Q.Explain the fact that in aryl alkyl ethers

The key to this problem is recognizing that each step in the sequence is a classic organic transformation, and the final product’s structure is determined by the directing effects of substituents already on the ring. Let’s walk through it step by step.

1. First step: Aromatization of n-hexane

   n-Hexane ($\mathrm{C_6H_{14}}$) is treated with $\mathrm{V_2O_5}$ at 20 atm and 773 K. This is a typical catalytic reforming or dehydrogenation condition. Vanadium pentoxide acts as a catalyst to cyclize and dehydrogenate the straight-chain alkane into an aromatic ring. The product $\mathrm{[P_1]}$ is benzene ($\mathrm{C_6H_6}$).

   Why? Under high temperature and pressure, alkanes with six or more carbons can undergo dehydrocyclization — losing hydrogen atoms and forming a stable aromatic ring. No other functional groups are introduced here.

2. Second step: Friedel–Crafts alkylation

   Benzene ($\mathrm{[P_1]}$) reacts with ethyl chloride ($\mathrm{C_2H_5Cl}$) in the presence of anhydrous $\mathrm{AlCl_3}$ under heat ($\triangle$). This is a classic Friedel–Crafts alkylation: the $\mathrm{AlCl_3}$ generates an ethyl carbocation (or an ethyl–$\mathrm{AlCl_3}$ complex) that attacks the benzene ring.

   The product $\mathrm{[P_2]}$ is ethylbenzene ($\mathrm{C_6H_5CH_2CH_3}$).

   Note: Alkylation gives a mono-substituted product because the ethyl group is an activating, ortho/para-directing group, but under these conditions further alkylation is minimal (or we assume mono-substitution as the intended step).

3. Third step: Nitration of ethylbenzene

   Ethylbenzene ($\mathrm{[P_2]}$) is treated with concentrated nitric acid and sulfuric acid (a nitrating mixture) at 333 K. This is an electrophilic aromatic substitution. The nitronium ion ($\mathrm{NO_2^+}$) is the attacking species.

   The ethyl group is an ortho/para-directing group (it donates electron density via hyperconjugation and inductive effect). Therefore, the nitro group will preferentially attach at the ortho or para positions relative to the ethyl group. …

Concept & Intuition

Electrophilic substitution on benzene proceeds via a positively charged intermediate (the arenium ion). Substituents that donate electrons (by resonance or induction) stabilize this intermediate and activate the ring, making it more reactive than benzene itself. Substituents that withdraw electrons destabilize the intermediate and deactivate the ring. The key is to compare the net electronic effect of each group:

• –OCH₃ (methoxy): Strongly activating. The oxygen lone pairs donate into the ring by resonance, overwhelming the weak inductive withdrawal.
• –CH₃ (methyl): Mildly activating. It donates electrons by hyperconjugation and the +I effect.
• –H (benzene): The reference point — no substituent effect.
• –CF₃ (trifluoromethyl): Strongly deactivating. The three fluorine atoms pull electrons inductively, and there is no resonance donation to compensate.

Thus the order is: anisole > toluene > benzene > trifluoromethylbenzene.

Step-by-step reasoning

1. Identify the substituent effects

   • III (–OCH₃): The oxygen lone pairs are conjugated with the π-system, creating strong resonance donation (the methoxy group is ortho/para-directing and strongly activating).
   • I (–CH₃): The methyl group donates via hyperconjugation (C–H σ bonds overlap with the ring π-orbitals) and a small +I effect. This is a mild activator.
   • II (–H): No substituent — reactivity is the baseline.
   • IV (–CF₃): The highly electronegative fluorines pull electron density through σ-bonds (strong –I effect). No resonance donation is possible, so the ring becomes electron-poor and deactivated.

2. Rank the activating power

   Strongest activator → weakest activator → deactivator:

   –OCH₃ > –CH₃ > –H > –CF₃.

3. Translate to reactivity order …

Concept and Intuition

This problem tests your knowledge of four classic organic reactions of benzene derivatives. Each reaction is distinguished by the reagent and conditions, which dictate the product. Instead of memorizing blindly, think about why each condition leads to a specific outcome:

• Diazonium salts are excellent leaving groups; in warm water, they simply hydrolyze to give phenol.
• Bromination of phenol can be controlled: in a non-polar solvent (CS₂) at low temperature, only one bromine substitutes at the para position (due to steric and electronic control). In aqueous solution, the reaction is much more vigorous, leading to trisubstitution.
• Reimer–Tiemann reaction uses chloroform and strong base to install a formyl group ortho to the phenol – a classic example of a carbene intermediate.

Let’s go through each match step by step.

