In S_{N}2 reactions,the correct order of reac | vedclass

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Question

(B) The rate of $S_{N}2$ reactions is inversely proportional to the steric hindrance around the electrophilic carbon atom.As the number of alkyl group

Vedclass · Accepted Answer

$CH_3Cl > CH_3CH_2Cl > (CH_3)_2CHCl > (CH_3)_3CCl$

Vedclass · Answer

(B) The rate of $S_{N}2$ reactions is inversely proportional to the steric hindrance around the electrophilic carbon atom.As the number of alkyl groups attached to the carbon bearing the halogen increases,the steric crowding increases,which hinders the approach of the nucleophile.The order of reactivity for alkyl halides in $S_{N}2$ reactions is: $Methyl \ halide > Primary \ (1^{\circ}) \ halide > Secondary \ (2^{\circ}) \ halide > Tertiary \ (3^{\circ}) \ halide$.Comparing the given compounds:$CH_3Cl$ (Methyl halide) > $CH_3CH_2Cl$ (Primary) > $(CH_3)_2CHCl$ (Secondary) > $(CH_3)_3CCl$ (Tertiary).Therefore,the correct order is $CH_3Cl > CH_3CH_2Cl > (CH_3)_2CHCl > (CH_3)_3CCl$.

In $S_{N}2$ reactions,the correct order of reactivity for the following compounds:

Similar Questions

Arrange the following in the correct order of reactivity towards $S_{N}2$ reaction:
$I$: $1$-Chloropropane
$II$: $2$-Chloropropane
$III$: $1$-Iodopropane
$IV$: $2$-Iodopropane

Which is the correct order of the following alkyl halides for an $S_{N}2$ reaction?

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Arrange the following compounds in order of decreasing rate of hydrolysis for $S_{N}1$ reaction:
$(I) \ C_6H_5-CH_2-Br$
$(II) \ p-CH_3-C_6H_4-CH_2-Br$
$(III) \ p-CH_3CH_2-C_6H_4-CH_2-Br$
$(IV) \ p-(CH_3)_2CH-C_6H_4-CH_2-Br$

Assertion : $CHCl_3$ is stored in dark bottles.
Reason : $CHCl_3$ is oxidised in dark.

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