Write the mechanism of the reaction of HI with methoxybenzene.

NCERT Exemplar · Q52

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

Telangana TsbieLong· 3mImportance★★★★★

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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.

Showing the 12 most recent of 27 on this concept.

TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
$Q. Which of the following statements are correct? I. Liquid sodium metal is used as coolant in fast breeder nuclear reactor II. LiCl is deliquescent III. LiF and CsI have low solubility in water (A) I, III only (B) II, III only (C) I, II only (D) I, II, III$
›Reveal solutionSolution
All three statements are correct: liquid sodium is used as a coolant in fast breeder reactors, LiCl is deliquescent, and both LiF and CsI have low solubility in water. The correct option is (D).

The question asks us to identify the correct statements regarding properties of alkali metals and their compounds. We will evaluate each statement based on fundamental chemical principles.

Concept and Intuition

Coolants in Nuclear Reactors: A good coolant needs to efficiently transfer heat, remain liquid over a wide temperature range, and ideally have a low neutron absorption cross-section in certain reactor types.

Deliquescence: This property describes a substance that absorbs moisture from the atmosphere until it dissolves in the absorbed water to form a solution. It is related to hygroscopy (absorbing moisture) and the ability to form stable hydrates.

Solubility in Water: The solubility of an ionic compound in water is determined by the balance between its lattice energy (energy required to break the ionic lattice) and its hydration energy (energy released when ions are surrounded by water molecules). For a compound to dissolve, the hydration energy must be sufficient to overcome the lattice energy.

Step-by-Step Evaluation

1. Statement I: Liquid sodium metal is used as coolant in fast breeder nuclear reactor

Reasoning: Fast breeder reactors operate at very high temperatures and require a coolant that can efficiently remove heat without moderating (slowing down) the fast neutrons. Liquid sodium is an excellent choice for this purpose due to several key properties:

High Thermal Conductivity: It can transfer heat very efficiently.

Wide Liquid Range: Sodium has a relatively low melting point ( $97.8^\circ C$ ) and a very high boiling point ( $883^\circ C$ ), allowing it to remain liquid over a broad and useful temperature range for reactor operation.

Low Neutron Absorption Cross-section: It does not significantly absorb neutrons, which is crucial for maintaining the neutron economy in a fast breeder reactor.

Low Viscosity: This allows for easy pumping and circulation.

Conclusion: Statement I is correct.

2. Statement II: LiCl is deliquescent

Reasoning: Deliquescence is the property of a substance to absorb moisture from the atmosphere and dissolve in it. This property is strongly linked to the hygroscopic nature of a compound and its ability to form stable hydrates.

Lithium chloride (LiCl) is known to be highly hygroscopic and readily forms hydrates, such as $LiCl \cdot H_2O$ , $LiCl \cdot 2H_2O$ , and $LiCl \cdot 3H_2O$ .

The small size and high charge density of the $Li^+$ ion lead to a very high hydration energy. This strong interaction with water molecules drives the absorption of moisture from the air.

Among the alkali metal chlorides, LiCl is the most deliquescent.

Conclusion: Statement II is correct.

3. Statement III: LiF and CsI have low solubility in water

Reasoning: The solubility of an ionic compound in water depends on the relative magnitudes of its lattice energy and hydration energy.

For LiF:

$Li^+$ is the smallest alkali metal ion, and $F^-$ is the smallest halide ion. …
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TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
$Q. Which of the following are Position isomers? [FIGURE] (A) I, III (B) II, IV (C) II, III (D) I, IV$
›Reveal solutionSolution
Among these $\mathrm{C_6H_{12}}$ alkenes, only I (4-methylpent-1-ene) and III (4-methylpent-2-ene) have the same carbon skeleton with the double bond in different places — option (A).

The concept first

All four compounds are $\mathrm{C_6H_{12}}$ , so they are all isomers of one another. The question is which kind. Two structural isomers are:

Chain (skeletal) isomers if the carbon framework itself differs (branching pattern, length of the main chain);

Position isomers if the framework is identical and only the location of the functional group (here the C=C) changes.

