Question

$\left(X\right)$ on treatment with sodium hydroxide followed by the addition of silver nitrate gives white precipitate at room temperature which is soluble in ${NH}_{4}OH$. $\left(X\right)$ can be:

• A. chlorobenzene
• B. ethyl bromide
• C. benzyl chloride
• D. vinyl chloride

Answer 1

The correct option is C benzyl chloride

Silver nitrate solution can be used to find out which halogen is present in a suspected halogenoalkane. $NaOH$ substitutes the $C{l}^{-}$ from haloalkane.

In Chlorobenzene and Vinyl chloride, the $Cl$ is in conjugation with benzene ring and does not give alcohol.

In Benzylchloride, the $Cl$ is not directly linked to benzene and thus react as haloalkane and gives white precipitate of $AgCl$ with $AgN{O}_3$ which is soluble in $N{H}_{4}OH$.

Ethyl bromide gives yellow precipitate of $AgBr$ which is insoluble in $N{H}_{4}OH$ at room temperature.