For the molecule CH_3(a)-CH_2(b)-CH_2(c)-CH_2 | vedclass

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Question

(C) The reactivity of hydrogens towards free radical chlorination depends on the stability of the intermediate radical formed after hydrogen abstracti

Vedclass · Accepted Answer

$b > c > a > d$

Vedclass · Answer

(C) The reactivity of hydrogens towards free radical chlorination depends on the stability of the intermediate radical formed after hydrogen abstraction.Stability of radicals follows the order: $3^o > 2^o > 1^o$.Additionally,the $-I$ effect of the fluorine atom destabilizes the radical formed on the adjacent carbons,with the effect decreasing as the distance from the $F$ atom increases.For $CH_3(a)-CH_2(b)-CH_2(c)-CH_2(d)-F$:- Hydrogen $d$ is on the carbon attached to $F$ (most destabilized by $-I$ effect).- Hydrogen $c$ is on the $\beta$-carbon.- Hydrogen $b$ is on the $\gamma$-carbon.- Hydrogen $a$ is on the $\delta$-carbon (least affected by $-I$ effect).However,considering the radical stability,$b$ and $c$ are $2^o$ radicals,while $a$ and $d$ are $1^o$ radicals.Comparing the $-I$ effect,the reactivity order is $b > c > a > d$.

For the molecule $CH_3(a)-CH_2(b)-CH_2(c)-CH_2(d)-F$,arrange the hydrogens $a, b, c, d$ in decreasing order of their reactivities towards free radical chlorination.

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