Why is the rate of reaction of polar EAS favored by inductive effects and the rate of less polar EAS reactions through hyperconjugation?

Question

1. Reactions involving strong electrophiles (nitration) are said to be governed by Inductive effects thus the order of nitration is
D>C>B>A
2.Chlorination is a less polar reaction thus chlorination said the be governed by hyper conjugation.
A>B>C>D

My question is, why?

The factors controlling EAS in both the cases will be the same for both the substitutes.
Hyperconjugation and inductive effects will both operate without distinction in both the cases.

So why is it that
The rate of reaction of polar EAS favored by inductive effects and the rate of less polar EAS reactions through hyperconjugation.

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SOURCE A guidebook to mechanism Peter Sykes

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EDIT I have been asking around for an answer to this question.

No one has been able to answer it conclusively , the primary school of thought is that

In Highly polar reactions , inductive effects dominate as more electron density is available , hyperconjugation on the other hand is just redistribution of electrons.

First of all is this correct ?

Even if it's correct but how does it explain the opposing rate orders for nitration and bromination?

Answer 1

I may be oversimplifying things.

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A look at electron density

Let's pick up the extreme cases of the substituted benzenes . $$\ce{A -> ArCH3}$$
$$\ce{D -> Ar(CH3)_3}$$
Case A shows the maximum hyperconjugation.
Case D on the other has the maximum +I effect.
For a moment let's imagine that the electrons are coins.
Case D , has more coins, more methyl groups (let's say 3).
Case A, on the other hand has 1 coin , hyperconjugation would simply mean that I am simply moving that coin in circles. It's simply redistribution of electrons.it doesn't actually increase the density.

So the first point which I'm going to use to answer the question is that -

• Inductive effects increases the electron density, hyperconjugation simply redistributes the electron density.

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A look at Coulomb's law

$$\ce{ F \propto Q_1*Q_2}$$

• More polar reactants will have a greater charge on them as compared with less polar reactants.
• Since inductive effects concentrate electron density at one area the negative charge concentration will be greater.

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HOMO/LUMO

Unoccupied orbitals are high in energy and occupied orbitals are low in energy. For the best overlap the energy gap between the HOMO and LUMO must be small.

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Rate Order for more polar reactions

In more polar reactions the electrostatic forces of attractions play the primary role as the polar reactants is highly charged. As established earlier ,inductive effects lead to a electron density thus a greater pull.
Homo/Lumo interactions are secondary as the overlapping happens after a collision occurs.

• The electrostatic forces are pulling the benzene and reactant. That would by itself increases the number of collisions.

• A greater electron density leads to the better stabilization of the +tive charge on the reactant.

inductive effects support more polar reactions

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Rate order for less polar reactions

The electrostatic forces of attraction are lower in the case as as the reactant is less charged.
So the control of the reaction is primarily through Homo/Lumo interactions.

• Unoccupied orbitals of the reactant are high in energy.( They are Unoccupied).

• To bridge the HOMO/LUMO gap the Lumo of benzene carbon must be high in energy. Non bonded electrons or the electrons which are delocalized through hyperconjugation are high in energy as they are held by only one atom.

• So the HOMO/Lumo gap reduces in the case of hyperconjugation thus hyperconjugation supports less polar reactions.

Answer 2

Nitration (strong electrophile, "hard" Lewis acid) has a low activation energy, so effects on the transition state will have a smaller relative contribution to the rate controlling step.

Chlorination (weaker electrophile, "softer" Lewis acid, more polarizable) has a higher activation energy and slower reaction rate. The transition state is more important in rate determination.

An aromatic compound has a "hard" face and a "soft" face: the sigma bonds (inductive effect) are more stable, less polarizable. The pi system is less stable, lower energy (diagramatically higher energy level!), more polarizable, and more directly connected to the hyperconjugation effect.

Nitration (very polar, "hard" cation) is influenced by inductive effects that make the aromatic ring more negative in the order D>C>B>A. The rate of nitration increases as you go from A to D. Compared to nitration, the transition state in chlorination is more important and thus more stabilized by hyperconjugation. In chlorination, hyperconjugation decreases in the series A>B>C>D, which gives a higher activation energy and slower rate of reaction as you go from A to D.

Answer 3

Chlorine is a weak electrophile, while the nitronium ion is a strong electrophile. Methyl groups are +I groups and they are electron releasing groups.

The NO2 group will attack a negatively charged carbon attack readily. The +I effect is permanent, and its magnitude decreases with increase in distance, and therefore, the nitronium ion will attack the carbon nearest from the methyl groups.

The Chlorine will not get attracted to to a negatively charged carbon atom as strongly as NO2 will. In hyperconjugation. the charge is delocalized i.e., it is distributed throughout the structure, in various resonance forms. Hence, the effect of hyperconjugation is weak. The Chlorine atom. therefore, will get attracted to to a weakly charged carbon atom, determined by hyperconjugation.

The hyperconjugation won't attract NO2, as the charge is delocalized, and not one specific carbon atom, or there is no carbon atom that is more charged than the others.