If an organic compound undergoes oxidation, any oxygen atom incorporated into the organic molecule is taken to have come from a water molecule, which also forms $2\text{H}^+ + 2\text{e}^-$. Any hydrogen atom removed may be regarded as being lost as $\text{H}^+ + \text{e}^-.$ These alterations may be shown by the two equations below. $\text{H}_2\text{O} \rightarrow [\text{O}] + 2\text{H}^+ + 2\text{e}^-$ $[\text{H}] \rightarrow \text{H}^+ + \text{e}^-$ Compound X is oxidised by heating under reflux with hot, acidified potassium dichromate(VI) for one hour. The half-equation for the reduction reaction is shown. $\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{e}^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}$ At these conditions, one mole of potassium dichromate(VI) oxidises three moles of X. What could X be?
- Apropanal
- Bpropan-1-ol
- Cpropan-1,2-diol
- Dpropan-1,3-diol