Comments on: Electrophilic Aromatic Substitution – The Mechanism

By: James Ashenhurst

James Ashenhurst — Thu, 17 Apr 2025 16:19:30 +0000

In reply to Mostafa.

Strictly speaking, no, since the electrophile is not providing the source of the electrons, the aromatic ring is.

It’s common to say things like “metals are attacked by strong acid” for example, when in reality it’s the metal providing the electrons and the acid is accepting them. It doesn’t really matter so long as you understand where the electrons are coming from.

By: Mostafa

Mostafa — Thu, 17 Apr 2025 02:49:50 +0000

Hello,
In some books they describe electrophilic aromatic substitution as the electrophile attack the benzene ring. However, they still draw the arrow from the pie electrons to the electrophile. Is it correct to state that the electrophile is attacking the nucleophile in this case? Thank you.

By: Kimathi Kendi

Kimathi Kendi — Mon, 25 Nov 2024 13:20:23 +0000

Hello thanks for the information…using curly curves to show the mechanism of electrophyllic substitution of benzene,show how the products of the following processes are formed: 1: chlorination
2: nitration
3: friedel craft alkylation

By: James Ashenhurst

James Ashenhurst — Tue, 07 Feb 2023 18:12:27 +0000

In reply to zey. There are two ortho positions and only one para position. So if it were completely random (and there were no meta) the yield would be 66%. The fact that the yield is 57% shows that the methyl group has some steric influence. The effect becomes much greater as the alkyl group is increased in size.

By: zey

zey — Tue, 07 Feb 2023 14:18:40 +0000

hello, thank you first! i have a question, in the example of methyl benzene, why did the reaction give the product where nitro group is in ortho position in a greater yield than the one where it is in para position?

By: James Ashenhurst

James Ashenhurst — Wed, 31 Aug 2022 19:11:34 +0000

Have you seen the section on EAS from “Modern Physical Organic Chemistry” (p. 608) by Anslyn? The description is quite clear. In most cases the reaction will be second-order overall, but when there is a bulky group (e.g. t-butyl) the rate of the reverse reaction can be significant. In those cases the reverse reaction starts to have about the same barrier as the rate of deprotonation, which leads to secondary isotope effects.

By: Ankit

Ankit — Fri, 26 Aug 2022 17:59:18 +0000

Hey! Great post however a few questions come to my mind from the texts I have. They have been great and clear but how primary isotopic effects and secondary isotopic effects are affected by the bulkiness of the electrophile have not been described very clearly, and other texts don’t seem have any information of them.

It mentions that primary isotopic effect is seen when the loss of an alpha-hydrogen happens in the slower than the formation of the Wheland-complex. It says this is only the case when the electrophile is a significantly bulkier group.

I’d also like to know how charge-transfer complexes are formed from pi-complexes, but that is a little off-topic and at the risk of digressing I wont talk too much about them (I dont understand them well enough anyway)

By: Satti

Satti — Mon, 23 May 2022 18:11:48 +0000

Assalam o alaikum.very good explanation sir.i enjoyed a lot reading it because it is giving me strong concepts.much helpful!!
I appreciate sir.

By: James Ashenhurst

James Ashenhurst — Mon, 30 Sep 2019 19:17:48 +0000

In reply to Piyush. There is only one intermediate - the carbocation. Are you referring to the two transition states?

By: Piyush

Piyush — Sun, 29 Sep 2019 16:34:19 +0000

The energy of second intermiediate is more than first in the given energy diagram? ….