Aryl ethers

An aryl ether has the general structure $Ar\text{–O–}R$: an alkoxy group $\text{–OR}$ attached directly to an aromatic ring carbon. The parent and most-tested example is anisole, $C_6H_5\text{–O–CH}_3$. Two ideas govern everything you need for this topic.

1. How it is made. You cannot simply react an aryl halide with an alkoxide the way you would for a dialkyl ether — aryl halides resist nucleophilic substitution because the ring's conjugation strengthens the C–X bond and blocks backside attack. Instead, the oxygen nucleophile must come from the aromatic side and the leaving group from the alkyl side. Sodium phenoxide, generated by treating phenol with base, is reacted with a primary alkyl halide:

where: $C_6H_5ONa$ is sodium phenoxide (the nucleophile, attacking through oxygen), $CH_3I$ is the alkyl halide (the electrophile, here methyl iodide), and $NaI$ is the salt by-product. This is ordinary $S_N2$ chemistry at the alkyl carbon, so the alkyl halide must be primary or methyl — a hindered secondary or tertiary alkyl halide favours elimination instead and ruins the yield.

2. How it reacts once made. The oxygen atom's lone pair is not just sitting there — it conjugates into the aromatic ring by resonance. This pushes extra electron density onto specific ring carbons and has two linked consequences: the ring becomes strongly activated (it reacts with electrophiles faster than benzene does, often needing no catalyst at all), and the incoming electrophile is directed almost exclusively to the ortho and para positions relative to the –OR group, because those are exactly the carbons the resonance structures load with extra electron density.

where: the trace of iron (as $FeCl_3$ generated in situ) is only needed to polarise $Cl_2$ enough to start the attack — far milder than the conditions benzene itself would need for the same halogenation, because anisole's ring is already electron-rich. The same pattern — mild conditions, ortho/para mixture — holds for nitration ($HNO_3/H_2SO_4$) and Friedel–Crafts alkylation or acylation ($RCOCl$ or $RX$ with $AlCl_3$).

The goal is to explain, at the mechanism level, why Williamson's synthesis of an aryl ether only works in one direction — phenoxide attacking an alkyl halide — and never the reverse of an aryl halide being attacked by an alkoxide, so that a student can justify the choice of starting materials on an exam rather than just recalling the rule.

Step 1. Identify the two possible disconnections of the $Ar\text{–O–}R$ bond. Route A: sodium phenoxide $(C_6H_5O^-)$ attacks an alkyl halide $(R\text{–}X)$ at the alkyl carbon. Route B: sodium alkoxide $(RO^-)$ attacks an aryl halide $(Ar\text{–}X)$ at the ring carbon bonded to the halogen.

Step 2. Test Route A against the $S_N2$ mechanism. The alkyl carbon in $R\text{–}X$ is $sp^3$-hybridised, sterically open on its backside, and the C–X bond is a normal, polarisable single bond. Phenoxide's oxygen lone pair attacks this carbon from the side opposite the leaving group, displacing $X^-$ in a single concerted step. Every requirement of $S_N2$ is satisfied, so Route A proceeds cleanly when $R$ is primary or methyl.

Step 3. Test Route B against the same mechanism and show it fails. The ring carbon in $Ar\text{–}X$ is $sp^2$-hybridised and locked flat in the plane of an aromatic ring; there is no accessible backside for a nucleophile to approach from, so classical $S_N2$ is geometrically impossible. Worse, the halogen's lone pair conjugates into the ring (the same resonance effect that activates aryl ethers), which pulls electron density toward the C–X bond and strengthens it, making it harder to break even by other pathways. $S_N1$ is equally hopeless because the resulting aryl cation, if it ever formed, would be destabilised by sitting in an $sp^2$ orbital orthogonal to the aromatic ring's delocalised pi system, unable to share the ring's stability.

Step 4. Conclude the direction. Since Route B is blocked by both geometry and electronics, the only viable path to an aryl ether is Route A: an alkoxide or phenoxide reacting at an alkyl halide's carbon, never at an aryl halide's ring carbon. This is why every aryl-ether synthesis problem on this syllabus specifies sodium phenoxide plus an alkyl halide, and why writing the reverse combination is treated as a fundamental error rather than a minor slip.

Problem. Starting from phenol, prepare anisole, then treat the anisole with bromine in the presence of a trace of iron. Identify the major and minor products and justify the site selectivity using resonance.

Step 1. Generate sodium phenoxide from phenol and base: $C_6H_5OH + NaOH \rightarrow C_6H_5ONa + H_2O$.

Step 2. Run Williamson's synthesis with methyl iodide: $C_6H_5ONa + CH_3I \rightarrow C_6H_5OCH_3\ (\text{anisole}) + NaI$. This is the $S_N2$ step justified above — methyl iodide's carbon is unhindered, so the reaction is fast and clean.

Step 3. React anisole with bromine and a trace of iron (which generates $FeBr_3$ in situ as a mild Lewis-acid catalyst): $C_6H_5OCH_3 + Br_2 \xrightarrow{Fe} o\text{-BrC}_6H_4OCH_3 + p\text{-BrC}_6H_4OCH_3 + HBr$. Only a trace of catalyst and mild conditions are needed, unlike unsubstituted benzene, because the ring is already activated.

Step 4. Justify the ortho/para outcome with resonance. When $Br^+$ attacks ortho or para to $\text{–OCH}_3$, one resonance structure of the resulting arenium ion places the positive charge directly on the ring carbon bearing oxygen, and the oxygen lone pair neutralises that charge by forming a double bond to it — a stabilisation unavailable if $Br^+$ had attacked meta. Between the two allowed sites, para is usually the major product because it is farther from the bulky $\text{–OCH}_3$ group and so meets less steric resistance during the attack. Answer: the reaction gives a mixture of ortho- and para-bromoanisole, with para predominating, because –OCH3 is a resonance-activating, ortho/para-directing group.

• Writing an aryl-ether synthesis as aryl halide plus sodium alkoxide instead of sodium phenoxide plus alkyl halide — the mechanism section above shows exactly why the reverse direction cannot work.
• Treating –OR as deactivating because oxygen is electronegative — the inductive electron-withdrawal is real but weak, and it is outweighed by the much stronger resonance (mesomeric) donation of the lone pair into the ring, which is what actually controls reactivity and orientation.
• Predicting meta substitution by analogy with –NO2 or –COOH — those are deactivating, meta-directing groups; –OR behaves oppositely and is always ortho/para-directing.
• Forgetting steric hindrance when asked for the major (not merely possible) product — para often wins over ortho purely because it sits farther from the bulky –OR group, even though both are equally favoured electronically over meta.

Ask the tutor: if you want to see the full arenium-ion resonance structures drawn out for ortho, para, and meta attack on anisole side by side, ask the AI tutor to walk through them with you.