Question

(a) Compound $\left(A\right)$ with molecular formula $C_2{H}_{4}O$ reduces Tollen's reagent. $\left(A\right)$ on treatment with $HCN$ gives compound $\left(B\right)$. Compound $\left(B\right)$ on hydrolysis with an acid gives compound $\left(C\right)$ with molecular formula $C_3{H}_6{O}_3$ which is an optically active compound. Compound $\left(A\right)$ on reduction with $N_2{H}_4/C_2{H}_{5}ONa$ gives a hydrocarbon $\left(D\right)$ of molecular formula $C_2{H}_6$. Identify $\left(A\right),\left(B\right),\left(C\right)$ and $\left(D\right)$ and explain the reactions.
(b) Ionic conductance at infinite dilution of ${Al}^{3+}$ and ${SO}_4^{2-}$ are $189{ohm}^{-1}{cm}^{2}gm$ ${equiv}^{-1}$ and $160{ohm}^{-1}{cm}^{2}gm$ ${equiv}^{-1}$. Calculate equivalent and molar conductance of the electrolyte at infinite dilution.

Answer 1

(a) Compound $A$ is Acetaldehyde
Compound $B$ is Acetaldehyde Cyanohydrin
Compound $C$ is Lactic acid
Compound $D$ is Pyruvic acid
(d) The electrolyte
${Al}_2{\left({SO}_4\right)}_3{\lambda }_{\infty }{Al}_2{\left({SO}_4\right)}_3=\frac{1}{3}{\lambda }_{\infty }{Al}^{3+}+{\lambda }_{\infty }{SO}_4^{2-}$
equivalent conductance at infinite dilution
${\lambda }_{\infty }{Al}_2{\left({SO}_4\right)}_3=\frac{189}{3}+\frac{160}{2}=63+80=143{cm}^{2}gm.{equiv}^{-1}$
molar conductance at infinite dilution
${\mu }_{\infty }{Al}_2{\left({SO}_4\right)}_3=2\times 189=858{cm}^2{mol}^{-1}$