Statement: Ethyl bromide reacts with alcoholic KOH to form C2H4. Analysis: - Ethyl bromide (C2H5Br) reacts with alcoholic KOH, which acts as a strong base.- The reaction mechanism is E2 (bimolecular elimination), where the base abstracts a proton from the beta carbon, and the bromine ...
<p>To solve the question, we need to analyze each statement regarding the reactions of ethyl bromide (C2H5Br) and determine which one is incorrect. </p><p>1. <strong>Statement 1</strong>: When C2H5Br reacts with alcoholic KOH, it forms ethene (C2H4).