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Amines Set-6 MCQ Online Practice Test | Class 12 | Quiz Tweek

Amines Set-6

Practice Important MCQs for Amines — Chemistry, Class 12.

The chapter on Amines is crucial for both CBSE boards and competitive exams because it links core ideas of organic chemistry—structure, basicity, reactions like diazotization, nitration control, and alkylation—with measurable equilibrium parameters (like $pK_b$ and degree of protonation). Questions from this chapter frequently test both conceptual understanding (resonance/inductive/steric effects) and calculation-based reasoning, making it high-scoring for exam preparation.

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Q1. A 0.10 M solution of methylamine $(

\mathrm{CH_3NH_2}

) $has$ pK_b=3.35 $. At$ pH=9.0 $, what fraction of methylamine exists as methylammonium ion$ (

\mathrm{CH_3NH_3^+}

)$?

(A)

$2.2\%$

(B)

$97.8\%$

(C)

$50.0\%$

(D)

$4.5\%$

Q2. Arrange the following amines in order of decreasing basic strength in aqueous solution: (i) methylamine $(

\mathrm{CH_3NH_2}

) $, (ii) pyridine$ (

\mathrm{C_5H_5N}

) $, (iii) aniline$ (

\mathrm{C_6H_5NH_2}

)$.

(A)

$\mathrm{C_6H_5NH_2} > \mathrm{C_5H_5N} > \mathrm{CH_3NH_2}$

(B)

$\mathrm{C_5H_5N} > \mathrm{CH_3NH_2} > \mathrm{C_6H_5NH_2}$

(C)

$\mathrm{CH_3NH_2} > \mathrm{C_6H_5NH_2} > \mathrm{C_5H_5N}$

(D)

$\mathrm{CH_3NH_2} > \mathrm{C_5H_5N} > \mathrm{C_6H_5NH_2}$

Q3. A 1:1 molar mixture of methylamine $(

\mathrm{CH_3NH_2}

) $and aniline$ (

\mathrm{C_6H_5NH_2}

) $is treated with one equivalent of benzyl bromide$ (

\mathrm{C_6H_5CH_2Br}

)$ in dry ether at 25°C for a short time (kinetic control). The major product will be:

(A)

N‑benzylmethylamine, $\mathrm{PhCH_2NHCH_3}$

(B)

N‑benzylaniline, $\mathrm{PhCH_2NHPh}$

(C)

An equimolar mixture of N‑benzylmethylamine and N‑benzylaniline

(D)

Predominant formation of over‑alkylated tertiary amine $\mathrm{N(CH_3)(CH_2Ph)_2}$

Q4. Assertion (A): In aqueous solution the basicity order of aliphatic amines is dimethylamine $((\mathrm{CH_3})_2\mathrm{NH}) >$ methylamine $(\mathrm{CH_3NH_2}) >$ trimethylamine $((\mathrm{CH_3})_3\mathrm{N})$ .
Reason (R): Trimethylamine is the weakest base in water because the inductive $(+I)$ effect of additional alkyl groups withdraws electron density from nitrogen, making the lone pair less available for protonation.

(A)

Both A and R are true and R is the correct explanation of A.

(B)

Both A and R are true but R is NOT the correct explanation of A.

(C)

A is true but R is false.

(D)

A is false but R is true.

Q5. Assertion (A): Direct nitration of aniline $(

\mathrm{C_6H_5NH_2}

) $with concentrated$ HNO_3 $/$ H_2SO_4 $at$ 0 $–$ 5^\circ\mathrm{C} $gives predominantly$ p $‑nitroaniline as the isolable product. Reason (R): Under strongly acidic nitrating conditions the amino group is protonated to form the anilinium ion$ (

\mathrm{C_6H_5NH_3^+}

)$, which is a meta‑directing substituent; therefore, if nitration proceeds under such conditions it should favour the meta isomer.

(A)

Both A and R are true and R is the correct explanation of A.

(B)

A is false but R is true.

(C)

A is true but R is false.

(D)

Both A and R are false.

Q6. At 25°C methylamine ( $\mathrm{CH_3NH_2}$ ) has pKb = 3.36. What percentage of methylamine molecules are present in the protonated form ( $\mathrm{CH_3NH_3^+}$ ) at pH = 7.00?

(A)

50%

(B)

0.023%

(C)

99.98%

(D)

99.9%

Q7. An equimolar mixture of methylamine ( $\mathrm{CH_3NH_2}$ , pKb = 3.36) and aniline ( $\mathrm{C_6H_5NH_2}$ , pKb = 9.40) is dissolved in water buffered at pH = 6.0. What is the ratio $[\mathrm{CH_3NH_3^+}]:[\mathrm{C_6H_5NH_3^+}]$ of their protonated forms?

(A)

1 : 26

(B)

26 : 1

(C)

100 : 1

(D)

1 : 100

Q8. Benzylamine ( $\mathrm{C_6H_5CH_2NH_2}$ ) is treated with $\mathrm{NaNO_2}$ and HCl at $0$ – $5^\circ\mathrm{C}$ and then the reaction mixture is warmed to room temperature. The major isolable organic product is:

(A)

Aniline, $\mathrm{C_6H_5NH_2}$

(B)

Nitrobenzene, $\mathrm{C_6H_5NO_2}$

(C)

Benzenediazonium chloride, $\mathrm{C_6H_5N_2^+Cl^-}$

(D)

Benzyl alcohol, $\mathrm{C_6H_5CH_2OH}$

Q9. Arrange the amines methylamine ( $\mathrm{CH_3NH_2}$ ), dimethylamine ( $\mathrm{(CH_3)_2NH}$ ) and trimethylamine ( $\mathrm{(CH_3)_3N}$ ) in order of increasing basicity (least → greatest) in (i) aqueous solution and (ii) gas phase.

