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: Understanding the Concept:
This is a multi-step conversion involving reduction, halogenation, and dehydrohalogenation.
Step 2: Formula Application:
1. $\mathrm{CH_3COOH} \xrightarrow{\mathrm{LiAlH_4}} \mathrm{CH_3CH_2OH}$ ($X$ = Ethanol). 2. $\mathrm{CH_3CH_2OH} \xrightarrow{\mathrm{PCl_5}} \mathrm{CH_3CH_2Cl}$ ($Y$ = Chloroethane). 3. $\mathrm{CH_3CH_2Cl} \xrightarrow{\text{alc. KOH}} \mathrm{CH_2=CH_2}$ ($Z$ = Ethene).
Step 3: Explanation:
Step 1 reduces the acid to a primary alcohol. Step 2 replaces the hydroxyl group with a chlorine atom. Step 3 involves $\beta$-elimination to form an alkene.
Step 4: Final Answer:
The final product Z is Ethene.