Accomplish the following conversions: (i) Nitrobenzene to benzoic acid (ii) Benzene to m-bromophenol (iii) Ben

Exercises · 9.8

Q. Accomplish the following conversions:

(i) Nitrobenzene to benzoic acid

(ii) Benzene to m-bromophenol

(iii) Benzoic acid to aniline

(iv) Aniline to 2,4,6-tribromofluorobenzene

(v) Benzyl chloride to 2-phenylethanamine

(vi) Chlorobenzene to p-chloroaniline

(vii) Aniline to p-bromoaniline

(viii) Benzamide to toluene

(ix) Aniline to benzyl alcohol.

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Concept understanding — Aromatic Synthesis Route

Aromatic Synthesis Route – First Principles

Imagine you are a chef who has been given a plain wooden board and told to carve a specific shape out of it. You can cut away wood, but you cannot add wood back. That is exactly the problem in aromatic synthesis: you start with a simple, cheap aromatic ring (like benzene) and you need to attach specific groups at specific positions. The ring itself is already there — you cannot rearrange its carbon skeleton. So the entire challenge is where to put the next group, and how to get it there.

The "route" is the sequence of reactions you choose. The order matters enormously because the groups already on the ring control where the next group will go. A wrong order can give you the wrong isomer, or force you into a dead end.

The Core Idea: The Ring Directs

Every substituent already on a benzene ring has a directing effect — it tells the next incoming group to go to certain positions. There are two families:

Ortho/para directors (e.g., –OH, –NH₂, –CH₃, –Cl) — they push the next group to positions 2 and 4 (ortho and para).
Meta directors (e.g., –NO₂, –CN, –CHO, –SO₃H) — they push the next group to position 3 (meta).

Watch out

A common mistake is to think you can just "add any group in any order." The ring is not passive — it has a memory of what is already attached. If you ignore directing effects, you will get a mixture of products, often with the wrong isomer as the major one.

The Precise Statement

An aromatic synthesis route is a planned sequence of electrophilic aromatic substitution (EAS) reactions, chosen so that each new substituent is introduced at the correct position relative to the existing ones. The route must account for:

Directing effects of all current substituents.
Activation/deactivation — some groups make the ring more reactive (activators), some make it less reactive (deactivators). You cannot do a reaction on a strongly deactivated ring without special conditions.
Order of introduction — sometimes you must introduce a meta director first, then an ortho/para director, or vice versa, to get the desired final pattern.

A Concrete Example: Making 4-Nitrobenzoic Acid

You want a benzene ring with –COOH at position 1 and –NO₂ at position 4 (para to each other).

–COOH is a meta director. –NO₂ is also a meta director. If you put –COOH first and then nitrate, the –COOH will send the –NO₂ to the meta position (3), not para (4). That gives the wrong isomer.

Correct route:

Nitrate benzene first → nitrobenzene ( –NO₂ is meta directing).
Then oxidise the methyl group (if you started with toluene) or use a different method to introduce –COOH. But wait — –NO₂ deactivates the ring strongly. So you cannot easily do Friedel-Crafts acylation on nitrobenzene.

So the actual correct route is different:

Start with toluene (methylbenzene). The –CH₃ is an ortho/para director and an activator.
Nitrate toluene → you get a mixture of ortho and para nitrotoluene. Separate the para isomer.
Oxidise the –CH₃ to –COOH using KMnO₄. The –NO₂ survives this oxidation.

Note

The order here is: introduce the ortho/para director first ( –CH₃ ), then nitrate to get para, then convert the –CH₃ to –COOH. If you had tried to put –COOH first, you would have a meta director that would send –NO₂ to the wrong place.

The General Strategy

When planning a route, ask yourself in order:

What is the final substitution pattern? (1,2- ; 1,3- ; 1,4- ; etc.)
Which groups are ortho/para directors and which are meta directors?
Can I introduce the meta director first, then the ortho/para director? (Often yes, because meta directors deactivate the ring, making further substitution harder — so you want to do the deactivating step last if possible.)
If I need a 1,3 pattern, I usually put a meta director first, then an ortho/para director. If I need a 1,4 pattern, I usually put an ortho/para director first, then a meta director (because the ortho/para director will send the next group to para, and the meta director will then be at the correct position). …

Why this formula?

