MCQ
$NaOH/{H^ + }$ reacts with
- A${C_6}{H_5}OC{H_3}$
- B$C{H_3}OH$
- ✓$C{H_3} - \mathop C\limits^{\mathop {||}\limits^O } - C{H_3}$
- D${C_2}{H_5}OH$
✓
Answer
Correct option: C.
$C{H_3} - \mathop C\limits^{\mathop {||}\limits^O } - C{H_3}$
c
$2C{H_3} - CO - C{H_3}\xrightarrow{{dil\,NaOH}}$ $\mathop {\begin{array}{*{20}{c}}
{\begin{array}{*{20}{c}}
{OH\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,O} \\
{\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,||}
\end{array}} \\
{C{H_3} - C - C{H_2} - C - C{H_3}} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{C{H_3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}
\end{array}}\limits_{(Diacetone\,\,\,alcohol)} $
$2C{H_3} - CO - C{H_3}\xrightarrow{{dil\,NaOH}}$ $\mathop {\begin{array}{*{20}{c}}
{\begin{array}{*{20}{c}}
{OH\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,O} \\
{\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,||}
\end{array}} \\
{C{H_3} - C - C{H_2} - C - C{H_3}} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{C{H_3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}
\end{array}}\limits_{(Diacetone\,\,\,alcohol)} $
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