Write the mechanism of the reaction of HI with methoxybenzene.

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Q. Write the mechanism of the reaction of HI with methoxybenzene.

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Concept understanding — Williamson Ether Synthesis

Williamson Ether Synthesis: From Intuition to Mechanism

Imagine you want to build a simple bridge between two carbon chains — an oxygen atom linking them together. That bridge is an ether ( $R-O-R'$ ). The Williamson ether synthesis is the most reliable way to build that bridge in a lab.

The Core Idea

You have two pieces: an alkoxide ion ( $RO^-$ ) and an alkyl halide ( $R'X$ ). The alkoxide is a strong nucleophile — it loves positive charge. The alkyl halide has a carbon attached to a halogen (like Cl, Br, I) that is slightly positive because the halogen pulls electrons away.

When you mix them, the alkoxide attacks that slightly positive carbon, kicks out the halide ion, and forms a new $C-O$ bond. The result? An ether.

R-O^- + R'-X \longrightarrow R-O-R' + X^-

That's the entire reaction in one line. But the devil is in the details — especially which alkyl halide you choose.

The Mechanism (SN2)

This is a classic SN2 reaction — one step, no intermediates. The alkoxide approaches the carbon from the opposite side of the halogen. As the $C-O$ bond forms, the $C-X$ bond breaks. The halide leaves as a stable anion.

Because it's SN2, the reaction is sensitive to steric hindrance. The carbon being attacked must be accessible.

Watch out

If the alkyl halide is tertiary (3°), the reaction will not work via SN2. The bulky carbon blocks the backside attack. Instead, the alkoxide will act as a base and cause elimination (forming an alkene). You'll get no ether.

The Practical Rule

Alkyl halide	Works?	Why
Methyl ( $CH_3X$ )	Yes	Least hindered, fastest SN2
Primary (1°)	Yes	Clean SN2
Secondary (2°)	Sometimes	Works if not too bulky; elimination competes
Tertiary (3°)	No	Elimination dominates
Aryl (e.g., bromobenzene)	No	SN2 impossible on $sp^2$ carbon

Tip

To make an ether like $R-O-R'$ , always use the less hindered alkyl halide and the more hindered alkoxide. For example, to make $CH_3CH_2-O-CH(CH_3)_2$ , use $CH_3CH_2O^-$ (primary alkoxide) + $(CH_3)_2CHBr$ (secondary halide) — not the other way around.

How to Choose the Alkoxide

You can't just buy alkoxide ions in a bottle. You make them by reacting an alcohol with a strong base like sodium hydride ( $NaH$ ) or sodium metal.

ROH + NaH \longrightarrow RO^-Na^+ + H_2

The alkoxide is then used immediately with the alkyl halide.

A Common Exam Trap …

Why this formula?

Williamson Ether Synthesis: Why the Key Principles Hold

The Williamson Ether Synthesis is a classic method to prepare ethers. The core reaction is:

\text{R-O}^- + \text{R'-X} \rightarrow \text{R-O-R'} + \text{X}^-

Where:

$\text{R-O}^-$ is an alkoxide ion (strong nucleophile)
$\text{R'-X}$ is an alkyl halide (electrophile)
$\text{X}^-$ is a halide ion (leaving group)

Let's break down why this works — the reasoning behind the key principles.

1. Why an Alkoxide (Not an Alcohol) is Needed

The Problem with Alcohols

Alcohols ( $\text{R-OH}$ ) are weak nucleophiles. The oxygen has a partial negative charge, but the O–H bond is strong. If you mix an alcohol with an alkyl halide, the reaction is extremely slow or doesn't happen at all.

The Solution: Deprotonation

By treating the alcohol with a strong base (like $\text{NaH}$ , $\text{Na}$ , or $\text{KOH}$ ), you remove the proton:

\text{R-OH} + \text{NaH} \rightarrow \text{R-O}^- \text{Na}^+ + \text{H}_2

The alkoxide ion ( $\text{R-O}^-$ ) has a full negative charge on oxygen. This makes it:

A much stronger nucleophile (higher electron density)
More reactive toward the electrophilic carbon in the alkyl halide

Key takeaway: The alkoxide's full negative charge is what drives the reaction — it's not just about having oxygen, but about having a charged, electron-rich oxygen.

