Question:
A compound on reaction with chloroform in a strong alkaline medium produces isocyanoethane (ethyl isocyanide). Identify the compound.
A compound on reaction with chloroform in a strong alkaline medium produces isocyanoethane (ethyl isocyanide). Identify the compound.
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The carbylamine test converts a primary amine, R-NH2, into an isocyanide, R-NC, using chloroform and alcoholic KOH.
Updated On: Jul 3, 2026
- Amino methane
- Amino ethane
- Nitromethane
- Nitroethane
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The Correct Option is B
Solution and Explanation
Step 1: This reaction is the carbylamine test, unique to primary amines. A primary amine, alcoholic \(KOH\), and chloroform together give an isocyanide (carbylamine) along with potassium chloride and water.
Step 2: The general equation is \[R-NH_2 + CHCl_3 + 3KOH \rightarrow R-NC + 3KCl + 3H_2O\]
Step 3: The product named is isocyanoethane, \(C_2H_5-NC\), so \(R\) must be \(C_2H_5\), the ethyl group.
Step 4: The starting amine carrying this ethyl group is \(C_2H_5-NH_2\), ethylamine, also called aminoethane.
Step 5: Nitro compounds do not undergo this test since they have no free \(-NH_2\) group.
\[\boxed{\text{Aminoethane (ethylamine)}}\]Was this answer helpful?
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