Organic Compounds Containing Nitrogen - MCQs with Explanations

1. Direct nitration of aniline is not a feasible process because

• (a) the reaction cannot be stopped at the mononitration stage
• (b) a mixture of o, m and p-nitroaniline is always obtained
• (c) nitric acid oxidises most of aniline to give oxidation products along with only a small amount of nitrated products
• (d) all of the above.

Correct Answer: (c)

Explanation: Direct nitration of aniline is not feasible because the strong oxidising property of nitric acid causes the oxidation of the sensitive amino group, leading to tarry oxidation products.

2. Aniline when diazotised in cold and then treated with dimethyl aniline gives a coloured product. Its structure would be

• (a) $CH_3NH-\text{C}_6H_4-N=N-\text{C}_6H_4-NHCH_3$
• (b) $CH_3-\text{C}_6H_4-N=N-\text{C}_6H_4-NH_2$
• (c) $(CH_3)_2N-\text{C}_6H_4-N=N-\text{C}_6H_5$
• (d) $(CH_3)_2N-\text{C}_6H_4-NH_2$

Correct Answer: (c)

Explanation: This is a coupling reaction where the benzene diazonium chloride reacts with N,N-dimethylaniline to form an azo dye.

3. An optically inactive amine (A) $C_4H_{11}N$ on treatment with $HNO_2$ gives an alcohol (B). The compound (B) on heating with conc. $H_2SO_4$ at 453 K gives an alkene (C). The (C) on treatment with HBr gives an optical active compound (D) having molecular formula $C_4H_9Br$. Identify (A).

• (a) $CH_3CH_2CH(NH_2)CH_3$
• (b) $CH_3CH_2CH_2CH_2NH_2$
• (c) $CH_3NHCH_2CH_2CH_3$
• (d) $C_2H_5NHC_2H_5$

Correct Answer: (b)

Explanation: n-Butylamine (A) reacts with $HNO_2$ to give n-butanol (B), which dehydrates to 1-butene (C). 1-butene reacts with HBr to form 2-bromobutane (D), which is optically active.

4. Aniline reacts with phosgene to form

• (a) Benzene diazonium chloride
• (b) Chlorobenzene
• (c) Benzamide
• (d) Phenyl isocyanate

Correct Answer: (d)

Explanation: The reaction of aniline with phosgene ($COCl_2$) results in the formation of phenyl isocyanate ($C_6H_5NCO$).

5. In the chemical reaction, Aniline $\xrightarrow[HCl, 278 K]{NaNO_2} A \xrightarrow{HBF_4} B$, the compounds A and B respectively are

• (a) nitrobenzene and chlorobenzene
• (b) nitrobenzene and fluorobenzene
• (c) phenol and benzene
• (d) benzene diazonium chloride and fluorobenzene.

Correct Answer: (d)

Explanation: Diazotisation of aniline gives benzene diazonium chloride (A), which then reacts with $HBF_4$ to eventually yield fluorobenzene (B).

6. The refluxing of $(CH_3)_2NCOCH_3$ with acid gives

• (a) $2CH_3NH_2 + CH_3COOH$
• (b) $2CH_3OH + CH_3CONH_2$
• (c) $(CH_3)_2NH + CH_3COOH$
• (d) $(CH_3)_2NCOOH + CH_4$

Correct Answer: (c)

Explanation: Acidic hydrolysis (refluxing) of N,N-dimethylacetamide yields dimethylamine and acetic acid.

7. The product (D) of the following reaction $CH_3Cl \xrightarrow{KCN} (A) \xrightarrow{H_2O/H^+} (B) \xrightarrow{NH_3} (C) \xrightarrow{\Delta} (D)$ is

• (a) $CH_3CH_2NH_2$
• (b) $CH_3CN$
• (c) $HCONH_2$
• (d) $CH_3CONH_2$

Correct Answer: (d)

Explanation: The sequence is: $CH_3Cl \rightarrow CH_3CN (A) \rightarrow CH_3COOH (B) \rightarrow CH_3COONH_4 (C) \rightarrow CH_3CONH_2$ (D, acetamide).

8. Acetamide is treated with the following reagents separately. Which one of these would yield methyl amine?

• (a) $NaOH-Br_2$
• (b) Sodalime
• (c) Hot conc. $H_2SO_4$
• (d) $PCl_5$

Correct Answer: (a)

Explanation: Hofmann bromamide degradation of acetamide with $Br_2$ and NaOH yields methylamine.

