Question

Ammoniacal silver nitrate forms a white precipitate easily with

• A. CH\(_3\)C = CH
• B. CH\(_3\)C = CCH\(_3\)
• C. CH\(_3\)CH = CH\(_2\)
• D. CH\(_2\) = CH\(_2\)

Hint:

Ammoniacal silver nitrate reacts with terminal alkenes (those with a double bond at the end of the chain) to form a white precipitate.

Answer 1

The Correct Option is A

Ammoniacal silver nitrate is used to test for alkynes with terminal hydrogen atoms, which forms a white precipitate due to the formation of silver acetylides. Let us evaluate each option to determine which compound reacts with ammoniacal silver nitrate:

• CH₃C≡CH: This compound is known as propyne. It is a terminal alkyne because there is a hydrogen atom attached to the terminal carbon of the alkyne group. Terminal alkynes readily react with ammoniacal silver nitrate to form a white precipitate.
• CH₃C≡CCH₃: Known as 2-butyne, it is an internal alkyne because the alkyne linkage is between two carbon atoms, both of which are attached to other carbon atoms. Internal alkynes do not form a precipitate with ammoniacal silver nitrate.
• CH₃CH=CH₂: This compound is propene, an alkene, which does not react with ammoniacal silver nitrate because the selective reaction occurs only with terminal alkynes.
• CH₂=CH₂: Known as ethene, this is also an alkene, similar to the previous compound, and will not react with ammoniacal silver nitrate.

Thus, the compound CH₃C≡CH, being a terminal alkyne, will form a white precipitate when reacted with ammoniacal silver nitrate.

Answer 2

To determine which hydrocarbon readily forms a white precipitate with ammoniacal silver nitrate, we need to understand the reaction mechanism involved. Ammoniacal silver nitrate, also known as Tollen's reagent, is primarily used for distinguishing aldehydes from ketones and can react with acetylenic (alkynic) hydrogens to form a precipitate.

An acetylenic hydrogen is a hydrogen atom attached to a terminal carbon of an alkyne group (C≡C-H). The silver ion (Ag⁺) in the Tollen's reagent can replace the hydrogen attached to the terminal alkyne carbon, forming a silver acetylide precipitate.

Among the given options, let's identify which one contains a terminal alkyne:

• CH₃C≡CH: Contains a terminal alkyne group.
• CH₃C≡CCH₃: Contains an internal alkyne group; no terminal hydrogen is present.
• CH₃CH=CH₂: Contains an alkene group, not an alkyne.
• CH₂=CH₂: Also contains an alkene group, not an alkyne.

Thus, CH₃C≡CH is the only compound with a terminal alkyne, allowing it to form a white precipitate with ammoniacal silver nitrate.

Therefore, the correct answer is CH₃C≡CH.