Aromatic Synthesis Route: Understanding the Why Behind the Key Principles

In organic chemistry, an aromatic synthesis route refers to a sequence of reactions designed to construct or modify an aromatic ring (typically benzene or its derivatives). The "key formulas" here are not single equations but rather rules and principles that govern reactivity and orientation. Let's break down the reasoning behind the most critical ones.

1. The 4n+2 Hückel Rule — Why Aromaticity Exists

Formula: A planar, cyclic, conjugated molecule is aromatic if it has $(4n+2)$ π electrons, where $n = 0,1,2,\dots$

Why this holds (the derivation):

In a cyclic conjugated system, the π electrons occupy molecular orbitals (MOs) that form a ring.
The energy levels of these MOs are given by the Frost circle (or polygon rule):
- For a regular polygon with $N$ vertices (atoms), inscribe it in a circle with one vertex at the bottom.
- The energy of each MO corresponds to the vertical coordinate of each vertex.
For benzene ( $N=6$ $N = 6$ ), the MOs split into:
- 1 low-energy bonding orbital
- 2 degenerate bonding orbitals
- 2 degenerate antibonding orbitals
- 1 high-energy antibonding orbital
Key insight: The 6 π electrons fill the 3 bonding MOs completely. This gives a closed-shell, highly stable configuration — the aromatic stabilization energy (~150 kJ/mol for benzene).
For $N=4$ (cyclobutadiene), the MO pattern gives 2 degenerate non-bonding orbitals — filling with 4 electrons creates an open-shell, antiaromatic (unstable) system.

Takeaway: The $(4n+2)$ rule is not arbitrary — it emerges from the symmetry of cyclic π systems and the filling of bonding MOs.

2. Electrophilic Aromatic Substitution (EAS) — The Reactivity Formula

General reaction:

\text{Ar-H} + \text{E}^+ \xrightarrow{\text{catalyst}} \text{Ar-E} + \text{H}^+

Why this is the only viable route for aromatic rings:

Aromatic rings are electron-rich (due to the π cloud) but resistant to addition — addition would break aromaticity.
Mechanism reasoning:
1. The electrophile $\text{E}^+$ attacks the ring, forming a σ-complex (arenium ion) — this step is slow (rate-determining).
2. The σ-complex is non-aromatic (4 π electrons in the ring) — it is high-energy and unstable.
3. To regain aromaticity, the complex loses a proton ( $\text{H}^+$ ) — this step is fast and thermodynamically driven.
Why substitution, not addition: Addition would permanently destroy aromaticity; substitution restores it.

Key formula: The rate law is $\text{Rate} = k[\text{Ar-H}][\text{E}^+]$ — first order in both, because the slow step involves both reactants.

3. Orientation Rules — Why Substituents Direct Where the Next Group Goes

Rule:

Activating groups (e.g., $-\text{OH}, -\text{NH}_2, -\text{OCH}_3$ ) direct to ortho/para positions.
Deactivating groups (e.g., $-\text{NO}_2, -\text{CN}, -\text{CHO}$ ) direct to meta positions.

Why this happens (resonance + inductive reasoning):

For ortho/para directors:

The substituent has a lone pair or π bond that can donate electrons into the ring via resonance.
Draw the resonance structures of the σ-complex for attack at ortho, meta, and para:
- Ortho attack: The positive charge can be delocalized onto the substituent (e.g., $-\text{OH}$ becomes $=\text{OH}^+$ ). This stabilizes the intermediate.
- Para attack: Similar stabilization — charge delocalized to the substituent.
- Meta attack: The positive charge cannot reach the substituent — less stable.
Result: Ortho/para intermediates are lower in energy → faster reaction.

For meta directors:

The substituent is electron-withdrawing (by induction or resonance, e.g., $-\text{NO}_2$ ).
Draw resonance for ortho attack: The positive charge is placed directly on the carbon bearing the withdrawing group — this is highly destabilizing (like putting a + charge next to a + pole).
For meta attack: The positive charge is never on the carbon with the withdrawing group — relatively more stable.
Result: Meta attack is the least destabilized → preferred. …

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Key Idea: Aromatic Synthesis Route

Each conversion uses standard functional-group transformations (reduction, diazotisation, Sandmeyer, Balz-Schiemann, etc.) while respecting directing effects.

