Q.Write chemical reaction for the preparation of phenol from chlorobenzene.

The concept first

All three statements orbit one theme: how the nitrogen lone pair of an aromatic amine behaves, and how we control it.

Diazonium chemistry. Aniline $+$ $\mathrm{NaNO_2/HCl}$ gives $\mathrm{C_6H_5N_2^+Cl^-}$. The ion is stabilised by resonance with the ring — which is precisely why the aromatic diazonium salt can be isolated at all, whereas aliphatic ones decompose instantly. But even so, it is only kinetically stable in ice-cold conditions. Above about $278$ K, the excellent leaving group $\mathrm{N_2}$ departs and water attacks, giving phenol.

Protecting the amino group. Aniline is so strongly activated that direct nitration is a mess: the strongly acidic nitrating mixture protonates the $-\mathrm{NH_2}$ to $-\mathrm{NH_3^+}$ (a meta-directing deactivator), and oxidation side-reactions occur. The classical fix is to acetylate aniline to acetanilide (N-phenylethanamide). Now the nitrogen lone pair is shared with the acetyl carbonyl, so the ring is still activated but much less so, the amine can't be protonated, and nitration gives clean para-nitroacetanilide, which is then hydrolysed back. Understanding why we bother acetylating is exactly what makes statement B obviously true.

Hinsberg's reagent. Benzenesulphonyl chloride, $\mathrm{C_6H_5SO_2Cl}$, distinguishes the three classes of amine: $1^\circ$ amines give a sulphonamide with an acidic N–H, soluble in alkali; $2^\circ$ amines give a sulphonamide with no N–H, insoluble in alkali; $3^\circ$ amines do not react. Note the $\mathrm{SO_2}$ — that is the fingerprint.

Step-by-step

Statement A — "Aniline forms a stable benzene diazonium chloride at 285 K".

Diazotisation is carried out at $273$–$278$ K ($0$–$5\,^\circ\mathrm{C}$). $285$ K is $12\,^\circ\mathrm{C}$ — above the safe window. At that temperature the salt decomposes:

So it is not stable at 285 K. A is incorrect. ✗ …

The concept first

In Friedel–Crafts alkylation, a Lewis acid generates a carbocation:

That carbocation is an electrophile; it is attacked by the aromatic ring's $\pi$ electrons. So the reaction rate is governed by one thing: how much electron density the ring can offer.

Substituents fall into two camps:

• Activating (ring-enriching): $-$CH$_3$ and other alkyls ($+I$, hyperconjugation), $-$OH, $-$OR, $-$NH$_2$ ($+R$). They raise the rate.
• Deactivating (ring-depleting): $-$NO$_2$, $-$CHO, $-$COR, $-$COOH, $-$CN, $-$SO$_3$H ($-I$ and $-R$). They lower the rate — and in the special case of Friedel–Crafts they lower it so much that the reaction fails entirely. (Nitrobenzene is even used as a solvent for Friedel–Crafts reactions, precisely because it will not react.)

Step-by-step

Step 1 — Rank the substituents by their electronic effect.

• (A) Benzene — no substituent; the baseline.
• (B) Nitrobenzene, $-$NO$_2$ — one of the strongest deactivators known ($-I$ and $-R$). Ring badly electron-poor.
• (C) Acetophenone, $-$COCH$_3$ — a carbonyl; $-I$ and $-R$, strongly deactivating. Also, its lone pair can complex with AlCl$_3$, poisoning the catalyst.
• (D) Toluene, $-$CH$_3$ — electron-donating by $+I$ and hyperconjugation. Ring electron-rich.

Step 2 — Apply the rule. Rate of electrophilic substitution increases with ring electron density: …

The concept first: one diazonium salt, four different name reactions

Aryl diazonium salts are a crossroads. From the same $\mathrm{C_6H_5N_2Cl}$ you can go in many directions, and each direction has its own name and its own reagent. Examiners test whether you can tell them apart, so learn them as a set, not one at a time:

The distinguishing mark of Sandmeyer is the cuprous (Cu(I)) salt, $\mathrm{Cu_2X_2}$, in the corresponding halogen acid.