9. Primary amines on heating with $CS_2$ followed by excess of mercuric chloride yields isothiocyanates. The reaction is called

• (a) Hofmann mustard oil reaction
• (b) Perkin reaction
• (c) Fries reaction
• (d) Diels-Alder reaction.

Correct Answer: (a)

Explanation: The formation of alkyl isothiocyanates (with a mustard-like smell) from primary amines is the Hofmann mustard oil reaction.

10. Which of the following reactions will not give a primary amine?

• (a) $CH_3CONH_2 \xrightarrow{Br_2/KOH}$
• (b) $CH_3CN \xrightarrow{LiAlH_4}$
• (c) $CH_3NC \xrightarrow{LiAlH_4}$
• (d) $CH_3CONH_2 \xrightarrow{LiAlH_4}$

Correct Answer: (c)

Explanation: Reduction of an isocyanide ($CH_3NC$) yields a secondary amine (dimethylamine), while the other options yield primary amines.

11. Which of the following reagents will be useful to distinguish between aniline and benzylamine?

• (a) Dilute HCl
• (b) $C_6H_5SO_2Cl$ and $OH^-/H_2O$
• (c) $HONO$ then $\beta$-naphthol
• (d) $AgNO_3$ in $H_2O$

Correct Answer: (c)

Explanation: Aniline (aromatic 1° amine) forms a diazonium salt that gives a coloured azo dye with $\beta$-naphthol, unlike benzylamine.

12. Which of the following is the strongest base?

• (a) $C_6H_5NH_2$
• (b) $p-CH_3-\text{C}_6H_4-NH_2$
• (c) $m-CH_3-\text{C}_6H_4-NH_2$
• (d) $C_6H_5CH_2NH_2$

Correct Answer: (d)

Explanation: In benzylamine, the lone pair on nitrogen is not in resonance with the benzene ring, making it more available for donation compared to aniline derivatives.

13. Identify the compound X in the following reactions: $CH_3NO_2 \xrightarrow[Excess]{Cl_2/NaOH} X \xleftarrow[CHCl_3]{HNO_3}$

• (a) $ClCH_2NO_2$
• (b) $Cl_2CHNO_2$
• (c) $Cl_3CNO_2$
• (d) $CH_3Cl$

Correct Answer: (c)

Explanation: Both reactions yield chloropicrin ($Cl_3CNO_2$), used as a tear gas and insecticide.

14. Which of the following statements is correct?

• (a) $C_2H_5NH_3^+OH^-$ is acidic.
• (b) $C_2H_5NH_2$ is a weaker base than $(C_2H_5)_3N$ in gas phase.
• (c) $C_2H_5NH_2$ is less basic than $NH_3$.
• (d) $C_2H_5NH_2$ forms salts with strong bases.

Correct Answer: (b)

Explanation: In the gas phase, tertiary amines are more basic than primary amines due to the +I effect of three alkyl groups.

15. In the Hoffmann's method for separation of 1º, 2º and 3º amines, the reagent used is

• (a) acetyl chloride
• (b) benzene sulphonyl chloride
• (c) diethyl oxalate
• (d) nitrous acid.

Correct Answer: (c)

Explanation: Hoffmann's separation method uses diethyl oxalate to distinguish amines based on the formation of oxamides or oxamic esters.

16. Aniline in a set of the following reactions yielded a coloured product Y. Aniline $\xrightarrow[273-278 K]{NaNO_2/HCl} X \xrightarrow{N,N\text{-dimethylaniline}} Y$. The structure of 'Y' would be

• (a) $C_6H_5-N=N-\text{C}_6H_4-N(CH_3)_2$
• (b) $C_6H_5-NH-N(CH_3)_2$
• (c) $C_6H_5-N=N-\text{C}_6H_4-NH_2$
• (d) $C_6H_5-N(CH_3)_2$

Correct Answer: (a)

Explanation: Diazotised aniline (X) couples with N,N-dimethylaniline at the para-position to form the azo compound (Y).

17. Which of the following statements about primary amines is false?

• (a) Alkyl amines are stronger bases than aryl amines.
• (b) Alkyl amines react with nitrous acid to produce alcohols.
• (c) Aryl amines react with nitrous acid to produce phenols.
• (d) Alkyl amines are stronger bases than ammonia.

