Question

Identify a reagent from the following list which can easily distinguish between $1 - butyne$ and $2 - butyne$
A) Bromine, $CC{l_4}$.
B) ${H_2}$, Lindlar catalyst
C) Dilute ${H_2}S{O_4},HgS{O_4}$
D) ammoniacal $C{u_2}C{l_2}$ solution

Answer 1

Hint: $1 - butyne$ and $2 - butyne$ are alkynes. It has a triple bond between carbon atoms. The molecular formula of $1 - butyne$ and $2 - butyne$ are ${C_4}{H_6}$.General alkyne formula is ${C_n}{H_{2n - 2}}$.In the presence of ammonical $C{u_2}C{l_2}$solutions as a catalyst formation of $1 - butyne$.

Complete step by step answer:
$1 - butyne$ will easily react with ammoniacal $C{u_2}C{l_2}$but $2 - butyne$ will not give any product. Hence, ammoniacal $C{u_2}C{l_2}$can easily distinguish between $1 - butyne$ and $2 - butyne$.
$1 - butyne$ is terminal alkyne so it reacts to give specific lists.
Silver nitrate and ammonium hydroxide $1 - butyne$ gives precipitate with Tollens reagent while $2 - butyne$ does not react.

$C{u_2}C{l_2} + N{H_4}OH$, $1 - butyne$ gives red precipitate while $2 - butyne$ does not react at all.
In $1 - butyne$, acidic hydrogen is present while in $2 - butyne$ acidic hydrogen is not present so $1 - butyne$ gives red precipitate with ammoniacal solution while $2 - butyne$ does not give these precipitate.

Test for $1 - butyne$
$2C{H_3} - C{H_2} - C \equiv CH + C{u_2}C{l_2} + 2N{H_4}OH \to 2C{H_3} - C{H_2} - C \equiv C - Cu \downarrow + 2N{H_4}Cl + 2{H_2}O$

Test for $2 - butyne$
$C{H_3} - C \equiv C - C{H_3} + C{u_2}C{l_2} + 2N{H_4}OH \to $No reaction
Cuprous chloride test=
Acidic alkynes form red precipitate of cuprous acetylides upon addition of alkyne to an ammoniacal solution of cuprous chloride.
For example, test for alkyne like as propyne-
\[C{H_3} - C \equiv CH + \mathop {Cu}\limits^ + \to Cu - C \equiv C - C{H_3} + \mathop H\limits^ + \]

Test for ethyne (Acetylene)
$CH \equiv CH + 2\mathop {Cu}\limits^ + \to Cu - C \equiv C - Cu + 2\mathop H\limits^ + $
These tests are not given by non-terminal alkynes ($R - C \equiv C - R$) and alkynes. Therefore, these reactions can be used to distinguish between alkenes and alkynes and also terminal alkynes such as $1 - butyne$ and non-terminal alkynes such as $2 - butyne$.

So, the correct answer is Option D.

Note: We have to remember that the first one was the dimerization of an alkyne on passing it through ammoniacal $C{u_2}C{l_2}$.
Glaser coupling is an identification test for terminal alkynes. In this reaction, a terminal alkyne reacts with ammoniacal cupric chloride which on subsequent oxidation in air gives diene.