NaCl reaction formula

Hi Guys,

I'm in need of some help desperately. I need to know what happens when you mix NaI with (CH3)2CO and then mix it with C4H9Cl. I think it forms NaCl, but I don't know how! I'm trying to work what the formula is for it.

Any help would be greatly appreciated! I'm new to this chemistry stuff...

- Row.

[Paul Robbins]

when you mix NaI with (CH3)2CO

At this stage you're dissolving the first reactant (sodium iodide) in acetone.

and then mix it with C4H9Cl.

Now you're bringing the reactants (sodium iodide and 1-chlorobutane) together in acetone.

When this happens, you'll get an SN2 reaction. The Cl on the chlorobutane will be replaced by I to produce 1-iodobutane.

As you said, NaCl will be formed. Unlike NaI, NaCl is insoluble in acetone and it will appear as a white precipitate.

C₄H₉Cl + NaI ----> C₄H₉I + NaCl_(s)