Step-by-step: test each option

1. $\mathrm{C_6H_5N_2Cl} \xrightarrow{\mathrm{Cu_2Br_2/HBr}} \mathrm{C_6H_5Br}$ — cuprous bromide in HBr, giving bromobenzene. This is exactly the Sandmeyer prescription. ✓
2. $\mathrm{C_6H_5N_2Cl} \xrightarrow{\mathrm{Cu/HBr}} \mathrm{C_6H_5Br}$ — the product is the same, but the reagent is copper powder, not the cuprous salt. This is the Gattermann reaction — a close cousin, deliberately placed here as the trap. ✗
3. $\mathrm{C_6H_5N_2Cl} \xrightarrow{\mathrm{H_2O}} \mathrm{C_6H_5OH}$ — this is simple hydrolysis to phenol (which is why diazonium salts must be kept cold). Not Sandmeyer. ✗ …

Concept & Intuition

This problem tests your ability to predict the outcome of a multi‑step synthesis on an aromatic ring. The three steps are classic transformations:

1. Williamson ether synthesis – converts a phenol into an aryl alkyl ether.
2. Friedel–Crafts acylation – introduces an acyl group (here, acetyl) ortho/para to the activating ether group.
3. Clemmensen reduction – reduces the carbonyl (C=O) to a methylene (CH₂), giving an alkyl side chain.

The key insight: the ether group (–OC₂H₅) is a strong activating and ortho/para‑directing substituent. So the acylation will occur at the para position (less steric hindrance than ortho). The final reduction then turns the –COCH₃ into –CH₂CH₃, leaving the ether untouched. The product is a 1,4‑disubstituted benzene with –OC₂H₅ and –C₂H₅.

Step‑by‑Step Reasoning

1. Step (i): NaOH, C₂H₅I The starting material is phenol (C₆H₅OH). NaOH deprotonates it to phenoxide (C₆H₅O⁻), which then attacks ethyl iodide (C₂H₅I) in an Sₙ2 reaction. This is a Williamson ether synthesis, giving ethoxybenzene (phenetole):

2. Step (ii): CH₃COCl, anhyd. AlCl₃ This is a Friedel–Crafts acylation. The –OC₂H₅ group is a strong ortho/para‑director because the oxygen lone pairs donate electron density into the ring.
   • The ortho positions are sterically hindered by the ethoxy group.
   • The para position is more accessible. Therefore, the acetyl group (–COCH₃) attaches para to the ethoxy group, forming 4‑ethoxyacetophenone:

3. Step (iii): Zn‑Hg, conc. HCl …

The reaction described is a classic Friedel–Crafts alkylation. In this reaction, an alkyl halide reacts with benzene in the presence of a Lewis acid catalyst (here, anhydrous AlCl₃) to form an alkylbenzene. The key mechanistic step is the generation of a carbocation from the alkyl halide. That carbocation then attacks the benzene ring.

The trap in this specific case is that the n-propyl carbocation (CH₃CH₂CH₂⁺) is primary and therefore relatively unstable. Under the reaction conditions, it rapidly rearranges to the more stable secondary carbocation (CH₃CH⁺CH₃) via a 1,2-hydride shift. This rearranged carbocation is what actually attacks benzene, giving isopropyl benzene as the major product.

Let’s walk through the reasoning step by step.

1. Generation of the electrophile. Anhydrous AlCl₃ coordinates with the chlorine atom of n-propyl chloride, weakening the C–Cl bond. This leads to the formation of a carbocation and an AlCl₄⁻ complex:

2. Carbocation rearrangement. The primary carbocation (CH₃CH₂CH₂⁺) is not very stable. A hydrogen atom from the second carbon (along with its bonding pair) shifts to the positively charged carbon. This 1,2-hydride shift converts the primary carbocation into a more stable secondary carbocation:

3. Electrophilic attack on benzene. The rearranged secondary carbocation acts as the electrophile. It attacks the electron-rich benzene ring, forming a sigma complex (arenium ion): $\text{C}_6\text{H}_6 + \text{CH}_3\text{CH}^+\text{CH}_3 \rightarrow \text{C}_6\text{H}_6^+(\text{CH}(\text{CH}_3)_2)$ …

The concept first

When a ring carries two substituents, the more powerfully activating one decides where the electrophile goes.

• $\mathrm{-OH}$: strongly activating via its lone pair ($+R \gg -I$), ortho/para directing.
• $\mathrm{-CHO}$: deactivating ($-R, -I$), meta directing. The phenol group is far stronger, so it wins. Conveniently, the positions ortho/para to the OH of salicylaldehyde are also meta to the CHO — the two groups' preferences agree, which is why the reaction is so clean.

Step-by-step

1. Number the ring. Salicylaldehyde is 2-hydroxybenzaldehyde: $\mathrm{CHO}$ at C-1, $\mathrm{OH}$ at C-2.
2. Find the positions activated by the OH (at C-2).
   • ortho to C-2 → C-1 (blocked by the CHO) and C-3 ✓
   • para to C-2 → C-5 ✓ So the two available activated sites are C-3 and C-5.
3. Cross-check against the CHO (at C-1). Meta to C-1 → C-3 and C-5 — exactly the same two carbons. Both directing effects therefore point to C-3 and C-5. …