[sharbeldam]

1-butene reacts with Cl2 with excess NaI, we should first get chloronium ion between carbon 1 and 2, then on which carbon will the I- attack? the primary or secondary? as I have read somewhere, that the more electronegative halogen will be bonded on the more substituted carbon. wouldn't that mean that the iodine will be on the primary carbon at the end since it's less electronegative than Chlorine?

Thanks in advance.

[spirochete]

Nucleophile attacks the more substituted carbon, which is better able to support positive charge. Which one will be the nucleophile here?

[sharbeldam]

Well we have two nucleophiles after the first chlorine attacks, the I- and the Cl-, if we had water i know that OH- will attack the more substituted carbon, but what about halogens? here we have I- and Cl-, what is the the rule? which halogen will attack second and on which carbon? it's confusing cause Chlorine is more electronegative but Iodine is more nucleophilic in a polar protic solvent...

[spirochete]

Your logic is good. The rest is somewhat an exercise in mind reading. By saying "excess" they basically want you to assume that iodide will be the nucleophile.

When you said "OH-", probably H2O would be the nucleophile in that case. But that doesn't relate to your question directly, just a random minor suggestion. Halogenation of an alkene makes an acid by product, so OH- couldn't be involved. It would be too basic to form.

Halohydrin formation (where H2O is the nucleophile) is another example where relative nucleophilicity doesn't seem to matter. Water is the nucleophile, even though the halide is stronger, because it is present in excess.

[sharbeldam]

So to sum it up, when I have an excess salt like NaCl or NaI, the halogen in the salt will attack the more substituted carbon in the second step ?

[spirochete]

Correct.