In your series, all of the alkyl benzenes will have roughly the same +I inductive effect. Where they differ is with regard to the resonance effect. Hyperconjugated structures such as those drawn below for ethylbenzene are often invoked to explain these differences.
There are two hydrogens in ethylbenzene that are capable of donating electrons into the aromatic ring by hypercojugation. In isopropyl benzene there is only one and there are none in *tert-*butylbenzene. Hence ethylbenzene should be the most electron donating of the compounds in your series due to resonance involving hyperconjugation.
Realize that these effects are relatively small, for example the relative rates for the electrophilic nitration of various aromatic compounds is:
Benzene has no substituent (other than hydrogen), so neither resonance or inductive effects play a role, it would be the slowest. The expected order in your series would therefore be:
In your series, all of the alkyl benzenes will have roughly the same +I inductive effect. Where they differ is with regard to the resonance effect. Hyperconjugated structures such as those drawn below for ethylbenzene are often invoked to explain these differences.
There are two hydrogens in ethylbenzene that are capable of donating electrons into the aromatic ring by hypercojugation. In isopropyl benzene there is only one and there are none in *tert-*butylbenzene. Hence ethylbenzene should be the most electron donating of the compounds in your series due to resonance involving hyperconjugation.
Realize that these effects are relatively small, for example the relative rates for the electrophilic nitration of various aromatic compounds is:
Benzene has no substituent (other than hydrogen), so neither resonance or inductive effects play a role, it would be the slowest. The expected order in your series would therefore be:
But what about the para position. As you are saying that for the steric hinderance the ortho site can be blocked but the para site is not hindered.More
I cannot imagine the para position to be extremely hindered by something that is happening on the other side of the molecule. But you loose two reactive sites which is bad for reaction speed.More
Actually the final answer lies in experimental values. These questions come in Indian Competetive exams are very complex to answer on basis on theory. Sometimes back I came across a question where in a species 2 resonance effect was dominating over 16 hyper conjugative effects.More
@jerepierre Is the relative rate reduced because of 2 sites blocked by sterics or 2 less hydrogens available for hyperconjugation? I come back to my earlier comment that the first-formed pi-complex shouldnt really be influenced by sterics. We are talking about some relatively small differences.More
I think @Jori is on to something here. Looking at the referenced table, toluene actually nitrates preferentially at the ortho position (o:56%, m:3.5%, p:40%). The steric bulk of t-butyl slows down nitration at the ortho positions. Looking at the relative rates, it seems like the difference is about what you would expect from eliminating two potential sites of reactivity.More
In your series, all of the alkyl benzenes will have roughly the same +I inductive effect. Where they differ is with regard to the resonance effect. Hyperconjugated structures such as those drawn below for ethylbenzene are often invoked to explain these differences.
There are two hydrogens in ethylbenzene that are capable of donating electrons into the aromatic ring by hypercojugation. In isopropyl benzene there is only one and there are none in *tert-*butylbenzene. Hence ethylbenzene should be the most electron donating of the compounds in your series due to resonance involving hyperconjugation.
Realize that these effects are relatively small, for example the relative rates for the electrophilic nitration of various aromatic compounds is:
benzene=1, toluene=24, tert-butylbenzene=15.7 (reference, page 1060)
Benzene has no substituent (other than hydrogen), so neither resonance or inductive effects play a role, it would be the slowest. The expected order in your series would therefore be:
ethylbenzene > isopropylbenzene > tert-butylbenzene > benzene
In your series, all of the alkyl benzenes will have roughly the same +I inductive effect. Where they differ is with regard to the resonance effect. Hyperconjugated structures such as those drawn below for ethylbenzene are often invoked to explain these differences.
There are two hydrogens in ethylbenzene that are capable of donating electrons into the aromatic ring by hypercojugation. In isopropyl benzene there is only one and there are none in *tert-*butylbenzene. Hence ethylbenzene should be the most electron donating of the compounds in your series due to resonance involving hyperconjugation.
Realize that these effects are relatively small, for example the relative rates for the electrophilic nitration of various aromatic compounds is:
benzene=1, toluene=24, tert-butylbenzene=15.7 (reference, page 1060)
Benzene has no substituent (other than hydrogen), so neither resonance or inductive effects play a role, it would be the slowest. The expected order in your series would therefore be:
ethylbenzene > isopropylbenzene > tert-butylbenzene > benzene
More
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