(A)

(i) $\mathrm{(CH_3)_3N} < \mathrm{CH_3NH_2} < \mathrm{(CH_3)_2NH}$ ; (ii) $\mathrm{CH_3NH_2} < \mathrm{(CH_3)_2NH} < \mathrm{(CH_3)_3N}$

(B)

(i) $\mathrm{CH_3NH_2} < \mathrm{(CH_3)_3N} < \mathrm{(CH_3)_2NH}$ ; (ii) $\mathrm{(CH_3)_3N} < \mathrm{(CH_3)_2NH} < \mathrm{CH_3NH_2}$

(C)

(i) $\mathrm{(CH_3)_2NH} < \mathrm{CH_3NH_2} < \mathrm{(CH_3)_3N}$ ; (ii) $\mathrm{(CH_3)_2NH} < \mathrm{CH_3NH_2} < \mathrm{(CH_3)_3N}$

(D)

(i) $\mathrm{(CH_3)_3N} < \mathrm{(CH_3)_2NH} < \mathrm{CH_3NH_2}$ ; (ii) $\mathrm{CH_3NH_2} < \mathrm{(CH_3)_3N} < \mathrm{(CH_3)_2NH}$

Q10. Which two-step sequence is the most appropriate to convert nitrobenzene ( $\mathrm{C_6H_5NO_2}$ ) to N,N-dimethylaniline ( $\mathrm{C_6H_5N(CH_3)_2}$ ) giving high yield while minimising side reactions?

(A)

Reduce $\mathrm{NO_2}$ to $\mathrm{NH_2}$ using $\mathrm{Fe/HCl}$ , then alkylate with excess $\mathrm{CH_3I}/\mathrm{K_2CO_3}$

(B)

Reduce $\mathrm{NO_2}$ to $\mathrm{NH_2}$ by catalytic hydrogenation ( $\mathrm{H_2/Pd}$ ), then perform Eschweiler–Clarke methylation using formaldehyde and formic acid ( $\mathrm{HCHO + HCOOH}$ )

(C)

Reduce $\mathrm{NO_2}$ to $\mathrm{NH_2}$ by $\mathrm{SnCl_2/HCl}$ , then react with excess $\mathrm{CH_3Cl}$ in presence of $\mathrm{AlCl_3}$

(D)

Convert $\mathrm{NO_2}$ to diazonium ( $\mathrm{N_2^+}$ ) and then treat with $\mathrm{CH_3MgBr}$ followed by hydrolysis

Q11. How many constitutional isomers of amines are possible with molecular formula $C_4H_{11}N$ (count primary, secondary and tertiary amines; ignore stereoisomers)?

(A)

(B)

(C)

(D)

Q12. Given $pK_b(\text{methylamine}) = 3.35$ and $pK_b(\text{aniline}) = 9.42$ , at $pH = 7$ the ratio $[\mathrm{BH}^+]/[\mathrm{B}]$ for each amine is closest to which of the following, and which amine is predominantly protonated?

(A)

Methylamine: $[\mathrm{BH}^+]/[\mathrm{B}]\approx 4.5\times10^{3}$ ; Aniline: $[\mathrm{BH}^+]/[\mathrm{B}]\approx 3.8\times10^{-3}$ — methylamine predominantly protonated

(B)

Methylamine: $[\mathrm{BH}^+]/[\mathrm{B}]\approx 3.8\times10^{-3}$ ; Aniline: $[\mathrm{BH}^+]/[\mathrm{B}]\approx 4.5\times10^{3}$ — aniline predominantly protonated

(C)

Methylamine: $[\mathrm{BH}^+]/[\mathrm{B}]\approx 4.5$ ; Aniline: $[\mathrm{BH}^+]/[\mathrm{B}]\approx 2.3\times10^{-3}$ — both partly protonated

(D)

Both methylamine and aniline are mostly protonated at $pH=7$ because $pH$ is below both $pK_b$ values

Q13. You have $50.0\ \text{mL}$ of $0.200\ \mathrm{M}$ methylamine ( $pK_b = 3.35$ ). What volume (in mL) of $0.100\ \mathrm{M}$ HCl must be added to convert $90.0\%$ of methylamine to its protonated form? (Assume volumes are additive.)

(A)

$20.0\ \text{mL}$

(B)

$90.0\ \text{mL}$

(C)

$50.0\ \text{mL}$

(D)

$110.0\ \text{mL}$

Q14. Treatment of optically active $(R)$ - $\mathrm{C_6H_5CH(NH_2)CH_3}$ (i.e. $(R)$ -1-phenylethylamine) with $\mathrm{NaNO_2/HCl}$ at $0$ – $5^\circ\mathrm{C}$ followed by warming typically gives which major organic product(s) and what happens to stereochemistry?

(A)

No reaction; optical activity retained

(B)

$(R)$ -1-phenylethanol only (retention of configuration)

(C)

Predominantly the corresponding alkene (styrene derivative) with inversion of configuration

(D)

A racemic mixture of 1-phenylethanol (with evolution of $N_2$ ), because the benzylic diazonium decomposes to a planar carbocation before capture by water

Q15. Assertion (A): Aniline does not undergo direct nitration with $HNO_3/H_2SO_4$ ; it is usually converted to acetanilide (acetylated) before nitration.
Reason (R): In the nitrating mixture aniline is oxidized and polymerizes (forms tars), so acetylation is performed to prevent such oxidation.

(A)

Both A and R are true and R is the correct explanation for A.

(B)

Both A and R are true but R is NOT the correct explanation for A.

(C)

A is true but R is false.

(D)

A is false but R is true.

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