2. Why the Alkyl Halide Must Be Primary (or Methyl)

The Mechanism: $\text{S}_\text{N}2$ is the Only Path

The Williamson synthesis proceeds exclusively via an $\text{S}_\text{N}2$ mechanism (bimolecular nucleophilic substitution). This means:

The nucleophile attacks the carbon from the backside
The leaving group departs from the opposite side
The reaction is concerted (one step, no intermediates)

Why Primary Halides Work Best

In $\text{S}_\text{N}2$ reactions, the rate depends on steric hindrance:

Alkyl Halide Type	Steric Hindrance	$\text{S}_\text{N}2$ Reactivity
Methyl ( $\text{CH}_3\text{X}$ )	Minimal	Very fast
Primary ( $\text{RCH}_2\text{X}$ )	Low	Fast
Secondary ( $\text{R}_2\text{CHX}$ )	Moderate	Slow
Tertiary ( $\text{R}_3\text{CX}$ )	High	Does not occur

Why Tertiary Halides Fail

With a tertiary halide, the bulky alkyl groups block the backside attack. Instead, the alkoxide (a strong base) will eliminate a proton from the halide, forming an alkene:

\text{R-O}^- + \text{R'}_3\text{C-X} \rightarrow \text{R-OH} + \text{alkene} + \text{X}^-

This is an E2 elimination — not the desired ether formation.

Key takeaway: The Williamson synthesis works only when the alkyl halide is primary or methyl because $\text{S}_\text{N}2$ requires an unhindered backside.

3. Why the Leaving Group Must Be Good

The Role of the Halide

The halide ( $\text{X}^-$ ) must be a good leaving group — meaning it can stabilize the negative charge after departure.

Halide	Leaving Group Ability	Reason
$\text{I}^-$	Excellent	Large, polarizable, weak base
$\text{Br}^-$	Good	Moderate size, weak base
$\text{Cl}^-$	Fair	Smaller, stronger base
$\text{F}^-$	Poor	Small, strong base, holds tightly

Why Fluoride Fails

Fluoride is a strong base and a poor leaving group. The C–F bond is very strong, and $\text{F}^-$ does not depart easily. So alkyl fluorides are unreactive in Williamson synthesis.

Key takeaway: The leaving group must be weakly basic and polarizable — iodide and bromide are ideal.

--- …

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The reaction of HI with methoxybenzene (anisole) proceeds via acid-catalyzed cleavage of the C–O bond, not by electrophilic substitution on the ring. The key is that HI protonates the oxygen, then the iodide ion attacks the methyl carbon (SN2), yielding iodomethane and phenol as the final products.

The reaction of hydrogen iodide with methoxybenzene (anisole) is a classic example of ether cleavage under acidic conditions. Most students instinctively think of electrophilic aromatic substitution when they see an aromatic ring with a strong acid, but that’s a trap here. The real action happens at the methoxy group, not on the ring.

Why does this happen? The oxygen in anisole has lone pairs that are basic. In the presence of a strong acid like HI, the oxygen gets protonated first. This turns the –OCH₃ group into a much better leaving group — instead of an alkoxide (which is a poor leaving group), you now have a neutral methanol molecule attached to the ring. You might object that methanol itself isn’t a great leaving group either — the real trick is that iodide ion (I⁻) is an excellent nucleophile and a weak base, so it attacks the methyl carbon in an SN2 fashion. The aromatic ring is too bulky and too stable to be attacked directly — the methyl group is the soft target.

Let’s walk through the mechanism step by step.

Protonation of the oxygen The lone pair on the oxygen of methoxybenzene abstracts a proton from HI. This gives a protonated anisole intermediate — an oxonium ion.

\ce{C6H5-O-CH3 + H-I -> C6H5-\overset{+}{O}(H)-CH3 + I^-}

The oxygen now bears a positive charge, making the C–O bond much more polarised and weaker.

Nucleophilic attack by iodide on the methyl carbon The iodide ion, present in solution, acts as a strong nucleophile. It attacks the methyl carbon from the back side (SN2 mechanism). The C–O bond breaks heterolytically, with the electron pair going to the oxygen (neutralising its charge).

\ce{C6H5-\overset{+}{O}(H)-CH3 + I^- -> C6H5-OH + CH3-I}

This step is the key cleavage event. The products are phenol and iodomethane.