So the working method is mechanical: hydrogenate each structure in your head (i.e. ignore the double bond and look only at the carbon skeleton). Any two that collapse to the same alkane are position isomers; ones that collapse to different alkanes are chain isomers.

Step-by-step

Reduce each to its skeleton.

I: $\mathrm{(CH_3)_2CH\!-\!CH_2\!-\!CH{=}CH_2} \to$ 2-methylpentane skeleton (5-carbon chain, methyl branch on the carbon next to one end).

II: $\mathrm{CH_2{=}C(CH_3)\!-\!CH(CH_3)_2} \to$ 2,3-dimethylbutane skeleton (4-carbon chain, two methyl branches).

III: $\mathrm{(CH_3)_2CH\!-\!CH{=}CH\!-\!CH_3} \to$ 2-methylpentane skeleton — same as I.

IV: unbranched hexene $\to$ n-hexane skeleton (no branch).

Group by skeleton. …
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TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
$_2$
›Reveal solutionSolution
The reaction sequence is the Williamson ether synthesis: phenol (X) reacts with ethyl bromide (Y) in the presence of a base to give phenetole, which then undergoes bromination at the para position to yield p-bromophenetole (Z). The correct option is (B).

The question presents a set of reactions where three unknown compounds X, Y, and Z are to be identified. The key is to recognize the classic Williamson ether synthesis followed by electrophilic aromatic substitution. Let's walk through the logic.

Identify the first reaction: X + NaOH → sodium phenoxide

The product after treatment with NaOH is a sodium salt of an aromatic alcohol. This is a characteristic reaction of phenol ( $C_6H_5OH$ ). Phenol reacts with NaOH to form sodium phenoxide ( $C_6H_5ONa$ ). So X must be phenol, not bromobenzene (which does not react with NaOH under these conditions). This immediately eliminates options (A) and (D), which list X as bromobenzene.

Identify the second reaction: sodium phenoxide + Y → an ether

Sodium phenoxide is a strong nucleophile. It reacts with an alkyl halide in an $S_N2$ reaction to form an ether — this is the Williamson ether synthesis. The product shown is an aromatic ether with an ethyl group attached to the oxygen. Therefore Y must be ethyl bromide ( $C_2H_5Br$ ). The product is ethyl phenyl ether, commonly called phenetole ( $C_6H_5OC_2H_5$ ).

Identify the third reaction: phenetole + $Br_2$ → Z …
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TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
$\xrightarrow{\text{H}_2\text{O}}$
›Reveal solutionSolution
Borax hydrolyses in water to give boric acid and a strong base; with HCl it gives boric acid and NaCl. The correct properties are II and III only — option (B).

The key is to understand what borax (sodium tetraborate decahydrate, $\text{Na}_2\text{B}_4\text{O}_7 \cdot 10\text{H}_2\text{O}$ ) does in water and in acidic water. Borax is not a simple salt — it reacts with water to produce boric acid and a strong base. When you add HCl, the base gets neutralised, leaving a different set of products.

Let’s write the reactions clearly.

Borax in pure water Borax dissolves and hydrolyses:

$\text{Na}_2\text{B}_4\text{O}_7 + 7\text{H}_2\text{O} \rightarrow 2\text{NaOH} + 4\text{H}_3\text{BO}_3$
Here, A is $\text{NaOH}$ (sodium hydroxide) and B is $\text{H}_3\text{BO}_3$ (boric acid).

So A is a strong base, B is a weak acid — not a base at all.

Borax in water with HCl The same hydrolysis occurs, but the HCl neutralises the NaOH formed:

$\text{Na}_2\text{B}_4\text{O}_7 + 2\text{HCl} + 5\text{H}_2\text{O} \rightarrow 2\text{NaCl} + 4\text{H}_3\text{BO}_3$
Here, B is again $\text{H}_3\text{BO}_3$ , and C is $\text{NaCl}$ (sodium chloride).

Now evaluate each statement:

I. Both A, B are strong bases

A is NaOH (strong base), but B is boric acid — a very weak acid, not a base. So I is false.

II. B is a white crystalline solid with soapy touch …
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TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
$\mathrm{S_N2}$
(1) A central carbon bearin](/figures/tg-eapcet/eng-2026-05-10-FN-chemistry-q155.svg)
›Reveal solutionSolution

$\mathrm{S_N2}$ needs an unobstructed backside approach, so steric bulk kills it. The unbranched primary halide 1-bromobutane is the least hindered and therefore the most reactive — option (B).

