Question

The correct order of reactivity of following compounds in nucleophilic substitution $\left(S_{N}2\right)$ is:

• A. $A>B>C>D$
• B. $D>C>B>A$
• C. $C>A>D>B$
• D. $C>A>B>D$

Answer 1

The correct option is A $A>B>C>D$

$C{H}_2=C{H}_2{B}_r>C{H}_{3}C{H}_{2}Br>$

Condition for reactivity order of $S{N}^2$

i) ally bromide accelerated by $\pi -$ conjunction in-transition sate because of that nudeophilic substitution take faster in allyl bromide rather than alkly bromide.

ii) Reactivity order in $S{N}^2$ is methyl $1>i^o>2^O>3^o$ some time $3^o$ alkyl holides are are so crowded that they do not genrally react by an $S{N}^{2}mec{h}^m.$