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Arrange the following compounds in the order of their increasing reactivity towards nucleophilic substitution:

Substituted benzyl chlorides

The answer to this question is $[3<2<1<4]$. We have to compare the positive charge density on the carbons to which leaving group is connected. In $(1)$, the phenyl ring shows negative inductive effect but my teacher did not mention anything for that. In $(3)$ and $(4)$ the connected groups ($\ce{OCH3}$ and $\ce{NO2}$) show +M and -M respectively. But, they also show -I effect. Why are we ignoring the inductive effect?

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    $\begingroup$ You're kind of overcomplicating it. The rate-determining step is attack of a nucleophile on the electrophilic C=O. The more electrophilic it is, the faster the reaction. You don't need to care about the inductive effect of the phenyl group, because all of them have phenyl groups. The only thing that matters is the 4-substituent. $\endgroup$ – orthocresol Jun 4 at 12:05
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    $\begingroup$ The person who decided it was OK to teach students that +ve and -ve are acceptable abbreviations is one of the worst people of the last century. + and - already mean positive and negative! $\endgroup$ – Zhe Jun 4 at 12:58
  • $\begingroup$ rate of reaction is proportional to the concentration of the nucleophile. But in this question, we don't know which nucleophile is being used. $\endgroup$ – bhuvanesh Jun 4 at 14:43
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Electronic effects govern in the order

Resonance/Mesomeric > Hyperconjugation > inductive

+M or +R dominate over inductive. That's because +M involves π electrons which are more reactive than the sigma electrons. + I operates through sigma electrons. So therefore +M effect has priority over +I.

Therefore the order of reactivity towards nucleophilic substitution is 4>1>2>3

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  • $\begingroup$ does that mean we completely ignore the inductive effect? $\endgroup$ – bhuvanesh Jun 5 at 4:09
  • $\begingroup$ No, it's just that +M must be considered first. That's because +M involves π electrons which are more reactive than the sigma electrons. + I operates through sigma electrons. $\endgroup$ – Yashwini Jul 14 at 18:20

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