Correct Answer: (c)

Explanation: Aryl primary amines react with nitrous acid at low temperatures to produce stable diazonium salts, not phenols directly.

18. What is the product when N-ethyl formamide is heated with $POCl_3$?

• (a) Ethyl cyanide
• (b) Propane nitrile
• (c) Ethyl carbylamine
• (d) Ethyl isocyanide

Correct Answer: (d)

Explanation: Dehydration of N-ethylformamide with $POCl_3$ in pyridine yields ethyl isocyanide.

19. The correct order of basicities of the following compounds is (I) Acetamidine, (II) Ethylamine, (III) Diethylamine, (IV) Acetamide.

• (a) $II > I > III > IV$
• (b) $I > III > II > IV$
• (c) $III > II > I > IV$
• (d) $I > II > III > IV$

Correct Answer: (b)

Explanation: Acetamidine (I) is most basic due to resonance stabilization of its cation. Among aliphatic amines, diethylamine (III) > ethylamine (II). Acetamide (IV) is least basic.

20. Which of the following compounds is the most basic in aqueous medium?

• (a) Acetamide
• (b) Benzamide
• (c) Acetamidine
• (d) Aniline

Correct Answer: (c)

Explanation: Acetamidine is the most basic because its conjugate acid is stabilized by two equivalent resonance structures.

21. Aniline is treated with bromine water to give compound 'X', which after diazotisation gives 'Y'. 'Y' on treatment with $Cu_2Cl_2$ and HCl gives 'Z'. Compound 'Z' is

• (a) o-bromochlorobenzene
• (b) p-bromochlorobenzene
• (c) 2,4,6-tribromophenol
• (d) 2,4,6-tribromochlorobenzene

Correct Answer: (d)

Explanation: Bromination of aniline gives 2,4,6-tribromoaniline (X). Subsequent diazotisation and Sandmeyer reaction yield 2,4,6-tribromochlorobenzene.

22. Which of the following is most basic in aqueous solution?

• (a) $CH_3NH_2$
• (b) $(CH_3)_2NH$
• (c) $(CH_3)_3N$
• (d) $NH_3$

Correct Answer: (b)

Explanation: For methylamines in aqueous solution, the basicity order is $(CH_3)_2NH > CH_3NH_2 > (CH_3)_3N > NH_3$ due to inductive, solvation, and steric effects.

23. Which amine amongst the following will answer positive the carbylamine test?

• (a) $C_6H_5-NH-CH_3$
• (b) $m-CH_3-\text{C}_6H_4-NH_2$
• (c) $C_6H_5-NH-C_4H_9$
• (d) $C_6H_5-N(C_2H_5)_2$

Correct Answer: (b)

Explanation: Only primary amines (aliphatic or aromatic) like m-toluidine give the carbylamine test.

24. Which 'A' gives red colour in the reactions? $A \xrightarrow{HNO_2} \text{Product} \xrightarrow{NaOH} \text{red colour}$

• (a) $CH_3CH_2NO_2$
• (b) $(CH_3)_2CHNO_2$
• (c) $(CH_3)_3CNO_2$
• (d) $C_6H_5NO_2$

Correct Answer: (a)

Explanation: Primary nitroalkanes react with $HNO_2$ to form nitrolic acids, which dissolve in NaOH to give a red solution.

25. The indicator that is obtained by coupling the diazonium salt of sulphanilic acid with N,N-dimethylaniline is

• (a) phenanthroline
• (b) methyl orange
• (c) methyl red
• (d) phenolphthalein.

Correct Answer: (b)

Explanation: The coupling reaction produces methyl orange, a common pH indicator.

26. Aniline is reacted with bromine water, then treated with $NaNO_2/HCl$, then $HBF_4$, then heated. The final product is

• (a) p-bromofluorobenzene
• (b) p-bromoaniline
• (c) 2,4,6-tribromofluorobenzene
• (d) 1,3,5-tribromobenzene.

Correct Answer: (c)

Explanation: Following the sequence: bromination $\rightarrow$ diazotisation $\rightarrow$ tetrafluoroborate formation $\rightarrow$ decomposition yields 2,4,6-tribromofluorobenzene.

