A) \[C{{H}_{2}}==CHCOOH\]
B) \[C{{H}_{2}}CHClCOOH\]
C) \[C{{H}_{3}}C{{H}_{2}}CN\]
D) \[C{{H}_{3}}C{{H}_{2}}OH\]
Correct Answer: A
Solution :
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.
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