Q.How will you convert

Concept: Electrophilic aromatic substitution (nitration + reduction) and chain-extension via the cyanide route.

Reasoning:

1. Benzene -> Aniline: Benzene is nitrated using conc. $\text{HNO}_3$/conc. $\text{H}_2\text{SO}_4$ to give nitrobenzene. This is then reduced (e.g., Sn/HCl or catalytic hydrogenation) to aniline.

2. Benzene -> N,N-Dimethylaniline: First prepare aniline as above. Then treat aniline with excess methyl iodide ($\text{CH}_3\text{I}$) in the presence of a base (like $\text{NaHCO}_3$) - this is an $\text{S}_\text{N}2$ reaction on the nucleophilic nitrogen. …

Let's break each conversion down by understanding the chemistry behind it.

(i) Benzene into aniline

Concept: Aniline is aminobenzene ($C_6H_5NH_2$). Benzene has no leaving group and is electron-rich, so we cannot directly substitute a hydrogen with $NH_2$ using simple nucleophilic substitution. Instead, we use electrophilic aromatic substitution to first introduce a nitro group ($-NO_2$), which is then reduced in place to an amino group.

1. Nitration of benzene Treat benzene with a nitrating mixture of concentrated $HNO_3$ and concentrated $H_2SO_4$. The electrophile is the nitronium ion ($NO_2^+$).

This gives nitrobenzene.

2. Reduction of nitrobenzene to aniline Reduce the nitro group to an amino group. Common reducing agents:
   • Sn / conc. HCl (then neutralize with NaOH)
   • Fe / conc. HCl
   • Catalytic hydrogenation ($H_2$ / Pd-C)

(ii) Benzene into N,N-dimethylaniline

Concept: N,N-dimethylaniline is $C_6H_5N(CH_3)_2$. We first make aniline (as above), then alkylate the amino group twice. Because the amine keeps reacting as long as methyl iodide is available, the amount of $CH_3I$ must be controlled at two moles — a large excess would push the reaction past the tertiary amine to the quaternary ammonium salt.

1. Prepare aniline from benzene (same as part i)

   Nitration → reduction.

2. Methylation of aniline

   Treat aniline with two moles (a controlled amount) of methyl iodide in the presence of a base (like $NaHCO_3$ or $K_2CO_3$) to neutralize the HI formed. The reaction proceeds via nucleophilic substitution at the nitrogen, one methyl at a time:

The base removes HI, keeping the amine free (unprotonated) so it can react; limiting the methyl iodide to two moles stops the sequence at the tertiary amine.

(iii) $Cl-(CH_2)_4-Cl$ into hexan-1,6-diamine …

Here is the clear, concept-first solution for each conversion.

(i) Benzene → Aniline

Method: Electrophilic Aromatic Substitution (Nitration) followed by Reduction

Benzene cannot be aminated directly — its electron-rich π-cloud repels nucleophiles, and there is no leaving group on the ring. So we install a nitro group by electrophilic substitution and then reduce it in place to $-NH_2$. (Note: the nitro group is never "substituted out" — it is chemically reduced to the amino group.)

Steps:

1. Nitration: Treat benzene with a nitrating mixture (conc. $HNO_3$ + conc. $H_2SO_4$) at 50–60°C.

   • Why? This introduces a nitro group ($-NO_2$) onto the ring, forming nitrobenzene.
   • Reaction: $C_6H_6 + HNO_3 \xrightarrow{H_2SO_4} C_6H_5NO_2 + H_2O$

2. Reduction: Reduce the nitro group to an amino group ($-NH_2$).

   • Reagent: Sn / conc. HCl (or Fe / HCl), followed by neutralization with NaOH.
   • Why? The nitro group is reduced to an amine.
   • Reaction: $C_6H_5NO_2 + 6[H] \xrightarrow{Sn/HCl} C_6H_5NH_2 + 2H_2O$

Final Product: Aniline ($C_6H_5NH_2$)

(ii) Benzene → N,N-Dimethylaniline

Method: Electrophilic Aromatic Substitution (Nitration) → Reduction → Nucleophilic Substitution (Alkylation at N)

We first make aniline (as above), then alkylate the amino group.

