Question

$C{{H}_{3}}C{{H}_{2}}COOH\xrightarrow[Fe]{C{{l}_{2}}}X\xrightarrow{alc.\,KOH}Y,$ the compound $Y$ is:

• A. $C{{H}_{2}}==CHCOOH$
• B. $C{{H}_{2}}CHClCOOH$
• C. $C{{H}_{3}}C{{H}_{2}}CN$
• D. $C{{H}_{3}}C{{H}_{2}}OH$

Answer 1

Answer: (A)

$C{{H}_{2}}==CHCOOH$

Answer 2

Propanoic acid, on treatment with halogen, in presence of catalyst, gives a-halo derivative. Thus, the complete reaction is as follows $C{{H}_{3}}.C{{H}_{2}}COOH\xrightarrow[Fe]{C{{l}_{2}}}\,C{{H}_{3}}-\overset{\begin{smallmatrix} Cl \\\ | \end{smallmatrix}}{\mathop{C}}\,H.COOH$ $\xrightarrow[\text{Dehydrodehalogenation}]{\text{alc}\text{.}\,\text{KOH}}\underset{\text{Acrylic}\,\,\text{acid}}{\mathop{C{{H}_{2}}==CH\cdot COOH}}\,$ Hence, the product is acrylic acid.