Animated Solution for Chemistry - Organic Chemistry: The product $A$ and $B$ formed in above reactions are

Saytzeff vs. Hofmann

Imagine you are an architect tasked with designing a molecule. You have a starting blueprint—a cyclohexane ring with a methyl group and a leaving group attached to the same carbon. Your goal is to create a double bond, but the tools you use will completely change the final structure. This problem is a beautiful demonstration of how reagents dictate the pathway and the outcome of elimination reactions.

The E1 Dehydration

In our first scenario, we start with $1$-methylcyclohexanol. We introduce it to $20\%{\text{ H}}_3{\text{PO}}_4$ and turn up the heat to $358\text{ K}$. This is the classic setup for an acid-catalyzed dehydration.

The acid's first job is to protonate the hydroxyl ($-\text{OH}$) group. Why? Because $-\text{OH}$ is a terrible leaving group, but once protonated, it becomes water (${\text{H}}_2\text{O}$), which is an excellent leaving group. As the water molecule departs, it takes its electrons with it, leaving behind a positively charged carbon. This forms a $3^{\circ }$ carbocation intermediate, which is highly stable due to hyperconjugation and the inductive effect of the surrounding alkyl groups.

Now, the molecule needs to stabilize itself by losing a proton (${\text{H}}^{+}$) from an adjacent carbon to form a double bond. Here, Saytzeff's rule takes the wheel. The rule states that the most substituted alkene is thermodynamically the most stable. Removing a proton from the ring creates a trisubstituted double bond, yielding $1$-methylcyclohexene. This is our major product, A.

The E2 Dehydrohalogenation

Now, let's look at the second reaction. We start with $1$-chloro-$1$-methylcyclohexane and react it with potassium tert-butoxide ($t{\text{-BuO}}^{-}{\text{K}}^{+}$).

Tert-butoxide is a strong base, which means it won't wait around for a carbocation to form. It wants to grab a proton immediately, driving the reaction via an $E_2$ mechanism. However, there is a massive catch: tert-butoxide is incredibly bulky. It's like trying to fit a large truck into a tiny parking space.

Due to severe steric hindrance, this bulky base cannot easily reach the internal protons of the cyclohexane ring. Instead, it takes the path of least resistance and abstracts a proton from the exposed, unhindered methyl group. This leads to the formation of the less substituted alkene, known as the Hofmann product. The resulting molecule is methylenecyclohexane, which is our major product, B.

By carefully analyzing the reagents, we've uncovered the hidden logic of these reactions. The acid-catalyzed dehydration favored thermodynamic stability, giving us the Saytzeff product ($1$-methylcyclohexene). In contrast, the bulky base in the dehydrohalogenation was governed by steric kinetics, yielding the Hofmann product (methylenecyclohexane). Matching these findings with our options, we confidently arrive at the correct answer.