# Reasons for dry conditions in substituition reactions with benzene

All substitution reactions of benzene must be carried out in dry conditions with a catalyst that produces a powerful electrophile.

This was a statement from my book. My question is, why must it be carried out in dry conditions?

What will occur if moisture is present?

• Are you sure that is what your textbook says? I am aware that there are solvent effects which make a reaction favorable kinetically or thermodynamically. But these effects vary depending on the intermediate or transition state complex. Also, I don't think that "substitution reaction" is specific enough. What textbook are you using? – CoffeeIsLife Mar 3 '14 at 11:10

Benzene and other aromatic hydrocarbons are immisible with water. So, probably there is no problem with benzene if we there is moisture.

In most of the substitution reactions which benzene undergo, requires Lewis acid catalyst, like ferric halide or aluminium halide. Most commonly used catalysts are Anhyd. $\ce{AlCl3}$ and $\ce{FeBr3}$, which become inactivated if they react with water. So, this requires dry conditions.

• But isn't that only for halogenation? – CoffeeIsLife Mar 3 '14 at 11:03
• @user97554. Benzene undergoes substitution reactions 1.Halogenation, 2.Sulphonation, 3.Nitration, 4.Alkylation, 5.Acylation. Almost all of them require lewis acid catalysts ($AlCl_3$ or $FeBr_3$). It is not only for halogenation. – Immortal Player Mar 3 '14 at 11:34

Electrophiles are reactive cations, they will react with water to form eg an alcohol (or some weak acid etc) and protons. No reaction with benzene.

With electrophilic aromatic substitution (EAS) you use a strong lewis acid to create an electrophile. $$\ce{AlCl3 + R-X \rightarrow AlCl3X- + R+}$$ $$\mathrm{R = Cl,\ Br,\ I,\ alky,\ aryl,\ RC(O),\ etc...}$$ $$\ce{X = Cl,\ Br,\ I, R-C(O)O,\ etc}...$$ These lewis acids (not the benzene) are incredibly sensitive to water which will irreversably react with the acid to form a hydrate. $$\ce{ACl3 + X H2O \rightarrow ACl3.X H2O}$$ $$\mathrm{For}\ \ce{AlCl3}: \ \mathrm{1\leq X \leq 9}$$ This hydrate is not as strong of a lewis acid as the anhydrous acid and no longer able to form electrophiles. That is why you must use anhydrous conditions with substitution reactions.

An electrophile goes after electron density, as in the two lone pairs in $\ce{H2O}$ plus poalrization. You then require conditions wherein water is inert (hypohalite halogenations, maybe samarium triflate), micelles exclude water, or the substituted benzene is so activated (dimethylaniline) that it doesn't care.

• I see a -1 here. Somebody has never done industrial chemistry wherein "good enough" is rarely challenged by "better, but more expensive." – Uncle Al Mar 3 '14 at 20:19
• Actually, I upvoted your answer ;) But preparative organic (photo)chemistry in 0.5M SDS is dreadful. Been there, done that back in 1993. Lots of sat. $\ce{CaCl2}$ was used to precipitate the laurylsulfate! – Klaus-Dieter Warzecha Mar 5 '14 at 7:16