What is the correct order of reactivity of ha | vedclass

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(B) The $S_{N}2$ reaction mechanism is highly sensitive to steric hindrance. The reactivity order for $S_{N}2$ is: $Primary > Secondary > Tertiary$. H

Vedclass · Accepted Answer

$(CH_3)_2CHX < CH_3CH_2X < CH_2=CH-CH_2X < PhCH_2X$

Vedclass · Answer

(B) The $S_{N}2$ reaction mechanism is highly sensitive to steric hindrance. The reactivity order for $S_{N}2$ is: $Primary > Secondary > Tertiary$. However,allylic $(CH_2=CH-CH_2X)$ and benzylic $(PhCH_2X)$ halides are highly reactive in $S_{N}2$ reactions due to the stabilization of the transition state by resonance. Comparing the given compounds: $1$. $(CH_3)_2CHX$ is a secondary halide (least reactive). $2$. $CH_3CH_2X$ is a primary halide. $3$. $CH_2=CH-CH_2X$ (allyl halide) and $PhCH_2X$ (benzyl halide) are more reactive than primary halides due to resonance stabilization of the transition state. Between allyl and benzyl,the benzyl group is generally more reactive due to extended conjugation. The correct order is: $(CH_3)_2CHX < CH_3CH_2X < CH_2=CH-CH_2X < PhCH_2X$.

What is the correct order of reactivity of halides for the $S_{N}2$ reaction?

Similar Questions

Which alkyl halide from the following pairs would you expect to react more rapidly by an $S_N2$ mechanism? Explain your answer.
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$(ii)$ $CH_3CH_2CH(Br)CH_3$ or $(CH_3)_3CBr$
$(iii)$ $CH_3CH(CH_3)CH_2CH_2Br$ or $CH_3CH_2CH(CH_3)CH_2Br$

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