How to determine which one is a better leaving group in SN2

1. determine by size of the group and the bond strength of C-X bond:
size: eg. I>Br>Cl>>F which I being the easiest to undergo SN2
and C-I < C-Br <C-Cl << C-F in terms of bond strength

2. yes

3. yes, because it can hardly change back to its acid, which is strong (i.e. more reactive). in other words, weak conjugate base -> stable species already.

4. you have to show whether that's an aqueous condition / in non-polar solvent.
for 1st condition, assume that's in aq. the reaction is not feasible because the
conc of Cl- is not high. The bond of C-Br is not that weak to let Cl- break
C-Br and form C-Cl.

for the 2nd condition, if the bromoethane is put in conc HCl, i guess the
reaction can happen as H+ conc is high, which feasible for formation of HBr.
This can weaken C-Br bond and let Cl- to attack the carbon easier.

correct me if i am wrong

2008-07-11 00:54:47 補充：
to 冷風:

or i should say in this way:
though KCl can dissolve in water easily, there's no chemical species to weaken C-Br bond, so Cl- cannot be easily attack to the nucleophilic carbon.

correct me if i am wrong.

You say the lower concentration of chloride ions in aqueous state???
I think your statement is wrong.
Since potassium chloride is ionic compound.
It can dissolve in water easily, so the conc. of chloride ions in water is quite high.

The final reaction would probably require heat to get it going, even if the conc. of HCl is high. This is because Cl is still a weak nucleophile than Br. HBr is produced as a result as mentioned and you would need 1eq. of base (eg. K2CO3) to 'mop up' the HBr in order for the reaction to go.