In the following sequence of reactions a compound \(A\), (molecular formula \(\left.\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{2}\right)\) with a straight chain structure gives a \(C_{4}\) carboxylic acid. \(A\) is : \(A \frac{{Li} {A} {H} {H}_{4}}{{H}_{3} {O}^{+}} \longrightarrow B \stackrel{\text { Oxidation }}{\longrightarrow} {C}_{4}-\) carboxylic acid

1. A \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{COO}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{CH}_{3}\)
2. B \(\begin{array}{*{20}{c}} {OH\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\ {|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\ {{\rm{C}}{{\rm{H}}_3} - {\rm{C}}{{\rm{H}}_2} - {\rm{CH}} - {\rm{C}}{{\rm{H}}_2} - {\rm{O}} - {\rm{CH}} = {\rm{C}}{{\rm{H}}_2}} \end{array}\)
3. C \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{COO}-\mathrm{CH}_{2}-\mathrm{CH}_{3}\)
4. D \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{O}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_{2}-\mathrm{OH}\)