Step-by-Step Matching

1. Reaction A: Diazonium salt with warm water

• Benzenediazonium chloride ($\mathrm{C_6H_5N_2^+Cl^-}$) is unstable in water. The $\mathrm{N_2^+}$ group is a very good leaving group, and upon heating, it is replaced by a hydroxyl group from water.
• Product: Phenol ($\mathrm{C_6H_5OH}$).
• So A → R (Phenol).

2. Reaction B: Phenol with $\mathrm{Br_2}$ in $\mathrm{CS_2}$ at 273 K

• In a non-polar solvent like carbon disulfide and at low temperature, bromination of phenol is monosubstitution. The –OH group is strongly activating and ortho/para-directing. However, the ortho positions are somewhat sterically hindered; the major product is the para isomer.
• Product: p-Bromophenol.
• So B → P (p-Bromophenol).

3. Reaction C: Phenol with $\mathrm{CHCl_3}$ and aqueous $\mathrm{NaOH}$, then $\mathrm{HCl}$

• This is the Reimer–Tiemann reaction. Chloroform in the presence of strong base generates dichlorocarbene ($:\mathrm{CCl_2}$), which attacks the electron-rich ortho position of the phenoxide ion. After hydrolysis with HCl, the product is an ortho-hydroxybenzaldehyde.
• Product: 2-Hydroxybenzaldehyde (salicylaldehyde). …

Step 1 — Normal behaviour of aniline

Free aniline's $-NH_2$ group has a lone pair that donates electron density into the ring by resonance, strongly activating the ring and directing electrophiles to the ortho and para positions.

Step 2 — What happens in strongly acidic medium

Nitration uses a mixture of concentrated $HNO_3$ and $H_2SO_4$, a highly acidic environment. Under these conditions, the nitrogen lone pair of aniline is protonated before nitration can occur, converting $-NH_2$ into $-NH_3^+$ (the anilinium ion).

Step 3 — Why anilinium directs meta …

Step 1 — Classify the two amines.

• Aniline, $\mathrm{C_6H_5-NH_2}$ — the nitrogen carries two hydrogens and one carbon → primary (1°) aromatic amine.
• N-methylaniline, $\mathrm{C_6H_5-NH-CH_3}$ — the nitrogen carries one hydrogen and two carbons → secondary (2°) amine.

So any reagent that distinguishes them must respond differently to 1° versus 2° amines. Anything that only probes the aromatic ring or the basicity of $-\mathrm{NH}-$ will react with both.

Step 2 — Test each reagent.

(A) Acetic anhydride. Acetylation needs an N–H bond. Aniline has two, N-methylaniline has one — both react:

Both give a substituted amide, so no distinction.

(B) $\mathrm{Br_2}$ water. The $-\mathrm{NH_2}$ and $-\mathrm{NHCH_3}$ groups are both strongly activating, ortho/para-directing groups. Both anilines therefore undergo rapid ring bromination to give a white precipitate of the 2,4,6-tribromo derivative. No distinction.

(C) Conc. HCl + anhydrous $\mathrm{ZnCl_2}$. This is Lucas reagent — it is a test for alcohols (1°/2°/3°), distinguishing them by how fast a turbidity of the alkyl chloride appears. It has nothing to do with telling two amines apart; both amines simply form soluble ammonium salts with the acid. Not applicable. …

Concept & Intuition

This problem tests your understanding of how functional groups direct electrophilic aromatic substitution, and how a reaction sequence can completely change the directing effect. The key is to track the activating/deactivating and directing nature of the substituent at each step. Phenol’s –OH group is strongly activating and ortho/para-directing. But the first step (Zn dust, heat) removes that –OH entirely, giving benzene. Then nitration adds a meta-directing nitro group. Finally, bromination under conditions that prevent side reactions (cold & dark, FeBr₃) will place bromine meta to the nitro group. A common pitfall is to forget that the –OH is gone after step 1, and to mistakenly apply phenol’s directing effects to later steps.

Step-by-step reasoning

1. Step 1: Reduction of phenol to benzene

   Phenol ($\ce{C6H5OH}$) is heated with zinc dust. This is a classic reduction that removes the hydroxyl group, replacing it with hydrogen. The product [X] is benzene ($\ce{C6H6}$).

   Why? Zinc dust at high temperature acts as a reducing agent, cleaving the C–O bond and adding H (from trace moisture or the Zn itself). No substituent remains to direct further reactions — we now have a plain benzene ring.

2. Step 2: Nitration of benzene to nitrobenzene

   Benzene [X] is treated with concentrated nitric acid and heat. This is standard electrophilic nitration. The product [Y] is nitrobenzene ($\ce{C6H5NO2}$).

