Consider the following reaction,CH_3-CH(D)-CH | vedclass

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(B) The reaction involves the abstraction of a hydrogen atom by a bromine radical $(\dot{Br})$.Stability of free radicals follows the order: $3^{\circ

Vedclass · Accepted Answer

$CH_3-CH(D)-\dot{C}(CH_3)-CH_3$

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(B) The reaction involves the abstraction of a hydrogen atom by a bromine radical $(\dot{Br})$.Stability of free radicals follows the order: $3^{\circ} > 2^{\circ} > 1^{\circ}$.In the given molecule $CH_3-CH(D)-CH(CH_3)-CH_3$,there are different types of hydrogen atoms available for abstraction.The abstraction of a hydrogen atom from the $C-3$ position leads to the formation of a $3^{\circ}$ free radical,$CH_3-CH(D)-\dot{C}(CH_3)-CH_3$,which is the most stable intermediate.Since the $C-H$ bond energy is lower than the $C-D$ bond energy,the hydrogen atom is abstracted preferentially over the deuterium atom.Thus,the major product $X$ is $CH_3-CH(D)-\dot{C}(CH_3)-CH_3$.

Consider the following reaction,$CH_3-CH(D)-CH(CH_3)-CH_3 + \dot{Br} \to X + HBr$. Identify the structure of major product $X$.

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