Animated Solution for Chemistry - Organic Chemistry: The major product obtained in the following reaction is

The journey of mastering organic chemistry is filled with fascinating battles between competing reaction pathways. Today, we are diving into a classic showdown: Substitution versus Elimination. When you look at a reaction, it's like reading a story where the reagents are the characters, and their structures dictate their actions. Let's unravel the mystery of 1-bromo-1,2-diphenylethane reacting with potassium tert-butoxide.

Imagine you are looking at the substrate: 1-bromo-1,2-diphenylethane. It's a secondary alkyl halide, meaning the carbon attached to the bromine is also attached to two other carbons. This makes it a prime candidate for both $S_{N}2$ substitution and E2 elimination.

But the real star of the show is our reagent: potassium tert-butoxide (${}^t\text{BuOK}$). This is not your average, everyday base. It is a massive, bulky molecule. Think of it like a giant trying to squeeze through a tiny door. Because of its sheer size, it suffers from immense steric hindrance. It simply cannot reach the electrophilic carbon to perform a nucleophilic attack.

Furthermore, we have heat ($\Delta$) added to the mix. Heat is the ultimate hype-man for elimination reactions. Why? Because elimination reactions create more molecules (increasing entropy), and higher temperatures favor processes that increase entropy. So, the stage is perfectly set for an E2 elimination.

Now, let's watch the E2 mechanism unfold. "E2" stands for Elimination Bimolecular. It's a concerted process, meaning everything happens in one beautiful, synchronized dance.

The bulky tert-butoxide ion, unable to attack the carbon, looks for an easier target: a proton on the adjacent carbon, known as the $\beta$-carbon.

As the base grabs the $\beta$-hydrogen, the electrons that once held that hydrogen swing down like a closing door to form a new $\pi$ bond between the $\alpha$ and $\beta$ carbons. Simultaneously, the bromine atom, which is an excellent leaving group, takes its electrons and departs as a bromide ion.

This entire process requires a specific 3D alignment called anti-periplanar geometry. The hydrogen being removed and the leaving group must be on opposite sides of the molecule to allow the orbitals to overlap perfectly and form the double bond.

As the dust settles from this concerted reaction, the single bonds have transformed into a double bond. The bromine is gone, and the $\beta$-hydrogen is gone.

What are we left with? A double bond between the two carbons, each still attached to a phenyl group. This molecule is 1,2-diphenylethene, commonly known as stilbene.

Because the bulky base completely shut down the substitution pathway, this alkene is our major product. It's a beautiful example of how steric hindrance can completely dictate the outcome of a chemical reaction.

I know these mechanisms can seem intimidating at first, but once you understand the "personality" of your reagents—like the bulky, elimination-loving tert-butoxide—predicting the products becomes second nature. Keep visualizing those electrons flowing, and you'll master organic chemistry in no time!