# Order of electrophilic substitution

What will be the order of reactivity towards electrophilic substitution in case of the following compounds:

benzene, ethyl benzene, isopropyl benzene, tert-butyl benzene

The answer at the end of the book says that ethyl benzene will be most reactive (the book doesn't explain the cause though it's a MCQ) but as according to inductive effect I think the tert-butyl group will have more +I effect on the ring. Doesn't it?

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

• Wouldn't the steric hindrance be of more importance than minor difference in hyperconjugating effect? I'm not sure if steric hindrance is even the issue here at all, but it seems very reasonable. Less available reaction sides reduce reaction speed, however I cannot confirm it as it isn't dealt with in Clayden et al. or Vollhardt & Schore (i.e. the introductory texts I own). I cannot imagine the difference in reactivity is very important here at all.
– Jori
Feb 24, 2015 at 17:35
• 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. Feb 24, 2015 at 17:38
• @Jori Yes, steric hindrance plays a role, but more so in determining the ortho\para ratio. Remember that a pi-intermediate is formed first and it should be less influenced by steric effects then the following sigma-intermediate (Wheland complex).
– ron
Feb 24, 2015 at 17:49
• 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. Feb 24, 2015 at 17:50
• @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 shouldn't really be influenced by sterics. We are talking about some relatively small differences.
– ron
Feb 24, 2015 at 17:54

Alkyl groups on benzene are ortho-para directing. The $\sigma$-conjugation effect of the alkyl group is not dependent on the type of carbon that is attached to is (methyl, primary, secondary or tertiary). Only the carbon connected directly to the ring can interact with it. More substituted alkyl groups will however hinder reaction at the ortho-site lowering the total reaction speed (e.g. for t-butyl benzene essentially no ortho attack can occur, it is too sterically hindered). Therefore the predicted order of reactivity from most reactive to least reactive is: ethylbenzene, isopropylbenzene, tert-butyl benzene, benzene.

• 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. Feb 24, 2015 at 17:15
• 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.
– Jori
Feb 24, 2015 at 17:23