[pantsboy]

Is I^- a strong enough nucleophile to cause 1-chloro-2-methylbutane to react via SN2?  Or is Cl^- to poor of LG for the reaction to happen?  I'm assuming it's SN2 because the solvent is aprotic (acetone).  Also would 1-chloropentane proceed quicker because is less substituted even though both are primary halides?



I'm just curious as to what is generally the 'cutoff' for reliable LG's?  Is it mainly dependent on other factors such as temperature, substrate, nucleophilicity, and dielectric strength of solvent?

[azmanam]

leaving group ability is inversely proportional to base strength.  Very stable anions make very weak bases.  Very stable anions also make very good leaving groups.  Cl- is the conjugate base of a very strong acid (HCl), thus Cl- is a very weak base.  Cl-, Br-, and I- are all very good leaving groups.  The SN2 will happen with no problem.  Fun fact of the day: The reaction of one alkyl halide with a halide anion in acetone (like the reaction of an alkyl chloride with sodium iodide in acetone) is called the Finkelstein.  Probably the coolest name for a reaction you'll see in the entire course

[Dan]

Iodide is a very good nucleophile, and chloride is a good leaving group, and the carbon is primary - so even in a protic solvent I'd still predict S_N2.

As Azmanam said, when considering how good a leaving group is, think about it's stability. The chloride ion is quite stable - consider how easily HCl dissociates - and so is a good LG. Note that this is also why iodide is a good leaving group as well - it is a good Nu and a good LG. As a result iodide can displace itsself in an S_N2 reaction - if an alkyl iodide has an asymmetric centre at the carbon bearing the iodine substituent, treatment with iodide can induce racemisation.

You are correct about 1-chloropentane reacting faster than 1-chloro-2-methylbutane. Bulky substituents adjacent to the (as well as at the) reactive centre will hinder the nucleophile - the classic example would be neopentyl compounds, which have a tertiary position adjacent to the reactive centre - e.g. 1-chloro-2,2-dimethylpropane - these react very slowly by S_N2 due to steric hinderance, and only very slowly by S_N1 because the carbocation intermediate (which usually rearranges in this situation) would still be primary.

Yeah, Finkelstein is a great name! Maybe only beaten by the Chichibabin reaction.

[tmartin]

Oh man you guys beat me to it twice while I typed this : (

I- can indeed displace the chloride.  This is commonly referred to as the Finkelstein reaction.  It works with a variety of primary and secondary alkyl halides (where the leaving group can be chloride or bromide, and in some cases a "pseudo-halide" like a mesylate or tosylaye).  The reaction does require heating, and utilizes differences in solubility.  NaI is soluble in acetone, whereas NaCl and NaBr are not, so they will crash out of solution.  Following Le Chatelier's Principle, this pushed the equilibrium toward exchange.

http://www.organic-chemistry.org/namedreactions/finkelstein-reaction.shtm

In general you can view halides (except fluorine) as a capable leaving group.  Also mesylates (-OMs) and tosylates (-OTs) are good leaving groups.  It does greatly depend upon conditions though, as -OH is in general a poor leaving group, it can be eliminated under different circumstances (generally not Sn2 reactions though).  One thing to look at for a leaving group is how well it can handle a negative charge, and the conditions of the reaction (acidic or basic).  The substrate is important in some cases, as tertiary carbons disfavor Sn2.