Nitrobenzene -> Benzoic acid Reduce $\ce{NO2}$ to $\ce{NH2}$ ( $\ce{Sn/HCl}$ ), then diazotise ( $\ce{NaNO2/HCl}$ , $0\text{-}5^\circ\text{C}$ ) and replace with $\ce{CN}$ (Sandmeyer: $\ce{CuCN}$ ). Hydrolyse $\ce{CN}$ to $\ce{COOH}$ ( $\ce{H3O+}$ , heat).
Benzene -> m-Bromophenol Nitrate benzene ( $\ce{HNO3/H2SO4}$ ) to get nitrobenzene. Since $\ce{-NO2}$ is a meta-director, brominate nitrobenzene directly ( $\ce{Br2/Fe}$ ) to give m-bromonitrobenzene. Reduce $\ce{-NO2}$ to $\ce{-NH2}$ ( $\ce{Sn/HCl}$ ), then diazotise ( $\ce{NaNO2/HCl}$ , $0$ - $5^\circ\text{C}$ ) and hydrolyse (warm with water) to replace $\ce{-N2+}$ with $\ce{-OH}$ , giving m-bromophenol.
Benzoic acid -> Aniline Convert $\ce{COOH}$ to $\ce{CONH2}$ (via $\ce{SOCl2}$ then $\ce{NH3}$ ). Hofmann rearrangement ( $\ce{Br2/NaOH}$ ) gives aniline directly (the amide loses its carbonyl carbon, which ends up as $\ce{Na2CO3}$ in the alkaline medium).
Aniline -> 2,4,6-Tribromofluorobenzene Treat aniline with excess $\ce{Br2}$ water (the free amino group is strongly activating, so tribromination occurs directly at 2,4,6 without needing protection) to get 2,4,6-tribromoaniline. Diazotise ( $\ce{NaNO2/HBF4}$ ) and heat (Balz-Schiemann) to replace $\ce{N2+}$ with $\ce{F}$ .
Benzyl chloride -> 2-Phenylethanamine Treat with $\ce{KCN}$ (SN2) -> benzyl cyanide. Reduce ( $\ce{LiAlH4}$ or $\ce{H2/Ni}$ ) to 2-phenylethanamine.
Chlorobenzene -> p-Chloroaniline Nitrate chlorobenzene ( $\ce{HNO3/H2SO4}$ ) - $\ce{Cl}$ is o/p-directing, so major product is p-chloronitrobenzene. Reduce $\ce{NO2}$ to $\ce{NH2}$ ( $\ce{Sn/HCl}$ ).
Aniline -> p-Bromoaniline Acetylate aniline (protects, reduces o-bromination). Brominate (electrophilic substitution gives p-bromoacetanilide). Deprotect ( $\ce{H3O+}$ ) -> p-bromoaniline.
Benzamide -> Toluene The amine route (LiAlH4 reduction of the amide to benzylamine, then diazotisation) does not work cleanly here, because benzylamine's amino group is on an aliphatic (benzylic) carbon, not on the ring itself - its diazonium salt is unstable and decomposes to give benzyl alcohol, not toluene (compare: $\ce{C6H5CH2NH2 + HNO2 -> C6H5CH2OH}$ ). The clean route instead removes the carbonyl carbon entirely and re-installs a methyl group by Friedel-Crafts alkylation: Hofmann-degrade benzamide to aniline (an aromatic amine), diazotise and deaminate with $\ce{H3PO2}$ to benzene, then methylate. $\ce{C6H5CONH2 ->[Br2/NaOH] C6H5NH2 ->[NaNO2/HCl] C6H5N2+ ->[H3PO2] C6H6 ->[CH3Cl/AlCl3] C6H5CH3}$
Aniline -> Benzyl alcohol …

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The key idea is to use the functional group transformations and directing effects of substituents on the benzene ring. Each conversion is achieved through a specific sequence of reactions: (i) reduction then diazotisation and substitution;

(ii) nitration, bromination, reduction, diazotisation, then hydrolysis;

(iii) amide formation, then Hofmann rearrangement to aniline directly;

(iv) bromination, diazotisation, then fluorination (Balz-Schiemann);

(v) nucleophilic substitution with cyanide followed by reduction;

(vi) nitration, then reduction;

(vii) acetylation, bromination, deprotection;

(viii) Hofmann rearrangement to aniline, diazotisation and deamination to benzene, then Friedel-Crafts methylation;

(ix) diazotisation, cyanation, hydrolysis, reduction to benzyl alcohol.

