Determination of the molecular formula of alcohols

Exercise 5

$0.704\;g $ of a saturated aliphatic primary mono-alcohol is quantitatively oxidized to carboxylic acid. This acid is titrated by exactly $20\;mL\;NaOH\;O.4 M $ 1) Determine the molecular formula of this alcohol! 2) Give the semi-structural formula knowing that it is optically active! 3) Represent the structural formula of the R enantiomer!

1) Titration: $n_{NaOH}$ $=$ $0.2\cdot 0.040$ $=$ $0.008 \;mol $ = $n_{acide}$ Oxidation: As the oxidation is quantitative, $n_{alcool}$ $=$ $n_{acide}$ $=$ $0.008 \;mol$ $M_{alcool}$ $=$ $\frac{0.704}{0.008}$ $=$ $88 \frac{g}{mol}$ General formula: Saturated aliphatic primary mono-alcohol: $C_nH_{2n+2}O$ $12n$ $+$ $2n$ $+$ $2$ $+$ $16$ $=$ $88$ $n=5$ Molecular formula: $C_5H_{12}O$ 2) Only possibility because of the optical activity: $CH_3CH_2C^*H(CH_3)CH_2OH$ 3) (R)-2-methylbutan-1-ol