Quote:
|
when you mix NaI with (CH3)2CO
|
At this stage you're dissolving the first reactant (sodium iodide) in acetone.
Quote:
|
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.
C
4H
9Cl + NaI ----> C
4H
9I + NaCl
(s)