Steps:

1. Nitration & Reduction (same as part i) to get aniline.

2. Controlled Methylation (Nucleophilic Substitution at N):

   • Treat aniline with two moles of methyl iodide ($CH_3I$) in the presence of a base (like $NaHCO_3$ or $K_2CO_3$).
   • Why? The lone pair on nitrogen attacks the electrophilic carbon of $CH_3I$ (an $S_N2$ reaction); the base mops up the HI so the amine stays free to react. Limiting the reagent to two moles stops the sequence at the tertiary amine (a large excess would carry on to the quaternary ammonium salt).
   • Reaction:
     • $C_6H_5NH_2 + CH_3I \rightarrow C_6H_5NHCH_3 + HI$
     • $C_6H_5NHCH_3 + CH_3I \rightarrow C_6H_5N(CH_3)_2 + HI$

Final Product: N,N-Dimethylaniline ($C_6H_5N(CH_3)_2$)

(iii) $Cl-(CH_2)_4-Cl$ → Hexan-1,6-diamine

Method: Nucleophilic Substitution (Bimolecular, $S_N2$) followed by Reduction

This is a classic chain-extension via cyanide substitution, then reduction. Count carbons first: the dichloride has 4, the target diamine has 6 — the two cyanide substitutions supply exactly the two extra carbons.

Steps:

1. Nucleophilic Substitution with Cyanide: …

🧪 Core Concept: Getting Nitrogen Onto Aromatic and Aliphatic Skeletons

• Benzene is electron-rich — it does not undergo nucleophilic substitution at all. To put an $-NH_2$ on the ring, first introduce a nitro group via electrophilic substitution, then reduce it in place ($-NO_2 \rightarrow -NH_2$). The nitro group is never "substituted out" — it is chemically reduced.
• Aliphatic dihalides undergo $S_N2$ at both ends — but before choosing a nucleophile, count carbons: if the target is longer than the starting material, the nucleophile itself must supply the extra carbons (cyanide does; ammonia does not).

✗ Common Mistake #1: Trying to directly aminate benzene

Wrong approach:

Benzene + NH₃ → Aniline ✗

Benzene does not react with nucleophiles like NH₃.

Why it fails:

Benzene's π-cloud repels nucleophiles. No leaving group, no reaction.

✓ Correct method (two steps):

1. Nitration

   $\text{Benzene} \xrightarrow{\text{HNO}_3/\text{H}_2\text{SO}_4} \text{Nitrobenzene}$

2. Reduction

   $\text{Nitrobenzene} \xrightarrow{\text{Sn/HCl or H}_2/\text{Pd}} \text{Aniline}$

Exam tip: Always write both steps with reagents. Partial credit is given for each correct step.

✗ Common Mistake #2: Uncontrolled methylation for N,N-dimethylaniline

Wrong approach:

Aniline + CH₃I (1 eq) → N,N-dimethylaniline ✗ (stops at N-methylaniline)

Aniline + CH₃I (large excess) → N,N-dimethylaniline ✗ (overshoots to the quaternary salt)

Why it fails:

• Aniline's nitrogen keeps reacting as long as methyl iodide is available: $1^\circ \rightarrow 2^\circ \rightarrow 3^\circ \rightarrow$ quaternary salt.
• One equivalent gives only mono-methylation; an uncontrolled excess pushes past the target to $C_6H_5N(CH_3)_3^+I^-$.

✓ Correct method:

1. First make aniline (as above).

2. Then methylate with a controlled two moles of $CH_3I$, with a mild base ($NaHCO_3$/$K_2CO_3$) to mop up the HI:

   $\text{Aniline} \xrightarrow{2CH_3I,\ \text{base}} \text{N,N-dimethylaniline}$

   Or better: use HCHO + HCOOH (Eschweiler–Clarke reaction) — clean dimethylation, no quaternization.

Exam tip: Write "2 moles of CH₃I, base" (or "HCHO/HCOOH") explicitly — the stoichiometry is what earns the mark.

✗ Common Mistake #3: Forgetting the carbon count in (iii)

Wrong approach:

Cl–(CH₂)₄–Cl + excess NH₃ → "hexan-1,6-diamine" ✗

Why it fails:

Direct ammonolysis replaces each Cl with $NH_2$ on the same skeleton: the product is $H_2N-(CH_2)_4-NH_2$ — butane-1,4-diamine, a 4-carbon compound. The target, hexan-1,6-diamine, has 6 carbons. No amount of ammonia, however large, can add the two missing carbons. (Direct ammonolysis of this dihalide also tends to cyclise — the mono-substituted intermediate $Cl-(CH_2)_4-NH_2$ closes rapidly to pyrrolidine — a second reason this route fails.)

✓ Correct method (chain extension, then reduction): …