# SN2 reaction

Discussion in 'General Science' started by boks, Nov 20, 2008.

1. ### boks Thread Starter Active Member

Oct 10, 2008
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0

Which of these five compounds will react easier after an S[sub]N[/sub]2 mechanism?

Since Cl is more electronegative than I, it is a better leaving group, so I would expect the answer to be B or D.

In B the halide is tertiary while in D it is primary, so I expect D to be the answer.

The correct answer is E. Why?

2. ### triggernum5 Senior Member

May 4, 2008
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Wouldn't your thoughts be more appropriate if the question had asked which compound is more likely to undergo an Sn2 rxn, rather than which will react more easily after the fact?

3. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
522
I didn't think we had done with relative strengths of acids yet?

However here goes.

Yes E is the correct answer.
This is due to the difference in solvation of the nuclephiles (anions). The smaller the anion the more densely solvated it is. The solvent molecules pack more tightly around the greater charge density of the Cl$^{-}$ shielding it from attack.

In fact the reactivity due to this is well marked increasing right to left with attacking group and downwards with leaving group over the table.

So F<<Cl<Br<I. It is very dificult to get Flourine to participate at all in these reactions.

By the way what are you studying for?