Why don't leaving groups act as nucleophiles in SN1 reactions?

[FeralisExtremum]

Using the example SN1 mechanism below:

So the leaving group (Br in this case) leaves and is now a Br- ion. Meanwhile, we've formed a carbocation. The next step is that water comes in and acts as the nucleophile, then eventually gets deprotonated and we have our end product.

My question is: Br- is a stronger nucleophile than H2O, so why doesn't the Br- simply re-attack the carbocation?

---

Members don't see this ad.

 

[Koalafied]

Using the example SN1 mechanism below:

So the leaving group (Br in this case) leaves and is now a Br- ion. Meanwhile, we've formed a carbocation. The next step is that water comes in and acts as the nucleophile, then eventually gets deprotonated and we have our end product.

My question is: Br- is a stronger nucleophile than H2O, so why doesn't the Br- simply re-attack the carbocation?

On paper, no.

A good nucleophile is typically compact and possesses a greater degree of basicity; and in your case, the bromide ion doesn't. Remember conjugate bases/acids? The bromide ion is a conjugate base of hydrobromic acid; a strong binary acid. The hydroxide ion is more compact and is more basic than the bromide ion. The hydroxide ions simply compete better than the bromide ions for the "attack."

 

[FeralisExtremum]

Water is certainly a better base than Br- (which explains why it deprotonates extra H on H2O in the final step), but we aren't referring to a deprotonation here. We're talking specifically about a nucleophilic attack on a carbocation, in which case we would be examining nucleophile strength, right?

 

[Koalafied]

Water is certainly a better base than Br- (which explains why it deprotonates extra H on H2O in the final step), but we aren't referring to a deprotonation here. We're talking specifically about a nucleophilic attack on a carbocation, in which case we would be examining nucleophile strength, right?

You do realize these reactions are dynamic and not statical as represented on static medium i.e. paper, screenshot, etc., right?

What exactly are you trying to ask here?

 

[FeralisExtremum]

I'm not sure what dynamic vs static has to do with my question. It's pretty straightforward:

My question is: Br- is a stronger nucleophile than H2O, so why doesn't the Br- simply re-attack the carbocation?

 

[Koalafied]

I'm not sure what dynamic vs static has to do with my question. It's pretty straightforward:

My question is: Br- is a stronger nucleophile than H2O, so why doesn't the Br- simply re-attack the carbocation?

Using the example SN1 mechanism below:

So the leaving group (Br in this case) leaves and is now a Br- ion. Meanwhile, we've formed a carbocation. The next step is that water comes in and acts as the nucleophile, then eventually gets deprotonated and we have our end product.

My question is: Br- is a stronger nucleophile than H2O, so why doesn't the Br- simply re-attack the carbocation?

On paper, no.

A good nucleophile is typically compact and possesses a greater degree of basicity; and in your case, the bromide ion doesn't. Remember conjugate bases/acids? The bromide ion is a conjugate base of hydrobromic acid; a strong binary acid. The hydroxide ion is more compact and is more basic than the bromide ion. The hydroxide ions simply compete better than the bromide ions for the "attack."

Then I believe I've already answered your question.

Honestly, it has been quite awhile. And in which case, I don't know how else to explain this.

 

[asne0407]

The better nucleophile is the stronger base which in this case is H2O (OH-) being the base is stronger than Br-.

 

[asne0407]

Another thing to think about...if Br was indeed a better nucleophile then water why would it leave and come back on, that would be a waste of energy..

 

[ChrisM07]

There are these notes floating around. You might want to take a look, they're called Ferali's Not...es...

Oh. Right. 😳

Anyways, I'll second asne0407. OH as a base is stronger than Br and will be a better nucleophile. Also, Br is larger than OH.

 

[IsaacYankem]

So the leaving group (Br in this case) leaves and is now a Br- ion. Meanwhile, we've formed a carbocation. The next step is that water comes in and acts as the nucleophile, then eventually gets deprotonated and we have our end product.

My question is: Br- is a stronger nucleophile than H2O, so why doesn't the Br- simply re-attack the carbocation?

Oxygen is the stronger nucleophile, it's far more electronegative than Br. Especially the Br- ion which has it's valence, and being a larger atom is relatively stable when compared to O

 

[FeralisExtremum]

That's what I'd thought too, but almost every source I've looked up ranks the halide ions as strong nucleophiles versus water as a weak nucleophile. e.g.:

 

[Koalafied]

That's what I'd thought too, but almost every source I've looked up ranks the halide ions as stronger nucleophiles (but weaker bases) than water. e.g.:

You do realize that water autoionizes in solution, right?

 

[FeralisExtremum]

You do realize that water autoionizes in solution, right?

The mechanism in question shows water - not its ions - acting directly as the nucleophile.

 

[Koalafied]

You do realize I am referring to a mechanism where water - not its ions - acts directly as the nucleophile, right?

I don't.

