Discussion Forum

 In the reaction,  $CH_{3}COOH$    $\underrightarrow{LiAlH_{4}}$  A   $\underrightarrow{PCl_{5}}$ B   $\underrightarrow{Alc.KOH}$  C .The product C is

Answer:

Explanation:

 This problem is based on successive reduction, chlorination and elimination reaction. To solve such a problem, use the function of the given reagents.

(i) LiAlH₄  causes a reduction

(ii) PCl₅ causes chlorination

(iii)  Alc. KOH causes elimination reaction

    $CH_{3}COOH$ $\underrightarrow{LiAlH_{4}}$ $CH_{3}CH_{2}OH$ (A) $\underrightarrow{PCl_{5}}$

    $CH_{3}CH_{2}Cl$ (B) $\xrightarrow[{Alc.KOH}]{-HCl}$ $CH_{2}=CH_{2}$ (C)