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JEE CHEMISTRY MCQ: ALDEHYDES & KETONES & CARBOXYLIC ACIDS

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Analyzing +4 | -1
In the reaction, CH3COOH→LiAlH4A→PCl5B→Alc.  KOHC C{H_3}COOH\xrightarrow{{LiAl{H_4}}}A\xrightarrow{{PC{l_5}}}B\xrightarrow{{Alc.\,\,KOH}}C\,. The product CC is
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Correct option: C
ethylene
Option C (ethylene) is correct because the reaction sequence involves reduction of carboxylic acid to alcohol, chlorination to alkyl halide, and then elimination to alkene.

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