Question

Which one of the ethane, ethene, and ethyne will form a white precipitate with an ammonical solution of silver nitrate.

Answer 1

Step 1:

The organic compounds having a single bond between two Carbon-Carbon atoms are called alkanes.

The general formula of saturated alkanes is $\left({\mathrm{C}}_{\mathrm{n}}{\mathrm{H}}_{2\mathrm{n}+2}\right)$.

The organic compounds having a double and triple bond present between two Carbon-Carbon atoms are called alkenes and alkynes.

The unsaturated alkenes and alkynes have the general formula are $\left({\mathrm{C}}_{\mathrm{n}}{\mathrm{H}}_{2\mathrm{n}}\right)$ and $\left({\mathrm{C}}_{\mathrm{n}}{\mathrm{H}}_{2\mathrm{n}-2}\right)$.

Step 2:

Treatment of Ammonical Silver nitrate $\left({\mathrm{AgNO}}_3\right)$:

Alkanes and alkenes do not react with an ammonical solution of silver nitrate.

Terminal alkynes give white precipitates with ammoniacal silver nitrate.

Ethyne reacts with Silver nitrate in the presence of Ammonium hydroxide forms

$$\begin{gathered} \mathrm{H}-\mathrm{C}\equiv \mathrm{C}-\mathrm{H}\stackrel{2{\mathrm{AgNO}}_3+2{\mathrm{NH}}_4\mathrm{OH}}{\to }\mathrm{Ag}-\mathrm{C}\equiv \mathrm{C}-\mathrm{Ag}+2{\mathrm{NH}}_4{\mathrm{NO}}_3+2{\mathrm{H}}_2\mathrm{O} \\ \left(\mathrm{Ethyne}\right)(\mathrm{Silver}\mathrm{acetylide})(\mathrm{Ammonium}(\mathrm{Water}) \\ \mathrm{nitrate}) \end{gathered}$$

Hence, Ethyne $\left({\mathrm{C}}_2{\mathrm{H}}_2\right)$ will form a white precipitate with an Ammonical solution of Silver nitrate.