DAT Destroyer 2014 OChem Question

[woosh]

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From one of the questions, a 3-chloropentane was reacted with NaI/Acetone.
The solutions says that the reaction would yield a 3-iodopentane via SN2 because the original molecule is a secondary halide, and polar aprotic solvent was used.

However, I was wondering why, since I- is a better leaving group than Cl-. Would the reaction really yield 3-iodopentane?

Thank you so much in advance!

 

[DatRav27]

Not sure if this is completely true but this is how I think of it. In the solution there are far more NaI molecules than 3-chloropentane. So essentially the Cl- and I- ions both compete for a spot on the pentane. Due to a far greater number of I- ions, the major product formed would be 3-iodopentane. Youre correct in saying that I- is better LG than Cl-.

If I am wrong, someone please correct me

 

[alligator10]

I- is both a better leaving group and nucliophile than cl- its very stable as a leaving group and its high electronegitivity and greater mass than cl makes it so great

 

[woosh]

Not sure if this is completely true but this is how I think of it. In the solution there are far more NaI molecules than 3-chloropentane. So essentially the Cl- and I- ions both compete for a spot on the pentane. Due to a far greater number of I- ions, the major product formed would be 3-iodopentane. Youre correct in saying that I- is better LG than Cl-.

If I am wrong, someone please correct me

I do get what you're saying.. but I'm not a hundred percent sure with your theory. Thanks though!

I- is both a better leaving group and nucliophile than cl- its very stable as a leaving group and its high electronegitivity and greater mass than cl makes it so great

Actually, in polar aprotic solvent, I- is the weaker nucleophile than Cl-.
quote (http://www.masterorganicchemistry.com/2012/06/18/what-makes-a-good-nucleophile/)