A) butanoic acid
B) 2-methyl butanoic acid
C) 1, 1, 1-trichloro-2-methyl butane
D) 1, 4-butane diol
Correct Answer: B
Solution :
\[C{{H}_{3}}=CH-C{{H}_{3}}+CHC{{l}_{3}}\xrightarrow[{}]{{}}\] \[C{{H}_{3}}-\overset{\begin{smallmatrix} CC{{l}_{3}} \\ | \end{smallmatrix}}{\mathop{C}}\,H-C{{H}_{2}}-C{{H}_{3}}\xrightarrow[hydrolysis]{NaOH}\] \[C{{H}_{3}}-\overset{\begin{smallmatrix} C{{(OH)}_{3}} \\ | \end{smallmatrix}}{\mathop{C}}\,H-C{{H}_{2}}-C{{H}_{3}}\xrightarrow{-{{H}_{2}}O}\] \[\underset{\text{2-methyl}\,\,\text{butanoic}\,\,\text{acid}}{\mathop{C{{H}_{3}}-\overset{\begin{smallmatrix} COOH \\ | \end{smallmatrix}}{\mathop{C}}\,H-C{{H}_{2}}C{{H}_{3}}}}\,\]
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