The decreasing order of reactivity towards dehydrohalogenation (E₁) reaction of the following compounds is :

Solution

Chemistry ma’am · 3 steps

In E1 the carbocation decides everything, so rank each chloride by how stable its cation is.

1. Slow step. In an E1 reaction the C–Cl bond first breaks on its own to give a carbocation, and this is the slow step. A proton is lost from the next carbon only after that. So the more stable the carbocation, the faster the reaction.

   Compound	Carbocation formed	Extra stabilisation	Rank
   D, 3-chlorobut-1-ene	2° allylic, right next to the C=C	resonance with the π bond, plus hyperconjugation	1
   B, 4-chlorobut-1-ene	1° at first, then a 1,2-H shift gives the 2° allylic cation	nearby π bond helps, and the shift leads to resonance	2
   C, 2-chlorobutane	2° alkyl	hyperconjugation only	3
   A, 1-chlorobutane	1° alkyl	very little	4

2. Why B beats C. The double bond in B is two carbons away from the cationic carbon. It helps that carbon lose Cl⁻, and a 1,2-hydride shift then turns the cation into the resonance-stabilised 2° allylic one. B still ranks below D, because D reaches that allylic cation directly without any primary stage.
3. So. The decreasing order of E1 reactivity is D > B > C > A, which is option (b).

The idea to keep. For any E1 or S_N1 ranking, first draw the carbocation each substrate gives. Allylic resonance beats plain alkyl groups. Also check whether a hydride shift can turn a poor cation into a better one.

Answer from the NTA final answer key. Solution by PrepWiser.