Q22Engineering Chemistry

Part A: Nucleophilic Substitution Bimolecular ($S_{N}2$) Mechanism

The $S_{N}2$ reaction is a fundamental class of nucleophilic substitution primarily observed in primary alkyl halides (e.g., methyl bromide, $C{H}_{3}Br$). The '2' stands for bimolecular, meaning the overall rate of the reaction is strictly dependent on the concentrations of both the substrate (the alkyl halide) and the attacking nucleophile simultaneously. Rate = $k[\text{Alkyl Halide}][\text{Nucleophile}]$

The Concerted One-Step Mechanism: Unlike the $S_{N}1$ reaction (which proceeds in two distinct steps with a carbocation intermediate), the $S_{N}2$ reaction is a concerted process. Everything happens simultaneously in a single, perfectly choreographed, continuous step.

1. Backside Attack: A strong, electron-rich nucleophile (like $O{H}^{-}$) is strongly attracted to the partially positive carbon atom attached to the electronegative halogen. Because the large halogen atom completely blocks access from the front, the nucleophile is physically forced to attack the carbon atom from the exact opposite side—the strictly $180^{\circ }$ "backside."

2. The Transition State: As the nucleophile crashes into the backside and begins to donate its electron pair to form a new covalent bond, the carbon atom simultaneously begins to sever its bond with the departing halogen leaving group. At the absolute peak of the energy barrier (the Transition State), the central carbon atom is momentarily pentacoordinate. It is weakly bonded to both the incoming nucleophile and the leaving halogen simultaneously. The three non-reacting substituent groups on the carbon atom are pushed perfectly flat into a planar arrangement, like the spokes of a wheel.

3. Completion and Inversion: The transition state is incredibly unstable and collapses instantly. The leaving group departs fully with its electrons, and the new bond with the nucleophile solidifies. The intense repulsive forces from the incoming nucleophile forcefully flip the three remaining substituent groups inside out, exactly like a high wind blowing an umbrella inside out. This profound stereochemical consequence is universally known as Walden Inversion.

Consequently, if the reaction occurs at a chiral center, an $S_{N}2$ reaction guarantees $100\%$ complete inversion of stereochemical configuration.

Part B: Anti-Markovnikov Addition of HBr (Peroxide Effect / Kharasch Effect)

Standard electrophilic addition of Hydrogen Bromide ($HBr$) to an unsymmetrical alkene (like propene, $C{H}_3-CH=C{H}_2$) strictly follows Markovnikov's rule, yielding 2-bromopropane because it proceeds via the most stable carbocation intermediate. However, if the exact same reaction is conducted in the presence of organic peroxides (like benzoyl peroxide), the regioselectivity entirely reverses. The bromine atom inexplicably attaches to the less substituted carbon atom, yielding 1-bromopropane. This is the Anti-Markovnikov addition or the Peroxide Effect.

This reversal occurs because the peroxide completely changes the fundamental reaction mechanism. It shifts from an ionic carbocation mechanism to a Free Radical Chain Mechanism.

The Free Radical Mechanism Steps:

• Initiation: The weak, highly unstable oxygen-oxygen bond in the organic peroxide ($R-O-O-R$) easily breaks homolytically under mild heat or light, generating two highly reactive alkoxy free radicals ($R{O}^{\bullet }$). $R-O-O-R\stackrel{\text{heat/light}}{\to }2R{O}^{\bullet }$ These aggressive alkoxy radicals immediately attack an $HBr$ molecule. They tear away the hydrogen atom to form stable alcohol ($ROH$), leaving behind a highly reactive Bromine free radical ($B{r}^{\bullet }$). $R{O}^{\bullet }+H-Br\to R-O-H+B{r}^{\bullet }$
• Propagation (The Reversal Step): The bromine radical ($B{r}^{\bullet }$) attacks the pi-bond of the unsymmetrical alkene (propene, $C{H}_3-CH=C{H}_2$). It can attack the central carbon or the terminal carbon. If it attacks the terminal carbon ($C{H}_2$), it generates a secondary ($2^{\circ }$) carbon free radical on the central carbon. $C{H}_3-CH=C{H}_2+B{r}^{\bullet }\to C{H}_3-C{H}^{\bullet }-C{H}_{2}Br\text{ (More stable }{2}^{\circ }\text{ radical)}$ If it attacked the central carbon, it would generate a primary ($1^{\circ }$) radical, which is much less stable. Therefore, the formation of the more stable secondary radical dictates the pathway, placing the bromine firmly on the terminal carbon.
• Chain Continuation: This newly formed secondary radical is highly reactive. It immediately attacks a fresh molecule of $HBr$, tearing away a hydrogen atom to satisfy its own valency, forming the final product (1-bromopropane) and generating a brand new Bromine radical to continue the chain reaction endlessly. $C{H}_3-C{H}^{\bullet }-C{H}_{2}Br+H-Br\to C{H}_3-C{H}_2-C{H}_{2}Br\text{ (1-bromopropane)}+B{r}^{\bullet }$

By forcing the intermediate to be a free radical rather than a carbocation, the peroxide dictates that the halogen attacks first (rather than hydrogen), leading to the precise reversal of Markovnikov's standard rule.