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

The world of organic chemistry is often a battlefield between competing reaction pathways. When you mix an alkyl halide with a strong base, you are essentially setting the stage for a classic showdown: Substitution ($S_{N}2$) versus Elimination ($E_2$).

In this problem, we are given 1-(2-bromoethyl)cyclohexene and treated with sodium methoxide ($\text{NaOMe}$) in methanol. Let's break down exactly why the reaction takes the path it does.

Imagine you are looking at the reactant molecule. We have a sturdy cyclohexene ring, and attached directly to one of the carbons of the double bond is a 2-bromoethyl side chain.

The reagent, sodium methoxide, is a strong base and a decent nucleophile. Because our leaving group (the bromine atom) is on a primary carbon, you might initially think, "Ah, this is a perfect setup for an $S_{N}2$ substitution!" But in chemistry, we must always look at the bigger picture.

Thermodynamic Control

To understand why elimination wins, we need to look for the most acidic proton. For an $E_2$ elimination, the base must abstract a $\beta$-proton (a proton on the carbon adjacent to the one holding the leaving group).

If we look at our side chain, the first carbon (the one attached to the ring) has two $\beta$-protons. But these aren't just any ordinary protons. They are allylic to the double bond inside the ring! This allylic position makes them unusually acidic because the resulting structure can stabilize itself through resonance.

Let's visualize the mechanism. The methoxide ion approaches and attacks one of these allylic protons. As the carbon-hydrogen bond begins to break, the electrons don't just sit there; they cascade down to form a new $\pi$ bond between the two carbons of the side chain.

Simultaneously, the carbon-bromine bond breaks, and the bromide ion departs. This all happens in one fluid, concerted motion—the hallmark of the $E_2$ mechanism.

The final product is 1-vinylcyclohexene. Notice the structure of this molecule. The newly formed double bond is separated from the ring's double bond by exactly one single bond. This is a conjugated diene.

Conjugation allows the $\pi$ electrons to delocalize over four carbon atoms, drastically lowering the overall energy of the molecule. This massive gain in thermodynamic stability is the ultimate driving force. It lowers the activation energy for the elimination pathway so much that it completely outcompetes the substitution reaction.

Whenever you see an opportunity to form a conjugated system, always bet on it being the major product!