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Old May 20th, 2007, 16:10
Row
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Default 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.
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Old January 10th, 2010, 13:24
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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.

C4H9Cl + NaI ----> C4H9I + NaCl(s)
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