
\(\begin{array}{*{20}{c}} {C{H_3} - CH - CH = C{H_2}} \\ {|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ {C{H_3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \end{array}\xrightarrow{{{O_3}/{H_2}{O_2}}}\)
\(\begin{array}{*{20}{c}} {C{H_3} - CH - COOH + HCOOH} \\ {|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ {C{H_3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \end{array}\)


નીપજ $(C)$ શું હશે ?
એસીટોએસીટીક એસ્ટર $\xrightarrow{NaOEt}$ $\xrightarrow{C{{H}_{3}}I}$ $\xrightarrow{NaOEt}$ $\xrightarrow{C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}Br}$
$A$, (પરમાણુ સૂત્ર $\left.{C}_{6} {H}_{12} {O}_{2}\right)$ સાથે સીધી સાંકળ શૃંખલા $C_{4}$ કાર્બોક્સિલિક એસિડ આપે છે. $A$ શું છે:
$A \frac{{Li} {A} {H} {H}_{4}}{{H}_{3} {O}^{+}} \longrightarrow B \stackrel{\text { Oxidation }}{\longrightarrow} {C}_{4}-$ કાર્બોક્સિલિક એસિડ