Amines Mock Test – Class 12 Chemistry
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Amines Mock Test – Class 12 Chemistry

Progressive Test — Guest First Round

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Amines – Progressive Test

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1. Examine the predicted carbylamine-test results below.

Row Amine Predicted result
P Formation of
Q Formation of
R Formation of
S Negative test

The valid rows are:

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2. Use the following passage.
A student protects aniline by acetylation and then brominates the protected compound. Analysis shows a major product P and a smaller amount of product Q. Both products contain one bromine atom. Hydrolysis of P gives -bromoaniline.
Which conclusion is most consistent with the experiment?

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3. Four operations are described below.
P: Prepare a substituted aromatic amine with the required ring-substitution pattern.
Q: Diazotise the amino group under cold acidic conditions.
R: Treat the diazonium species with nitrite under suitable copper-assisted conditions.
S: Obtain the corresponding nitroarene.
Which sequence is correct?

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4. The major stages in Gabriel phthalimide synthesis proceed in the order:

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5. Assertion: tert-Butylamine is a tertiary amine because the carbon atom bonded to is tertiary.
Reason: The degree of an amine is determined by the number of carbon groups directly bonded to nitrogen.

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6. Aniline, , is diazotised with conversion. Of the diazonium salt formed, is converted into the tetrafluoroborate salt. On heating, of that salt produces fluorobenzene, which is isolated in yield. Calculate the isolated fluorobenzene mass and the volume of nitrogen formed during thermal decomposition at STP. Use , , and at STP.

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7. The difference between a secondary amine and a diamine is that:

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8. The pair that represents constitutional isomers differing only in whether methyl is attached to nitrogen or to the aromatic ring is:

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9. Examine the formulas below.

Row Formula Classification
P Benzenediazonium chloride
Q Benzenediazonium tetrafluoroborate
R Benzenediazonium phenyl salt
S Benzenediazonium oxide

The valid rows are:

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10. Which comparison of carbon-count changes is correct?

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11. Assertion: -Methylaniline can be less basic than -methylaniline even though the methyl group is electron releasing in both compounds.
Reason: The ortho methyl group can hinder proton approach and hydration of the protonated amine.

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12. Compare the diazonium conversions below.

Row Diazonium replacement Carbon-count consequence
P replaced by One carbon added
Q replaced by Carbon count unchanged
R replaced by One carbon removed
S replaced by Aromatic ring destroyed

The valid rows are:

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13. A graph plots moles of phenol formed against moles of benzenediazonium chloride warmed in excess water. The hydrolysis is quantitative. The graph shows:

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14. Assertion: Diethylamine is a simple amine even though it is secondary.
Reason: The terms simple and secondary refer to different structural features.

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15. An aromatic solid is described below.
A solid aromatic compound contains both an ammonium group and a sulphonate group in the same molecule. It has a high melting point and behaves more like an ionic solid than a typical neutral aromatic compound.
Which explanation best accounts for these properties?

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16. Three compounds are listed below.
Compound P:
Compound Q:
Compound R:
The pair in which the nitrogen lone pair can interact directly with the benzene -system is:

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17. Read the reaction account below.
A chemist nitrates aniline directly using concentrated nitric acid and concentrated sulphuric acid. The product mixture contains ortho-, meta-, and para-nitroaniline. Another chemist first acetylates aniline and then performs nitration under controlled conditions.
Which prediction is most appropriate?

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18. Ethylamine gives ethyl isocyanide rather than propanenitrile in the carbylamine reaction because:

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19. When trimethylamine forms a bond with a methyl group to produce , the nitrogen centre changes from:

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20. Azo coupling usually favours the para position of an activated aromatic ring, when that position is free, because:

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21. A mixture contains diethylamine and triethylamine. Treatment with exactly of benzenesulphonyl chloride converts all the diethylamine into of insoluble sulphonamide. The separated triethylamine is then completely neutralised with hydrochloric acid. Using molar masses and , the original triethylamine amount and acid volume required are:

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22. A storage plan for a diazonium salt is described below.
A student prepares a cold solution of benzenediazonium chloride and divides it into two portions. Portion P is kept in an ice bath. Portion Q is evaporated toward dryness and then warmed. The student plans to store both portions for later use.
Which safety assessment is most appropriate?

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23. Assertion: Gabriel synthesis does not directly prepare a secondary amine.
Reason: Cleavage of an -alkylphthalimide releases a compound of the form .

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24. Nitrobenzene is converted into aniline in a molar ratio. What mass of aniline is obtained from of nitrobenzene at yield? Use molar masses for nitrobenzene and for aniline.