Let's walk through each conversion step by step, focusing on the why behind each choice.

(i) Nitrobenzene to benzoic acid

Concept: We need to replace the nitro group ( $-NO_2$ ) with a carboxyl group ( $-COOH$ ). The nitro group is a strong deactivator and meta-director, but we can first reduce it to an amino group ( $-NH_2$ ), which is activating and ortho/para-directing. Then, via diazotisation, we can replace the amino group with a cyano group ( $-CN$ ), which can be hydrolysed to a carboxylic acid.

Reduce nitrobenzene to aniline: Use Sn/HCl or Fe/HCl. The nitro group is reduced to an amino group.

$C_6H_5NO_2 \xrightarrow{Sn/HCl} C_6H_5NH_2$
Diazotise aniline: Treat with NaNO $_2$ and HCl at 0-5°C to form the diazonium salt.

$C_6H_5NH_2 \xrightarrow{NaNO_2/HCl, 0-5^\circ C} C_6H_5N_2^+Cl^-$
Replace diazonium with cyano group: Use CuCN (Sandmeyer reaction) or KCN with Cu catalyst.

$C_6H_5N_2^+Cl^- \xrightarrow{CuCN} C_6H_5CN$
Hydrolyse the nitrile to benzoic acid: Acidic or basic hydrolysis.

$C_6H_5CN \xrightarrow{H_3O^+, \Delta} C_6H_5COOH$

Watch out

A common mistake is to try direct oxidation of nitrobenzene to benzoic acid. That can never work: the group on the ring is a nitrogen substituent, and no oxidation can turn the $-NO_2$ nitrogen into the $-COOH$ carbon — benzoic acid has a seventh carbon that nitrobenzene simply does not have. The carboxyl carbon must be brought in from outside, which is exactly what the $CN^-$ of the Sandmeyer step supplies; hydrolysis then converts it to $-COOH$ .

(ii) Benzene to m-bromophenol

Concept: We need to introduce a bromine atom and a hydroxyl group ( $-OH$ ) in a meta relationship. Since $-OH$ is strongly activating and ortho/para-directing, we cannot brominate phenol directly to get meta-bromophenol (it would give ortho/para). Instead, we first introduce a meta-directing group (like $-NO_2$ ), then brominate, then convert the nitro group to hydroxyl.

Nitrate benzene: Use conc. HNO $_3$ and conc. H $_2$ SO $_4$ to get nitrobenzene.

$C_6H_6 \xrightarrow{HNO_3/H_2SO_4} C_6H_5NO_2$
Brominate nitrobenzene: The nitro group is meta-directing, so bromination gives m-bromonitrobenzene.

$C_6H_5NO_2 \xrightarrow{Br_2/FeBr_3} m-BrC_6H_4NO_2$
Reduce the nitro group to amino: Use Sn/HCl.

$m-BrC_6H_4NO_2 \xrightarrow{Sn/HCl} m-BrC_6H_4NH_2$
Diazotise and hydrolyse to phenol: Diazotise with NaNO $_2$ /HCl, then warm with water.

$m-BrC_6H_4NH_2 \xrightarrow{NaNO_2/HCl, 0-5^\circ C} m-BrC_6H_4N_2^+Cl^- \xrightarrow{H_2O, \Delta} m-BrC_6H_4OH$

Tip

The key insight: to get a meta relationship between two substituents where one is ortho/para-directing, you must first introduce a meta-directing group, then the second substituent, and finally convert the meta-directing group into the desired ortho/para-directing one.

