Question

In the elimination reactions, the reactivity of alkyl halides follows the sequence

• A. $R-F>R-I>R-Br>R-Cl$
• B. $R-F>R-Cl>R-Br>R-I$
• C. $R-I>R-Br>R-Cl>R-F$
• D. $R-I>R-F>R-Br>R-Cl$

Answer 1

Answer: (C)

For a general reaction
$R-L+B^{-}⟶$ alkene $+BH+L^{\ominus }$
where, $L^{\ominus }$ is the leaving group.
Rate of cleavage of $C-X$ bond is in order
$\to X\rightleftharpoons \to \oplus +X^{-}$
$\underset{\text{ fastest }}{I}>Br>Cl>\underset{\text{ slowest }}{F}$
So, the ease of the leaving group in alkyl halide is
$-I>-Br>-Cl>-F$
The bond strengths being in order
$C-F>C-Cl>C-Br>C-I$
and amongst the alkyl group the order of reactivity is
ter $>$ sec $>$ pri
But in the given question the alkyl group is the same. So, the reactivity of the alkyl halide is in the following order
$R-I>R-Br>R-Cl>R-F$