Question

The compound $C_3{H}_4$ has a triple bond, which is confirmed by its reaction with

• A. Baeyers reagent
• B. Bromine in $CC{l}_4$
• C. Ammonical silver nitrate
• D. All the above

Answer 1

The correct option is D All the above

Molecular formula $C_3{H}_4$ of a compound with a triple bond is
$C{H}_3-C\equiv CH$
Bromine in $CC{l}_4$ is used to test for the presence of unsaturation in compound. Both alkene and alkyne will reacting with the bromine in $CC{l}_4$, changing its color from an intense yellow to a colorless solution.
The Baeyer's test for unsaturation is for determining the presence of carbon-carbon double bonded and carbon-carbon trible bonded compounds.
Both these reagent cannot differentiate the double and triple bond in compound. So it won't confirm the triple bond in compound.
Terminal alkynes contain acidic hydrogen. These terminal alkynes react with ammoniacal solutions of silver nitrate to give white precipitate of silver alkynide.
$C{H}_3-C\equiv CH+\left[Ag\right(N{H}_3{)}_2{]}^{+}N{O}_3^{-}\to C{H}_3-C\equiv C^{-}A{g}^{+}\downarrow +N{H}_4^{+}N{O}_3^{-}+N{H}_3$