Animated Solution for Chemistry - Organic Chemistry: The alkene formed as a major product in the above elimination reaction is

A Bulky Behemoth

Imagine you are a tiny hydroxide ion (${\text{OH}}^{\ominus }$), floating in a hot solution, looking for a proton to abstract. You come across a massive, intimidating molecule. It's a quaternary ammonium salt. At its core is a nitrogen atom, bearing a positive charge, bonded to four different carbon groups: a methyl group, an ethyl group, an n-butyl group, and a bulky cyclohexyl ring.

This is not your everyday, simple alkyl halide. The leaving group here is a trialkylamine ($-{\text{NR}}_3$), which is incredibly bulky and a relatively poor leaving group compared to a nimble bromide or chloride ion.

When you heat this behemoth in the presence of a strong base, an elimination reaction is inevitable. Specifically, this is an $\text{E}2$ elimination. But because of the sheer size of the leaving group, the standard rules of the game are about to be flipped upside down.

Zaitsev vs. Hofmann

In a typical $\text{E}2$ elimination with a small leaving group, the reaction follows Zaitsev's Rule. This rule dictates that the major product will be the most substituted, and therefore most thermodynamically stable, alkene. The base carefully maneuvers to abstract the proton that leads to the most stable double bond.

However, our quaternary ammonium salt is a different beast. The massive $-{\text{NR}}_3^{\oplus }$ group creates a tremendous amount of steric hindrance. It's like trying to park a bus in a crowded alleyway. The base simply cannot reach the more hindered protons easily.

Enter Hofmann's Rule. Because of the steric clash in the transition state, the base takes the path of least resistance. It attacks the most exposed, least sterically hindered $\beta$-hydrogen it can find. As a result, Hofmann elimination yields the least substituted alkene as the major product.

To predict the outcome, we must become molecular detectives and hunt down all the $\beta$-hydrogens.

First, we identify the $\alpha$-carbons. These are the carbon atoms directly attached to the positively charged nitrogen.

Next, we look one bond further to find the $\beta$-carbons. Any hydrogen attached to a $\beta$-carbon is a $\beta$-hydrogen, a prime target for our hydroxide base. Let's analyze the four groups attached to the nitrogen:

1. The Methyl Group ($-{\text{CH}}_3$): This carbon is directly attached to the nitrogen, making it an $\alpha$-carbon. It has no adjacent carbons, meaning it has zero $\beta$-hydrogens. It cannot participate in the elimination.

2. The Ethyl Group ($-{\text{CH}}_2-{\text{CH}}_3$): The terminal methyl group is a $\beta$-carbon. It proudly presents three unhindered $\beta$-hydrogens. Let's call them ${\beta }_1(\text{H})$.

3. The n-Butyl Group ($-{\text{CH}}_2-{\text{CH}}_2-{\text{CH}}_2-{\text{CH}}_3$): The second carbon in this chain is a $\beta$-carbon, possessing two $\beta$-hydrogens, which we will label ${\beta }_2(\text{H})$.

4. The Cyclohexyl Ring: This ring has two identical $\beta$-carbons adjacent to the $\alpha$-carbon. Each of these $\beta$-carbons holds two $\beta$-hydrogens, giving us ${\beta }_3(\text{H})$.

Comparing the Alkenes

Now, let's simulate the elimination for each set of $\beta$-hydrogens and see what alkenes we produce.

If the base attacks ${\beta }_1(\text{H})$ on the ethyl group, the bond collapses, the bulky amine leaves, and we form ethene (${\text{CH}}_2={\text{CH}}_2$).

If the base attacks ${\beta }_2(\text{H})$ on the n-butyl group, the resulting double bond gives us but-1-ene (${\text{CH}}_2=\text{CH}-{\text{CH}}_2-{\text{CH}}_3$).

If the base attacks ${\beta }_3(\text{H})$ on the cyclohexyl ring, the double bond forms within the ring, yielding cyclohexene.

We have our three contenders. Now, we must classify them based on their substitution: - Ethene has zero alkyl groups attached to the double bond. It is unsubstituted. - But-1-ene has one alkyl group attached. It is monosubstituted. - Cyclohexene has two alkyl groups attached (the rest of the ring). It is disubstituted.

We bring back our guiding principle: Hofmann's Rule. The major product must be the least substituted alkene because it comes from the least sterically hindered transition state.

Comparing ethene, but-1-ene, and cyclohexene, the winner is crystal clear. Ethene is completely unsubstituted. The $\beta$-hydrogens on the ethyl group are the most exposed and easiest for the base to grab.

Therefore, the base overwhelmingly attacks the ethyl group, making ethene (${\text{CH}}_2={\text{CH}}_2$) the major product of this reaction!

This problem is a beautiful demonstration of how steric hindrance can completely override thermodynamic stability, forcing a chemical reaction down a completely different path. Always respect the size of your leaving group!