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(D) The $SN_1$ reaction rate depends on the stability of the carbocation formed in the rate-determining step.The carbocations formed are:$I$. $CH_3CH_

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$I < III < II < IV$

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(D) The $SN_1$ reaction rate depends on the stability of the carbocation formed in the rate-determining step.The carbocations formed are:$I$. $CH_3CH_2CH_2CH_2^+$ (Primary)$II$. $CH_3CH_2CH^+CH_3$ (Secondary)$III$. $(CH_3)_2CHCH_2^+$ (Primary,but branched)$IV$. $(CH_3)_3C^+$ (Tertiary)Tertiary carbocations are more stable than secondary,which are more stable than primary carbocations.Among the primary carbocations,$III$ is slightly more stable than $I$ due to the inductive effect of the methyl group.Therefore,the stability order is $I Since the rate of $SN_1$ reaction follows the stability order of carbocations,the reactivity order is $I < III < II < IV$.

The order of reactivity of the following compounds towards dilute aqueous $KOH$ in $SN_1$ reaction is
$I$. $CH_3CH_2CH_2CH_2Br$
$II$. $CH_3CH_2CHBrCH_3$
$III$. $(CH_3)_2CHCH_2Br$
$IV$. $(CH_3)_3CBr$

Similar Questions

Which of the following rules applies in the reaction:
$CH_3CHBrCH_2CH_3 \xrightarrow{alc. KOH}$
$(i)$ $CH_3CH = CHCH_3$ (major product)
$(ii)$ $CH_2 = CHCH_2CH_3$ (minor product)

Which of the following alkyl halides has the highest boiling point?

$CH_3-CH_2-CH_2-Cl$ $\xrightarrow{KOH_{(aq)}} X$ $\xrightarrow{H^{+}/\Delta} Y$ $\xrightarrow{Cl_2/H_2O} Z$ (Major); $Z$ is

The following reaction is supposed to take place through the $S_N1$ mechanism: $CH_3-C(Ph)(OH)-CH_2-CH_3 \xrightarrow{HI} \text{products}$. If the configuration of the substrate is $D$,then the configuration of the products will be:

Most reactive compound toward $S_N1$ is

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The order of reactivity of the following compounds towards dilute aqueous $KOH$ in $SN_1$ reaction is$I$. $CH_3CH_2CH_2CH_2Br$$II$. $CH_3CH_2CHBrCH_3$$III$. $(CH_3)_2CHCH_2Br$$IV$. $(CH_3)_3CBr$