Nucleophilicity and Leaving Groups?!?

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mikemohan0482

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My organic chem textbook tells me that nucleophilicity increases as you go down a column on the periodic table and thus I- will be a stronger nucleophile than Cl-. This makes sense to me.

Then in the leaving group section they say for the same types of reactions (SN2), leaving group stability is inversely related to nucleophilicity and basicity and this makes sense as well because the less nucleophilic an ion, the less it will resist being displaced.

What I don't get is that they then state I- is a better leaving group than Cl-. How can I- be both a stronger nucleophile and a better leaving group? Cl- should be the better leaving seeing as how it is the more stable anion and weaker base.

What's going on here? Anyone please?

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Hmmm.. well, first of all do you have the Kaplan books? They have a pretty good explanation in there. Otherwise, let me do my best to explain and then you shall owe me 3 virgins.

In a PROTIC solution (Sn1), LARGE atoms are better nucleophiles because they are more polarizable.. this is also related to how good a leaving group they would be as well.

In an APROTIC solution (Sn2), nucleophiles are related to their basicity (electron donating). So you'd get F- as the mega-nucleophile.

----

As I said, in terms of leaving groups, think "bigger is better" for Sn1. For Sn2's.. just make sure it's a DAMN strong nucleophile that can kick off the leaving group.
 
I- is a better leaving group than Cl- because it has a smaller Electroneg. So it won't be holding on as tight and then will be more willing to "let go" than Cl-. At least that is how I think about it. Good Luck
 
I thought Iodide was less stable than chloride. This should make cl- a better leaving group than I-. This is why HI is more reactive but less acidic than HF right?

edit: I have HI and HF backwards here... don't want to mess someone up. This was on one of last years exams.

HF = more reactive
HI = far more acidic
 
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No, Cl is a worse leaving group than I-.. too tightly packed.. small atomic radii. Whereas I- is a big ass son of a bitch.

In terms of nucleophility Cl- would be good for Sn2 reactions because it's small and says "Bitch, I'm coming in the backdoor."
 
ahh yes...'tis true. I was imagining the halogen period upside down for some reason. Another way to think about it is recognize that iodide has a larger area over which to spread the negative charge.
 
Okay so I think the missing link in my understanding was realizing nucleophilicity varies with the solvent used. In protic solvents (SN1/E1), large anions will be more nucleophilic despite their often decreased basicity because they can shed solvent more easily (more polarizable). In aprotic (SN2/E2), all you need to look at is how basic they are.

One final detail I need to clear up is then with leaving groups. So in aprotic solutions since solubilization won't be an issue, the most stable anions will be best (eg. I-), right?

What I'm not sure of though is what the situation is in (SN2/E2) protic solutions. Because solubilization is a stabilizing force that affects small molecules like F- the most (making them weak nucleophiles), shouldn't they be the best leaving groups in this situation?
 
Originally posted by mikemohan0482
One final detail I need to clear up is then with leaving groups. So in aprotic solutions since solubilization won't be an issue, the most stable anions will be best (eg. I-), right?

What I'm not sure of though is what the situation is in (SN2/E2) protic solutions. Because solubilization is a stabilizing force that affects small molecules like F- the most (making them weak nucleophiles), shouldn't they be the best leaving groups in this situation?

In an aprotic solution F- would be the best. The key word is Aprotic. It's not being buffered by a bunch of protons whereas in PROTIC solutions I- would be better.

Also Sn1 is in PROTIC solution whereas Sn2 is APROTIC. Makes sense if you think about it. Sn1 just has to create a stable cation and will like to be STABILIZED by a PROTIC solution. Whereas Sn2 is a ONE step process so the better NUCLEOPHILE is something that's SMALL and can attack from behind with STRONG force.

Hope this makes sense.
 
Ok,

For Leaving Groups:

For either Sn1, Sn2, E1, or E2:
Let's say "LG" is the leaving group.

In any of those 4 reactions, the LG is taking electrons along with it. In the transition state (TS), the LG becomes partially negative. The group that can stabilize the TS best is the one that will have a lower Activation energy (and thus a faster reaction).

