Acetylene reacts with ammoniacal \[AgN{O_3}\] , forming:
A) silver acetylide
B) silver acetate
C) metal silver
D) silver mirror
Answer
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Hint:We know that acetylene reacts with ammoniacal \[AgN{O_3}\] and the reaction is known as Tollens’s Test. In the following Tollens test, we get a white precipitate. So we have to find which of the following options white precipitate is.
Complete step by step answer
We know that in the Tollens test acetylene reacts with ammoniacal \[AgN{O_3}\] to form silver acetylide.
We can show the Tollens reaction by the following reaction:
\[{C_2}{H_2} + 2AgN{O_3} + 2N{H_4}OH \to {C_2}A{g_2} + 2N{H_4}N{O_2} + 2{H_2}O\]
From the following reaction, we know that silver acetylene \[\left( {{C_2}A{g_2}} \right)\] is formed as a white precipitate.
We must know that the Tollens test is used to determine the presence of a triple bond in hydrocarbons.
Therefore, we can conclude that the correct answer to the question is option A.
Additional Information:
We know that the Tollens reagent is used to determine the presence of the aldehyde functional group and also a ketone functional group. We know that the Tollens reagent consists of A reagent consisting of a solution of silver nitrate, ammonia, and some sodium hydroxide (to maintain a basic pH of the reagent solution).
We know that a positive test with Tollens reagent is indicated by the precipitation of elemental silver, producing a characteristic "silver mirror" on the inner surface of the reaction vessel.
Let us see the uses of Tollens' reagent which apply a silver mirror to the glassware; for example, the inside of an insulated vacuum flask. It is also used as the reducing agent is glucose (an aldehyde) for such applications. We must know that the clean glassware is required for a high-quality mirror.
Note:We know that acetylene consists of a triple bond, so the reaction with ammoniacal \[AgN{O_3}\] will give silver acetylide because silver acetylide also contains a triple bond in its structure. The remaining options do not contain a triple bond in their structure.
Complete step by step answer
We know that in the Tollens test acetylene reacts with ammoniacal \[AgN{O_3}\] to form silver acetylide.
We can show the Tollens reaction by the following reaction:
\[{C_2}{H_2} + 2AgN{O_3} + 2N{H_4}OH \to {C_2}A{g_2} + 2N{H_4}N{O_2} + 2{H_2}O\]
From the following reaction, we know that silver acetylene \[\left( {{C_2}A{g_2}} \right)\] is formed as a white precipitate.
We must know that the Tollens test is used to determine the presence of a triple bond in hydrocarbons.
Therefore, we can conclude that the correct answer to the question is option A.
Additional Information:
We know that the Tollens reagent is used to determine the presence of the aldehyde functional group and also a ketone functional group. We know that the Tollens reagent consists of A reagent consisting of a solution of silver nitrate, ammonia, and some sodium hydroxide (to maintain a basic pH of the reagent solution).
We know that a positive test with Tollens reagent is indicated by the precipitation of elemental silver, producing a characteristic "silver mirror" on the inner surface of the reaction vessel.
Let us see the uses of Tollens' reagent which apply a silver mirror to the glassware; for example, the inside of an insulated vacuum flask. It is also used as the reducing agent is glucose (an aldehyde) for such applications. We must know that the clean glassware is required for a high-quality mirror.
Note:We know that acetylene consists of a triple bond, so the reaction with ammoniacal \[AgN{O_3}\] will give silver acetylide because silver acetylide also contains a triple bond in its structure. The remaining options do not contain a triple bond in their structure.
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