In the reaction, CH _ 3 COOH LiAlH _ 4 → A PCl _ 5 → B Alc. KOH → C...

${LiAlH}_{4}$ is a reducing agent and reduces carboxylic acid to alcohol.

${\text{CH}}_{3} COOH \overset{{\text{LiAlH}}_{4}}{\to} {\text{CH}}_{3} - {\text{CH}}_{2} - \text{OH}$
(A)

$\overset{{\text{PCl}}_{5}}{\to} {\text{CH}}_{3} - {\text{CH}}_{2} - \text{Cl} \mathop{\to}_{\text{KOH}}^{\text{alc}} {\text{CH}}_{2} = {\text{CH}}_{2}$