(iii) Benzoic acid to aniline

Concept: We need to replace the carboxyl group ( $-COOH$ ) with an amino group ( $-NH_2$ ), losing the carbonyl carbon in the process. The classic method is the Hofmann rearrangement of the amide derived from benzoic acid - this converts the amide directly to aniline in one step, with no intermediate alcohol stage.

Convert benzoic acid to benzamide: Treat with NH $_3$ or first make the acid chloride (with SOCl $_2$ or PCl $_5$ ), then react with NH $_3$ .

$C_6H_5COOH \xrightarrow{SOCl_2} C_6H_5COCl \xrightarrow{NH_3} C_6H_5CONH_2$
Hofmann rearrangement: Treat benzamide with Br $_2$ and NaOH. The amide loses CO $_2$ and forms aniline directly.

$C_6H_5CONH_2 \xrightarrow{Br_2/NaOH} C_6H_5NH_2$

R-CONH_2 + Br_2 + 4NaOH \rightarrow R-NH_2 + Na_2CO_3 + 2NaBr + 2H_2O

(iv) Aniline to 2,4,6-tribromofluorobenzene

Concept: We need to introduce three bromine atoms at the 2,4,6 positions (ortho and para to the amino group) and then replace the amino group with fluorine. The amino group is strongly activating and ortho/para-directing, so bromination of aniline gives 2,4,6-tribromoaniline. Then, diazotisation followed by treatment with fluoroboric acid (HBF $_4$ ) gives the fluorobenzene derivative (Balz-Schiemann reaction).

Brominate aniline: Treat with excess Br $_2$ in water. The amino group directs bromine to ortho and para positions, giving 2,4,6-tribromoaniline.

$C_6H_5NH_2 + 3Br_2 \rightarrow 2,4,6-Br_3C_6H_2NH_2 + 3HBr$
Diazotise: Treat with NaNO $_2$ /HCl at 0-5°C.

$2,4,6-Br_3C_6H_2NH_2 \xrightarrow{NaNO_2/HCl} 2,4,6-Br_3C_6H_2N_2^+Cl^-$
Replace diazonium with fluorine (Balz-Schiemann reaction): Add HBF $_4$ to form the diazonium fluoroborate, then heat to decompose it.

$2,4,6-Br_3C_6H_2N_2^+Cl^- \xrightarrow{HBF_4} 2,4,6-Br_3C_6H_2N_2^+BF_4^- \xrightarrow{\Delta} 2,4,6-Br_3C_6H_2F + N_2 + BF_3$

Watch out

Direct fluorination of aniline is not possible; the Balz-Schiemann reaction is the standard method to introduce fluorine onto an aromatic ring.

(v) Benzyl chloride to 2-phenylethanamine

Concept: We need to increase the carbon chain by one carbon and introduce an amino group. The classic method is nucleophilic substitution of the chloride with cyanide, followed by reduction of the nitrile to a primary amine.

Nucleophilic substitution with cyanide: Treat benzyl chloride with KCN (or NaCN) in ethanol-water.

$C_6H_5CH_2Cl \xrightarrow{KCN} C_6H_5CH_2CN$
Reduce the nitrile to a primary amine: Use LiAlH $_4$ or catalytic hydrogenation (H $_2$ /Ni) or Na/ethanol.

$C_6H_5CH_2CN \xrightarrow{LiAlH_4} C_6H_5CH_2CH_2NH_2$

Tip

This is a classic chain-elongation method: alkyl halide -> nitrile -> primary amine (with one extra carbon).

(vi) Chlorobenzene to p-chloroaniline

Concept: We need to introduce an amino group at the para position relative to chlorine. Chlorine is ortho/para-directing but deactivating. Direct nitration of chlorobenzene gives a mixture of ortho and para nitrochlorobenzene, which can be separated. Then reduce the nitro group to amino.

Nitrate chlorobenzene: Use conc. HNO $_3$ and conc. H $_2$ SO $_4$ . The major product is p-nitrochlorobenzene (along with some ortho).

$C_6H_5Cl \xrightarrow{HNO_3/H_2SO_4} p-O_2NC_6H_4Cl$ (major)
Separate the para isomer (by fractional distillation or crystallisation).
Reduce the nitro group: Use Sn/HCl or Fe/HCl.

