Organic Compounds Containing Nitrogen — NEET UG practice

57 questions

Practice NEET UG Organic Compounds Containing Nitrogen questions free — each with a detailed solution, graded instantly. Nothing is saved; log in to track your accuracy and build a streak.

Sample questions with solutions

Q1 · 2026

The following two reactions give the same foul smelling product ZZ.

C2H5ClXZC_2H_5Cl \xrightarrow{X} Z

XX and ZZ, respectively, are :

  • A.

    X=AgCNX=AgCN; Z=C2H5NCZ=C_2H_5NC

  • B.

    X=KCNX=KCN; Z=C2H5CNZ=C_2H_5CN

  • C.

    X=AgCNX=AgCN; Z=C2H5CNZ=C_2H_5CN

  • D.

    X=KCNX=KCN; Z=C2H5NCZ=C_2H_5NC

Answer: A
  1. When an alkyl halide reacts with AgCNAgCN, the reaction proceeds through the nitrogen atom of the cyanide ion (because Ag+Ag^+ makes the CNC-N bond more covalent, and nitrogen becomes the more reactive nucleophilic site), forming an isocyanide.
C2H5Cl+AgCNC2H5NC+AgClC_2H_5Cl + AgCN \rightarrow C_2H_5-NC + AgCl
  1. This isocyanide, ethyl isocyanide (C2H5NCC_2H_5NC), is well known for its strong, unpleasant (foul) smell.

  2. Therefore, since both reactions in the question give the same foul-smelling product, XX must be AgCNAgCN and ZZ must be C2H5NCC_2H_5NC.

  3. In contrast, KCNKCN reacts through the carbon atom of the cyanide ion, giving a nitrile (C2H5CNC_2H_5CN), which does not have a foul odor.

  4. Hence, the answer is option A.

Q2 · 2026

Two products XX and YY are formed in the following reaction sequence.

The suitable method that can be used for the separation of products X and Y is :

  • A.

    Fractional distillation

  • B.

    Sublimation

  • C.

    Differential extraction

  • D.

    Continuous extraction

Answer: A
  1. In many aromatic substitution reactions, a mixture of ortho (o-) and para (p-) isomers is formed as the two major products.

  2. These isomers have different boiling points because the more symmetrical para isomer packs and interacts differently than the ortho isomer.

Given the boiling points,

o-isomer: b.p.=222C,p-isomer: b.p.=238C\text{o-isomer: b.p.} = 222^{\circ}C, \qquad \text{p-isomer: b.p.} = 238^{\circ}C
  1. Since these boiling points are sufficiently different, the mixture can be separated by carefully heating it and collecting each component as it evaporates within its own boiling range, typically under reduced pressure.

  2. This technique is known as fractional distillation.

  3. Hence, the answer is option A.

Q3 · 2026

The major product Z formed in the following sequence of reactions is

  • A.

    C2H5NO2C_2H_5NO_2

  • B.

    C2H5N=NOHC_2H_5-N=N-OH

  • C.

    C2H5OHC_2H_5OH

  • D.

    C2H5NH2C_2H_5NH_2

Answer: C
  1. An aliphatic primary amine reacts with nitrous acid (HNO2HNO_2, generated in situ from NaNO2+HClNaNO_2 + HCl) to form an unstable diazonium salt.
C2H5NH2+HNO2[C2H5N2+]Cl+2H2OC_2H_5NH_2 + HNO_2 \rightarrow [C_2H_5-N_2^+]Cl^- + 2H_2O
  1. Unlike aromatic diazonium salts, aliphatic diazonium salts are highly unstable, even at low temperature, and decompose immediately with loss of nitrogen gas.

  2. The resulting ethyl carbocation quickly reacts with water present in the reaction mixture to give an alcohol.

C2H5N2+C2H5++N2H2OC2H5OHC_2H_5-N_2^+ \rightarrow C_2H_5^+ + N_2\uparrow \xrightarrow{H_2O} C_2H_5OH
  1. This brisk evolution of nitrogen gas along with alcohol formation is actually used as a test to identify primary aliphatic amines.

  2. Hence, the major product ZZ is ethanol (C2H5OHC_2H_5OH), so the answer is option C.

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