Lets not get nebulous with regards to what medium this reaction is being held in.

So in what type of solution is this reaction being held in, then? (e.g. aqueous, etc.)

 

[FeralisExtremum]

This is just a straightforward SN1 alkyl halide reaction, so unfortunately the medium isn't really specified in any of the mechanisms I've looked up (since it's SN1, whatever solvent it is must be polar protic). My assumption would be that the solvent and the nucleophile are the same here, though. Perhaps the resulting hydronium ions from the steps above stabilizes the Br- ions?

 

[Koalafied]

This is just a straightforward SN1 alkyl halide reaction, so unfortunately the medium isn't really specified in any of the mechanisms I've looked up (since it's SN1, whatever solvent it is must be polar protic). My assumption would be that the solvent and the nucleophile are the same here, though. Perhaps the resulting hydronium ions from the steps above stabilizes the Br- ions?

I know. So what's the confusion then?

The basic concepts being asked/tested at hand are as follow:

• Understanding pKa
• Understanding conjugate acid/base
• Understanding rate limiting step
• Understanding polarizability
• Understanding basic trends i.e. electronegativity, atomic radius, basicity, and acidity
• Understanding autoionization
• Understanding rearrangement, if any
• Understanding leaving groups
  Just to name a few.

The word nucleophile takes root from Greek: "nucleo" nucleus, "phile" lover of, hence lover of nucleus.

Then again, nucleophiles are not that important in an Sn1 RXN; only in Sn2 as far as you're concerned for the scope of the DAT.

You do realize that an effective solvent is needed in order to proceed past the rate limiting step of an Sn1 RXN, right? Do you know which kind?

 

[FeralisExtremum]

The confusion is why the halide ion does not act immediately as a nucleophile after leaving, since it is a stronger nucleophile that the water molecule. You seem to be questioning why being able to answer that question is relevant, which is a valid point for the purposes of the DAT, but doesn't actually answer the question itself.

The effective solvent, as mentioned above, would be polar protic for SN1. But I'm not sure how that plays in (if at all) to the main question here.

 

[helloworld9]

Hey Ferali,

I did some searching around and thought I found the right answer so many times, but it was simple. The leaving group leaves (because in a polar protic solvent like water, the carbocation will be more stable if the bromine left) and then the bromine basically gets neutralized by hydrogen bonding. You might be wondering how that's possible because usually H-bonds need F, O, or N atoms but according to this site:

"Increasing stabilization of the nucleophile by the solvent results in decreasing reactivity. Thus, polar protic solvents will stabilize the chloride and bromide ions through the formation of hydrogen bonds to these smaller anions."

This will leave the carbocation wide open for attack from a better nucleophile such as water that is readily available (also because the rate law does not depend on the nucleophile (Rate=K[electrophile]), water is simply waiting around ready to attack once the bromine leaves.

I'll include a link to the page here:
http://www.organic-chemistry.org/namedreactions/nucleophilic-substitution-sn1-sn2.shtm

Hope this helps answer your question! But that's what I would say, is that the bromine probably would attack it...if its hands weren't caught up being surrounded by water molecules.
Another good read here:
http://www.masterorganicchemistry.com/2012/12/04/deciding-sn1sn2e1e2-the-solvent/

 

[FeralisExtremum]

Hey Ferali,

I did some searching around and thought I found the right answer so many times, but it was simple. The leaving group leaves (because in a polar protic solvent like water, the carbocation will be more stable if the bromine left) and then the bromine basically gets neutralized by hydrogen bonding. You might be wondering how that's possible because usually H-bonds need F, O, or N atoms but according to this site:

"Increasing stabilization of the nucleophile by the solvent results in decreasing reactivity. Thus, polar protic solvents will stabilize the chloride and bromide ions through the formation of hydrogen bonds to these smaller anions."

This will leave the carbocation wide open for attack from a better nucleophile such as water that is readily available (also because the rate law does not depend on the nucleophile (Rate=K[electrophile]), water is simply waiting around ready to attack once the bromine leaves.

I'll include a link to the page here:
http://www.organic-chemistry.org/namedreactions/nucleophilic-substitution-sn1-sn2.shtm

Hope this helps answer your question! But that's what I would say, is that the bromine probably would attack it...if its hands were caught up being surrounded by water molecules.
Another good read here:
http://www.masterorganicchemistry.com/2012/12/04/deciding-sn1sn2e1e2-the-solvent/

That answered it, thank you so much!

 

[Koalafied]

The confusion is why the halide ion does not act immediately as a nucleophile after leaving, since it is a stronger nucleophile that the water molecule. You seem to be questioning why being able to answer that question is relevant, which is a valid point for the purposes of the DAT, but doesn't actually answer the question itself.

The effective solvent, as mentioned above, would be polar protic for SN1. But I'm not sure how that plays in (if at all) to the main question here.

http://en.wikipedia.org/wiki/Leaving_group