Chemistry Multiple Choice Questions on “Qualitative Analysis of Organic Compounds”.
1. Carbon and hydrogen are detected by heating the compound with which of the following?
a) Copper (II)oxide
b) Iron(II)oxide
c) Iron(III)oxide
d) Copper(I)oxide
Answer: a
Clarification: Carbon and hydrogen are detected by heating the compound with copper (II) oxide. In this way, the carbon that is present in the compound will be oxidized to carbon dioxide and hydrogen to water. Carbon dioxide is tested by limewater which develops turbidity, and hydrogen is tested with anhydrous copper sulphate, which turns blue.
2. Which compound gets precipitated in the detection of carbon and hydrogen?
a) Copper
b) Carbon dioxide
c) Calcium carbonate
d) Copper sulphate
Answer: c
Clarification: Calcium carbonate (CaCO3) gets precipitated during the detection of carbon and hydrogen. Firstly carbon will react with copper (II) oxide to form copper and carbon dioxide. Then, hydrogen will react with the same copper (II) oxide to form copper and water. After this, the presence of carbon dioxide is tested by reacting it with calcium hydroxide and the product obtained is calcium carbonate, which gets precipitated and water.
3. Identify the element that cannot be detected by Lassaigne’s test.
a) Nitrogen
b) Fluorine
c) Sulfur
d) Phosphorous
Answer: b
Clarification: Fluorine cannot be detected by lassaigne’s test. Even though lassaigne’s test is used for the detection of halogens, all halogens except fluorine can be detected. This is because, in lassaigne’s test, the sodium extract is treated with silver nitrate. Only in the case of fluorine, the silver fluoride formed is soluble, unlike the others which is insoluble. Thus, the precipitate will not be formed and so, this method cannot be used for fluorine detection.
4. Potassium can replace sodium in lassaigne’s test.
a) True
b) False
Answer: a
Clarification: Potassium, like sodium is electropositive in nature. In lassaigne’s test, the elements present in the compound are converted from their covalent form to their ionic form by fusing the compound with sodium metal. Since, potassium has similar characteristics of electro positivity as sodium and since potassium is highly reactive, it can be used instead of sodium in lassaigne’s test.
5. What is Lassaigne’s test extract called as?
a) Fusion extract
b) Sodium fusion extract
c) Lassaigne extract
d) Sodium extract
Answer: b
Clarification: Lassaigne’s test extract is called as sodium fusion extract. The cyanides, sulphides and halides of sodium will be formed. These will be extracted from the fused mass by boiling it with distilled water. Hence, the name of the extract is sodium fusion extract.
6. In the test for nitrogen, the sodium fusion extract is acidified with which of the following?
a) Dilute sulphuric acid
b) Dilute hydrochloric acid
c) Concentrated hydrochloric acid
d) Concentrated sulphuric acid
Answer: d
Clarification: In the test for nitrogen, sodium cyanide first reacts with iron (III) sulphate and forms sodium hexacyanoferrate (II). On heating with concentrated sulphuric acid, some iron (II) ions are oxidized to iron (III)ions which react with sodium hexacyanoferrate (II) to produce iron (III) hexacyanoferrate (II), which is Prussian blue in color. Moreover, nitrogen atoms are soluble in concentrated sulphuric acid.
7. What is the color of the precipitate obtained in the test for sulphur?
a) White
b) Black
c) Violent
d) Blue
Answer: b
Clarification: In the test for sulphur, the sodium fusion extract is acidified with acetic acid and lead acetate is added to it. Once this reaction takes place, a black precipitate is formed. This black precipitate is lead acetate, indicating the presence of sulphur.
8. In case of both nitrogen and sulphur existence, Prussian blue is still the color of the end product.
a) True
b) False
Answer: b
Clarification: In case, both nitrogen and sulphur are present in an organic compound, sodium thiocyanate is formed. The color formed is blood red and not Prussian blue. This is because, in this case, there are no free cyanide ions.
Na + C + N+ S → NaSCN
Fe3+ + SCN– → [Fe (SCN)] 2+ (blood red)
9. A X color precipitate, which is Y in ammonium hydroxide indicates presence of chlorine. Identify X and Y.
a) X = yellowish, Y = soluble
b) X = yellow, Y = insoluble
c) X = white, Y = insoluble
d) X = white, Y = soluble
Answer: d
Clarification: During the detection of chlorine, when the organic compound reacts with sodium, it forms sodium chloride. This sodium chloride gives the white precipitate of silver nitrate with silver nitrate solution. This white precipitate is also soluble in ammonium hydroxide.
10. Which is the oxidizing agent used in the test for phosphorous?
a) Hydrogen peroxide
b) Potassium nitrate
c) Sodium peroxide
d) Nitric acid
Answer: c
Clarification: The phosphorous present in the organic compound is oxidized to phosphate by the oxidizing agent sodium peroxide. This solution is then boiled with nitric acid and treated with ammonium molybdate. As a result, a yellow precipitate is formed, indicating the presence of phosphorous.