18A.4-18A.7 - Benzene And Phenol Reactions

Why Benzene Substitutes Rather Than Adds

Benzene is not just "cyclohexatriene with three double bonds". Its six pi electrons are delocalised over the whole ring, giving a stable aromatic system. That stability changes the reaction pattern: benzene tends to substitute one hydrogen atom for another group, rather than add across the ring and permanently destroy delocalisation.

The electrophile, $\ce{E+}$, is electron deficient. The benzene ring uses its delocalised pi electrons to form a new bond to $\ce{E}$, then a proton is lost so the delocalised ring is restored. The product is a substituted benzene, not an addition product.

Combustion is the exception to that substitution pattern. Benzene burns in oxygen in air, but it has a high carbon-to-hydrogen ratio, so in an ordinary air flame incomplete combustion can form carbon particles (soot), making it smoky.

$\ce{2C6H6 + 15O2 -> 12CO2 + 6H2O}$

This balanced equation represents complete combustion. The observation of a smoky flame is the useful practical cue: compared with many smaller aliphatic hydrocarbons, benzene is carbon rich, so soot formation is more noticeable.

Common-error contrast

Do not explain benzene reactions by saying "benzene has three double bonds, so it adds like an alkene." Alkenes have a localised $\ce{C=C}$ pi bond that readily donates an electron pair to an electrophile. Benzene has delocalised pi electron density, and reactions that keep or restore aromatic delocalisation are favoured.

The Benzene Reaction Set

Benzene reactions are best learned as reagent-condition-product sets. The route is easier if you first identify the electrophile that must be generated.

The catalyst is needed because benzene is not electron rich enough, in its stable delocalised form, to react rapidly with weak or unactivated electrophiles. The catalyst helps produce a much stronger electrophile.

For bromination, the halogen carrier accepts a lone pair from bromine and polarises the $\ce{Br-Br}$ bond strongly:

$\ce{Br2 + AlBr3 -> Br+ + AlBr4-}$

For nitration, sulfuric acid protonates nitric acid and helps generate the nitronium ion:

$\ce{HNO3 + H2SO4 -> NO2+ + HSO4- + H2O}$

For Friedel-Crafts reactions, aluminium chloride accepts chloride and produces an electrophile with $\ce{AlCl4-}$ as the counter-ion:

$\ce{RCl + AlCl3 -> R+ + AlCl4-}$
$\ce{RCOCl + AlCl3 -> RCO+ + AlCl4-}$

These are simplified electrophile-generation equations, but they show the assessed idea: the reagent alone is not usually enough; the catalyst creates or strengthens the electrophile.

Worked example: choose the product and electrophile

Question: Benzene reacts with ethanoyl chloride, $\ce{CH3COCl}$, in the presence of anhydrous aluminium chloride. Name the reaction family, identify the electrophile and write the organic product.

The reagent is an acyl chloride, so this is Friedel-Crafts acylation. The electrophile is the acylium ion, $\ce{CH3CO+}$. One hydrogen on benzene is replaced by the acyl group:

$\ce{C6H6 + CH3COCl -> C6H5COCH3 + HCl}$

The product is phenylethanone, often also called acetophenone.

For a yield calculation, use the balanced equation to convert reactant amount to the theoretical product mass, then multiply by yield/100. For example, $6.24\,\mathrm{g}$ benzene with $M_r=78.0$ is $0.0800\,\mathrm{mol}$. The 1:1 nitration equation gives a theoretical nitrobenzene mass of $0.0800\times123.0=9.84\,\mathrm{g}$. At 80.0% yield, the isolated mass is $9.84\times0.800=7.87\,\mathrm{g}$ to three significant figures. The yield factor changes the recovered mass; it does not change the equation coefficients.

Guided practice

Complete the table.

Electrophilic Substitution Mechanism

The electrophilic substitution mechanism has the same core logic for halogenation, nitration and Friedel-Crafts reactions. The details of electrophile generation change, but the benzene ring always follows the same two-stage pattern.

Stage 1: the benzene ring forms a bond to the electrophile.

$\ce{C6H6 + E+ -> [C6H6E]+}$

In a drawn mechanism, the first curly arrow starts from the benzene pi system and points to the electrophile. The intermediate is sometimes called an arenium ion or sigma complex. It is positively charged and has temporarily lost full aromatic delocalisation because one ring carbon now has an extra sigma bond.

Stage 2: a base removes $\ce{H+}$ from the carbon that has gained the electrophile.

$\ce{[C6H6E]+ -> C6H5E + H+}$

In a drawn mechanism, the curly arrow for this step starts from the $\ce{C-H}$ bond and returns the electron pair to the ring. That restores the delocalised pi system. A nearby negative ion, such as $\ce{AlBr4-}$, $\ce{HSO4-}$ or $\ce{AlCl4-}$, accepts the proton and helps regenerate the catalyst or acid.

Worked example: nitration of benzene

Generate the electrophile:

$\ce{HNO3 + H2SO4 -> NO2+ + HSO4- + H2O}$

Then substitute onto benzene:

$\ce{C6H6 + NO2+ -> [C6H6NO2]+}$

Overall:

$\ce{C6H6 + HNO3 -> C6H5NO2 + H2O}$

The sulfuric acid is a catalyst in the overall nitration process because it is used to generate $\ce{NO2+}$ and is regenerated when the sigma complex loses a proton.

Worked example: bromination of benzene

Generate the electrophile:

$\ce{Br2 + AlBr3 -> Br+ + AlBr4-}$

Then substitute onto benzene:

$\ce{C6H6 + Br+ -> [C6H6Br]+}$

The aluminium bromide catalyst is regenerated in the final step. The reaction gives bromobenzene and hydrogen bromide.