The correct decreasing order of SN^2 reactivi | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) In $SN^2$ reactions,the reactivity depends primarily on steric hindrance. The general order of reactivity for alkyl halides is $Methyl > Primary (

Vedclass · Accepted Answer

$II > IV > III > I$

Vedclass · Answer

(B) In $SN^2$ reactions,the reactivity depends primarily on steric hindrance. The general order of reactivity for alkyl halides is $Methyl > Primary (1^\circ) > Secondary (2^\circ) > Tertiary (3^\circ)$. $(II)$ $CH_3-Br$ is a methyl halide,which has the least steric hindrance and is most reactive. $(IV)$ $CH_3-CH_2-Br$ is a primary $(1^\circ)$ alkyl halide. $(III)$ $CH_3-CH(Br)-CH_3$ is a secondary $(2^\circ)$ alkyl halide. $(I)$ $CH_2=CH-Br$ is a vinyl halide,which is the least reactive because the $C-Br$ bond has partial double bond character due to resonance,making it very difficult to break. Therefore,the correct decreasing order of reactivity is $II > IV > III > I$.

The correct decreasing order of $SN^2$ reactivity of the following alkyl halides is:
$(I)$ $CH_2=CH-Br$
$(II)$ $CH_3-Br$
$(III)$ $CH_3-CH(Br)-CH_3$
$(IV)$ $CH_3-CH_2-Br$

Similar Questions

The reaction of a mixture of two alkyl halides with sodium in ether gives $2-$methylpropane. The alkyl halides are:

$A$ student was carrying out a Grignard reaction between $PhMgBr$ and ethyl benzoate. She ran out of anhydrous ether just after the Grignard reagent was made. Which of the following solvents can still be used to dissolve the ethyl benzoate for its reaction with already formed $PhMgBr$?

Which among the following compounds undergoes $SN^2$ reaction most rapidly?

Arrange the following three chlorides in decreasing order of their reactivity towards $S_{N^1}$ reaction.

Which of the following,on heating with aqueous $KOH$,produces acetaldehyde?

Vedclass Test Series

Exam Paper Generator

Online Exam Module

The correct decreasing order of $SN^2$ reactivity of the following alkyl halides is: $(I)$ $CH_2=CH-Br$ $(II)$ $CH_3-Br$ $(III)$ $CH_3-CH(Br)-CH_3$ $(IV)$ $CH_3-CH_2-Br$