
B. Decreased reaction rate with ethyl because steric hindrance of the leaving group has increased.
C. Increased reaction rate with ethyl because steric hindrance of the leaving group has decreased.
D. Decreased reaction rate with ethyl because steric hindrance of the leaving group has decreased.

C and D are obviously incorrect. B seemed reasonable because of the "steric hindrance" keyword - when hindrance is the main factor in substitution reactions, increased hindrance leads to a lower reaction rate. But apparently the answer is A.

Spoiler: Explanation

Choice A is correct. The ethyl substituent is larger than the methyl substituent, so the triethyl phosphine is bulkier than the trimethyl phosphine. This makes the triethyl phosphine a better leaving group. The rate of the reaction increases with the ethyl substituents, making choice A the best answer. Whether the reaction mechanism is associative or dissociative, the leaving group affects the reaction rate.

This feels unsatisfying. I did notice that I didn't give much weight to the leaving group's stability, but that was because the question specified "steric hindrance". That phrasing implies that steric hindrance effects dominate leaving group stability effects, which would lead to overall decreased reaction rate.

But suppose I assume that the mechanism is dissociative because it's octahedral, which is contextually ugh-hair-tearing but chemically reasonable. Okay, the leaving group stability is now the most important factor. But why is Et3P a better leaving group than Me3P? I thought that Et3P would be worse at leaving than Me3P since gas phase basicities of amines increase with alkyl substitution, and comparing phosphines to amines seemed reasonable since they are in the same family. But I looked up the electronegativity of phosphorus and now I could see the argument for Et3P being a better leaving group... but I thought that knowing exact electronegativity values was not the MCAT's style. Is there a solvent effect or something?

IDK. I can understand their explanation, but the question still feels like a total toss-up (which makes me sad and not sleep well).

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[amy_k]

I think they may be trying to through you off by saying "steric hindrance" rather than "bulkiness", but leaving group stability increases with size because larger, bulkier leaving groups have a lower density of electrons, so Et3P is a better leaving group than Me3P.

 

[chemtopper]

This seemed like a straightforward problem, but I guess not.

TBR Chemistry Ch 9 Passage IX Q62

The passage describes an experiment to study the chelating effect on transition metal complex substitutions. The reactant is an octahedral complex with three sites bound to a certain polymer and the other three sites occupied by three identical phosphines. C4H13N3 is added, and the passage states that the three amine functional groups will replace the three phosphine ligands, creating product polymer-M-C4H13N3.

62. If in one reaction the R group of the phosphine is methyl, what is observed when the methyl is replaced by ethyl?
A. Increased reaction rate with ethyl because steric hindrance of the leaving group has increased.
B. Decreased reaction rate with ethyl because steric hindrance of the leaving group has increased.
C. Increased reaction rate with ethyl because steric hindrance of the leaving group has decreased.
D. Decreased reaction rate with ethyl because steric hindrance of the leaving group has decreased.

C and D are obviously incorrect. B seemed reasonable because of the "steric hindrance" keyword - when hindrance is the main factor in substitution reactions, increased hindrance leads to a lower reaction rate. But apparently the answer is A.

Spoiler: Explanation

Choice A is correct. The ethyl substituent is larger than the methyl substituent, so the triethyl phosphine is bulkier than the trimethyl phosphine. This makes the triethyl phosphine a better leaving group. The rate of the reaction increases with the ethyl substituents, making choice A the best answer. Whether the reaction mechanism is associative or dissociative, the leaving group affects the reaction rate.

This feels unsatisfying. I did notice that I didn't give much weight to the leaving group's stability, but that was because the question specified "steric hindrance". That phrasing implies that steric hindrance effects dominate leaving group stability effects, which would lead to overall decreased reaction rate.

But suppose I assume that the mechanism is dissociative because it's octahedral, which is contextually ugh-hair-tearing but chemically reasonable. Okay, the leaving group stability is now the most important factor. But why is Et3P a better leaving group than Me3P? I thought that Et3P would be worse at leaving than Me3P since gas phase basicities of amines increase with alkyl substitution, and comparing phosphines to amines seemed reasonable since they are in the same family. But I looked up the electronegativity of phosphorus and now I could see the argument for Et3P being a better leaving group... but I thought that knowing exact electronegativity values was not the MCAT's style. Is there a solvent effect or something?

IDK. I can understand their explanation, but the question still feels like a total toss-up (which makes me sad and not sleep well).

Check the stability of the complex with bulky groups:
(C2H5)3P is much bulkier than (CH3)3P and presence of three such groups in the complex increases the steric hindrance on the complex.Here reaction rate increases because the complex is very unstable with triethylphosphine .The new complex formed has more rings due to tridentate ligand (presence of three amine groups) and is more stable.Removal of bulky ligands decreases the steric hindrance around the metal complex and this is the main cause of increase in the rate of the reaction.

 

[Odi]

@amy_k is right. You're over thinking the concept being applied.

Always think of leaving groups in terms of bases. A good leaving group is a weak base which should make intuitive sense; because a strong base is binding tightly to other atoms. A weak base on the other hand doesn't bind itself as tight and for the most part stays dissociated in solution making it a great LG.