The atoms/compounds that can stabilize that extra negative charge are those that can DELOCALIZE and HOLD/PULL that negative charge. Delocalization comes in two flavors (RESONANCE and SIZE OF THE ANION). Now obviously, no one ever heard of resonance during the TS, but it helps lower the TS.

HOLDING/PULLING negative charge towards the nucleus (positive charge) doesn't allow the electrons to leave and go back to the Carbon that it's leaving the electrons with. Therefore the HIGHER the ELECTRONEGATIVITY (EN-) of the LG, the better it is as a LG. Also, the higher EN- of the LG, the more positive the carbon becomes, making it "hotter" for a Nucleophile to attack it.

So, for DELOCALIZATION, the more RESONANCE (like in COO-, OTs, OTf, etc) the more stable that negative charge is, therefore the better LG (for TS and product). Also, notice how the atoms holding that negative charge is a very EN- atom (Oxygen), thereby making them even better leaving groups.
Also, the SIZE of the anion allows for that extra negative charge to be spread out all over the anion (electrons aren't static), therefore in ANY SOLVENT, I- > Br- > Cl- > F- (notice that EN- is practically negligible). HOWEVER, everyone knows that radius of atom DECREASES across a period (F is smaller than O). That size difference is negligible, and EN- wins out.
Also, larger atoms like I have a longer bond distance to C (a smaller atom). This larger distance (compared to F) makes it easier to break the bond. Usually atoms with similar size (like C and F) make stronger bonds.

So BIGGER ANION, MORE RESONANCE, AND MORE EN- tend to produce best LEAVING GROUPS
The same reasoning applies to Bases. The best Acids are those that have the BEST LG's.

Quick note on solvents...if the solution is POLAR PROTIC, F- LG is OK, b/c it'll be stabilized via Hydrogen bonds. However, that stabilization is still way lower than the delocalization of electrons in a big atom.

NUCEOPHILICITY

First off, Nucleophiles (Nu) are different from Bases, however good Bases tend to be good Nu.
A GOOD BASE is opposite of a good LG. So, smaller anion size (F- > Cl- > Br- > I-), minimal resonance, low EN- (C- > N- > O- > F-). So it's solely based on how much NEGATIVE CHARGE is on the anion.
Therefore a BULKY BASE (with a many alkyl groups that donate e- to anion) is a better BASE (taking away Hydrogen only)...t-butyl oxide > methoxide (OCH3) > OH-
The opposite trend makes the better Nu (OH- > -OCH3 > -O(CH3)3
This is true for any solvent.

Now, about the halogens.
There are two factors: SOLVENT AND POLARIZABILITY

SOLVENT:

POLAR PROTIC VS. POLAR APROTIC

POLAR PROTIC solvents will surround smaller anions best, therefore making F- a weaker nucleophile than I-. The partial positive Hydrogen makes hydrogen bonds with anions that have a more "concentrated" negative charge. In bigger anions, the charge is dispersed so it's harder to interact.
Therefore, in Polar Protic Solvents,
I- > Br- > Cl - > F-

In POLAR APROTIC Solvents, ALL Nucleophiles become stronger because there's no solvation (nothing to stop it from attacking) because no H-bonds form. However, this time, I- is not that much stronger than F-.
In polar aprotic solvents, the trend is not that clear-cut. The trend may even be opposite.

Another BIG FACTOR is POLARIZABILITY. Larger elements have larger, more diffuse, and more polarizable electron clouds. These electron clouds allow for more effective orbital overlap in the SN2 TS. Therefore, lower TS energy and thus a faster reaction.
The 5p orbital of Iodine is easily able to distort it's electron cloud TOWARDS the electropositive sp3 hybridized Carbon atom. The less polarizable 2p orbital of flourine cannot interact as well.

Some of the stuff was taken from Vollhardt/Schore textbook.

Hope this helped 😀
 
I can make a book out of this thread. Reading all this o-chem is making me dizzy. I think I've bitched enough about o-chem for the night!
 
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