A mono-alcohol $A$ has a density of $0.79 \frac {kg} {L}$. a) By reaction of $9.5\;mL$ of A with an excess of sodium are collected $1.4\;L\; H_2$ (STP). Find the molecular formula of $A$.
$m_A$ $=$ $0.79\cdot 9.5$ $ =$ $ 7.5\; g$ $n_{H_2}$ $=$ $\frac{1.4}{22.4}$ $=$ $0.0635\; mol$ 2 mole A produce une mole $H_2$ $n_A$ $=$ $2\cdot 0.0635$ $=$ $0.127\; mol$ $M_A$ $=$ $\frac{m_A}{n_A}$ $=$ $60\;\frac{g}{mol}$ Formula of A: $C_nH_{2n+2}O$ $12n+2n+2+16$ $=$ $60$ so $n=3$ Molecular formula: $C_3H_7OH$
b) Establish equations of reactions of all isomers of $A$ with an excess of potassium permanganate in an acid medium. Calculate each time the ratio $\frac{n_A}{n_MnO_4^-}$
1st isomer: propan-1-ol
-1st step:
Propan-1-ol $\Rightarrow$ propanal:
$CH_3CH_2CH_2OH$ $-$ $2e^-$ $\rightarrow$ $CH_3CH_2CHO$ $+$ $2H^+$ | $\cdot 5$
$MnO_4^-$ $+$ $5e^-$ $+$ $8H^+$ $\rightarrow$ $Mn^{2+}$ $+$ $4H_2O$ | $\cdot 2$
$5CH_3CH_2CH_2OH$ $+$ $2MnO_4^-$ $+$ $6H^+$ $\rightarrow$ $5CH_3CH_2CHO$ $+$ $2Mn^{2+}$ $+$ $8H_2O$
- 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:
$\frac{n_{alcool}}{n_{permanganate}}=\frac{5}{4}$
2nd isomer: propan-2-ol
Propan-2-ol $\Rightarrow$ propanone:
$CH_3CHOHCH_3$ $-$ $2e^-$ $\rightarrow$ $CH_3COCH_3$ $+$ $2H^+$ | $\cdot 5$
$MnO_4^-$ $+$ $5e^-$ $+$ $8H^+$ $\rightarrow$ $Mn^{2+}$ $+$ $4H_2O$ | $\cdot 2$
$5CH_3CHOHCH_3$ $+$ $2MnO_4^-$ $+$ $6H^+$ $\rightarrow$ $5CH_3COCH_3$ $+$ $2Mn^{2+}$ $+$ $8H_2O$
Ratio:
$\frac{n_{alcool}}{n_{permanganate}}$ $=$ $\frac{5}{2}$
c)At a second sample of $9.5\;mL$ water was added to obtain $100\;mL$ solution. $10\;mL$ of this solution are acidified, and then titrated with a $0.2\;M$ potassium permanganate solution . The mixture turns pink after addition of $25.4\;mL$ of this solution. What alcohol is it and why?
In $10\; mL$: $n_A$ $=$ $0.0127\; mol$ In $25.4\; mL$: $n_{MnO_4^-}$ $=$ $0.0254\cdot 0.2$ $=$ $0.00508 mol$ $\frac{n_A}{n_{MnO_4^-}}$ $=$ $2.5$ $=$ $\frac{5}{2}$ It is propan-2-ol !
