Animated Solution for Chemistry - Organic Chemistry: The organic chloro compound, which shows complete stereochemical inversion during an ${\text{S}}_{\text{N}}2$ reaction is

Imagine a fortress under siege. The front gate is heavily guarded by a massive, electron-rich leaving group (like a chlorine atom). For the invading army—our nucleophile—a frontal assault is impossible due to electrostatic repulsion. The only way in is through the back door. This is the essence of the ${\text{S}}_{\text{N}}2$ (Substitution Nucleophilic Bimolecular) mechanism.

Because the nucleophile must attack from the side exactly opposite to the leaving group, the existing bonds on the carbon atom are forced to flip to the other side as the new bond forms. This phenomenon is beautifully described as Walden inversion, much like an umbrella turning inside out during a violent windstorm.

Steric Hindrance

For this backside attack to be successful, the path to the central carbon must be clear. If the carbon is surrounded by bulky alkyl groups, they act like a physical shield, blocking the nucleophile's approach. This physical crowding is known as steric hindrance.

Therefore, the reactivity of alkyl halides towards ${\text{S}}_{\text{N}}2$ reactions strictly follows the order of least crowding to most crowding:

Let's evaluate the candidates provided in the problem to see which one offers the clearest path for our nucleophile:

1. $({\text{C}}_2{\text{H}}_5{)}_2\text{CHCl}$: This is a secondary ($2^{\circ }$) alkyl halide. It has two bulky ethyl groups attached to the alpha carbon. It will experience significant steric hindrance. 2. $({\text{CH}}_3{)}_3\text{CCl}$: This is a tertiary ($3^{\circ }$) alkyl halide. With three methyl groups, the backside is completely blocked. It will practically never undergo an ${\text{S}}_{\text{N}}2$ reaction. 3. $({\text{CH}}_3{)}_2\text{CHCl}$: Another secondary ($2^{\circ }$) alkyl halide. While less hindered than the first option, it still presents a barrier to the nucleophile. 4. ${\text{CH}}_3\text{Cl}$: This is methyl chloride. The central carbon is bonded only to three tiny hydrogen atoms.

Methyl chloride (${\text{CH}}_3\text{Cl}$) offers virtually zero steric hindrance. The nucleophile has a wide-open runway to execute a perfect backside attack. Because it exclusively follows the ${\text{S}}_{\text{N}}2$ pathway without any competition from ${\text{S}}_{\text{N}}1$ (which would lead to racemization), it undergoes complete stereochemical inversion.

Thus, the correct answer is undeniably ${\text{CH}}_3\text{Cl}$.