Acid dissociation:

H-A -> H+ & A-

Is no different than:

H-LG -> H+ & LG-

The same rules that apply to acids and conjugate bases apply to leaving groups. As the size of the anion (conjugate base) increases, the strength of the acid increases because dissociation increases; think acid strength HI > HBr > HCl because I- is the largest anion which makes it the best leaving group--because of it's low electron density it forms weaker bonds.

Same application to Me3P versus Et3P.

Et3P is a larger leaving group and thus a weaker base; so it will dissociate better than Me3P allowing for a faster reaction.

The correct answer A says steric hindrance "of the leaving group has increased" which is the same thing as saying "the size of the leaving group has increased."

The scenario you're thinking of is if the steric hindrance of a non-leaving group substituent had increased, which in that case you'd be correct to pick B.

 

[whistle47]

I think they may be trying to through you off by saying "steric hindrance" rather than "bulkiness", but leaving group stability increases with size because larger, bulkier leaving groups have a lower density of electrons, so Et3P is a better leaving group than Me3P.

yeah, the phrasing really threw me off because usually steric hindrance is used in a very different context. oh well. it's all about the least incorrect answer right???

Check the stability of the complex with bulky groups:
(C2H5)3P is much bulkier than (CH3)3P and presence of three such groups in the complex increases the steric hindrance on the complex.Here reaction rate increases because the complex is very unstable with triethylphosphine .The new complex formed has more rings due to tridentate ligand (presence of three amine groups) and is more stable.Removal of bulky ligands decreases the steric hindrance around the metal complex and this is the main cause of increase in the rate of the reaction.

this is a nice perspective. I think it might have been what the question writers had in mind. the phrasing of the answer choices actually makes sense since steric hindrance of the ethyl groups is what increases the dissociation rate of the leaving groups.

@amy_k is right. You're over thinking the concept being applied.

Always think of leaving groups in terms of bases. A good leaving group is a weak base which should make intuitive sense; because a strong base is binding tightly to other atoms. A weak base on the other hand doesn't bind itself as tight and for the most part stays dissociated in solution making it a great LG.

Acid dissociation:

H-A -> H+ & A-

Is no different than:

H-LG -> H+ & LG-

The same rules that apply to acids and conjugate bases apply to leaving groups. As the size of the anion (conjugate base) increases, the strength of the acid increases because dissociation increases; think acid strength HI > HBr > HCl because I- is the largest anion which makes it the best leaving group--because of it's low electron density it forms weaker bonds.

Same application to Me3P versus Et3P.

Et3P is a larger leaving group and thus a weaker base; so it will dissociate better than Me3P allowing for a faster reaction.

The correct answer A says steric hindrance "of the leaving group has increased" which is the same thing as saying "the size of the leaving group has increased."

The scenario you're thinking of is if the steric hindrance of a non-leaving group substituent had increased, which in that case you'd be correct to pick B.

well, phosphines aren't particularly basic in water. plus, Et3N has a higher pKa than Me3N. polarizability/softness makes sense when comparing the conjugate bases of hydrogen halides but I don't think thinking about base strength is as useful here.

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Lessons learned:
1. be careful of "incorrectly" used keywords/keyphrases
2. don't assume sn1/sn2 or associative/dissociative mechanism without solid reasoning
3. reactant instability, not just product stability, contributes towards reaction rate
4. phosphate is less electronegative than carbon and is much softer than nitrogen

thanks everyone 🙂

 

[erythrocyte666]

Whoa, organometallics is like my favorite chemistry!
Yeah, since it's an octahedral complex, the mechanism will most likely be dissociative (however, the number of electrons contributed by the metal and ligands should be 18 to further confirm dissociative mechanism...though for square planar complexes like of Pt or Rh you can have dissociative even with 16 electrons - check out the hydroformylation reaction mechanism with Wilkinson's catalyst). Because the mechanism is dissociative, ligand dissociation is the rate-limiting step, so having very bulky ligands will allow this to occur more quickly. In other words, having bulky ligands makes the complex inherently unstable, raising its energy, and hence the activation energy barrier to ligand dissociation is not as high. A good bit of research has been done on the Tolman cone angles of phosphines and their effects on reaction rates. Triphenylphosphine would be even faster since it has a significantly larger cone angle than triethylphosphine, which has a larger cone angle than trimethylphosphine.
Also, because phosphorus has lower electronegativity than nitrogen, phosphine is a STRONGER field ligand than amine (despite the fact that it is more diffuse than nitrogen). However, here you have a kappa-3 triamine chelate replacing individual phosphines. A chelate like that will have extremely high binding affinity to the metal due to the proximity and entropic effect (the amine chelators are nearby and you don't have the entropy decrease inherently required for association of 2 ligands).
Also, I'm curious what this complex looks like since several other factors like the electron donation (or pi-backdonation, depending on the number of electrons available on the metal as well as the electronegativity of atoms present in the other ligands) capability of the other ancillary ligands and their location relative to each other (known as the trans effect vs. trans influence) make a huge difference in how the ligand substitution proceeds.