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The increasing order of reduction of alkyl halides with zinc and dilute $\mathrm{HCl}$ is
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The correct answer is:
$\mathrm{R}-\mathrm{Cl} < \mathrm{R}-\mathrm{Br} < \mathrm{R}-$ I
$\mathrm{R}-\mathrm{Cl} < \mathrm{R}-\mathrm{Br} < \mathrm{R}-$ I
The reactivity of halogens with alkanes follows the order: $\mathrm{F}_2>\mathrm{Cl}_2>\mathrm{Br}_2>\mathrm{I}_2$. Further, the reactivity of reduction of alkyl halides with $\mathrm{Zn}$ \& dilute $\mathrm{HCl}$ increases as the strength of $\mathrm{C}-\mathrm{X}(\mathrm{X}=\mathrm{F}, \mathrm{Cl}, \mathrm{Br}, \mathrm{I})$ bond decreases. Hence, reduction of alkyl halide with $\mathrm{Zn}$ and dilute $\mathrm{HCl}$ follows the order :
$$
\mathrm{R}-\mathrm{I}>\mathrm{R}-\mathrm{Br}>\mathrm{R}-\mathrm{Cl} \text {. }
$$
$$
\mathrm{R}-\mathrm{I}>\mathrm{R}-\mathrm{Br}>\mathrm{R}-\mathrm{Cl} \text {. }
$$
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