$p-O_2NC_6H_4Cl \xrightarrow{Sn/HCl} p-H_2NC_6H_4Cl$

Watch out

Direct amination of chlorobenzene is not possible under normal conditions. The nitration-reduction route is standard.

(vii) Aniline to p-bromoaniline

Concept: We need to introduce a bromine atom at the para position relative to the amino group. The amino group is strongly activating and ortho/para-directing. Direct bromination of aniline gives 2,4,6-tribromoaniline (all three positions are activated). To get mono-bromination at the para position, we must first protect the amino group (e.g., by acetylation) to reduce its activating power, then brominate, then deprotect.

Protect the amino group by acetylation: Treat aniline with acetic anhydride (or acetyl chloride) to form acetanilide.

$C_6H_5NH_2 \xrightarrow{(CH_3CO)_2O} C_6H_5NHCOCH_3$
Brominate acetanilide: The acetamido group is still ortho/para-directing but less activating, so bromination gives mainly p-bromoacetanilide.

$C_6H_5NHCOCH_3 \xrightarrow{Br_2/CH_3COOH} p-BrC_6H_4NHCOCH_3$
Deprotect: Hydrolyse the amide back to the amine using dilute HCl or NaOH.

$p-BrC_6H_4NHCOCH_3 \xrightarrow{H_3O^+, \Delta} p-BrC_6H_4NH_2$

Tip

Acetylation is a common protecting group strategy to control the degree and position of electrophilic substitution on aniline.

(viii) Benzamide to toluene …

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Aromatic Synthesis Route — One Clear Method per Conversion

Method Used: Retrosynthetic Disconnection + Functional Group Interconversion (FGI)

For each conversion, identify the key functional-group transformation and the directing effects involved, then write the forward synthesis step by step.

(i) Nitrobenzene → Benzoic acid

Method: Reduction → Diazotisation → Sandmeyer (CN) → Hydrolysis

Reduce nitrobenzene to aniline: $C_6H_5NO_2 \xrightarrow{Sn/HCl} C_6H_5NH_2$
Diazotise at 273–278 K: $C_6H_5NH_2 \xrightarrow{NaNO_2/HCl} C_6H_5N_2^+Cl^-$
Sandmeyer with CuCN: $C_6H_5N_2^+Cl^- \xrightarrow{CuCN} C_6H_5CN$
Hydrolyse the nitrile: $C_6H_5CN \xrightarrow{H_3O^+,\ \Delta} C_6H_5COOH$

Key result: $\boxed{C_6H_5NO_2 \rightarrow C_6H_5COOH}$

(ii) Benzene → m-bromophenol

Method: install a meta director first, brominate, then convert it to $-OH$

Nitrate benzene: $C_6H_6 \xrightarrow{HNO_3/H_2SO_4} C_6H_5NO_2$
Brominate nitrobenzene — $-NO_2$ is a meta director, so bromine enters meta: $C_6H_5NO_2 \xrightarrow{Br_2/FeBr_3} m\text{-}BrC_6H_4NO_2$
Reduce the nitro group: $m\text{-}BrC_6H_4NO_2 \xrightarrow{Sn/HCl} m\text{-}BrC_6H_4NH_2$
Diazotise and hydrolyse: $m\text{-}BrC_6H_4NH_2 \xrightarrow{NaNO_2/HCl,\ 273\text{–}278\ K} m\text{-}BrC_6H_4N_2^+Cl^- \xrightarrow{H_2O,\ \Delta} m\text{-}BrC_6H_4OH$

Key result: $\boxed{C_6H_6 \rightarrow m\text{-}BrC_6H_4OH}$

(iii) Benzoic acid → Aniline

Method: amide formation → Hofmann bromamide degradation

$C_6H_5COOH \xrightarrow{SOCl_2} C_6H_5COCl \xrightarrow{NH_3} C_6H_5CONH_2$
$C_6H_5CONH_2 \xrightarrow{Br_2/NaOH} C_6H_5NH_2$

Key result: $\boxed{C_6H_5COOH \rightarrow C_6H_5NH_2}$

(iv) Aniline → 2,4,6-tribromofluorobenzene

Method: direct tribromination → diazotisation → Balz–Schiemann (F)

