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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:

Similar Questions

The order of reactivity of the given haloalkanes towards nucleophiles is:

$CCl_4$ and $CHCl_3$ can be distinguished by .....

$CH_3-CH_2-CH(Br)-CH_3 \xrightarrow{Alc. KOH} X$ (major)
$CH_3-CH_2-CH(NMe_3^+) - CH_3 \xrightarrow{EtONa, \Delta} Y$ (major)
Product $(X)$ and $(Y)$ respectively are:

$(i) \, (CH_3)_2CH-CH_2Br \xrightarrow{C_2H_5OH} (CH_3)_2CH-CH_2OC_2H_5 + HBr$
$(ii) \, (CH_3)_2CH-CH_2Br \xrightarrow{C_2H_5O^-} (CH_3)_2CH-CH_2OC_2H_5 + Br^-$
The reaction mechanisms for $(i)$ and $(ii)$ are respectively:

$CH_3-C(CH_3)=CH-CH_2-CH_2-Cl \xrightarrow[CaCO_3]{H_2O} (X)$
The major product $(X)$ of the above reaction is:

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