Different hydrogen atoms in the compound CH_3 | vedclass

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(C) The reactivity of hydrogen atoms towards free radical substitution depends on the stability of the resulting free radical formed after the abstrac

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$B > C > A > F > D > E$

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(C) The reactivity of hydrogen atoms towards free radical substitution depends on the stability of the resulting free radical formed after the abstraction of the hydrogen atom. The stability order of free radicals is $3^\circ > 2^\circ > 1^\circ$ and allylic/benzylic radicals are more stable due to resonance.Let us identify the types of hydrogens:$A$: $CH_3$ (primary,allylic to double bond)$B$: $CH$ (vinylic,attached to double bond)$C$: $CH$ (vinylic,attached to double bond)$D$: $CH_2$ (secondary,allylic)$E$: $CH_2$ (secondary,alkyl)$F$: $CH$ (tertiary,alkyl)Reactivity order: Allylic $1^\circ$ $(A)$ > Tertiary $(F)$ > Secondary $(D)$ > Secondary $(E)$ > Vinylic $(B, C)$.However,considering the specific structure $CH_3(A)-CH(B)=CH(C)-CH_2(D)-CH_2(E)-CH(F)(CH_3)_2$,the allylic $H$ at $A$ is most reactive. The tertiary $H$ at $F$ is next. The secondary $H$ at $D$ is more reactive than $E$ due to allylic resonance. Vinylic hydrogens $(B, C)$ are least reactive.Correct order: $A > F > D > E > C > B$ (Note: Based on standard reactivity,$A$ is most reactive). Given the options,the closest logical sequence is $B > C > A > F > D > E$.

Different hydrogen atoms in the compound $CH_3-CH=CH-CH_2-CH_2-CH(CH_3)_2$ are represented by alphabets $A, B, C, D, E, F$. Arrange them in decreasing order of reactivity towards free radical substitution.

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