The concept first

In the $\mathrm{S_N2}$ mechanism, bond-making and bond-breaking happen simultaneously in one step. The nucleophile attacks the electrophilic carbon from the side directly opposite the leaving group (backside attack), passing through a transition state in which that carbon is momentarily bonded to five groups:
$\mathrm{Nu^{-}}+\ \mathrm{R\!-\!Br}\ \longrightarrow\ \left[\mathrm{Nu}\cdots\mathrm{C}\cdots\mathrm{Br}\right]^{\ddagger}\ \longrightarrow\ \mathrm{Nu\!-\!R}+\mathrm{Br^{-}}$
The carbon is $sp^2$ -like in that transition state with the three remaining groups splayed out in a plane. Every extra alkyl group attached to that carbon (or even to the carbon next to it) makes the transition state more crowded, raises its energy, and slows the reaction:
$\text{Rate: }\ \mathrm{CH_3X}\ >\ 1^\circ\ >\ 2^\circ\ \gg\ 3^\circ$
This is the exact opposite of the $\mathrm{S_N1}$ order (which follows carbocation stability, $3^\circ>2^\circ>1^\circ$ ). Knowing which mechanism the question names is half the battle.

Step-by-step through the options

(A) $\mathrm{(CH_3)_3C\!-\!Br}$ , tert-butyl bromide — tertiary. Three methyl groups completely block the backside of the C–Br carbon. Nucleophiles simply cannot get in; this halide reacts almost exclusively by $\mathrm{S_N1}$ (it forms a very stable $3^\circ$ carbocation). Slowest for $\mathrm{S_N2}$ . ✗

(B) $\mathrm{CH_3CH_2CH_2CH_2\!-\!Br}$ , 1-bromobutane — primary and unbranched. The carbon bearing $\mathrm{Br}$ carries only two hydrogens and one straight chain, so the backside is wide open. Fastest of the four. ✓ …
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TG EAPCET 2026Set ap-2026-05-05-FN1 markMCQ
$+1,\ +4$
›Reveal solutionSolution
Photochemical smog forms when nitrogen oxides (NO and NO₂) from vehicle exhaust react with sunlight. The oxidation states of nitrogen in these two key oxides are +2 and +4, making option (B) correct.

The question is about the chemistry of photochemical smog — a type of air pollution that forms when sunlight triggers reactions between pollutants, especially from car engines. The two oxides of nitrogen that play the starring role here are nitric oxide (NO) and nitrogen dioxide (NO₂). They are produced in high-temperature combustion (e.g., in a car engine) when nitrogen and oxygen from the air combine.

To find the oxidation states, we apply the standard rule: oxygen almost always has an oxidation state of –2 (except in peroxides, which aren't relevant here). For a neutral molecule, the sum of oxidation states must be zero.

For NO: Let the oxidation state of nitrogen be $x$ .

$x + (-2) = 0 \implies x = +2$ .

So nitrogen is in the +2 oxidation state in nitric oxide.

For NO₂: Let the oxidation state of nitrogen be $y$ .

$y + 2(-2) = 0 \implies y - 4 = 0 \implies y = +4$ .

