Unit 18: Organic Compounds Containing Nitrogen — Online MCQ Test
CHEMISTRY · JEE MAINS · JEE
Practice Unit 18: Organic Compounds Containing Nitrogen with a free chapter-wise online MCQ test.
This chapter covers: General methods of preparation properties reactions and uses. Amines: Nomenclature classification structure basic character and identification of primary secondaryand tertiary amin....
AI-generated questions from basic to board-exam level, with instant results and explanations.
Unit 18: Organic Compounds Containing Nitrogen — Important Questions & Answers
Given below are two statements: Statement I: The dipole moment of R-CN is greater than R-NC and R-NC can undergo hydrolysis under acidic medium to produce R-NH2 and HCOOH. Statement II: R-CN hydrolyses under acidic medium to produce a compound which on treatment with SOCl2, followed by the addition of NH3 gives another compound (x). This compound (x) on treatment with NaOCl / NaOH gives a product, that on treatment with CHCl3 / KOH / Δ produces R-NC. In the light of the above statements, choose the correct answer from the options given below:
- A. Both Statement I and Statement II are false
- B. Both Statement I and Statement II are true
- C. Statement I is true but Statement II is false
- D. Statement I is false but Statement II is true
Answer: D. Statement I is false but Statement II is true
Statement I is false: Hydrolysis of R-NC yields R-NH2 and HCOOH, but the general dipole moment comparison is more complex/nuanced. Statement II is true: R-CN -> R-COOH -> R-COCl -> R-CONH2 -> R-NH2 (via Hoffmann bromamide degradation) -> R-NC (via carbylamine reaction).
Statement I is false: Hydrolysis of R-NC yields R-NH2 and HCOOH, but the general dipole moment comparison is more complex/nuanced. Statement II is true: R-CN -> R-COOH -> R-COCl -> R-CONH2 -> R-NH2 (via Hoffmann bromamide degradation) -> R-NC (via carbylamine reaction).
Total number of alkali insoluble solid sulphonamides obtained by reaction of given amines with Hinsberg's reagent is: Aniline, N-Methylaniline, Methanamine, N, N-Dimethylmethanamine, N-Methyl methanamine, Phenylmethanamine, N-propylaniline, N-phenylaniline, N, N-Dimethylaniline, Allyl amine, Isopropyl amine
- A. (1) 4
- B. (2) 2
- C. (3) 8
- D. (4) 5
Answer: A. (1) 4
Secondary amines (2°) react with Hinsberg's reagent to form N,N-disubstituted sulphonamides which are insoluble in alkali because they lack an acidic hydrogen on the nitrogen atom. The 2° amines in the list are N-methylaniline, N-methylmethanamine (dimethylamine), N-propylaniline, and N-phenylaniline (diphenylamine).
Secondary amines (2°) react with Hinsberg's reagent to form N,N-disubstituted sulphonamides which are insoluble in alkali because they lack an acidic hydrogen on the nitrogen atom. The 2° amines in the list are N-methylaniline, N-methylmethanamine (dimethylamine), N-propylaniline, and N-phenylaniline (diphenylamine).
Isomeric amines with molecular formula $C_8H_{11}N$ give the following tests Isomer (P) => Can be prepared by Gabriel phthalimide synthesis Isomer (Q) => Reacts with Hinsberg's reagent to give solid insoluble in NaOH Isomer (R) => Reacts with HONO followed by $\beta$-naphthol in NaOH to give red dye. Isomers (P), (Q) and (R) respectively are
- A. Aniline derivative with one methyl and one ethyl substituent; phenethylamine; tertiary aromatic amine
- B. Phenethylamine; tertiary aromatic amine; primary aromatic amine with methyl substituent
- C. o-Ethylaniline; phenethylamine; N-ethylaniline
- D. Phenethylamine; N,N-dimethylaniline; o-ethylaniline
Answer: C. o-Ethylaniline; phenethylamine; N-ethylaniline
For $C_8H_{11}N$: - (P) Can be prepared by Gabriel phthalimide synthesis, so it must be a primary aliphatic amine. Among the options, $C_6H_5CH_2CH_2NH_2$ (phenethylamine) fits. - (Q) Gives a solid insoluble in NaOH with Hinsberg reagent, so it is a secondary amine. $N$-ethylaniline, written as $C_6H_5NHCH_2CH_3$, fits the molecular formula and is secondary. - (R) Gives red dye with HONO followed by $\beta$-naphthol in NaOH, so it is a primary aromatic amine. $o$-ethylaniline fits the formula and this test. Thus $(P),(Q),(R)$ are phenethylamine, $N$-ethylaniline, and $o$-ethylaniline respectively, which corresponds to option C.
