Question

What is the action of following on ethyl iodide a) alc.KOH b) Zn, HCl

Answer 1

Explanation:

a) Ethyl iodide on reaction with alcoholic KOH gives ethene as a product.

$C_2H_5I+KOH(alc)\rightarrow C_2H_4+KI+H_2O$

b) When ethyl iodide reacts with HCl in presence of zinc it forms ethane as a final product.

The nascent hydrogen released due to reaction between HCl and zinc is reduces ethyl iodide to ethane

$C_2H_5I\overset{Zn/HCl}\rightarrow C_2H_6+HI$

Answer 2

Explanation:

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