Q30 / 180 · COMEDK 2023 (Morning Shift) · Chemistry
Ammoniacal silver nitrate forms a white precipitate easily with
- A$\mathrm{CH}_3 \mathrm{CH}=\mathrm{CH}_2$
- B$\mathrm{CH}_3 \mathrm{C}=\mathrm{CCH}_3$
- Correct answerC$\mathrm{CH}_3 \mathrm{C}=\mathrm{CH}$
- D$\mathrm{CH}_2=\mathrm{CH}_2$
Solution
Ammoniacal silver nitrate, also known as Tollens reagent, reacts with terminal alkynes to form a white precipitate of silver acetylide.
The general reaction is $R-C\equiv CH + [Ag(NH_3)_2]^+ OH^- \rightarrow R-C\equiv C-Ag \downarrow + 2NH_3 + H_2O$.
Among the given options, $\mathrm{CH}_3 C\equiv CH$ (propyne) is a terminal alkyne, meaning it has an acidic hydrogen atom attached to the triply bonded carbon.
The other compounds, $\mathrm{CH}_3 \mathrm{CH}=\mathrm{CH}_2$ (propene), $\mathrm{CH}_3 C\equiv CCH_3$ (but-2-yne, an internal alkyne), and $\mathrm{CH}_2=\mathrm{CH}_2$ (ethene), do not possess an acidic terminal hydrogen atom and therefore do not form a precipitate with ammoniacal silver nitrate.
Answer: $\mathrm{CH}_3 \mathrm{C}=\mathrm{CH}$
The general reaction is $R-C\equiv CH + [Ag(NH_3)_2]^+ OH^- \rightarrow R-C\equiv C-Ag \downarrow + 2NH_3 + H_2O$.
Among the given options, $\mathrm{CH}_3 C\equiv CH$ (propyne) is a terminal alkyne, meaning it has an acidic hydrogen atom attached to the triply bonded carbon.
The other compounds, $\mathrm{CH}_3 \mathrm{CH}=\mathrm{CH}_2$ (propene), $\mathrm{CH}_3 C\equiv CCH_3$ (but-2-yne, an internal alkyne), and $\mathrm{CH}_2=\mathrm{CH}_2$ (ethene), do not possess an acidic terminal hydrogen atom and therefore do not form a precipitate with ammoniacal silver nitrate.
Answer: $\mathrm{CH}_3 \mathrm{C}=\mathrm{CH}$
Hence, the correct answer is option: 'C'.