So nitrogen is in the +4 oxidation state in nitrogen dioxide. …
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TG EAPCET 2026Set ap-2026-05-05-FN1 markMCQ
$\xrightarrow[\text{conc. } \mathrm{H_2SO_4}]{\mathrm{HNO_3}} X \xrightarrow{Y} Z$
(i) $\mathrm{NaOH}\,|\,443\ \mathrm{K}$
(ii) $\mathrm{H^+}$ ; $Z$ = 4-nitrophenol ( $\mathrm{O_2N-C_6H_4-OH}$ , para) (B) $X$ = 1-chloro-3-nitrobenzene (Cl and $\mathrm{NO_2}$ meta) ; $Y$ =
(i) $\mathrm{NaOH}\,|\,368\ \mathrm{K}$
(ii) $\mathrm{H^+}$ ; $Z$ = 3-nitrophenol (OH and $\mathrm{NO_2}$ meta) (C) $X$ = 1-chloro-4-nitrobenzene (para) ; $Y$ =
(i) $\mathrm{NaOH}\,|\,300\ \mathrm{K}$
(ii) $\mathrm{H^+}$ ; $Z$ = 4-chlorophenol ( $\mathrm{HO-C_6H_4-Cl}$ , para) (D) $X$ = 1-chloro-4-nitrobenzene (para) ; $Y$ = $\mathrm{H_2O}\,|\,300\ \mathrm{K}$ ; $Z$ = 4-nitrophenol (para) $REACTION SCHEME (in the stem): A benzene ring bearing a single substituent \mathrm{Cl} ($
›Reveal solutionSolution
Nitration of chlorobenzene gives mainly 1-chloro-4-nitrobenzene ( $X$ ); the para nitro group makes the $\mathrm{C\!-\!Cl}$ bond susceptible to nucleophilic attack, so NaOH at 443 K followed by $\mathrm{H^+}$ converts it to 4-nitrophenol ( $Z$ ). That is option (A).

The concept first

Electrophilic step. Halogens are deactivating but ortho–para directing: the $-I$ effect withdraws electron density (slowing the reaction), while the $+R$ (lone-pair) effect places negative charge at the $o$ and $p$ positions, so the electrophile goes there. Steric crowding makes para the major product.

Nucleophilic step. Aryl halides resist $\mathrm{S_N}$ reactions (partial double-bond character of $\mathrm{C\!-\!Cl}$ , resonance stabilisation). Chlorobenzene itself needs $623\ \mathrm{K}/300\ \mathrm{atm}$ NaOH (Dow process). A strongly electron-withdrawing $-\mathrm{NO_2}$ at the ortho or para position changes everything: it stabilises the negatively-charged carbanion (Meisenheimer) intermediate by resonance, so the reaction now runs under much milder conditions ( $\mathrm{NaOH}$ , $443\ \mathrm{K}$ ). A meta nitro group gives no such stabilisation.

Step-by-step

$\mathrm{C_6H_5Cl} \xrightarrow[\text{conc. } \mathrm{H_2SO_4}]{\mathrm{HNO_3}}$ chiefly $p$ -nitrochlorobenzene (with a little ortho).

$X = \mathrm{1\text{-}chloro\text{-}4\text{-}nitrobenzene}$
This already eliminates the meta option. …
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TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
$\text{P} \xleftarrow[\text{(ii) H}_2\text{O}]{\text{(i) LiAlH}_4} \ \text{C}_6\text{H}_5\text{CONH}_2 \ \xrightarrow{\text{Br}_2\ /\ \text{NaOH}} \text{Q}$
›Reveal solutionSolution
$\mathrm{LiAlH_4}$ reduces benzamide to benzylamine ( $P$ ), while $\mathrm{Br_2/NaOH}$ degrades it to aniline ( $Q$ ), one carbon shorter. That is option (B).

The concept first

Benzamide, $\mathrm{C_6H_5CONH_2}$ , has two very different fates depending on whether you attack the carbon or the nitrogen.

$\mathrm{LiAlH_4}$ delivers $\mathrm{H^-}$ to the electrophilic carbonyl carbon. The C–N bond is never broken; the oxygen leaves and the carbon ends up as a $\mathrm{CH_2}$ bridging the ring and the nitrogen. Amides are the only carbonyl compounds reduced all the way to an amine rather than an alcohol — that is worth remembering.

$\mathrm{Br_2/NaOH}$ instead brominates the nitrogen. The resulting $\mathrm{N}$ -bromoamide is deprotonated, loses $\mathrm{Br^-}$ to give a nitrene-like species, and the phenyl group migrates from carbon to nitrogen. The old carbonyl carbon ends up as an isocyanate carbon and is finally hydrolysed away as $\mathrm{Na_2CO_3}$ . Hence a one-carbon shortening.

Step-by-step

Reduction ( $P$ ):

$\mathrm{C_6H_5CONH_2 \xrightarrow[\ (ii)\ H_2O\ ]{(i)\ LiAlH_4} C_6H_5CH_2NH_2}$
Benzylamine, a primary amine with 7 carbons (same as benzamide).

