Animated Solution for Chemistry - Organic Chemistry: The product formed in the first step of the reaction of $\begin{array}{cc}& \text{Br}\\ & |\\ {\text{CH}}_3-{\text{CH}}_2-& \text{CH}-{\text{CH}}_2\\ & |\\ & \text{CH}-{\text{CH}}_3\\ & |\\ & \text{Br}\end{array}$ with excess ${\text{Mg/Et}}_2\text{O}$ ($\text{Et}={\text{C}}_2{\text{H}}_5$) is

Welcome to a classic organic chemistry puzzle that tests your ability to see past deceptive drawings and apply fundamental reaction mechanisms. At first glance, the molecule presented in the question might look like a complex, branched nightmare. However, the secret to mastering organic chemistry lies in systematically identifying the longest continuous carbon chain.

Let's take a closer look at the starting material. The structure is drawn with a vertical bend, but if we trace the longest continuous path of carbon atoms, we find a straight chain of six carbons.

This reveals the true identity of our molecule: 2,4-dibromohexane. It is a simple dihalide with bromine atoms attached to the second and fourth carbon atoms. Examiners often use these folded representations to test your IUPAC nomenclature and structural visualization skills.

Now, we introduce the reagents: Magnesium ($\text{Mg}$) in the presence of dry ether (${\text{Et}}_2\text{O}$). This is the textbook recipe for creating a Grignard reagent.

When an alkyl halide reacts with magnesium metal, the magnesium atom undergoes an oxidative insertion directly into the carbon-halogen bond.

The dry ether is not just a passive bystander; it acts as a crucial solvent. The lone pairs on the oxygen atoms of the ether molecules coordinate with the electron-deficient magnesium, stabilizing the highly reactive Grignard complex.

Here is where the critical detail of the question comes into play: the word "excess".

If we had only one equivalent of magnesium, we would have to worry about which bromine atom reacts first (though in this symmetrical-like case, it wouldn't matter much). However, because we have an excess of magnesium, there is no competition. Every single carbon-bromine bond in the molecule will undergo the insertion reaction.

Magnesium inserts itself into the $\text{C}-\text{Br}$ bond at carbon-2, and simultaneously, another magnesium atom inserts into the $\text{C}-\text{Br}$ bond at carbon-4.

The result of this dual insertion is a di-Grignard reagent. Both bromine atoms have been successfully replaced by $\text{MgBr}$ groups.

This perfectly matches the structure given in option (d). The question specifically asks for the product formed in the first step, and this di-Grignard reagent is exactly that.

While we have found our answer, elite students always ask: What happens next?

Grignard reagents are powerful nucleophiles. In a molecule with two such reactive centers in close proximity (a 1,3-relationship), an intramolecular Wurtz-type coupling is highly likely to occur upon further heating or reaction progression. The nucleophilic carbon attached to one magnesium would attack the other carbon, eliminating ${\text{MgBr}}_2$ and forming a stable three-membered cyclopropane ring (specifically, 1-ethyl-2-methylcyclopropane).

Always keep these subsequent steps in mind, as JEE frequently designs questions that require you to traverse multiple stages of a reaction pathway!