A) \[HOC{{H}_{2}}-\underset{C{{H}_{3}}}{\mathop{\underset{|}{\mathop{CH}}\,}}\,-COOH\]
B) \[C{{H}_{3}}-\overset{OH}{\mathop{\overset{|}{\mathop{\underset{C{{H}_{3}}}{\mathop{\underset{|}{\mathop{C}}\,}}\,}}\,}}\,-COOH\]
C) \[C{{H}_{3}}-\overset{OH}{\mathop{\overset{|}{\mathop{CH}}\,}}\,-C{{H}_{3}}\]
D) \[C{{H}_{3}}C{{H}_{2}}\overset{O}{\mathop{\overset{||}{\mathop{C}}\,}}\,-\overset{O}{\mathop{\overset{||}{\mathop{C}}\,}}\,-OH\]
Correct Answer: A
Solution :
\[{{(C{{H}_{3}})}_{2}}C=O\xrightarrow{{{H}^{+}},HCN}C{{H}_{3}}-\underset{C{{H}_{3}}}{\mathop{\overset{OH}{\mathop{\underset{|}{\overset{|}{\mathop{C}}}\,}}\,}}\,-CN\] \[\xrightarrow{{{H}_{3}}{{O}^{+}}}C{{H}_{3}}-\underset{C{{H}_{3}}}{\mathop{\overset{OH}{\mathop{\underset{|}{\overset{|}{\mathop{C}}}\,}}\,}}\,-COOH\xrightarrow[\Delta ]{{{H}_{2}}S{{O}_{4}}}\] \[C{{H}_{2}}=\underset{C{{H}_{3}}}{\mathop{\underset{|}{\mathop{C}}\,}}\,-COOH\] \[PC{{l}_{5}}\xrightarrow{{}}C{{H}_{2}}=\underset{C{{H}_{3}}}{\mathop{\underset{|}{\mathop{C}}\,}}\,-COCl\xrightarrow[{{H}_{2}}{{O}_{2}}.O{{H}^{-}}]{B{{H}_{3}}.\,THF}\] \[C{{H}_{3}}-\overset{OH}{\mathop{\underset{\underset{(\text{minor})}{\mathop{C{{H}_{3}}}}\,}{\mathop{\underset{|}{\overset{|}{\mathop{C}}}\,}}\,}}\,-COOH+HOC{{H}_{2}}-\underset{\underset{(major)}{\mathop{C{{H}_{3}}}}\,}{\mathop{\underset{|}{\mathop{C}}\,}}\,H-COOH\]
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