If chloroform is treated with KOH followed by the addition of 1,6-diamino hexane, hexane-1,6-di isocyanide is formed. How many formula units of KCI are formed per molecule of diamine ?Correct answer is '6'. Can you explain this answer?

Question

Answer

Reaction Overview
When chloroform (CHCl3) is treated with potassium hydroxide (KOH), a dehydrohalogenation reaction occurs, resulting in the formation of potassium chloride (KCl) and an intermediate, potassium formate (HCOOK). This intermediate reacts further with 1,6-diaminohexane to yield hexane-1,6-diisocyanide.
Step-by-Step Reaction Mechanism
- Chloroform Treatment:
- CHCl3 + KOH → KCl + HCOOK + H2O
- For each molecule of chloroform, one molecule of KCl is produced.
- Formation of Isocyanide:
- The intermediate formate reacts with 1,6-diaminohexane.
- Each amino group (-NH2) from the diamine can react with the formate to form isocyanide.
Number of KCl Units Produced
- Total KCl Formation:
- The reaction of chloroform with KOH produces 1 mole of KCl.
- However, for each nitrogen atom in the 1,6-diaminohexane, there is a reaction that can lead to the release of KCl.
- Diamine Structure:
- 1,6-diaminohexane has 2 amino groups.
- Hexane-1,6-diisocyanide formation involves multiple iterations of the reaction, leading to the release of additional KCl.
Final Calculation
- Each amino group reacts to produce KCl.
- Considering the complete stoichiometry and the possibility of multiple cycles of reaction, 3 moles of KCl are released for each amino group reacted.
- Therefore, for 2 amino groups in 1,6-diaminohexane, 2 x 3 = 6 units of KCl are formed.
Conclusion
In total, 6 formula units of KCl are produced per molecule of diamine when chloroform is treated with KOH and subsequently reacts with 1,6-diaminohexane.

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