I am unable to understand how to decide reactivity of alkyl halides based on inductive effect in a nucleophilic substitution reaction. I have 2 explanation for deciding. Consider 2 alkyl halides:

  • (A) 2-chloro-2-methylbutane
  • (B) 2-chloro-3-methylbutane

First explanation:

In (A) we have a $3^\circ$ carbon so if we break the $\ce{C--Cl}$ bond a $3^\circ$ carbocation will be formed, which is stabilized by +Inductive effect of surrounding groups. Therefore we can say that as breaking of $\ce{C--X}$ bond in (A) leads to the formation of a more stable carbocation than that formed in (B){$2^\circ$ carbocation}. And due to this the reactivity will be more of (A).

Second explanation:

As in (A) $3^\circ$ carbocation is formed, it will get more stabilized than $2^\circ$ carbocation formed in (B). So this means that $2^\circ$ carbocation in (B) becomes more positive than $3^\circ$ carbocation in (A), making it easier for a nucleophile to attack on (B)'s carbocation. This would make (B) more reactive.

Which one of them is correct?


2 Answers 2


Assuming your concern is on $\mathrm{S_N1}$ type nucleophilic substitution reactions, you are in the correct path with your first explanation. However, everything has gone south from that point on. As you found out, the stability of carbocations in $\mathrm{S_N1}$ reactions are: $3^\circ \gt 2^\circ \gt 1^\circ$. However, reaction rates of $\mathrm{S_N1}$ reactions do not depend on electron density (or lack of it) on carbocation, because rate determine step is rate of carbocation formation step:

Reaction paths

As depicted in the diagram, $3^\circ$-carbocation from 2-chloro-2-methylbutane forms faster than $2^\circ$-carbocation from 2-chloro-3-methylbutane because the stability of $3^\circ$-carbocation is much larger than that of $2^\circ$-carbocation as you predicted in your first explanation. Also, you need to understand that in these reactions, 1,2-hydride (or methide) transfer to give more stable $3^\circ$-carbocation from $2^\circ$-carbocation is possible if there are available $\alpha$-hydrogen or $\alpha$-methyl group in the sought $2^\circ$-carbocation, alike in the case of 2-chloro-3-methylbutane here. Regardless of this rearrangement, the rate of carbocation formation of 2-chloro-2-methylbutane is still faster than that of 2-chloro-3-methylbutane, because it is strictly the stability of $3^\circ$- versus $2^\circ$-carbocations.

The second reaction (attack of the nucleophile to available carbocation) is very fast compared to chloride dissociation, thus it is not affecting the overall rate of the reaction. Thus, the product formation from 2-chloro-2-methylbutane is faster than the product formation from 2-chloro-3-methylbutane, and hence, 2-chloro-2-methylbutane is more reactive than 2-chloro-3-methylbutane.

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    $\begingroup$ @Ameya M: Good catch. $\endgroup$ Commented Apr 27, 2020 at 9:31

In tertiary alkyl halides, carbocation formation occurs and reaction proceeds via SN1 Mechanism. Formation of carbocation is a slow step. It requires energy,(endothermic). Carbocation as well as the leaving group (halide ion) is stabilised by solvation by a polar protic solvent Carbocation formation is ofcourse the Rate Determining step which means that stability of carbocation determines the rate at which reaction will proceed and not better positive centre for nucleophilic attack. Nucleophilic attack will be very fast and won't affect the rate of reaction


And in the said secondary alkyl halide it depends on nucleophile whether it will proceed via SN1 or SN2 mechanism as secondary alkyl halides are less hindered and use of a strong nucleophile will favour SN2 mechanism.


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