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(B) The reaction of alkenes with $HBr$ proceeds via the formation of a carbocation intermediate,which is the rate-determining step. The reactivity is

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$b > a > c$

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(B) The reaction of alkenes with $HBr$ proceeds via the formation of a carbocation intermediate,which is the rate-determining step. The reactivity is directly proportional to the stability of the carbocation formed.$1$. For $(b)$,the carbocation formed is $CH_3O-C^+(CH_3)-CH_3$,which is highly stabilized by resonance due to the lone pair on the oxygen atom ($+M$ effect).$2$. For $(a)$,the carbocation formed is $(CH_3)_2C^+-CH_3$,which is stabilized by $9$ hyperconjugative $\alpha$-hydrogens.$3$. For $(c)$,the carbocation formed is $CH_3OCH_2-C^+(CH_3)-CH_3$. The $CH_3OCH_2-$ group exerts an electron-withdrawing inductive effect ($-I$ effect),which destabilizes the carbocation.Thus,the order of stability of the carbocations is $(b) > (a) > (c)$. Consequently,the order of reactivity towards $HBr$ is $(b) > (a) > (c)$.

Arrange the following compounds in the decreasing order of their reactivity towards $HBr$:
$(a)$ $(CH_3)_2C=CH_2$
$(b)$ $CH_3O-C(CH_3)=CH_2$
$(c)$ $CH_3OCH_2-C(CH_3)=CH_2$

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Arrange the following compounds in the decreasing order of their reactivity towards $HBr$:$(a)$ $(CH_3)_2C=CH_2$$(b)$ $CH_3O-C(CH_3)=CH_2$$(c)$ $CH_3OCH_2-C(CH_3)=CH_2$