what will treatment of ammonia with excess of ethyl iodide yeild ??????
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Solution
as per given, reaction takes place as follows,
NH3+C2H5I→C8H20N+NH3+C2H5I→C8+C8H20N+C8H20N
is the chemical formula of tetraethylammonium iodide
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Explanation:
From the above reaction, we can see that when ethyl iodide in excess is made to react with ammonia, the product formed would be tetraethylammonium iodide.
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