In the following reaction sequence, [C] is
(a) Cl-C6H4-CH2-CH2-C6H4-Cl
(b) CH3-C6H4-C6H4-CH3
(c) Cl-

Q: Why doesn't the $Cl_2/h\nu$ attack the benzene ring directly?

Electrophilic aromatic substitution on the benzene ring requires a Lewis acid catalyst (like $FeCl_3$ or $AlCl_3$) to generate a strong $Cl^+$ electrophile. Sunlight ($h\nu$), however, causes homolytic cleavage of $Cl_2$ to form chlorine radicals, which specifically target the benzylic position due to the high stability of the resulting benzylic radical.

Q: Why does the Wurtz reaction only couple at the $CH_2Cl$ group and not the ring $Cl$?

Aryl chlorides (where Cl is directly attached to the benzene ring) are highly unreactive towards nucleophilic substitution and radical formation compared to alkyl or benzylic chlorides. This is because the lone pairs on the aryl chlorine participate in resonance with the ring, giving the C-Cl bond a partial double bond character, making it very difficult to break.

Q: What exactly is the intermediate formed right before Product A?

The intermediate is a diazonium salt, specifically $p$-toluenediazonium chloride. It is formed when the primary amine reacts with nitrous acid (generated in situ from $NaNO_2$ and $HCl$) at $0-5^\circ C$.

Q: Can we use aqueous ether instead of dry ether in the last step?

Absolutely not! Sodium metal reacts violently and explosively with water to form sodium hydroxide and hydrogen gas. Therefore, the ether solvent must be strictly anhydrous (dry) to ensure the Wurtz reaction proceeds safely and successfully.

Q: What would happen if we used excess $Cl_2$ in step 2?

If excess chlorine gas was used under prolonged sunlight, the free radical substitution would continue until all three hydrogens on the benzylic methyl group were replaced, ultimately forming a $-CCl_3$ (trichloromethyl) group instead of a $-CH_2Cl$ group.

Question

In the following reaction sequence, [C] is

Options:

(A) Cl-C6H4-CH2-CH2-C6H4-Cl

(B) CH3-C6H4-C6H4-CH3

(C) Cl-C6H4-CH2-C6H4-CH2-Cl

(D) CH2(Cl)-C6H4-C6H4-CH2(Cl)

Sigma Percentile · Accepted Answer

### The Starting Line: Diazotization

Welcome to a classic multi-step organic synthesis journey! We begin our adventure with **$p$-toluidine**, an aromatic molecule featuring both an amino group ($-NH_2$) and a methyl group ($-CH_3$) attached to a benzene ring. Our mission is to navigate through three distinct chemical transformations to uncover the final major product, $[C]$.

Our first set of reagents is a chilling combination of sodium nitrite ($NaNO_2$) and hydrochloric acid ($HCl$) maintained strictly between $0-5^\circ C$. This is the unmistakable signature of **diazotization**. Under these ice-cold conditions, the primary aromatic amine is converted into a highly reactive diazonium salt. The $-NH_2$ group transforms into the $-N_2^+Cl^-$ group, setting the stage for the next dramatic shift.

### The Sandmeyer Transformation

Without missing a beat, the diazonium salt is treated with cuprous chloride ($Cu_2Cl_2$) and $HCl$. This is the renowned **Sandmeyer reaction**. The diazonium group is an exceptionally good leaving group because it departs as stable nitrogen gas ($N_2$).

As the nitrogen bubbles away, a chlorine atom seamlessly takes its place on the aromatic ring. This elegant substitution yields our first intermediate, **Product $[A]$**, which is $p$-chlorotoluene.

### Sunlight and Radicals: The Benzylic Attack

Now, the plot thickens. We take $p$-chlorotoluene and expose it to chlorine gas ($Cl_2$) bathed in sunlight ($h
u$). The presence of ultraviolet light is a massive clue: we are no longer in the realm of ionic reactions; we have entered the chaotic world of **free radical substitution**.

But where will the chlorine radical attack? The benzene ring or the methyl group? Because it's a radical mechanism, it seeks the path of maximum stability. Abstracting a hydrogen from the methyl group creates a **benzylic radical**, which is beautifully stabilized by resonance with the adjacent benzene ring. Consequently, the substitution occurs exclusively at the side chain, replacing one hydrogen with a chlorine atom. This gives us **Product $[B]$**, $p$-chlorobenzyl chloride.

### The Grand Finale: Wurtz Coupling

For our final act, we introduce Product $[B]$ to sodium metal ($Na$) in a bath of dry ether. If you've studied your hydrocarbon synthesis, you'll immediately recognize the **Wurtz reaction**. This reaction is famous for taking two alkyl (or benzylic) halides and coupling them together to form a longer carbon chain.

In this scenario, two molecules of $p$-chlorobenzyl chloride approach the sodium metal. The sodium eagerly strips away the benzylic chlorine atoms, leaving behind two benzylic $-CH_2^\bullet$ radicals. These two radicals quickly find each other and couple, forming a strong new carbon-carbon single bond ($-CH_2-CH_2-$).

**Crucial Note:** The chlorine atom attached directly to the benzene ring remains completely untouched during this process because aryl halides are notoriously unreactive towards Wurtz coupling due to their partial double bond character.

The resulting masterpiece is **1,2-bis(4-chlorophenyl)ethane**, which perfectly matches option (a).

Animated Solution for Chemistry - Organic Chemistry: In the following reaction sequence, [C] is

Select Answer:

Visualized Solution

$p -toluidine$

$Diazotization$

$Sandmeyer Reaction$

$Free Radical Halogenation$

$Benzylic Substitution$

$Wurtz Reaction$

$Final Coupling$

$The Way Forward$

The Sigma Insight: Haloalkanes & Haloarenes

The Starting Line

The Sandmeyer Transformation

Sunlight and Radicals

The Grand Finale

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