Brominate aniline directly — the free $-NH_2$ is so activating that bromine water substitutes all three ortho/para positions at once (no protection step; protection would moderate the ring and stop at mono-bromination): $C_6H_5NH_2 \xrightarrow{Br_2/H_2O} 2,4,6\text{-}Br_3C_6H_2NH_2$
Diazotise at 273–278 K: $2,4,6\text{-}Br_3C_6H_2NH_2 \xrightarrow{NaNO_2/HCl} 2,4,6\text{-}Br_3C_6H_2N_2^+Cl^-$
Balz–Schiemann: $2,4,6\text{-}Br_3C_6H_2N_2^+Cl^- \xrightarrow{HBF_4} 2,4,6\text{-}Br_3C_6H_2N_2^+BF_4^- \xrightarrow{\Delta} 2,4,6\text{-}Br_3C_6H_2F$

Key result: $\boxed{C_6H_5NH_2 \rightarrow 2,4,6\text{-}Br_3C_6H_2F}$

(v) Benzyl chloride → 2-phenylethanamine

Method: chain extension with cyanide → reduction

$C_6H_5CH_2Cl \xrightarrow{KCN\ (alc.)} C_6H_5CH_2CN$
$C_6H_5CH_2CN \xrightarrow{LiAlH_4\ (\text{or } H_2/Ni)} C_6H_5CH_2CH_2NH_2$

Key result: $\boxed{C_6H_5CH_2Cl \rightarrow C_6H_5CH_2CH_2NH_2}$

(vi) Chlorobenzene → p-chloroaniline

Method: nitration (Cl is o/p-directing) → separation → reduction

$C_6H_5Cl \xrightarrow{HNO_3/H_2SO_4} p\text{-}ClC_6H_4NO_2$ (major, with some ortho)
Separate the para isomer (crystallisation/distillation)
$p\text{-}ClC_6H_4NO_2 \xrightarrow{Sn/HCl} p\text{-}ClC_6H_4NH_2$

Key result: $\boxed{C_6H_5Cl \rightarrow p\text{-}ClC_6H_4NH_2}$

(vii) Aniline → p-bromoaniline

Method: protection → mono-bromination → deprotection

Acetylate (to moderate the ring and favour para): $C_6H_5NH_2 \xrightarrow{(CH_3CO)_2O} C_6H_5NHCOCH_3$
Brominate: $C_6H_5NHCOCH_3 \xrightarrow{Br_2/CH_3COOH} p\text{-}BrC_6H_4NHCOCH_3$
Hydrolyse: $p\text{-}BrC_6H_4NHCOCH_3 \xrightarrow{H_3O^+,\ \Delta} p\text{-}BrC_6H_4NH_2$

Key result: $\boxed{C_6H_5NH_2 \rightarrow p\text{-}BrC_6H_4NH_2}$

(viii) Benzamide → Toluene

Method: Hofmann degradation → deamination to benzene → Friedel–Crafts methylation

$C_6H_5CONH_2 \xrightarrow{Br_2/NaOH} C_6H_5NH_2$ …

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Common Mistakes in Aromatic Synthesis Routes (and How to Avoid Them)

These conversions test your understanding of directing effects, protecting groups, reduction/oxidation sequences, and order of reactions. Here are the most frequent errors students make — and how to fix them.

(i) Nitrobenzene → Benzoic acid

✗ Mistake: Trying to oxidise the nitro group directly

The $-NO_2$ nitrogen can never become the $-COOH$ carbon — benzoic acid's carboxyl group is a seventh carbon that nitrobenzene does not have, and no oxidation of a nitrogen substituent can create it. The carboxyl carbon must be brought in via $CN^-$ (Sandmeyer with CuCN, then hydrolysis).

✓ Correct approach

Reduce $-NO_2$ to $-NH_2$ (Sn/HCl), diazotise, replace with $-CN$ (Sandmeyer, CuCN), then hydrolyse $-CN$ to $-COOH$ .

Key idea: the functional group must be exchanged via diazonium chemistry, not oxidised.

(ii) Benzene → m-bromophenol

✗ Mistake: Brominating first, then hydroxylating

Bromobenzene's $-Br$ is ortho/para-directing, so subsequent chemistry lands para, giving p-bromophenol.

