[Winga]

What is the driving force to make leaving groups to leave?

e.g. SN2 reactions, the leaving groups leave because of the backside attack of nucleophiles. Then, how about SN1 reactions?

[786mine]

S_N1 three-steps with first slowest so rate depends only on alkyl halide concentration, product is racemic since carbocation intermediate is achiral

[Winga]

I mean...

e.g. alkyl iodide, what is the driving force to lose iodide in SN1 reaction?

Alkyl iodide is much stable than the carbocation, right?

[786mine]

Wrong. I just did my free radical chapter (chapter 10 in solomons) and it said there that Idodine radicals are much the least stable of all the halides. Even if you did have an alkyl iodide, the iodine would break off instanteously and a carbocanine would be formed. i hope this helps? if not ask away.

[Mitch]

Wrong. I just did my free radical chapter (chapter 10 in solomons) and it said there that Idodine radicals are much the least stable of all the halides. Even if you did have an alkyl iodide, the iodine would break off instanteously and a carbocanine would be formed. i hope this helps? if not ask away.

There are several wrong statements in there, but I'll only pick the most blatant.

Radical formation and carbocation formation are not related! One is a homolytic bond clevage the other is a heterolytic bond clevage. People answering questions without a thorough understanding of the subject matter has been increasingly problematic.  

[movies]

Are you familiar with the Curtin-Hammett principle?  I think that may explain the situation.

[dexangeles]

the strength of the attacking nucleophile is important in SN2 mechanism, but it is  hardly a contributing factor in SN1 because the carbocation intermediate is formed so easily, with reference to Hammond's Postulate. The stability of the carbocation (which translates to lower Ea to form the intermediate) is the driving force for SN1 mechanism.

in the case of alkyl iodide, we must take note that iodide is a strong nucleophile, thus it wouldn't be a very good leaving group. considering ejecting iodide from the alkane structure thru SN2 wouldn't be feasible because the new C-Nu bond formed must be more energetic than that of C-I. by rejecting the SN2 route, SN1 route (by default) is viable. The activation energy for this reaction would be the bond energy of the C-I bond to cause heterolysis of the C-I bond. The driving force for SN1 route would thus be probability (in my opinion).

correct me if i am wrong. organic chemistry is hardly a strong area of study for me.

1. Doesn't Hammond's postulate deal with whether the transition is early or late?  This meaning that the transition state's stability is decided on whether it's structure is closer to the reactant, or the product.

2.  Isn't the carbocation the intermediate in the reaction?  I always thought that formation of the carbocation is the slow step because it is not as stable as the original compound and it takes an extreme amount of energy to force this process.  The carbocation is in a higher potential energy than the alkyl halide in example.  This show's that it is not as stable as the original structure.

3.  As far as I know, Iodine is the best leaving group, especially amongst nucleophiles.  Amongst halogens, in example, Iodine forms the weakest bond.  Fluorine is the strongest nucleophile, and forms the strongest bond.  It is solvation that hinders Fluorine to express it's anionic properties.  But with substitution in the prescence of crown ethers such as 18-crown-6, fluorine is able to show the strength of it's nucleophilicity compared to other halogens.


So, wouldn't the final product be the deciding factor in this reaction?  As long as the final product is more stable, thus in a lower potential as the originating reactant, and with the help of added energy, the reaction should be fissible.  The carbocation is not really present that long for after formation, the reaction goes so fast that the final product is actually obeserved right away.

Just my two cents

please correct me if I am wrong too

[movies]

Fanning the flames.

Hammond's Postulate:

"If two states, as, for example, a transition station and an unstable intermediate, occur consecutively during a reaction process and have nearly the same energy content, their interconversion will involve only a small reorganization of molecular structure."

The take home message:

For exothermic reactions, the transition state will resemble the reactants.  For endothermic reactions, the transition state will resemble the products.

Discuss!

[dexangeles]

Fanning the flames.

Hammond's Postulate:

"If two states, as, for example, a transition station and an unstable intermediate, occur consecutively during a reaction process and have nearly the same energy content, their interconversion will involve only a small reorganization of molecular structure."

The take home message:

For exothermic reactions, the transition state will resemble the reactants.  For endothermic reactions, the transition state will resemble the products.

Discuss!

I'm sorry if you thought there is some heat going on, I just thought it was a good discussion    no flames here  

C-I exhibits the lowest bond energy among the halogens. Given the acid strength of H-X (X: halogen) increases down the group, the basicity of halide anions decreases down the group.

but shouldn't this mean as you go down the periodic table, the better the leaving group.  I am confused now cause I swear my book kept stating that with leaving groups its I>Fr>Cl>F

CH₃I should go SN2 right, a methyl carbocation should be very unstable and would not go SN₁.  This is as far as I know with SN1 tertiary>secondary>primary>methyl and it's the opposite with SN2.

[movies]

I was just fanning the flames by tossing some info in there without answering anything.  It's a good discussion!

By the way, CH₃I definitely reacts S_N2 style.

Anyone want to attempt a molecular orbital explanation?

[Donaldson Tan]

ok. i went to read my organic chemistry textbook and realise i made a big theoretical mistake,

[corey2]

SN2 reactions proceeds first with addition, generating a negative charge transient, than with an elimination, resulting in inversion of the chiral centre involved with the substitution (if there is any).

SN1 reactions proceeds with an elimination, generating a carbocation intermediate (not a transient, a carbocation can live from milliseconds to seconds), than with the attack of the nucleophile.

[Donaldson Tan]

when you heat tertiary alkyl halide, an equilibrium is formed between the activated carbocation and the original state. the driving force for SN1 reaction would be the disruption of the equilibrium brought about by the presense of nucleophiles in the system. The disruption by nucleophiles can be kinetically prevented by protic solvents, as the solvent molecules 'hydrate' the nucleophile, making them bulkier and less mobile.

in the case of E1, the same equilibrium is setup. E1 is favoured in presence of strong bases because they can abstract H more effectively. The driving force of the E1 mechanism is that the carbocation +ve charge is neutralised by the bonding electron pair from a nearby C-H bond. This results in the ejection of H+ which readily reacts with the bases. E1 can be disrupted if the bases are allowed to act as nucleophiles. Aprotic solvents will help to stabilise the carbocation intermediate but not 'hydrate' the nucleophile. This enhances the drive for E1.

[corey2]

True, thanx for the correction.  

[dexangeles]

I guess heat itself drives the halogen off.  With the addition of heat, more compounds would have enough activation evergy to form the carbocation.  And as Geodome stated, the presence of a nucleophile only disrupts the equilibrium balance.  That makes a lot of sense, learned a new thing....thanks