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

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The reactivity of hydrogen atoms towards free radical substitution depends on the stability of the resulting free radical formed after hydrogen abstraction.Stability order of free radicals: $3^\circ > 2^\circ > 1^\circ > 	ext{vinylic}$.Let's analyze the types of hydrogen atoms:$(A)$ is $1^\circ$ allylic (adjacent to double bond).$(B)$ is vinylic (directly on double bond),which is least reactive.$(C)$ is $2^\circ$ allylic (adjacent to double bond).$(D)$ is $2^\circ$ alkyl.$(E)$ is $3^\circ$ alkyl.$(F)$ is $1^\circ$ alkyl.Comparing stability: $2^\circ$ allylic $(C) > 1^\circ$ allylic $(A) > 3^\circ$ alkyl $(E) > 2^\circ$ alkyl $(D) > 1^\circ$ alkyl $(F) > 	ext{vinylic } (B)$.Wait,re-evaluating: The reactivity order is $C > A > E > D > F > B$. Thus,the correct option is $A$.

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

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