Animated Solution for Chemistry - Organic Chemistry: Which one of the following is likely to give a precipitate with $AgN{O}_3$ solution?

A Quest for Carbocations

Whenever you encounter a question asking which compound gives a precipitate with silver nitrate ($AgN{O}_3$), your mind should immediately jump to one core concept: carbocation stability.

The silver ion ($A{g}^{+}$) has an incredibly high affinity for chloride ions ($C{l}^{-}$). It acts as a chemical thief, trying to pull the chloride away from the organic molecule. When it succeeds, it leaves behind a positively charged carbon atom—a carbocation—and forms a highly insoluble white precipitate of silver chloride ($AgCl$).

The golden rule here is simple: The easier it is to form this carbocation, the faster the precipitate forms. Therefore, the entire problem boils down to breaking the carbon-chlorine bond in each option and evaluating which one leaves behind the most stable positive charge.

The Unstable Misfits

Let's test our options one by one.

First, we look at option (a), vinyl chloride ($C{H}_2=CH-Cl$). If we remove the chloride ion, we get a vinyl carbocation ($C{H}_2=C{H}^{+}$). Notice where the positive charge is sitting—it is on a double-bonded carbon. A double-bonded carbon is $sp$ hybridized, making it quite electronegative. An electronegative atom inherently hates bearing a positive charge! Because of this, the vinyl carbocation is highly unstable and will practically never form under these conditions.

Next, let's examine options (b) and (c): carbon tetrachloride ($CC{l}_4$) and chloroform ($CHC{l}_3$). Removing a chloride from them gives us the trichloromethyl ($CC{l}_3^{+}$) and dichloromethyl ($CHC{l}_2^{+}$) carbocations, respectively. Are these stable? Absolutely not.

Chlorine is highly electronegative. It pulls electron density away through the $-I$ inductive effect. Imagine a carbon atom already struggling with a positive charge, and then having two or three chlorines pulling even more electrons away from it! This intensifies the positive charge and makes these carbocations extremely unstable.

The Tertiary Butyl Carbocation

Finally, let's look at option (d), tert-butyl chloride ($(C{H}_3{)}_{3}C-Cl$). When this molecule loses its chloride ion, it forms a tertiary butyl carbocation ($(C{H}_3{)}_3{C}^{+}$). Let's see what makes this one special.

This central carbon is surrounded by three methyl groups. These groups are electron-donating, providing a strong $+I$ inductive effect. Even better, there are 9 $\alpha$-hydrogens available for hyperconjugation!

Hyperconjugation allows the electrons from adjacent $C-H$ sigma bonds to delocalize into the empty p-orbital of the carbocation. This extensive electron delocalization beautifully stabilizes the positive charge. It is like having nine friends helping you carry a heavy load.

Because the tertiary butyl carbocation is exceptionally stable, tert-butyl chloride will very readily undergo ionization. It will happily give up its chloride to the silver ion, instantly forming a thick white precipitate of silver chloride.

Therefore, option (d) is our correct answer. Understanding these nuances of intermediate stability is the absolute key to mastering $S_{N}1$ reaction mechanisms!