Reactivity order of the following alkyl halides towards dehydrohalogenation is :
(A) 
(B) 
(C) 
(D) 
Choose the correct answer from the options given below :
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A) > (B) > (C) > (D)
In dehydrohalogenation, a haloalkane heated with alcoholic KOH loses a hydrogen from the β-carbon and the halogen from the α-carbon, and an alkene is formed. The C–X bond must break in this step, so the weaker the C–X bond, the faster the reaction. The C–X bond becomes weaker as the halogen gets larger: C–F is the strongest and C–I the weakest.
\( \displaystyle \ce{CH3CH2CH(X)CH3} \xrightarrow{\text{alc. KOH},\ \Delta} \ce{CH3CH=CHCH3 + KX + H2O} \)
The four compounds are 2-halobutanes with the same carbon skeleton, so only the halogen differs:
R–I (A) > R–Br (B) > R–Cl (C) > R–F (D)
So (A) comes first, (B) second, (C) third and (D) last.
Shortcut: The iodide (A) is the most reactive and the fluoride (D) the least; only Option (1) starts with (A) and ends with (D).