27. 3-nitroaniline is subjected to: (i) $NaNO_2/HCl$, (ii) KI, (iii) Cu powder. The final product will be

• (a) 3,3'-diaminobiphenyl
• (b) 3-iodoaniline
• (c) 3-nitroiodobenzene
• (d) 3,3'-dinitrobiphenyl.

Correct Answer: (d)

Explanation: (i) and (ii) convert 3-nitroaniline to 3-nitroiodobenzene. (iii) is the Ullmann reaction, coupling two molecules to form 3,3'-dinitrobiphenyl.

28. In the diazotization of aryl amines, an excess of hydrochloric acid is used primarily to

• (a) suppress the concentration of free aniline available for coupling.
• (b) suppress hydrolysis of phenol
• (c) ensure a stoichiometric amount of nitrous acid.
• (d) neutralise the base liberated.

Correct Answer: (a)

Explanation: Excess acid ensures that all aniline is converted to salt form, preventing premature coupling between the diazonium salt and free aniline.

29. Which one of the following nitrophenols is the strongest acid?

• (a) 2-nitro-4,6-dimethylphenol
• (b) 4-nitro-2,6-dimethylphenol
• (c) 3-nitrophenol
• (d) 2-nitrophenol

Correct Answer: (a)

Explanation: In ortho-nitrophenols with bulky substituents, steric inhibition of resonance can increase acidity by preventing intramolcular hydrogen bonding.

30. An organic compound (A) on reduction gives (B). (B) with $CHCl_3/KOH$ gives (C). (C) on reduction gives N-methylaniline. (A) is

• (a) methylamine
• (b) nitromethane
• (c) aniline
• (d) nitrobenzene

Correct Answer: (d)

Explanation: Nitrobenzene (A) $\rightarrow$ Aniline (B) $\rightarrow$ Phenyl isocyanide (C) $\rightarrow$ N-methylaniline.

31. Mark the incorrect reaction.

• (a) $C_6H_5N_2Cl \xrightarrow{H_3PO_2} C_6H_6$
• (b) $C_6H_5N_2Cl \xrightarrow{CuCl/HCl} C_6H_5Cl$
• (c) $C_6H_5N_2Cl \xrightarrow{NaOH} C_6H_5OH$
• (d) All are incorrect.

Correct Answer: (b)

Explanation: Reaction (b) is the Sandmeyer reaction, and it is a standard way to produce chlorobenzene from a diazonium salt.

32. The end product of the reaction $C_2H_5NH_2 \xrightarrow{HNO_2} [A] \xrightarrow{PCl_5} [B] \xrightarrow{NH_3} [C]$ is

• (a) Ethyl alcohol
• (b) Ethylamine
• (c) Methylamine
• (d) Acetamide.

Correct Answer: (b)

Explanation: $C_2H_5NH_2 \rightarrow C_2H_5OH (A) \rightarrow C_2H_5Cl (B) \rightarrow C_2H_5NH_2$ (C, ethylamine).

33. Which of the following is true regarding the conversion of acetamide into methylamine?

• (a) It is a Hofmann ammonolysis synthesis.
• (b) It is an industrial preparation of amine.
• (c) Reaction is useful for descent in a series.
• (d) Both (a) and (b).

Correct Answer: (c)

Explanation: The Hofmann degradation reaction involves the descent of the series because the resulting amine has one carbon atom less than the parent amide.

34. Molecular formula $C_3H_9N$ represents three amines (1°, 2°, 3°). Which type of isomerism is exhibited?

• (a) Position
• (b) Chain
• (c) Metamerism
• (d) Functional

Correct Answer: (d)

Explanation: 1°, 2°, and 3° amines with the same molecular formula are functional isomers of each other.

35. Mark the factor that controls the basicity of amines.

• (a) Availability of lone pair
• (b) Resonance effect
• (c) Hyperconjugation
• (d) -I effect

Correct Answer: (a)

Explanation: The basicity of an amine is primarily determined by the availability of the lone pair of electrons on the nitrogen atom.

36. Which of the following describes 'amino' group as a substituent in electrophilic aromatic substitution?

• (a) Weakly activating and o/p-directing
• (b) Strongly activating and o/p-directing
• (c) Weakly deactivating, meta-directing
• (d) Strongly activating, meta-directing

Correct Answer: (b)

Explanation: The amino group is a strongly activating and ortho/para-directing group due to the +R effect of the lone pair.

