Which of the following will give a precipitate with an aqueous solution of copper (I) chloride?

Analysis

The question asks us to identify which of the given hydrocarbons reacts with an aqueous solution of copper(I) chloride (specifically ammoniacal copper(I) chloride) to form a precipitate.

Concept Explanation

This is a classic test for terminal alkynes.

• Terminal alkynes are hydrocarbons with a triple bond at the end of the carbon chain (e.g., $R-C \equiv C-H$).
• The hydrogen atom attached to the triply-bonded carbon is acidic.
• When a terminal alkyne reacts with ammoniacal copper(I) chloride (or ammoniacal silver nitrate), the acidic hydrogen is replaced by the metal, forming a metal acetylide precipitate.
• Internal alkynes (where the triple bond is not at the end) and alkenes do not have this acidic hydrogen, so they do not react with these reagents to form a precipitate.

Step-by-Step Solution

1. Analyze the options: • A. But-2-ene ($CH_3-CH=CH-CH_3$): This is an alkene. It does not react with copper(I) chloride. • B. But-1-ene ($CH_2=CH-CH_2-CH_3$): This is an alkene. It does not react with copper(I) chloride. • C. But-2-yne ($CH_3-C \equiv C-CH_3$): This is an internal alkyne. The triple bond is between the second and third carbons. There is no hydrogen atom attached to the triply-bonded carbons, so it cannot form a precipitate. • D. But-1-yne ($CH \equiv C-CH_2-CH_3$): This is a terminal alkyne. The triple bond is at the first carbon, meaning there is an acidic hydrogen ($H-C \equiv C-CH_2-CH_3$). This hydrogen reacts with the copper(I) ions to form a reddish-brown precipitate of copper(I) butynide.

2. Conclusion: Only But-1-yne possesses the terminal acidic hydrogen required to react with the reagent to form a precipitate.

Exam Tip

Remember the "Terminal Rule": If you see a question about forming precipitates with ammoniacal $AgNO_3$ (Tollens' reagent) or $CuCl$, always look for the molecule that ends in "-1-yne". If it's an alkene or an internal alkyne (like but-2-yne), it will not react!