Question:
Which from following reaction results in azo coupling?
Which from following reaction results in azo coupling?
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Azo coupling usually means:
\[
\text{diazonium salt} + \text{phenol / aniline} \rightarrow \text{azo dye}
\]
Updated On: May 14, 2026
- \(\text{C}_6\text{H}_5\text{NH}_2 \xrightarrow{\text{HNO}_2} \dots\)
- \(\text{C}_6\text{H}_5\text{N}_2\text{Cl} + \text{C}_6\text{H}_5\text{OH} \xrightarrow{\text{OH}^-} \dots\)
- \(\text{C}_6\text{H}_5\text{N}_2\text{Cl} \xrightarrow{\text{HBr}_4} \dots\)
- \(\text{C}_6\text{H}_5\text{N}_2\text{Cl} \xrightarrow{\frac{\text{Cu (Powder)}}{\text{HCl}}} \dots\)
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The Correct Option is B
Solution and Explanation
Concept:
Azo coupling reaction involves coupling of an aromatic diazonium salt with an activated aromatic ring such as phenol or aniline. This reaction gives azo compounds containing the \(-\text{N}=\text{N}-\) linkage. ip
Step 1: Recall the requirement for azo coupling.
Azo coupling occurs when:
• an aryl diazonium salt is present
• and it reacts with phenol or aniline in suitable medium e} ip
Step 2: Check the given options.
Option (B) contains benzenediazonium chloride reacting with phenol in alkaline medium. This is the standard azo coupling reaction. ip Hence, the correct answer is:
\[ \boxed{(B)\ \text{C}_6\text{H}_5\text{N}_2\text{Cl} + \text{C}_6\text{H}_5\text{OH} \xrightarrow{\text{OH}^-} \dots} \]
Azo coupling reaction involves coupling of an aromatic diazonium salt with an activated aromatic ring such as phenol or aniline. This reaction gives azo compounds containing the \(-\text{N}=\text{N}-\) linkage. ip
Step 1: Recall the requirement for azo coupling.
Azo coupling occurs when:
• an aryl diazonium salt is present
• and it reacts with phenol or aniline in suitable medium e} ip
Step 2: Check the given options.
Option (B) contains benzenediazonium chloride reacting with phenol in alkaline medium. This is the standard azo coupling reaction. ip Hence, the correct answer is:
\[ \boxed{(B)\ \text{C}_6\text{H}_5\text{N}_2\text{Cl} + \text{C}_6\text{H}_5\text{OH} \xrightarrow{\text{OH}^-} \dots} \]
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