37. Which of the following will not give coloured dye with benzenediazonium chloride?

• (a) Phenol
• (b) Isopropyl alcohol
• (c) Aniline
• (d) N,N-dimethylaniline

Correct Answer: (b)

Explanation: Only aromatic compounds with activating groups undergo coupling; aliphatic alcohols like isopropyl alcohol do not.

38. The acid which contain -COOH group is

• (a) picric acid
• (b) salicylic acid
• (c) sulphanilic acid
• (d) p-toluenesulphonic acid.

Correct Answer: (b)

Explanation: Salicylic acid is 2-hydroxybenzoic acid and contains the carboxyl (-COOH) group.

39. The correct structure of zwitter ion of sulphanilic acid is

• (a) $H_2N-\text{C}_6H_4-SO_3H$
• (b) $H_3N^+-\text{C}_6H_4-SO_3^-$
• (c) $H_2N-\text{C}_6H_4-SO_3^-$
• (d) $H_3N^+-\text{C}_6H_4-SO_3H$

Correct Answer: (b)

Explanation: In the zwitterionic form, the acidic sulphonic group donates a proton to the basic amino group in the same molecule.

40. A pair of compounds and a test name is given. In which option the test cannot differentiate the pair?

• (a) Benzyl amine and dimethyl amine: Nitrous acid test
• (b) Aniline and ethyl amine: Azo-dye test
• (c) Nitromethane and methylcyanide: Mulliken-Barker test
• (d) Diethyl amine and triethyl amine: Mustard oil test.

Correct Answer: (d)

Explanation: The mustard oil test is specific for 1° amines; it cannot differentiate between 2° and 3° amines as neither reacts.

41. Consider the sequence $X \xrightarrow{KCN} B \xrightarrow{H_2/Ni} CH_3CH_2NH_2 \xrightarrow{HNO_2, 0^\circ C} C_2H_5OH$. Find starting material X.

• (a) $CH_3Br$
• (b) $CH_3CH_2MgBr$
• (c) $CH_3Cl$
• (d) $CH_3CH_2Br$

Correct Answer: (c)

Explanation: The sequence is: $CH_3Cl \xrightarrow{KCN} CH_3CN \xrightarrow{H_2/Ni} CH_3CH_2NH_2 \xrightarrow{HNO_2} CH_3CH_2OH$.

42. Ethyl isocyanide on hydrolysis in acidic medium generates

• (a) ethyl amine salt and methanoic acid
• (b) propanoic acid and ammonium salt
• (c) ethanoic acid and ammonium salt
• (d) methyl amine salt and ethanoic acid.

Correct Answer: (a)

Explanation: Acidic hydrolysis of ethyl isocyanide yields ethylamine and formic acid.

43. For the conversion $CH_3CH(OH)CH_3 \rightarrow CH_3C(CH_3)CH_2NH_2$, find the sequence of reagents.

• (a) $K_2Cr_2O_7/H^+; NH_3; ThO_2/\Delta$
• (b) $K_2Cr_2O_7/H^+; CH_3MgCl; NH_3$
• (c) $KMnO_4/H^+; HCN; H_3O^+$
• (d) $K_2Cr_2O_7/H^+; HCN; H_2/Ni$

Correct Answer: (d)

Explanation: Propan-2-ol is oxidised to acetone, which then reacts with HCN to form a cyanohydrin, which is finally reduced to an amine.

44. The reaction $ArN_2Cl \xrightarrow{Cu/HCl} ArCl + N_2 + CuCl$ is named as

• (a) Sandmeyer reaction
• (b) Gatterman reaction
• (c) Stephen reaction
• (d) Carbylamine reaction.

Correct Answer: (b)

Explanation: The use of copper powder and HCl to convert diazonium salt to aryl chloride is the Gatterman reaction.

45. Which reagent can provide distinction between aliphatic and aromatic $-NH_2$ group?

• (a) Benzenediazonium chloride
• (b) Benzenesulphonyl chloride
• (c) $CHCl_3/KOH$
• (d) $CH_3COCl$

Correct Answer: (a)

Explanation: Aromatic amines undergo coupling reactions with diazonium salts to form coloured dyes, whereas aliphatic amines do not.