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

The key idea is that chlorobenzene is less reactive than alkyl halides toward nucleophilic substitution due to resonance stabilisation of the C–Cl bond. Under harsh conditions, it undergoes nucleophilic aromatic substitution via a benzyne intermediate.

Step 1: Chlorobenzene is heated with aqueous sodium hydroxide at high temperature (about 300 °C) and high pressure.

Step 2: The hydroxide ion attacks the aromatic ring, forming a benzyne intermediate (via elimination of HCl).

Step 3: Water adds across the benzyne triple bond, yielding phenol after tautomerisation.

The balanced reaction is: …

The reaction you are asking about — the conversion of chlorobenzene to phenol — is a classic example of a nucleophilic aromatic substitution that proceeds through a benzyne intermediate. This is not the usual electrophilic substitution you might expect for an aromatic ring. Instead, the chlorine atom is displaced by a hydroxide ion, but the mechanism is unusual because the aromatic ring is normally resistant to direct nucleophilic attack.

Why does this happen? Chlorobenzene has a carbon–chlorine bond that is slightly polarised, but the lone pairs on chlorine can donate into the ring, making the C–Cl bond stronger than in an alkyl halide. A direct $\mathrm{S_N2}$ attack is impossible because the ring’s $\pi$ system blocks the backside approach. An $\mathrm{S_N1}$ pathway would require forming a phenyl cation, which is highly unstable. So, nature takes a different route: under extreme conditions (high temperature and pressure), a strong base abstracts a proton from the position ortho to chlorine, leading to elimination of HCl and formation of a highly reactive benzyne intermediate. The hydroxide ion then attacks this triple bond, and after tautomerisation and acidification, phenol is obtained.

Let’s walk through the reaction step by step.

1. Reaction conditions

   Chlorobenzene is heated with aqueous sodium hydroxide ($\ce{NaOH}$) at a temperature of about $300\,^\circ\mathrm{C}$ and a pressure of around $200\ \mathrm{atm}$. These harsh conditions are necessary to force the reaction to occur.

2. Formation of the benzyne intermediate

   The hydroxide ion acts as a strong base. It abstracts a proton from the carbon atom ortho to the chlorine. The electrons from the broken C–H bond move to form a second $\pi$ bond between the ortho carbon and the carbon bearing the chlorine. Simultaneously, the C–Cl bond breaks, and the chlorine leaves as $\ce{Cl-}$. This creates a benzyne — a benzene ring with an extra triple bond between two adjacent carbons.

   The reaction can be written as:

3. Nucleophilic attack on benzyne

   The benzyne intermediate is extremely reactive because the triple bond is strained (the bond angles are far from the ideal $180^\circ$ for a triple bond). The hydroxide ion attacks one of the sp-hybridised carbons of the triple bond, adding a $\ce{OH}$ group and generating a negative charge on the adjacent carbon. This forms a phenoxide ion.

   $\ce{\text{benzyne} + OH- -> C6H5O-}$ …

Method: Nucleophilic Aromatic Substitution (Addition-Elimination via Benzyne Mechanism)

This is not an electrophilic aromatic substitution — it is a nucleophilic aromatic substitution that proceeds through a benzyne intermediate. This is a key exception to the usual EAS reactions of benzene.

Steps of the Reaction

1. Formation of the benzyne intermediate

   Chlorobenzene is treated with a strong base (like molten NaOH at high temperature, ~623 K). The hydroxide ion abstracts a hydrogen ortho to the chlorine, and the chlorine leaves as chloride ion. This creates a highly strained benzyne (aryne) intermediate — a benzene ring with a triple bond.

2. Nucleophilic attack on benzyne

   The hydroxide ion (from NaOH) attacks one of the sp-hybridized carbons of the benzyne triple bond. The triple bond is unsymmetrical, so attack occurs at the less hindered carbon (usually the one para to the original chlorine position, but both ortho and para products can form).

3. Protonation to form phenol

   The resulting carbanion (phenoxide ion) is protonated by water (or acid work-up) to give phenol.

Chemical Reaction

Then, acidification: …

Here are the common mistakes students make when writing the reaction for the preparation of phenol from chlorobenzene, along with how to avoid each.

1. Forgetting the Harsh Reaction Conditions

• The Mistake: Students often write only the reactants and products, omitting the high pressure (300 atm) and high temperature (350–370°C). They also forget to mention the aqueous NaOH (not just solid NaOH).
• Why it matters: This is a classic example of nucleophilic aromatic substitution (SNAr) , which requires extreme conditions because the benzene ring is electron-rich and resists nucleophilic attack. Without these conditions, the reaction simply won’t happen.
• How to avoid: Always write the full reaction as:

Then, acidify to get phenol:

2. Writing the Wrong Mechanism (Saying it’s Electrophilic Substitution)

• The Mistake: Because the chapter is titled “Electrophilic Aromatic Substitution,” students mistakenly write an electrophilic mechanism (e.g., $\ce{Cl+}$ attacking the ring). In reality, this is a nucleophilic substitution.
• Why it matters: The chlorine atom is attached to an sp² carbon, making it less reactive toward nucleophiles. The reaction proceeds via a benzyne intermediate (elimination-addition), not via a carbocation.
• How to avoid: Memorise the benzyne mechanism:
  1. Strong base ($\ce{OH-}$) abstracts a hydrogen ortho to the chlorine.
  2. A triple bond forms (benzyne intermediate).
  3. $\ce{OH-}$ attacks the benzyne, giving phenoxide ion.
  4. Acidification yields phenol.

3. Forgetting the Acidification Step

• The Mistake: Students stop at $\ce{C6H5-ONa}$ (sodium phenoxide) and write it as the final product.
• Why it matters: The question explicitly asks for phenol ($\ce{C6H5-OH}$), not its salt. In the exam, writing only the salt will lose marks.
• How to avoid: Always add the second step:

(Use dilute $\ce{HCl}$ or $\ce{CO2}$ + water.)

4. Using the Wrong Base or Reagent

• The Mistake: Writing $\ce{NaOH}$ (solid) or $\ce{KOH}$ (alcoholic) instead of aqueous NaOH.
• Why it matters: Aqueous medium provides the $\ce{OH-}$ ions needed for nucleophilic attack. Alcoholic KOH would lead to elimination (forming benzene) instead of substitution.
• How to avoid: Write “aq. NaOH” or “NaOH (aq)” clearly.

5. Ignoring the By-Products

• The Mistake: Writing only $\ce{C6H5-Cl + NaOH -> C6H5-OH + NaCl}$. …