Oxidation of alcohols by strong oxidants in acid medium

Exercise 1

1) Establish the reaction equations of all isomers $C_3H_7OH$ with an excess of potassium permanganate in an acid medium. Calculate each time the ratio $\frac{n_A}{n_MnO_4^-}$

1) 1st isomer: propan-1-ol - 1st step: Propan-1-ol $\Rightarrow$ propanal: $CH_3CH_2CH_2OH$ $-$ $2e^-$ $\rightarrow$ $CH_3CH_2CHO+2H^+$ $MnO_4^-$ $+$ $5e^-$ $+$ $8H^+$ $\rightarrow$ $Mn^{2+}$ $+$ $4H_2O$ $5CH_3CH_2CH_2OH$ $+$ $2MnO_4^-$ $+$ $6H^+$ $\rightarrow$ $5CH_3CH_2CHO$ $+$ $2Mn^{2+}$ $+$ $8H_2O$ - 2nd step: Propanal $\Rightarrow$ propanoic acid: $CH_3CH_2CHO$ $-$ $2e^-$ $+$ $H_2O$ $\rightarrow$ $CH_3CH_2COOH$ $+$ $2H^+$ $MnO_4^-$ $+$ $5e^-$ $+$ $8H^+$ $ \rightarrow$ $Mn^{2+}$ $+$ $4H_2O$ $5CH_3CH_2CHO$ $+$ $2MnO_4^-$ $+$ $6H^+$ $\rightarrow$ $5CH_3CH_2COOH$ $ +$ $ 2Mn^{2+}$ $+$ $3H_2O$ Ratio: $\frac{n_{alcohol}}{n_{permanganate}}$ $=$ $\frac{5}{4}$ 2nd isomer: propan-2-ol Propan-2-ol $\Rightarrow$ propanone: $CH_3CHOHCH_3$ $-$ $2e^-$ $ \rightarrow$ $CH_3COCH_3$ $+$ $2H^+$ $MnO_4^-$ $+$ $5e^-$ $+$ $8H^+$ $ \rightarrow $ $Mn^{2+}$ $+$ $4H_2O$ $5CH_3CHOHCH_3$ $+$ $2MnO_4^-$ $+$ $6H^+$ $\rightarrow$ $5CH_3COCH_3$ $+$ $2Mn^{2+}$ $+$ $8H_2O$ Ratio: $\frac{n_{alcohol}}{n_{permanganate}}$ $=$ $\frac{5}{2}$

2) Take a $9.5\;mL$ of alcohol of the formula $ C_3H_7OH$, add water to obtain a $100\; mL $ solution. $10\; mL$ of this solution are acidified, and then a $0.2 \;M$ potassium permanganate solution is added gradually. The mixture turns pink after addition of $25.4\; mL$ of this solution. What alcohol is it and why?

2) In $10\; mL$: $n_A=0.0127\; mol$ Dans $25.4\; mL$: $n_{MnO_4^-}$ $=$ $0.0254\cdot 0.2$ $ =$ $ 0.00508 mol$ $\frac{n_A}{n_{MnO_4^-}}$ $=$ $ 2.5$ $ =$ $ \frac{5}{2}$ It is propan-2-ol !