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(C) The rate of solvolysis follows an $S_N1$ mechanism,which depends on the stability of the carbocation intermediate formed after the departure of th

Vedclass · Accepted Answer

$2 > 3 > 1$

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(C) The rate of solvolysis follows an $S_N1$ mechanism,which depends on the stability of the carbocation intermediate formed after the departure of the leaving group $(Br^-)$.$(1)$ $CH_2=C(CH_3)^+$ is a vinylic carbocation,which is highly unstable due to the $sp$ hybridization of the positively charged carbon.$(2)$ $1-methylcyclohexyl$ cation is a tertiary $(3^{\circ})$ carbocation,which is highly stable due to hyperconjugation and inductive effects.$(3)$ $CH_3-CH^+-CH_2-CH(CH_3)_2$ is a secondary $(2^{\circ})$ carbocation,which is less stable than a tertiary carbocation but more stable than a vinylic carbocation.Therefore,the order of stability of the carbocations is $(2) > (3) > (1)$.Thus,the rate of solvolysis is $(2) > (3) > (1)$.

Rank the following compounds in order of decreasing rate of solvolysis with aqueous ethanol (fastest $\to$ slowest):
$(1)$ $CH_2=C(CH_3)Br$
$(2)$ $1-bromo-1-methylcyclohexane$
$(3)$ $CH_3-CH(Br)-CH_2-CH(CH_3)_2$

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Rank the following compounds in order of decreasing rate of solvolysis with aqueous ethanol (fastest $\to$ slowest):$(1)$ $CH_2=C(CH_3)Br$$(2)$ $1-bromo-1-methylcyclohexane$$(3)$ $CH_3-CH(Br)-CH_2-CH(CH_3)_2$