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Question 57

In the reaction, $$CH_3COOH \xrightarrow{LiAlH_4} A \xrightarrow{PCl_5} B \xrightarrow{\text{Alc. KOH}} C$$, the product C is:

Solution

Reduction of $$\text{CH}_3\text{COOH}$$ by $$\text{LiAlH}_4$$ yields ethanol ($$\text{A}: \text{CH}_3\text{CH}_2\text{OH}$$).

Reaction of ethanol with $$\text{PCl}_5$$ produces ethyl chloride ($$\text{B}: \text{CH}_3\text{CH}_2\text{Cl}$$).

Treatment of ethyl chloride with alc. $$\text{KOH}$$ causes dehydrohalogenation to form ethylene ($$\text{C}:\text{CH}_2=\text{CH}_2$$).

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