Question

Propyne and propene can be distinguished by:
A. Conc. $H_{2}S{O}_4$
B. $B{r}_2$ in $CC{l}_4$.
C. Dil. $KMn{O}_4$
D. $AgN{O}_3$ in ammonia.

Answer 1

Hint: Silver nitrate in ammonia is actually Tollen’s reagent. It also reacts with acidic hydrogens to give precipitates.

Complete step by step answer:
To get the answer, let us check each reaction one by one:
Conc. $$H_{2}S{O}_4$$:
Propyne on reaction with conc. $H_{2}S{O}_4$ produces a ketone.

Propene on reaction with conc. $H_{2}S{O}_4$ produces alcohol.

Ketone and alcohol can’t be distinguished at this condition without adding any other reagent. Hence, it is not the correct option.

$B{r}_2]in[CC{l}_4$:
Propyne on reaction with $B{r}_2$ in $CC{l}_4$ produces a vicinal dibromo alkene whereas propene on reaction with $B{r}_2$ in $CC{l}_4$ produces a vicinal dibromo alkane.

Hence, the two compounds can’t be distinguished by this process.

Dil. $KMn{O}_4$:
Propynes on reaction with Dil. $KMn{O}_4$ produce -ketoacid.
Propene on reaction with Dil. $KMn{O}_4$ produce a vicinal diol.

Hence, the two compounds cannot be distinguished by this method also.
$AgN{O}_3$ in ammonia:
Propyne on reaction with $AgN{O}_3$ in ammonia produces a white precipitate, while propene does not show any reaction with the Tollen’s reagent.

This is due to the fact that Tollen’s reagent reacts with an acidic hydrogen forming a precipitate like in a propyne. The hydrogen which is replaced is very acidic due to high electronegativity of an sp hybridized carbon.

Hence, this reagent can be used to distinguish between a propyne and a propene.

Therefore, the correct answer is (D) $AgN{O}_3$ in ammonia.

Note: When an alkyne is reacted with conc. Sulphuric acid, ketone is formed by the Markovnikov addition of -OH, followed by the rearrangement of the alcohol formed from enol form to keto.