Question

Beilstein test is used for the detection of which element in an organic compound?
A.N
B.Cl
C.Na
D.P

Answer 1

Hint:Beilstein test accounts for the qualitative analysis of organic halide present in the given sample of organic compound. But the test is not used to detect fluorine or fluorides.

Complete answer:
The Beilstein test is a chemical test for the detection of organic halides in organic compounds. It is a qualitative test which means this test is only used to check whether presence or absence of organic halides, this test does not account for the quantity of organic halides present in the given organic compound sample. It was developed by Friedrich Konrad Beilstein.
To perform this test, materials required are copper wire, Bunsen burner and samples of organic compounds to be tested. Firstly heat up the copper wire with the help of Bunsen burner in the presence of air or oxygen which leads to oxidation of copper wire by oxygen and formation of copper oxide will be there on the surface of copper wire. Since oxidation is an endothermic process, that’s why Bunsen burner is used. Now dip the heated copper wire into the given sample of organic compound. After dipping, again heat up the copper wire which now contains both samples of organic compound and copper oxide. If heating of this shows a green color flame it corresponds to a positive test which means the sample of organic compound contains organic halide but if it does not heat up with green flame then the sample of organic compound does not contain organic halide.
The disadvantage of this test is that it cannot be used for the sample containing polychloroprene because on heating it can produce highly toxic chloro dioxins.

Thus, the correct option is B.
Note:
Sodium fusion test is commonly used to determine qualitatively organic halide. This test converts the organic compound to inorganic salts like sodium halide. Addition of silver nitrate solution causes any halide to precipitate as the respective silver halide. For example: $$\text{NaCl}+\text{AgN}{\text{O}}_3\to \text{AgCl}\left(\text{whiteppt}\right)+\text{NaN}{\text{O}}_3$$ .