Identify the products [A] and [B], respectively in the following reaction: Chlorobenzene (i) NaOH, 623 K, 300 atm; (ii)…

Step 1: Chlorobenzene + NaOH (623 K, 300 atm) → sodium phenoxide, then H+ → [A]

This is the Dow process / alkaline fusion for industrial phenol manufacture: $\ce{C6H5Cl + 2NaOH ->[ 623K, 300 atm ] C6H5ONa + NaCl + H2O}$

Acidification with $\ce{H+}$: $\ce{C6H5ONa + HCl -> C6H5OH + NaCl}$

[A] = Phenol ($\ce{C6H5OH}$)

Step 2: Phenol + Na2Cr2O7/H2SO4 → [B]

$\ce{Na2Cr2O7/H2SO4}$ is a strong oxidizing agent. Oxidation of phenol goes to benzoquinone (p-benzoquinone = cyclohexa-2,5-diene-1,4-dione):

$\ce{C6H5OH ->[ Na2Cr2O7/H2SO4 ] }$ benzoquinone

The -OH is oxidized and a second $\ce{C=O}$ is introduced at the para position.

[B] = Benzoquinone (1,4-benzoquinone)

Step 3: Common mistakes

• Students may think $\ce{Na2Cr2O7}$ just oxidises -OH to -CHO or -COOH. For phenol (aromatic), the oxidation gives quinone, not an aldehyde.
• Option (b) is wrong — Na2Cr2O7 does not give sodium phenoxide.

Key Points to Remember:

• Chlorobenzene → phenol requires severe conditions (623 K, 300 atm NaOH) — this is SNAr on unactivated ring.
• Phenol oxidized by $\ce{Na2Cr2O7/H2SO4}$ (or $\ce{KMnO4}$) gives benzoquinone.
• Benzoquinone is an important compound used in photographic development.