Question

\( \mathrm{CH_3COOH \xrightarrow{LiAlH_4} X \xrightarrow{PCl_5} Y \xrightarrow{Alcoholic\ KOH} Z} \). Find product \( Z \) in above reaction.

• A. Ethyne
• B. Ethanol
• C. Ethene
• D. Ethanal

Hint:

Remember the sequence: carboxylic acid \( \rightarrow \) primary alcohol by \( \mathrm{LiAlH_4} \), alcohol \( \rightarrow \) alkyl chloride by \( \mathrm{PCl_5} \), and alkyl chloride \( \rightarrow \) alkene by alcoholic KOH.

Answer 1

The Correct Option is C

Step 1: Reduction of acetic acid by \( \mathrm{LiAlH_4} \).
Acetic acid, \( \mathrm{CH_3COOH} \), is reduced by \( \mathrm{LiAlH_4} \) to give the corresponding primary alcohol. Therefore, compound \( X \) is ethanol, \( \mathrm{CH_3CH_2OH} \).
\[ \mathrm{CH_3COOH \xrightarrow{LiAlH_4} CH_3CH_2OH} \] Step 2: Reaction of ethanol with \( \mathrm{PCl_5} \).
Alcohols react with \( \mathrm{PCl_5} \) to form alkyl chlorides. Hence, ethanol is converted into ethyl chloride, \( \mathrm{CH_3CH_2Cl} \). So, compound \( Y \) is ethyl chloride.
\[ \mathrm{CH_3CH_2OH \xrightarrow{PCl_5} CH_3CH_2Cl} \] Step 3: Reaction of ethyl chloride with alcoholic KOH.
Alcoholic KOH causes dehydrohalogenation of alkyl halides. In this reaction, one molecule of HCl is eliminated from ethyl chloride, producing an alkene. Thus, ethyl chloride gives ethene, \( \mathrm{CH_2{=}CH_2} \).
\[ \mathrm{CH_3CH_2Cl \xrightarrow{Alcoholic\ KOH} CH_2{=}CH_2} \] Step 4: Conclusion.
So, the final product \( Z \) obtained in the reaction sequence is ethene.
Final Answer:Ethene.