Question

Ethyl alcohol on heating with $HI$ in presence of $ZnC{l}_2$ yields:

• A. ethane
• B. ethyl Iodide
• C. ethylene
• D. methane

Answer 1

The correct option is B ethyl Iodide

This is an example of a substitution reaction
ethanol in presence of $ZnC{l}_2$ with $HI$ undergoes carbocation formation which undergoes an attack by $I-$nucleophile to ethyl iodide. $I-$attacks the substrate since its active concentrated is more n its a stronger nucleophile.