Proton transfer (if needed) The phenol formed is already neutral, so no further deprotonation is required. The reaction stops here. …

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Common Mistakes in Williamson Ether Synthesis (HI + Methoxybenzene)

This question tests your understanding of ether cleavage — specifically, how HI reacts with an aryl methyl ether (methoxybenzene). Students often confuse this with the formation of ethers (the actual Williamson Ether Synthesis). Let's break down the key errors.

Mistake 1: Confusing Ether Formation with Ether Cleavage

The error: Students write the mechanism for making an ether (e.g., $CH_3ONa + C_6H_5Br \rightarrow C_6H_5OCH_3$ ) instead of breaking it.

Why it happens: The question says "Williamson Ether Synthesis" in the heading, but the actual task is the reverse reaction — acid-catalyzed cleavage.

How to avoid:

Read carefully: "Reaction of HI with methoxybenzene" means cleavage, not synthesis.
Remember: HI (a strong acid) protonates the ether oxygen, then $I^-$ attacks.
The product is phenol ( $C_6H_5OH$ ) and iodomethane ( $CH_3I$ ), not an ether.

Mistake 2: Forgetting the Aromatic Ring Doesn't Undergo $S_N2$

The error: Students show $I^-$ attacking the aryl carbon (the carbon attached to the benzene ring), writing $C_6H_5I$ as a product.

Why it happens: They treat the aryl-oxygen bond like an alkyl-oxygen bond.

How to avoid:

Aryl halides do not form easily via $S_N2$ — the $sp^2$ carbon is too hindered and the $\pi$ system resists backside attack.
The correct attack is on the methyl carbon ( $CH_3$ ), which is $sp^3$ and unhindered.
Result: $C_6H_5OH$ (phenol) + $CH_3I$ (iodomethane).

Mistake 3: Writing the Wrong Mechanism Steps

The error: Students skip protonation or show direct $S_N2$ on the neutral ether.

Why it happens: They forget that HI provides both $H^+$ (to activate the ether) and $I^-$ (the nucleophile).

How to avoid — write these steps in order:

Protonation of the ether oxygen by $H^+$ from HI:

C_6H_5OCH_3 + H^+ \rightarrow C_6H_5O^+HCH_3

Nucleophilic attack by $I^-$ on the methyl carbon (not the aryl carbon):

I^- + C_6H_5O^+HCH_3 \rightarrow C_6H_5OH + CH_3I

Deprotonation (if needed): The phenol product may lose $H^+$ in basic workup, but in acidic HI, it remains as phenol.

Mistake 4: Ignoring the Role of the Aromatic Ring

The error: Students treat methoxybenzene like a simple aliphatic ether (e.g., $CH_3OCH_3$ ).

Why it happens: They don't consider resonance stabilization of the aryl-oxygen bond.

How to avoid:

The lone pair on oxygen is delocalized into the benzene ring, making the $C_6H_5-O$ bond stronger than a typical $C-O$ bond.
This is why $I^-$ attacks the methyl group — breaking the $CH_3-O$ bond is easier than breaking the $C_6H_5-O$ bond. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
$(CH_3)_3CONa + CH_3Br \rightarrow (CH_3)_3COCH_3$
›Reveal solutionSolution
Aryl halides do not undergo ordinary nucleophilic substitution the way alkyl halides

do, so bromobenzene + sodium methoxide will not simply hand you anisole. Answer: (D).

Concept and Intuition

(A) $(CH_3)_3CONa + CH_3Br \rightarrow (CH_3)_3COCH_3$ : this is Williamson ether synthesis. The rule of thumb is that the alkyl halide should be unhindered (methyl or primary) for a clean $S_N2$ ; the bulk of the alkoxide doesn't matter much since it's the nucleophile, not the electrophile. Methyl bromide is a perfect $S_N2$ substrate, so this reaction proceeds cleanly to give methyl tert-butyl ether. Correct.

(B) $(C_2H_5)_2O \xrightarrow[\Delta]{excess\ HI} 2C_2H_5I + H_2O$ : with excess hot HI, a symmetrical ether is cleaved completely, both alkyl-oxygen bonds broken, to give two equivalents of alkyl iodide. Correct textbook fact.

