
In Lassaigne’s test, if both $N$ and $S$ are present in the organic compound, they are converted to:
A.$N{a_2}S$ and $NaCN$
B.$NaSCN$
C.$N{a_2}S{O_3}$ and $NaCN$
D.$N{a_2}S$ and $NaCNO$
Answer
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Hint: The presence of nitrogen, sulphur and halogens can be detected with the help of Lassaigne’s test in which sodium metal is heated with the compound to convert the compound into ionic salt which is then boiled with distilled water to extract the desired product from the sodium fusion extract. There are various different tests to detect the presence of nitrogen, sulphur and halogens in a compound.
Complete step by step answer:
The Lassaigne’s test is a test used for detecting the presence of nitrogen, sulphur and halogens in a compound. In this test, sodium metal is heated in a fusion tube with the organic compound in which we have to detect the presence of nitrogen, sulphur and halogens and thus after the setup is done and reaction takes place, the sodium metal converts all the elements into their ionic form. These ionic salts are formed and then extracted from the fused mass through the process of boiling the mass with distilled water and then the desired product is filtered out. This is known as the sodium fusion extract.
Once the extract is filtered out it is treated with various reagents and elements and it undergoes different tests to indicate the presence of nitrogen, sulphur and halogens.
For the above question,
We know that when $N$ is present in the compound the test gives,
$ Na + C + N \to NaCN $
And when $S$ is present in the compound the test gives,
$ 2Na + S \to N{a_2}S $
According to the question, if both $N$ and $S$ are present in the compound, they are converted into,
$ Na + C + N + S \to NaSCN $
Thus, the correct option is (B) $NaSCN$.
Note:
Nitrogen is an important odourless gas which is detected with the appearance of Prussian blue colour after the extract is boiled with $FeS{O_4}$ and concentrated ${H_2}S{O_4}$ and sulphur is indicated by the presence of violet colour on treating the extract with sodium nitroprusside and when the extract is treated with $HN{O_3}$ and $AgN{O_3}$ then halogens are detected in the presence of various indicators.
Complete step by step answer:
The Lassaigne’s test is a test used for detecting the presence of nitrogen, sulphur and halogens in a compound. In this test, sodium metal is heated in a fusion tube with the organic compound in which we have to detect the presence of nitrogen, sulphur and halogens and thus after the setup is done and reaction takes place, the sodium metal converts all the elements into their ionic form. These ionic salts are formed and then extracted from the fused mass through the process of boiling the mass with distilled water and then the desired product is filtered out. This is known as the sodium fusion extract.
Once the extract is filtered out it is treated with various reagents and elements and it undergoes different tests to indicate the presence of nitrogen, sulphur and halogens.
For the above question,
We know that when $N$ is present in the compound the test gives,
$ Na + C + N \to NaCN $
And when $S$ is present in the compound the test gives,
$ 2Na + S \to N{a_2}S $
According to the question, if both $N$ and $S$ are present in the compound, they are converted into,
$ Na + C + N + S \to NaSCN $
Thus, the correct option is (B) $NaSCN$.
Note:
Nitrogen is an important odourless gas which is detected with the appearance of Prussian blue colour after the extract is boiled with $FeS{O_4}$ and concentrated ${H_2}S{O_4}$ and sulphur is indicated by the presence of violet colour on treating the extract with sodium nitroprusside and when the extract is treated with $HN{O_3}$ and $AgN{O_3}$ then halogens are detected in the presence of various indicators.
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