An organic compound (x) was heated with \(\mathrm{NaOH}\) solution and then acidified with dilute \(\mathrm{HNO}_{3} \cdot\) Addition of silver nitrate solution gave a white precipitate completely soluble in ammonium hydroxide. The compound (x) is (a) \(\mathrm{p}\) -chlorotoluene (b) vinyl chloride (c) \(\mathrm{p}\) - dichlorobenzene (d) benzyl chloride

For each given compound, we need to identify the functional groups that can react with \(\mathrm{NaOH}\) and silver nitrate. Here are the given compounds and their functional groups: - (a) \(\mathrm{p}\) -chlorotoluene (aromatic and halogen, no reactive functional group) - (b) vinyl chloride (alkene and halogen, no reactive functional group) - (c) \(\mathrm{p}\) - dichlorobenzene (aromatic and halogen, no reactive functional group) - (d) benzyl chloride (benzylic halide and halogen, reactive functional group)

Now, let's analyze the expected reactions for each functional group and see if they fit the given reaction conditions: - (a) Aromatic and halogen groups will not react with \(\mathrm{NaOH}\) and form a soluble complex with silver nitrate after acidification. - (b) Alkene and halogen groups will not react with \(\mathrm{NaOH}\) and form a soluble complex with silver nitrate after acidification. - (c) Aromatic and halogen groups will not react with \(\mathrm{NaOH}\) and form a soluble complex with silver nitrate after acidification. - (d) The benzylic halide in benzyl chloride can react with \(\mathrm{NaOH}\) to produce a carboxylic acid. After acidification and the addition of silver nitrate, carboxylic acids tend to form a white precipitate that is soluble in ammonium hydroxide.

Based on the reaction analysis, we can conclude that compound (d) benzyl chloride is the compound (x) that fits the given conditions of the problem. The white precipitate observed after treating with silver nitrate is silver carboxylate, which is soluble in ammonium hydroxide. So, the correct answer is: (d) benzyl chloride