(C) $(CH_3)_3COC_2H_5 \xrightarrow[\Delta]{HI} (CH_3)_3CI + C_2H_5OH$ : this is a mixed ether with one tertiary and one primary alkyl group. Cleavage proceeds via $S_N1$ at the carbon that gives the more stable carbocation (tertiary), so the tert-butyl group leaves as the iodide and the ethyl-oxygen fragment is released as ethanol. This is the standard textbook outcome. Correct.

(D) Aryl halides like bromobenzene have their C–X bond strengthened by resonance with the ring (partial double-bond character) and the carbon is $sp^2$ , blocking backside $S_N2$ attack. Nucleophilic substitution on an unactivated aryl halide by a simple alkoxide under ordinary conditions simply does not happen — it requires either very forcing conditions (high temperature and pressure, as in phenol manufacture from chlorobenzene) or a strong base capable of a benzyne (elimination-addition) pathway (e.g. $NaNH_2$ ). Plain $CH_3ONa$ at ordinary conditions will not convert bromobenzene to anisole. This statement is therefore NOT correct.

Step-by-Step Solution …
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AP EAPCET 2026Set ap-2026-05-20-AN1 markMCQ
$Y$
›Reveal solutionSolution
Tertiary C-H oxidation gives tert-butanol, then Williamson ether synthesis gives MTBE; the only remaining subtlety is citing the substituent prefixes in the correct alphabetical order.

Concept and Intuition

Tertiary C-H bonds are the weakest and most easily oxidised C-H bonds in an alkane (the resulting radical/cation is most stabilised), so a strong oxidant like $KMnO_4$ selectively converts isobutane's one tertiary hydrogen into a tertiary alcohol rather than attacking the primary methyl hydrogens. Converting that alcohol to its sodium alkoxide and reacting with a primary alkyl halide (Williamson ether synthesis, $S_N2$ at the primary carbon of $CH_3Br$ ) builds the ether cleanly, since $S_N2$ works well on primary halides.

Step-by-Step Solution

Isobutane $(CH_3)_3CH$ + $KMnO_4$ → oxidation at the sole tertiary C-H → X = tert-butyl alcohol, $(CH_3)_3C-OH$ (2-methylpropan-2-ol).

X + Na → sodium tert-butoxide, $(CH_3)_3C-O^-Na^+$ (Na displaces the O-H proton).

Sodium tert-butoxide + $CH_3Br$ → Williamson ether synthesis (the alkoxide's oxygen performs $S_N2$ on the primary carbon of methyl bromide) → Y = $(CH_3)_3C-O-CH_3$ , methyl tert-butyl ether. …
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AP EAPCET 2025Set ap-2025-05-20-AN1 markMCQ
$\mathrm{C_6H_5NH_2}$
›Reveal solutionSolution
Aniline → diazonium salt → phenol (X) → anisole (via Williamson ether synthesis) → 4-bromoanisole (Y, major product of electrophilic bromination directed para by the methoxy group).

Concept and Intuition

This is a multi-step synthesis chaining three classic reactions: diazotisation/hydrolysis (amine → phenol via the diazonium salt), Williamson ether synthesis (phenoxide + alkyl halide → aryl alkyl ether), and electrophilic aromatic bromination directed by a strongly activating, ortho/para-directing methoxy group. Since anisole's methoxy substituent is already present on the ring, the incoming $\mathrm{Br^+}$ (from $\mathrm{Br_2}$ in acetic acid) attacks preferentially para (and some ortho), and para is reported as the major product due to less steric crowding.

Step-by-Step Solution

$\mathrm{C_6H_5NH_2} \xrightarrow{\mathrm{NaNO_2/HCl},\ 273K} \mathrm{C_6H_5N_2^+Cl^-}$ (benzenediazonium chloride) — standard diazotisation of a primary aromatic amine at 0–5°C.

$\mathrm{C_6H_5N_2^+Cl^-} \xrightarrow{\mathrm{H_2O},\ \Delta}$ phenol ( $\mathrm{C_6H_5OH}$ ) + $\mathrm{N_2}$ + HCl — hydrolysis of the diazonium salt on warming. So X = phenol.