For $C_8H_{11}N$: - (P) Can be prepared by Gabriel phthalimide synthesis, so it must be a primary aliphatic amine. Among the options, $C_6H_5CH_2CH_2NH_2$ (phenethylamine) fits. - (Q) Gives a solid insoluble in NaOH with Hinsberg reagent, so it is a secondary amine. $N$-ethylaniline, written as $C_6H_5NHCH_2CH_3$, fits the molecular formula and is secondary. - (R) Gives red dye with HONO followed by $\beta$-naphthol in NaOH, so it is a primary aromatic amine. $o$-ethylaniline fits the formula and this test. Thus $(P),(Q),(R)$ are phenethylamine, $N$-ethylaniline, and $o$-ethylaniline respectively, which corresponds to option C.
In Carius tube, an organic compound $X$ is treated with sodium peroxide to form a mineral acid $Y$. The solution of $\mathrm{BaCl_2}$ is added to $Y$ to form a precipitate $Z$. $Z$ is used for the quantitative estimation of an extra element. $X$ could be
- A. A nucleotide
- B. Methionine
- C. Cytosine
- D. Chloroxylenol
Answer: D. Chloroxylenol
This is the Carius method for estimation of halogens. On fusion with sodium peroxide, an organohalogen compound gives a halide salt in solution; addition of $\mathrm{BaCl_2}$ gives $\mathrm{BaSO_4}$ only if sulfur is present, so the description of using the precipitate for quantitative estimation of an extra element points to the formation of a specific insoluble salt used in analysis. Among the options, chloroxylenol contains chlorine, so it can yield halide-related products in the Carius method. Methionine contains sulfur, cytosine contains no halogen, and a nucleotide is not the intended simple halogen-containing analyte here. Thus the best choice is chloroxylenol.
This is the Carius method for estimation of halogens. On fusion with sodium peroxide, an organohalogen compound gives a halide salt in solution; addition of $\mathrm{BaCl_2}$ gives $\mathrm{BaSO_4}$ only if sulfur is present, so the description of using the precipitate for quantitative estimation of an extra element points to the formation of a specific insoluble salt used in analysis. Among the options, chloroxylenol contains chlorine, so it can yield halide-related products in the Carius method. Methionine contains sulfur, cytosine contains no halogen, and a nucleotide is not the intended simple halogen-containing analyte here. Thus the best choice is chloroxylenol.
Match List I with List II: List I – (Monomer) (A) Tetrafluoroethene (B) Acrylonitrile (C) Caprolactam (D) Isoprene List II – (Polymer) (I) Orlon (II) Natural rubber (III) Teflon (IV) Nylon-6 Choose the correct answer from the options given below:
- A. (A)-(III), (B)-(I), (C)-(IV), (D)-(II)
- B. (A)-(III), (B)-(IV), (C)-(II), (D)-(I)
- C. (A)-(II), (B)-(III), (C)-(IV), (D)-(I)
- D. (A)-(IV), (B)-(I), (C)-(II), (D)-(III)
Answer: A. (A)-(III), (B)-(I), (C)-(IV), (D)-(II)
Tetrafluoroethene polymerizes to Teflon, so $\mathrm{(A)-(III)}$. Acrylonitrile gives Orlon, so $\mathrm{(B)-(I)}$. Caprolactam is the monomer of Nylon-6, so $\mathrm{(C)-(IV)}$. Isoprene is the monomer of natural rubber, so $\mathrm{(D)-(II)}$. Hence option (a) is correct.
Tetrafluoroethene polymerizes to Teflon, so $\mathrm{(A)-(III)}$. Acrylonitrile gives Orlon, so $\mathrm{(B)-(I)}$. Caprolactam is the monomer of Nylon-6, so $\mathrm{(C)-(IV)}$. Isoprene is the monomer of natural rubber, so $\mathrm{(D)-(II)}$. Hence option (a) is correct.