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GUJCET 2026 Chemistry Paper

Question : 1 of 40

Marks: +1, -0

\text{CH}_3\text{COOH} \xrightarrow{\text{LiAlH}_4} \text{X} \xrightarrow{\text{PCl}_5} \text{Y} \xrightarrow{\text{Alcoholic KOH}} \text{Z}

Z

' in the above reaction.

Solution:

Concept:

Carboxylic acid is reduced to alcohol, then converted to alkyl halide, and finally dehydrohalogenated to alkene.

Explanation:

\mathrm{CH}_3\mathrm{COOH} \xrightarrow{\mathrm{LiAlH}_4} \mathrm{CH}_3\mathrm{CH}_2\mathrm{OH}

(X, ethanol).

\mathrm{CH}_3\mathrm{CH}_2\mathrm{OH} \xrightarrow{\mathrm{PCl}_5} \mathrm{CH}_3\mathrm{CH}_2\mathrm{Cl}

(Y, ethyl chloride).

\mathrm{CH}_3\mathrm{CH}_2\mathrm{Cl} \xrightarrow{\text{alcoholic KOH}} \mathrm{CH}_2=\mathrm{CH}_2

(Z, ethene).

Answer:

Product Z is ethene, so the correct option is C.

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