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)

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.