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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

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$2 > 3 > 1$

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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):
$(1)$ $CH_2=C(CH_3)Br$
$(2)$ $1-bromo-1-methylcyclohexane$
$(3)$ $CH_3-CH(Br)-CH_2-CH(CH_3)_2$

Similar Questions

The correct reactivity order of the following compounds towards alkaline hydrolysis is:-
$I. \ p-CH_3O-C_6H_4-CH_2Cl$
$II. \ C_6H_5-CH_2Cl$
$III. \ p-CH_3-C_6H_4-CH_2Cl$
$IV. \ m-CH_3-C_6H_4-CH_2Cl$

In the given reaction
$CH_3-CH(OTs)-CH_2-CH_2-CH(OTs)-CH_3$ $\xrightarrow[(ii) \ KOH]{(i) \ SH^{\ominus} (one \ equivalent)}$ $[X]$
$[X]$ will be :

The graph which represents the following reaction is :
$(C_6H_5)_3C-Cl \xrightarrow{OH^{-} / \text{Pyridine}} (C_6H_5)_3C-OH$

The reagents required for the following two-step transformation are:

The decreasing order of reactivity of the compounds given below towards solvolysis under identical conditions is:

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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)$ $CH_3-CH(Br)-CH_2-CH(CH_3)_2$