Elementary analysis of carboxylic acids

Exercise 3

    

The reaction of a saturated monoalcohol $A$ on an unsaturated carboxylic acid $B$ forms an ester $E$. $3.7\;g$ of $B$ was obtained by hydrating $2.8\;g$ of an alkene $F$. Deduce the molecular formula of $B$ and possible condensed formulas. The controlled oxidation of $B$ gives a compound which reacts with DNPH but not with Fehling's solution. What is the condensed formula of $B$? $V=50\;cm^3$ aqueous solution containing $0.400\;g$ of $A$ consumes $17.5\;mL\;0.5\;M \;NaOH$ by titration. Find the condensed formula of $A$ Find the condensed formula of $E$ and its name.

1) Hydratation reaction: $C_nH_{2n}$ $+$ $H_2O$ $\rightarrow$ $C_nH_{2n+2}O$ Alcene + water $\rightarrow$ alcohol x mol alcene give x mol alcohol: $\frac{x\cdot M_{F}}{x\cdot M_{B}}$ $=$ $\frac{14n}{14n+18}$ $=$ $\frac{2.8}{3.7}$ so $n$ $=$ $4$ Molecular formula of $ B$: $C_4H_{10}O$ $B$ is a secondary alcohol: Condensed formula of $B$: $CH_3CHOHCH_2CH_3$ 2) Titration: At equivalence: $n_{acid}$ $=$ $n_{NaOH}$ $=$ $0.5\cdot 0.0175$ $=$ $0.00875$ $M_{acid}$ $=$ $\frac{0.400}{0.00875}$ $\approx$ $46 \frac{g}{mol}$ General formula: $C_nH_{2n}O_2$ $12n$ $+$ $2n$ $+$ $32$ $=$ $46$ $n$ $=$ $1$ Molecular formula: $CH_2O_2$ Condensed formula of $A$: $HCOOH$ $E$ is 1-methylpropylmethanoate $HCOOCH(CH_3)CH_2CH_3$