${\mathrm{CH}}_3\mathrm{CHO}\underset{Pt}{\overset{{\mathrm{H}}_2}{\to }}A\underset{573\mathrm{K}}{\overset{\mathrm{Cu}}{\to }}\mathrm{B}\stackrel{{\mathrm{NaBH}}_4}{\to }\mathrm{C}$ $\frac{\text{ Conc. }{\mathrm{H}}_2{\mathrm{SO}}_4}{443\mathrm{K}}\to D$$\underset{\text{ b) }{\mathrm{H}}_2{\mathrm{O}}_2/{\mathrm{OH}}^{-}}{\overset{\text{ a) }{\mathrm{BH}}_3/\text{ THF }}{\to }}E$  In the above conversion, the compound ethyl alcohol is