Question

Which of the compounds is soluble in conc. $H_{2}S{O}_4$, decolourises Bromine $CC{l}_4$ and give a white precipitate with $AgN{O}_3$ in $N{H}_{4}OH$ but does not give a precipitate with $AgN{O}_3$ in alcohol?

• A. $C{H}_{3}C{H}_{2}C{H}_{2}CH=C{H}_2$
• B. $C{H}_{3}CH\left(Cl\right)C{H}_{2}CH=C{H}_2$
• C. $C{H}_{3}C{H}_{2}C\equiv CH$
• D. $C{H}_{3}C{H}_{2}C{H}_{2}C{H}_{2}C{H}_3$

Answer 1

The correct option is C $C{H}_{3}C{H}_{2}C\equiv CH$

It is terminal alkyne $C{H}_{3}C{H}_{2}C\equiv CH$ which forms white precipitate with Tollen's reagent.
Ammoniacal silver nitrate solution is called Tollen's reagent which reacts with terminal alkynes to form white ppt.
$C{H}_{3}C{H}_{2}C\equiv CH+AgN{O}_3+N{H}_{4}OH\to C{H}_{3}C{H}_{2}C\equiv CAg\downarrow \text{(white ppt)}+N{H}_{4}N{O}_3+H_{2}O$
This reaction is not given with alcoholic silver nitrate solution.
Since $C{H}_{3}C{H}_{2}C\equiv CH$ is unsaturated compound, it decolorises bromine in carbon tetrachloride. Two molecules of bromine add to $C\equiv C$ bond.

Option C is correct.