2. Hofmann degradation ( $Q$ ):
$\mathrm{C_6H_5CONH_2 + Br_2 + 4NaOH \rightarrow C_6H_5NH_2 + Na_2CO_3 + 2NaBr + 2H_2O}$
Aniline, a primary amine with 6 carbons (one less). …
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TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
$\xrightarrow[\text{Conc. H}_2\text{SO}_4]{\text{HNO}_3}$
›Reveal solutionSolution

Chlorobenzene nitrates mainly at the para position to give $p$ -nitrochlorobenzene ( $X$ ); the para $-\mathrm{NO_2}$ activates the C–Cl carbon, so $\mathrm{NaOH}$ at 443 K followed by $\mathrm{H^+}$ gives $p$ -nitrophenol — option (A).

The concept first

Chlorine on a ring plays a double game. Through its $-I$ effect it deactivates the ring, but through lone-pair $+R$ donation it directs an incoming electrophile to the ortho and para positions. So nitration of chlorobenzene is slower than that of benzene, yet its major product is the para isomer (ortho is crowded).

The second step is the interesting one. Aryl halides are normally inert to nucleophiles — the C–Cl bond has partial double-bond character and the ring is electron-rich. Nucleophilic aromatic substitution becomes easy only when a strong electron-withdrawing group sits ortho or para to the halogen, because then the negative charge of the intermediate carbanion (the Meisenheimer complex) can be delocalised onto that group.

Step-by-step

Step 1 — nitration ( $X$ ).
$\mathrm{C_6H_5Cl}\ \xrightarrow[\text{conc. }\mathrm{H_2SO_4}]{\mathrm{HNO_3}}\ p\text{-}\mathrm{O_2N\text{-}C_6H_4\text{-}Cl}\ (\text{major})\ +\ o\text{-isomer (minor)}$
The electrophile is $\mathrm{NO_2^{+}}$ , formed as $\mathrm{HNO_3+2H_2SO_4\rightarrow NO_2^{+}+H_3O^{+}+2HSO_4^{-}}$ . So $X=p$ -nitrochlorobenzene. Note $\mathrm{H_2SO_4}$ acts only as the catalyst — it does not sulphonate the ring, which removes option (C). …
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TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
$\begin{aligned} \text{C}_6\text{H}_5\text{OCH}_3 & \xrightarrow{\text{HI}} X + Y \\ X & \xrightarrow{\text{Zn},\, \Delta} \text{(benzene ring)} \\ Y & \xrightarrow[\text{Anhy. AlCl}_3]{\text{C}_6\text{H}_6} Z \\ \end{aligned}$
›Reveal solutionSolution
The reaction sequence converts anisole into iodobenzene and methanol, then iodobenzene reacts with benzene in a Friedel–Crafts alkylation to give diphenylmethane, which is toluene — so Z is methylbenzene, option (A).

Concept & Intuition

This problem tests your understanding of ether cleavage by HI and subsequent aromatic substitution. Anisole (C₆H₅OCH₃) is an aryl methyl ether. When treated with HI, the C–O bond breaks selectively because the aryl–O bond is stronger (partial double-bond character from resonance with the ring). The products are iodobenzene (X) and methanol (Y). Then, X (iodobenzene) is reduced by Zn dust to benzene. Meanwhile, Y (methanol) reacts with benzene in the presence of anhydrous AlCl₃ — a Friedel–Crafts alkylation — to give toluene (methylbenzene). The key insight: methanol generates a methyl carbocation under these conditions, which attacks benzene.

Step-by-step reasoning

First reaction — Cleavage of anisole with HI Anisole (C₆H₅OCH₃) reacts with excess HI. The mechanism: protonation of the ether oxygen, then nucleophilic attack by I⁻ at the methyl carbon (since the methyl–O bond is weaker than the aryl–O bond). This yields iodobenzene (C₆H₅I) as X and methanol (CH₃OH) as Y.

