A) butanoic acid
B) 2-methyl butanoic acid
C) \[1,1,1-\]trichloro\[-2-\]methyl butane
D) \[1.4-\]butanediol
Correct Answer: B
Solution :
\[C{{H}_{3}}CH=CH-C{{H}_{3}}+CHC{{l}_{3}}\xrightarrow{{}}\] \[C{{H}_{3}}-\overset{CC{{l}_{3}}}{\mathop{\overset{|}{\mathop{CH}}\,}}\,-C{{H}_{2}}-C{{H}_{3}}\xrightarrow[\text{hydrol}]{\text{NaOH}}\] \[C{{H}_{3}}-\overset{C(OH)3}{\mathop{\overset{|}{\mathop{CH}}\,}}\,-C{{H}_{2}}-C{{H}_{3}}\xrightarrow[{}]{-{{H}_{2}}O}\] \[\underset{2-\text{methyl}\,\text{butanoic}\,\text{acid}}{\mathop{C{{H}_{3}}-\overset{COOH}{\mathop{\overset{|}{\mathop{CH}}\,}}\,-C{{H}_{2}}C{{H}_{3}}}}\,\]
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