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25. A coupling mixture contains (0.120 mathrm{mol}) of an arenediazonium salt and (0.100 mathrm{mol}) of phenol. Under buffered alkaline conditions, (90.0%) of the phenolic component is present as phenoxide at equilibrium. Because phenol and phenoxide rapidly interconvert, (80.0%) of the total phenolic component undergoes coupling, and the azo product is isolated in (75.0%) yield. If (M(mathrm{azo product})=198 mathrm{g,mol^{-1}}), calculate the isolated product mass and the amount of chemically unreacted diazonium salt.

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26. Complete conversion of of an unknown monofunctional amine gives its hydrochloride. The isolated hydrochloride weighs , and the isolation yield is . Assuming one mole of amine binds one mole of hydrochloric acid, the amine is:

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27. A graph shows water solubility on the vertical axis and carbon-chain length on the horizontal axis for a homologous series of straight-chain primary amines. The most reasonable overall trend is:

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28. Replacement of the diazonium group by a nitro group can be achieved at recognition level by treating a suitable arenediazonium species with:

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29. The systematic name and retained name of , respectively, are:

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30. The same resonance effect that lowers the basicity of aniline also causes the amino group to:

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31. The mixture contains methylamine and dimethylamine. Complete neutralisation requires of hydrochloric acid. When a separate identical mixture is treated with excess nitrous acid, of nitrogen is obtained at STP. Assuming at STP, the amounts of methylamine and dimethylamine and the mass percentage of methylamine are:

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32. Base P has , and base Q has . The ratio is:

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33. A compound is named -dimethylaniline. This notation indicates:

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34. Match each pair in Column I with the type of constitutional isomerism in Column II.

Column I Column II
P. Pentan-1-amine and -methylbutan-1-amine 1. Position isomerism
Q. Pentan-1-amine and pentan-2-amine 2. Class isomerism
R. Methylbutylamine and ethylpropylamine 3. Chain isomerism
S. Pentan-1-amine and ethylpropylamine 4. Metamerism

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35. A mixture of methylamine and ethylamine has a total mass of and requires exactly of hydrochloric acid for complete neutralisation. The amounts of methylamine and ethylamine are, respectively:

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36. Consider the following statements about diazonium-salt stability.
Statement I: Arenediazonium salts are relatively stable in cold aqueous solution.
Statement II: Aliphatic diazonium ions are generally less stable than aromatic diazonium ions.
Statement III: Dry diazonium salts should be handled cautiously because some may decompose explosively.
Statement IV: Heating always increases the storage stability of diazonium salts.
Select the applicable combination.

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37. Consider the following statements about common names of amines.
Statement I: Identical alkyl groups are indicated using prefixes such as “di” or “tri.”
Statement II: Different alkyl groups may be cited alphabetically before the word “amine.”
Statement III: Common nomenclature always requires -locants.
Select the applicable combination.

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38. Examine the predictions below.

Row Substrate and condition Prediction
P Aniline, alkyl chloride, Clean ordinary Friedel–Crafts alkylation
Q Aniline, acyl chloride, Clean ordinary Friedel–Crafts acylation
R Aniline and No interaction because nitrogen has no lone pair
S Aniline and Formation of a salt-like Lewis acid–base complex

The valid row is:

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39. An unknown amine has molecular formula . It gives a positive carbylamine test. With cold nitrous acid it evolves nitrogen and produces an alcohol that is oxidised to propanone. The unknown amine is:

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40. Equal-concentration solutions of weak amines P and Q have values and , respectively, at . Assuming the weak-base approximation is valid, the ratio is:

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41. Assertion: The molecular formula can represent primary, secondary, and tertiary amines.
Reason: Three carbon atoms satisfy the minimum carbon requirements for all three amine degrees.

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42. An aqueous extract contains of ethylammonium chloride and of excess hydrochloric acid. It is treated with of sodium hydroxide. Given and , the final is approximately:

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43. Equal initial amounts of benzenediazonium ion and an aliphatic diazonium ion are maintained at the same low temperature. A graph plots the fraction of each ion remaining against time. Which description is most reasonable?

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44. Assertion: Many azo compounds are strongly coloured.
Reason: Their aromatic rings and azo linkage form an extended conjugated system capable of absorbing visible light.

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45. A name written as “methylethylamine” is being converted to the preferred common-name order. The suitable repair is:

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46. Review these statements about simple and mixed amines.
Statement I: Diethylamine is a simple secondary amine.
Statement II: Methylethylamine is a mixed secondary amine.
Statement III: Ethyldimethylamine is a simple tertiary amine.
Select the applicable combination.

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47. A graph shows against moles of strong acid added during titration of a weak amine solution. Which feature is expected at the equivalence point?

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48. Free ethylamine is liberated from ethylammonium bromide when the salt is treated with:

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49. Hydrolysis of the isocyanate intermediate in Hoffmann degradation ultimately gives:

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50. Amide reduction and nitrile reduction are similar in that:

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