Phenol + NaOH → sodium phenoxide ( $\mathrm{C_6H_5O^-Na^+}$ ); phenoxide + $\mathrm{CH_3Br}$ → anisole ( $\mathrm{C_6H_5OCH_3}$ ) via $\mathrm{S_N2}$ (Williamson ether synthesis). …
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AP EAPCET 2024Set eng-2024-05-21-FN1 markMCQ
$\mathrm{Y \xleftarrow{NaOEt} CH_3CH_2CH_2CH_2Br \xrightarrow[(ii)\,H_2O]{(i)\,Mg/dry\ ether} X}$
›Reveal solutionSolution
Grignard + water gives the alkane (not the alcohol you'd get from a carbonyl); sodium ethoxide on a primary halide gives clean $S_N2$ substitution (an ether), not elimination.

Concept and Intuition

A Grignard reagent $R\text{-}MgX$ behaves like a carbanion $R^-$ . Any proton source acidic enough (even water, $pK_a\approx15.7$ , is far more acidic than the alkane conjugate acid) instantly protonates it: $R\text{-}MgX + H_2O \to R\text{-}H + Mg(OH)X$ . This is the classic way to convert a halide into the corresponding hydrocarbon — it is not how you make an alcohol (that needs the Grignard to attack a carbonyl carbon first).

Separately, when a nucleophile/base attacks an alkyl halide, the outcome ( $S_N2$ vs $E2$ ) depends on both the base's bulk and the substrate's branching. Ethoxide ( $\mathrm{NaOEt}$ ) is a comparatively small, strong base. Against a primary substrate with an easily accessible backside carbon (n-butyl bromide), steric hindrance to backside attack is minimal, so $S_N2$ substitution is the major pathway, not elimination.

Step-by-Step Solution

$\mathrm{CH_3CH_2CH_2CH_2Br} + Mg \xrightarrow{dry\ ether} \mathrm{CH_3CH_2CH_2CH_2MgBr}$ (Grignard reagent formed).

$\mathrm{CH_3CH_2CH_2CH_2MgBr} + H_2O \to \mathrm{CH_3CH_2CH_2CH_3} + Mg(OH)Br$ — protonolysis gives butane, so $X=\mathrm{CH_3CH_2CH_2CH_3}$ . …
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AP EAPCET 2024Set eng-2024-05-21-FN1 markMCQ
$\mathrm{(CH_3)_3CONa + CH_3CH_2Br \rightarrow X}$
›Reveal solutionSolution
$S_N2$ vs $E2$ is decided by whether backside attack at the carbon bearing the leaving group is sterically possible — here that means looking at the alkyl halide's own branching, not just the base's bulk.

Concept and Intuition

The classic trap in these paired reactions is to only look at how bulky the base/nucleophile is. What actually controls the outcome is whether the nucleophile can reach the back lobe of the C–LG bond:

If the carbon bearing the leaving group is primary and unhindered (ethyl bromide), $S_N2$ remains fast and dominant even with a bulky base like $t$ -butoxide, because the bulk of the base doesn't block approach to a completely open primary carbon nearly as much as branching at the substrate would.

If the carbon bearing the leaving group is tertiary (tert-butyl bromide), backside attack is essentially impossible regardless of which base is used — the only viable pathway is $E2$ (proton abstraction from a $\beta$ -carbon), so even a small, strong, ionic base like ethoxide gives elimination as the major product.

Step-by-Step Solution

Reaction 1: $(CH_3)_3CONa$ (base/nucleophile) + $\mathrm{CH_3CH_2Br}$ (primary substrate). The electrophilic carbon (in $CH_3CH_2Br$ ) is unhindered, so $t$ -butoxide's oxygen performs $S_N2$ substitution: $X = (CH_3)_3C\text{-}O\text{-}CH_2CH_3$ . …
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AP EAPCET 2023Set ap-2023-05-23-FN1 markMCQ
$\mathrm{CH_3CH_2C(CH_3)_2OCH_2CH_2CH_3}$
›Reveal solutionSolution
Williamson ether synthesis needs the bulky/aryl partner as the alkoxide and the unhindered partner as the primary halide, to avoid elimination or a non-reactive aryl-halide substitution.

Concept and Intuition

Williamson synthesis is an $\mathrm{S_N2}$ reaction between an alkoxide and an alkyl halide. Two structural traps show up here: (1) tertiary alkyl halides are poor $\mathrm{S_N2}$ substrates — with a strong nucleophile/base like an alkoxide they instead undergo E2 elimination; and (2) aryl halides (like bromobenzene) cannot undergo $\mathrm{S_N2}$ at all because the aryl carbon is sp2, in-plane, and shielded, and the C–X bond is strengthened by ring resonance.

