[NewtoAtoms]

[macman104]

[NewtoAtoms]

Thank you so much for your help Macman.

I have gone back to the drawing board with this question and due to your help have come to some more conclusions.

c. 1,3-Dichlorobutane + sodium iodide (1 mole) in acetone - As NaI is soluble in acetone, and the I- could substitute the 1-chloro- THIS IS MY NEW CHOICE!
d. 1,3-Diiodobutane + sodium chloride (1 mole) in acetone - After researching the productivity of NaCl in acetone I learned that it is insoluable, therefore this is also SCRATCHED
e. 3-Bromo-1-iodobutane + sodium chloride (1 mole) in acetone - I have researched leaving group priority and learned that Cl-<Br-<I- therefore I think it would be safe to say that in this reaction 1-iodo would leave first, therefore not giving the correct 3-chloro-1-iodobutane as a product... so this option is now SCRATCHED.

Therefore after more work, and I am so grateful for your help... I would have to propose that c is the correct answer.

[macman104]

Congrats, nice research!  I have actually done a similar reaction with 1-chloropropane and NaI in acetone.  It's pretty cool to see the solid NaCl come out of solution and it goes very cleanly (almost 100%). 

In addition, even if in e, we decided bromine and iodine were equivalent in leaving group ability, the iodine is primary and still favored over the secondary bromine.

[NewtoAtoms]

Thank you macman for taking the time to walk me through the question.
I AM SO GRATEFUL!!!

Have a great day, and thank-you for helping me learn.