Question

What will happen if ethyl iodide is treated with zinc and distilled

Answer 1

Explanation:

I know this much only

Answer 2

End Product is Propane

Explanation:

Subjecting haloalkane to a reducing medium such as Zn + Conc. HCl,$H_2$/Raney Nickel,

Zn-Cu couples,$C_2H_5OH$ will give a direct replacement of hydrogen with halogen.

This can be understood through a simple reaction:

$RX+2H -- >RH+HX$

Now as per the question,

$CH_3CH_2I+(Zn-Cu+C_2H_5OH)+H_2/Renay\ Nickel+Zn+Conc.\ HCl --> CH_3CH_2CH_3+HI$

Hence,we can say that the final products of ethyl iodide ,Zn-Cu couples and ethanol are Propane along with some by products.