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(D) The reaction of haloalkanes with alcoholic $KOH$ proceeds via the $E2$ elimination mechanism.In $CH_3CH_2CH_2Br$ (a primary alkyl halide),the pres

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$CH_3CH_2CH_2Br$

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(D) The reaction of haloalkanes with alcoholic $KOH$ proceeds via the $E2$ elimination mechanism.In $CH_3CH_2CH_2Br$ (a primary alkyl halide),the presence of the alkyl group provides a $+I$ effect,which facilitates the elimination of the leaving group.$CH_2=CHBr$ is a vinyl halide where the $C-X$ bond has partial double bond character,making it very unreactive towards elimination.$CH_3COCH_2CH_2Br$ contains an electron-withdrawing $-CO$ group ($-I$ effect),which destabilizes the transition state for elimination compared to a simple alkyl halide.Comparing $CH_3CH_2Br$ and $CH_3CH_2CH_2Br$,the latter has a slightly higher $+I$ effect,making it more reactive in this context.$CH_3CH_2CH_2Br + KOH \xrightarrow{	ext{alc.}} CH_3CH=CH_2 + KBr + H_2O$

Among the following,the most reactive towards alcoholic $KOH$ is:

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

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