   Why? The nitronium ion ($\ce{NO2^+}$) attacks the benzene ring. Since there is no activating or deactivating group yet, the first substitution occurs at any position equally, but the product is simply nitrobenzene. The nitro group is strongly deactivating and meta-directing for further substitutions.

3. Step 3: Bromination of nitrobenzene …

Rate of electrophilic substitution tracks the electron density of the ring:

• Isopropylbenzene (IV): the isopropyl group is electron-donating (+I, hyperconjugation) → activated, most reactive.
• Benzene (III): the unsubstituted reference. …

Concept

Friedel–Crafts alkylation is an electrophilic aromatic substitution, and its rate is set by how electron-rich the ring is. Electron-donating groups raise the electron density and activate the ring; electron-withdrawing groups lower it and deactivate the ring. The stronger the effect, the larger the change in reactivity.

Solution

1. D (nitrobenzene): $-\text{NO}_2$ withdraws electrons strongly by both resonance and induction — the ring is the most deactivated, so D is least reactive.
2. B (bromobenzene): $-\text{Br}$ is net deactivating (inductive withdrawal dominates), so B is less reactive than plain benzene.
3. A (benzene): the reference, with no substituent. …

The key here is to understand how the amino group ($-\text{NH}_2$) in aniline behaves. It is a strongly activating and ortho/para-directing group because the lone pair on nitrogen can donate into the benzene ring through resonance. That makes aniline extremely reactive toward electrophilic substitution — too reactive, in fact, for some conditions.

But the same lone pair also makes aniline basic. In the presence of a strong Lewis acid (like $\text{AlCl}_3$), the nitrogen donates its lone pair to the acid, forming a salt. That salt has a positive charge on nitrogen, which is a powerful electron-withdrawing group. The ring becomes deactivated, and the Friedel-Crafts reaction fails.

Let’s go through each option.

1. Nitration — Aniline undergoes nitration readily. In fact, it is so reactive that direct nitration with $\text{HNO}_3/\text{H}_2\text{SO}_4$ gives a mixture including meta-product due to oxidation of the amino group. To avoid this, the amino group is first protected (e.g., by acetylation to acetanilide), then nitrated. But the point is: nitration does occur, so this is not the answer.

2. Sulphonation — Aniline reacts with concentrated $\text{H}_2\text{SO}_4$ to form anilinium hydrogen sulphate at low temperature. On heating, it undergoes sulphonation to give ortho- and para-aminobenzenesulphonic acids. So sulphonation is possible.

3. Friedel-Crafts reaction — This is the one that fails. In a Friedel-Crafts alkylation or acylation, the catalyst is a Lewis acid like $\text{AlCl}_3$. Aniline, being a strong base, forms a complex with $\text{AlCl}_3$: …

Concept: Diazo coupling.

In a coupling reaction the benzenediazonium ion, C6H5-N2+, acts as a weak ELECTROPHILE and attacks a highly activated aromatic ring (phenol in mild alkali, or aniline in mildly acidic medium). The activated ring's pi cloud attacks the terminal N of the diazonium ion, a sigma (arenium) complex forms, and loss of H+ restores aromaticity, giving the azo dye Ar-N=N-Ar'. …

The reaction you’re asking about is the Gattermann–Koch reaction — a classic method to introduce an aldehyde group directly onto an aromatic ring. The key idea is that carbon monoxide (CO) and hydrogen chloride (HCl) combine, under the Lewis acid catalysis of anhydrous AlCl₃, to generate a reactive electrophile: the formyl cation ($\mathrm{HCO}^+$). This electrophile then attacks the electron-rich benzene ring in an electrophilic aromatic substitution.

Why does this work? AlCl₃ coordinates with CO, polarising it, and then HCl provides the proton, forming a complex that effectively acts as $\mathrm{HCO}^+$. This is a strong electrophile, capable of substituting a hydrogen on benzene. The product is benzaldehyde — no further substitution occurs because the aldehyde group is deactivating and meta-directing, so the reaction stops at mono-formylation.

Let’s walk through the mechanism step by step.

1. Generation of the electrophile Anhydrous AlCl₃ (a strong Lewis acid) coordinates with the oxygen of carbon monoxide, pulling electron density away from the carbon. This makes the carbon highly electrophilic. HCl then adds, forming a complex that can be thought of as $\mathrm{HCO}^+$ (formyl cation) stabilised by $\mathrm{AlCl_4^-}$:

2. Electrophilic attack on benzene The formyl cation attacks the benzene ring, forming a sigma complex (arenium ion). The positive charge is delocalised over the ring:

3. Deprotonation to restore aromaticity The $\mathrm{AlCl_4^-}$ ion (or a chloride ion) abstracts a proton from the sigma complex, regenerating the aromatic ring and yielding benzaldehyde: …