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

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$Q < P < R$

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(A) The reaction of alkenes with $HBr$ proceeds via the formation of a carbocation intermediate,which is the rate-determining step. The rate of reaction depends on the stability of the carbocation formed. When $H^+$ attacks the double bond,the carbocation is formed at the carbon adjacent to the heteroatom. In $(P)$,the carbocation is stabilized by the $+M$ effect of the oxygen atom. In $(R)$,the carbocation is stabilized by the $+M$ effect of the nitrogen atom. In $(Q)$,there is no such resonance stabilization,only inductive effects. Since nitrogen is less electronegative than oxygen,it donates its lone pair more effectively via resonance ($+M$ effect). Therefore,the carbocation formed from $(R)$ is the most stable,followed by $(P)$,and then $(Q)$. The order of stability of carbocations is: $(R) > (P) > (Q)$. Thus,the increasing order of the rate of reaction is: $Q < P < R$.

Rank the following in the increasing order of rate of reaction with $HBr$.

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