Reactions are usually in equilibrium unless a product of the 'forward reaction' leaves the solution (becomes a solid, or leaves as a gas). In this case, I- is a better leaving group than Cl- because it can host electron density more readily (bigger atomic radius) and thus this reaction takes place, but equilibrium is probably shifted left.
Reactions are usually in equilibrium unless a product of the 'forward reaction' leaves the solution (becomes a solid, or leaves as a gas). In this case, I- is a better leaving group than Cl- because it can host electron density more readily (bigger atomic radius) and thus this reaction takes place, but equilibrium is probably shifted left.
yea! What you are saying makes sense. But technically this reaction does happen, so the answer should be yes right? I is a better leaving group but it is also a better nucleophile. So it will force Cl to leave and make this reaction happen.
ok kaplan is stupid, it had this reaction and it was asking whether it would happen or not. whatever!! I think even though Cl- is a weak leaving group, it would still work out since I- is such a good nucelophile. Thanks!
Like you said, this is a bad question.
Iodine is the best leaving group and also the best nucleophile so the reaction will be in equilibrium with about half COCl and half COI (just guessing it will be equal).