Step-by-Step Solution

For X, $\mathrm{CH_3CH_2C(CH_3)_2{-}O{-}CH_2CH_2CH_3}$ : the tertiary-pentyl part must come in as the alkoxide (sodium tert-alkoxide), while the primary propyl part comes in as the halide (propyl bromide) — the primary halide undergoes clean $\mathrm{S_N2}$ with the bulky alkoxide, avoiding elimination.

For Y, phenyl propyl ether ( $\mathrm{C_6H_5{-}O{-}CH_2CH_2CH_3}$ ): since aryl halides (bromobenzene) cannot undergo nucleophilic substitution, the correct route is phenoxide (from phenol) attacking the primary alkyl halide (propyl bromide). …
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AP EAPCET 2022Set ap-2022-07-11-AN1 markMCQ
$-C(CH_3)_2Br$
›Reveal solutionSolution
Bulky sodium tert-butoxide is a classic E2-promoting, poor-SN2 base; on the primary bromide side chain it eliminates HBr to give a terminal alkene (vinyl group), while the more hindered tertiary bromide is unaffected — matching option (C).

Concept and Intuition

The identity of a base/nucleophile controls whether an alkyl halide undergoes substitution or elimination. Sodium tert-butoxide, $(CH_3)_3CO^-Na^+$ , is strongly basic (as a good alkoxide) but its oxygen is buried under three bulky methyl groups, making it a very poor nucleophile for SN2 (it cannot easily approach a carbon from the back side). This steric bulk is exactly why tert-butoxide is the textbook reagent of choice to force elimination (E2) rather than substitution even on primary alkyl halides, which would normally favour SN2 with a small nucleophile like hydroxide or ethoxide. For a primary bromide such as $-CH_2CH_2Br$ , tert-butoxide therefore removes a $\beta$ -hydrogen (E2) to give the terminal alkene $-CH=CH_2$ , releasing Br $^-$ .

Meanwhile, this same molecule also carries a tertiary, benzylic bromide, $-C(CH_3)_2Br$ , directly on the ring. In problems of this kind, the answer choices make clear that the reaction is selective for the primary position over the tertiary one under the given conditions/timescale — the tertiary center is left as the unreacted starting bromide in the major product.

Step-by-Step Solution

Identify the base: Me $_3$ CONa = sodium tert-butoxide — strong base, very weak/hindered nucleophile.

Recognize its standard textbook behaviour: promotes E2 elimination, especially valuable for driving primary alkyl halides toward alkenes instead of the substitution product they would normally favour with a small nucleophile.

Apply E2 to the primary bromide arm: $-CH_2-CH_2-Br \xrightarrow{-HBr} -CH=CH_2$ (a terminal vinyl group, the only possible elimination product from a 2-carbon primary chain). …
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AP EAPCET 2021Set eng-2021-08-25-AN1 markMCQ
$\underset{CH_3}{\overset{CH_3}{CH_3-C-ONa}} + CH_3Cl \longrightarrow\ ?$
›Reveal solutionSolution
A branched sodium alkoxide plus methyl chloride is a textbook Williamson ether synthesis: the alkoxide's own carbon skeleton is retained, and a new methyl group is added onto its oxygen.

Concept and Intuition

Williamson ether synthesis works best when the alkyl halide is primary/unhindered (favours clean $S_N2$ ) — methyl halides are ideal because they have no $\beta$ -hydrogens at all, so competing elimination is impossible regardless of how bulky the incoming alkoxide nucleophile is. The alkoxide's own carbon skeleton is untouched by the reaction (only its oxygen attacks); the methyl group from the alkyl halide simply attaches to that oxygen.

Step-by-Step Solution

Identify the nucleophile: the sodium alkoxide of the branched alcohol, $R_3C$ – $O^-\,Na^+$ (with $R_3C$ its full carbon skeleton, retained unchanged throughout the reaction).

Identify the electrophile: $CH_3Cl$ — small, unhindered, no possibility of elimination (no $\beta$ -H).

$S_N2$ attack: the alkoxide oxygen's lone pair displaces $Cl^-$ from the methyl carbon. …
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Q.Write the mechanism of the reaction of HI with methoxybenzene.

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