✓ Correct approach

Install the meta director first: nitrate benzene, brominate nitrobenzene (Br enters meta to $-NO_2$ ), reduce $-NO_2$ to $-NH_2$ , then diazotise and warm with water to convert $-N_2^+$ to $-OH$ .

Key idea: for a meta relationship, a meta-directing group must be on the ring before the second substituent arrives.

(iii) Benzoic acid → Aniline

✗ Mistake: Trying to reduce $-COOH$ directly to $-NH_2$

No such one-step reduction exists.

✓ Correct approach

Convert $-COOH$ to $-CONH_2$ (via $SOCl_2$ then $NH_3$ ), then Hofmann bromamide degradation ( $Br_2$ /NaOH) gives aniline — the carbonyl carbon is lost as carbonate.

(iv) Aniline → 2,4,6-tribromofluorobenzene

✗ Mistake: Protecting the amine before bromination

Acetylating first is exactly backwards for this target: protection deliberately moderates the ring so that bromination stops at the mono (mainly para) stage — that is its entire purpose in the p-bromoaniline synthesis of part (vii). Here we want all three positions substituted, and the free, fully activated $-NH_2$ delivers 2,4,6-tribromoaniline instantly in bromine water.

✗ Mistake: Attempting direct fluorination

Elemental fluorination is violent and unselective; fluorine goes on via the diazonium salt, never directly.

✓ Correct approach

Direct tribromination: aniline + excess $Br_2$ /water → 2,4,6-tribromoaniline (no protection step).
Diazotise ( $NaNO_2$ /HCl, 273–278 K), then Balz–Schiemann: form the diazonium fluoroborate with $HBF_4$ and heat → 2,4,6-tribromofluorobenzene.

Key idea: protect the amine only when you want controlled mono-substitution. For exhaustive substitution, leave it free.

(v) Benzyl chloride → 2-phenylethanamine

✗ Mistake: Using $NH_3$ directly on benzyl chloride

That gives benzylamine ( $C_6H_5CH_2NH_2$ ) — one carbon short of the target.

✓ Correct approach

$KCN$ (chain extension) → benzyl cyanide, then reduce ( $LiAlH_4$ or $H_2$ /Ni) → 2-phenylethanamine.

Key idea: to add a carbon, go through cyanide then reduce.

(vi) Chlorobenzene → p-chloroaniline

✗ Mistake: Trying to aminate the C–Cl bond directly with ammonia

The aryl carbon–chlorine bond does not undergo nucleophilic substitution with ammonia under ordinary conditions — and even if it did, that would put the nitrogen at the wrong position (where the Cl was) and lose the chlorine the target still needs.

✗ Mistake: Abandoning the given starting material

Routes that begin from aniline instead of chlorobenzene do not answer the question — the conversion asked for starts from chlorobenzene.

✓ Correct approach

Nitrate chlorobenzene ( $-Cl$ is ortho/para-directing; para is the major product), separate the para isomer, then reduce $-NO_2$ to $-NH_2$ with Sn/HCl. The ortho/para mixture is not a reason to reject this route — separating the crystalline para isomer is standard practice.

(vii) Aniline → p-bromoaniline

✗ Mistake: Brominating aniline directly

The free $-NH_2$ is so activating that even one equivalent of $Br_2$ gives 2,4,6-tribromoaniline instantly.

✓ Correct approach

Acetylate first (acetanilide), brominate (para major), then hydrolyse the amide → p-bromoaniline.

Key idea: this is the mirror image of part (iv) — protection is for mono-substitution, free amine is for exhaustive substitution.

(viii) Benzamide → Toluene

✗ Mistake: Reducing benzamide to benzylamine and then diazotising …

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Q.Accomplish the following conversions:

(i) Nitrobenzene to benzoic acid

(ii) Benzene to m-bromophenol

(iii) Benzoic acid to aniline

(iv) Aniline to 2,4,6-tribromofluorobenzene

(v) Benzyl chloride to 2-phenylethanamine

(vi) Chlorobenzene to p-chloroaniline

(vii) Aniline to p-bromoaniline

(viii) Benzamide to toluene …

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