# Detection of Foreign Elements in an Organic Compound

The presence of foreign elements in organic compounds can be detected by Lassaigne’s test. This test is also known as sodium extract test or sodium fusion test. It is the most reliable method for the detection of foreign compounds in organic compounds.

Preparation:

In Lassagines’ test, the given sample of organic compounds is heated with sodium metal in an ignition tube. Thus, heated ignition tube is plunged in an porcelain basin containing distilled water. The content in the basin is crushed, boiled and then filtered. The filtrate is called Lassaigne’s solution which is used for further test for the identification of element.

The Lassigne’s solution may contain the following compounds:

If nitrogen is present:

$$\ce{Na + C + N-> NaCN <=>Na^++ + CN^-}$$

If sulphur is present,

$$\ce{Na + S->Na2S->2Na^+ + S^– }$$

If both nitrogen and sulphur are present,

$$\ce{Na + CN + S->NaCNS<=>Na^+ + CNS^- }$$

If halogen is present,

$$\ce{Na + X->NaX <=>Na^+ + X^-}$$

Detection of nitrogen:

At first the sodium extract is made alkaline by adding NaOH. Then to the alkaline solution of sodium extract, freshly prepared FeSO4 solution is added. After then there is a formation of dirty green ppt. which is boiled from some time and then dissolved by adding dilute HCl. Finally, ferric chloride is added to the solution and the prussian blue color is formed which confirms the presence of nitrogen. The reactions involved are as follows:

Preparation of sodium extract:

$$\ce{Na + C + N-> \underset{\text{Sodium cyanide (Sodium extract)}}{\ce{NaCN}}}$$

Addition of ferrous sulphate and dilute hydrochloric acid:

$$\ce{FeSO4 + 2NaOH->Fe(OH)2 + Na2SO4}$$

$$\ce{Fe(OH)2 + 2NaCN->Fe(CN)2 + 2NaOH}$$

$$\ce{Fe(CN)2 + 4NaCN->\underset{\text{Sodium ferrocyanide}}{\ce{Na4Fe(CN)6}}}$$

$$\ce{2Cu + \underset{\text{Steam}}{\ce{H2O}} ->\underset{\text{Copper Oxide}}{\ce{CuO}} + H2}$$

Confirmation for presence of nitrogen:

$$\ce{3Na4Fe(CN)6 + 4FeCl3->\underset{\text{Ferric ferrocyanide (Prussian blue)}}{\ce{ Fe4[Fe(CN)6]3}} + 12NaCl}$$

Detection of both nitrogen and sulphur:

If nitrogen and sulphur both are present, blood red coloration is formed.

$$\ce{Na + C + N +S->NaCNS}$$

$$\ce{NaCNS + FeCl3->\underset{\text{Blood red}}{\ce{Fe(CNS)3}} + NaCl}$$

Detection of Sulphur

Sulphur is detected in either by lead acetate test or sodium nitroprusside test:

• Lead acetate test: The Lassaigne’s test is first acidified with acetic acid and then lead acetate solution is added. Formation of black precipitate confirms the presence of sulphur.

$$\ce{Na + S->Na2S}$$

$$\ce{Na2S +(CH3COO)2Pb->\underset{\text{Black}}{\ce{PbS ^ }} + CH3COONa}$$

• Sodium Nitroprusside test: To a small portion of sodium extract, sodium nitroprusside solution is added. Formation of violet coloration shows the presence of sodium sulphur in organic compound.

$$\ce{Na2S + \underset{\text{sodium nitroprusside}}{\ce{Fe(CN)5NO}} ->\underset{\text{Sodium sulphonitroprusside}}{\ce{Na4[Fe(CN)5NOS]}}}$$

Detection of halogens:

Few drops of concentrated HNO3 is added to the sodium extract. The solution is then boiled to remove any gas is produced. A few drops of AgNO3 solution is added to the Lassaigne’s test.

Formation of curdy white ppt soluble in ammonia solution confirms the presence of chlorine.

$$\ce{Na + Cl->NaCl}$$

$$\ce{NaCl + AgNO3-> \underset{\text{Curdy white}}{\ce{AgCl v }}}$$

Formation of pale yellow ppt. partially soluble in ammonia solution confirms the presence of bromine.

$$\ce{Na + Br->NaBr}$$

$$\ce{NaBr + AgNO3->\underset{\text{Pale yellow}}{\ce{AgBr v }} + NaNO3}$$

Formation of pale yellow precipitate shows the presence of iodine.

$$\ce{I + Na->NaI}$$

$$\ce{NaI + AgNO3->\underset{\text{Silver iodide}}{\ce{AgI v }} + NaNO3}$$

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