Animated Solution for Chemistry - Organic Chemistry: The decreasing order of reactivity towards dehydrohalogenation ($E_1$) reaction of the following compounds is :

The $E_1$ (Elimination Unimolecular) reaction is a classic two-step process in organic chemistry. The hallmark of this mechanism is its rate-determining step: the departure of the leaving group to form a carbocation intermediate.

Because this first step is the slowest, it acts as the bottleneck for the entire reaction. Therefore, the golden rule of $E_1$ reactions is simple: the more stable the carbocation intermediate, the faster the reaction.

To solve this problem, we must act as molecular detectives, ionizing each compound by removing the chloride ion and analyzing the stability of the resulting carbocations.

Let's break down the intermediates formed by each of the four compounds:

Compound (A): 1-chloropropane When the chloride ion leaves, it forms a primary ($1^{\circ }$) carbocation. Primary carbocations are notoriously unstable because they only have one adjacent alkyl group to provide electron density through the inductive effect and hyperconjugation. This will likely be our slowest reactor.

Compound (B): 1-chloro-2-butene The departure of the chloride ion here results in a primary allylic carbocation. While it is technically primary, it has a superpower: resonance. The adjacent double bond can delocalize the positive charge, spreading it across multiple carbon atoms. This resonance stabilization makes the allylic carbocation significantly more stable than a simple primary carbocation.

Compound (C): 2-chloropropane When this compound loses its chloride ion, a secondary ($2^{\circ }$) carbocation is formed. This carbocation is stabilized by the inductive effect and hyperconjugation from two adjacent methyl groups. It is comfortably more stable than the primary carbocation from compound (A).

Compound (D): 3-chloro-1-butene The loss of the chloride ion generates a secondary allylic carbocation. This is the jackpot! It enjoys the powerful resonance stabilization from the adjacent double bond (just like B), but it also has an extra alkyl group providing inductive stabilization. This dual-action stabilization makes it the most stable carbocation among all four.

Here is a fascinating catch that often trips up students. If you draw the resonance structures, you will realize that compounds (B) and (D) actually form the exact same delocalized allylic carbocation!

So, if the intermediate is the same, why do they react at different rates?

The answer lies in the transition state. Compound (D) is a secondary allylic halide, while (B) is a primary allylic halide. As the carbon-chlorine bond begins to break, the developing positive charge in the transition state is better stabilized by the adjacent methyl group in (D). Because the transition state for (D) is lower in energy, it ionizes faster than (B). Therefore, (D) reacts faster than (B).

Now, let's put it all together.

The secondary allylic system (D) is the most stable and reacts the fastest. Next is the primary allylic system (B), because resonance stabilization generally trumps simple inductive effects. Then comes the standard secondary carbocation (C), and the least stable is the primary carbocation (A).

The final decreasing order of reactivity is D > B > C > A.

This perfectly matches option (c). Always remember, when dealing with $E_1$ or $S_{N}1$ reactions, let carbocation stability be your ultimate guide!