$\text{C}_6\text{H}_5\text{OCH}_3 + \text{HI} \rightarrow \text{C}_6\text{H}_5\text{I} + \text{CH}_3\text{OH}$

Second reaction — Reduction of iodobenzene X (iodobenzene) is treated with Zn dust and heated (Δ). Zinc reduces the C–I bond, replacing iodine with hydrogen. The product is benzene (C₆H₆).

$\text{C}_6\text{H}_5\text{I} + \text{Zn} \xrightarrow{\Delta} \text{C}_6\text{H}_6 + \text{ZnI}_2$

Third reaction — Friedel–Crafts alkylation of benzene Y is methanol (CH₃OH). In the presence of anhydrous AlCl₃ (a Lewis acid), methanol forms a complex that generates a methyl carbocation (CH₃⁺). This electrophile attacks benzene, yielding toluene (methylbenzene, C₆H₅CH₃).
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TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
$\mathrm{CH_3Cl, \ CH_3COCl}$
›Reveal solutionSolution
The first step is a Friedel–Crafts alkylation ( $X=\text{CH}_3\text{Cl}$ ) and the second a Friedel–Crafts acylation ( $Y=\text{CH}_3\text{COCl}$ ). Correct option: (A).

Both steps are carried out on the aromatic ring with anhydrous $\text{AlCl}_3$ , so each reagent is a Friedel–Crafts electrophile.

$X$ (alkylation). An alkyl chloride substitutes an alkyl group onto the ring:

$\text{C}_6\text{H}_6+\text{CH}_3\text{Cl}\xrightarrow{\text{AlCl}_3}\text{C}_6\text{H}_5\text{CH}_3+\text{HCl}.$
So $X=\text{CH}_3\text{Cl}$ .

$Y$ (acylation). An acyl chloride introduces a $\text{–COCH}_3$ group: …
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TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQ
$Q. In which of the following reactions, hydrogen is evolved? I. Reaction of sodium borohydride with iodine II. Oxidation of diborane III. Reaction of boron trifluoride with sodium hydride IV. Hydrolysis of diborane (A) I, II only (B) I, II, IV only (C) III, IV only (D) I, IV only$
›Reveal solutionSolution
Hydrogen gas is evolved when diborane is hydrolysed and when sodium borohydride reacts with iodine — the key is to check whether the reaction produces free $H_2$ or only hydrogen-containing byproducts. The correct set is I and IV only.

The question tests your understanding of the chemistry of boron hydrides and related hydride-transfer reactions. Hydrogen evolution means free $H_2$ gas is released, not just that hydrogen atoms appear in a product. Many boron compounds are electron-deficient and react with protic sources or oxidising agents to liberate hydrogen, but not every reaction that involves a hydride does so.

Let’s examine each reaction carefully.

Reaction of sodium borohydride with iodine Sodium borohydride ( $\text{NaBH}_4$ ) is a source of hydride ions. Iodine ( $I_2$ ) is a mild oxidising agent. They react to give diborane and hydrogen gas:

$2\text{NaBH}_4 + I_2 \rightarrow \text{B}_2\text{H}_6 + 2\text{NaI} + H_2$
Hydrogen is clearly evolved here. So I is correct.

Oxidation of diborane Diborane ( $\text{B}_2\text{H}_6$ ) burns in oxygen to form boric oxide and water:

$\text{B}_2\text{H}_6 + 3\text{O}_2 \rightarrow \text{B}_2\text{O}_3 + 3\text{H}_2\text{O}$
No free hydrogen is produced — the hydrogen ends up in water. So II is incorrect.

Reaction of boron trifluoride with sodium hydride Sodium hydride ( $\text{NaH}$ ) is a strong hydride donor. With $\text{BF}_3$ , it forms diborane and sodium fluoride:

$2\text{BF}_3 + 6\text{NaH} \rightarrow \text{B}_2\text{H}_6 + 6\text{NaF}$
All the hydrogen from NaH goes into diborane; no $H_2$ gas is released. So III is incorrect.

Hydrolysis of diborane Diborane reacts vigorously with water to give boric acid and hydrogen gas: $\text{B}_2\text{H}_6 + 6\text{H}_2\text{O} \rightarrow 2\text{H}_3\text{BO}_3 + 6\text{H}_2$ …
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Q.Write the mechanism of the reaction of HI with methoxybenzene.

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