Detect the presence of nitrogen atom in a given organic compound 'O'
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The carbon and nitrogen present in the organic compound on fusion with sodium metal gives sodium cyanide (NaCN) soluble in water. This is converted in to sodium ferrocyanide by the addition of sufficient quantity of ferrous sulphate. Ferric ions generated during the process react with ferrocyanide to form prussian blue precipitate of ferric ferrocyanide.
Na + C + N → NaCN
6NaCN + FeSO4 → Na4[Fe(CN)6] + Na2SO4
Sodium ferrocyanide
Na4[Fe(CN)6] + Fe3+ → Fe4[Fe(CN)6]3 Ferric ferrocyanide
rishilaugh:
great thanks
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Good Question . Thanks for asking it.
Test for Nitrogen in Organic compound:
Step 1: Sodium metal is fused with organic compound containing Nitrogen and Crabon.it Gives Sodium cyanide which is soluble in water.
Na+ N+ C ----> NaCN.
Step 2:Sodium Cyanide is converted into Sodium ferrocynaide by adding ferrous sulphate.
6NaCN+FeSO4 ----> Na₄ [Fe(CN)₆] + Na₂SO₄
Step 3:
Ferric ions formed during this process react with ferrocynaide to form blue precipitate of Ferric ferrocyanide.
Na₄ [Fe(CN)₆] +Fe³⁺ ----> Fe₄[Fe(CN)₆]₃
Test for Nitrogen in Organic compound:
Step 1: Sodium metal is fused with organic compound containing Nitrogen and Crabon.it Gives Sodium cyanide which is soluble in water.
Na+ N+ C ----> NaCN.
Step 2:Sodium Cyanide is converted into Sodium ferrocynaide by adding ferrous sulphate.
6NaCN+FeSO4 ----> Na₄ [Fe(CN)₆] + Na₂SO₄
Step 3:
Ferric ions formed during this process react with ferrocynaide to form blue precipitate of Ferric ferrocyanide.
Na₄ [Fe(CN)₆] +Fe³⁺ ----> Fe₄[Fe(CN)₆]₃
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