Animated Solution for Chemistry - Organic Chemistry: The decreasing order of reactivity of the following organic molecules towards ${\text{AgNO}}_3$ solution is

Silver Nitrate and the $S_{N}1$ Pathway

When we see ${\text{AgNO}}_3$ reacting with an alkyl halide, we should immediately think of the $S_{N}1$ mechanism. Silver has a very high affinity for halogens, forming a stable precipitate of silver halide (like $\text{AgCl}$). This acts as a powerful driving force to break the carbon-halogen bond, leaving behind a carbocation intermediate.

The rate of this entire reaction is dictated by the stability of this carbocation. The more stable the carbocation, the faster it forms, and the more reactive the original molecule is. Therefore, our task is to ionize all four molecules and compare the stability of the resulting positive charges.

The Power of Aromaticity

When molecule (A) loses its chloride ion, it forms the cyclopropenyl cation. At first glance, a three-membered ring with a positive charge might seem strained and unstable. However, we must count the $\pi$ electrons.

The double bond provides $2\pi$ electrons, and the empty p-orbital on the carbocation completes the continuous cyclic conjugation. According to Hückel's rule ($4n+2$), setting $n=0$ gives us exactly $2\pi$ electrons. This makes the cyclopropenyl cation aromatic. Aromaticity provides a massive thermodynamic stabilization energy, making this carbocation exceptionally stable despite the ring strain.

Aromaticity Meets the Mesomeric Effect

Molecule (B) also forms a cyclopropenyl cation upon losing chloride, so it starts with the same baseline aromatic stability as (A). But it has a secret weapon: the methoxy ($-\text{OMe}$) group attached directly to the ring.

The oxygen atom has lone pairs of electrons. Through the positive mesomeric effect (+M), oxygen can donate a lone pair into the conjugated system, further delocalizing the positive charge. This extra electron density acts like a booster shield, making carbocation (B) even more stable than (A).

Standard Hyperconjugation

Molecule (C) is isopropyl chloride. When it ionizes, it forms an isopropyl cation, which is a secondary ($2^{\circ }$) aliphatic carbocation.

It is stabilized by hyperconjugation from the six $\alpha$-hydrogens on the two adjacent methyl groups. While hyperconjugation provides decent stability, it is a much weaker stabilizing force compared to the immense resonance energy of an aromatic ring. Therefore, (C) is significantly less stable than both (A) and (B).

The Destructive Inductive Effect

Finally, we look at molecule (D). It also forms a secondary carbocation, but it has a nitro ($-{\text{NO}}_2$) group attached to the adjacent carbon.

The nitro group is one of the most powerful electron-withdrawing groups in organic chemistry. Because it is separated from the positive charge by a sigma bond, it exerts a strong negative inductive effect (-I). It pulls electron density away from the already electron-deficient carbocation. This severe electrostatic repulsion severely destabilizes the intermediate, making (D) the least reactive molecule in the group.

Putting it all together, the stability of the carbocations dictates the reactivity order. Molecule (B) is the most stable due to aromaticity and the +M effect. Molecule (A) follows closely with pure aromaticity. Molecule (C) is next with standard hyperconjugation. Molecule (D) is the least stable due to the strong -I effect of the nitro group.

Thus, the decreasing order of reactivity is (B) > (A) > (C) > (D).