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Amines Set-5 MCQ Online Practice Test | Class 12 | Quiz Tweek

Amines Set-5

Practice Important MCQs for Amines — Chemistry, Class 12.

Amines are crucial in Class 12 Chemistry because they connect structure with basicity and reactivity, and they form the basis of several high-scoring chapters’ reactions (diazotization, substitution, protection/deprotection like acetylation, and N-containing functional group transformations). For board and competitive exams, understanding how substituents, resonance, inductive effects, and solvation/steric effects control basic strength and reaction outcomes is essential—exactly what this set tests.

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Q1. A 0.10 M solution of methylamine ( $CH_3NH_2$ ) has $pK_b = 3.36$ at 25°C. If the solution's pH is adjusted to $9.00$ , what percentage of methylamine exists as its protonated form $CH_3NH_3^+$ ?

(A)

2.25%

(B)

97.75%

(C)

50.0%

(D)

76.2%

Q2. Arrange the following amines in decreasing order of basicity in aqueous solution: benzylamine ( $C_6H_5CH_2NH_2$ ), aniline ( $C_6H_5NH_2$ ), $p$ -nitroaniline ( $p$ - $NO_2$ – $C_6H_4$ – $NH_2$ ).

(A)

$C_6H_5CH_2NH_2 > p\text{-}NO_2\text{-}C_6H_4\text{-}NH_2 > C_6H_5NH_2$

(B)

$C_6H_5NH_2 > C_6H_5CH_2NH_2 > p\text{-}NO_2\text{-}C_6H_4\text{-}NH_2$

(C)

$C_6H_5CH_2NH_2 > C_6H_5NH_2 > p\text{-}NO_2\text{-}C_6H_4\text{-}NH_2$

(D)

$p\text{-}NO_2\text{-}C_6H_4\text{-}NH_2 > C_6H_5CH_2NH_2 > C_6H_5NH_2$

Q3. Assertion (A): In aqueous solution dimethylamine $(CH_3)_2NH$ is a stronger base than trimethylamine $(CH_3)_3N$ .
Reason (R): Steric hindrance in trimethylamine reduces hydration (solvation) of its conjugate acid, thereby lowering its basicity in water.

(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

Q4. Which of the following reagent sequences converts aniline ( $C_6H_5NH_2$ ) to $p$ -bromoaniline ( $p$ - $Br$ – $C_6H_4$ – $NH_2$ ) selectively in good yield?

(A)

(i) Nitration: $HNO_3/H_2SO_4$ → (ii) Reduction: $Sn/HCl$ → (iii) Bromination: $Br_2/FeBr_3$

(B)

(i) Acetylation: $CH_3COCl$ (or $Ac_2O$ )/pyridine → (ii) Bromination: excess $Br_2$ (or $Br_2/FeBr_3$ ) → (iii) Hydrolysis: $NaOH$ (hot), then acidify

(C)

(i) Sulfonation: $H_2SO_4$ → (ii) Desulfonation (steam) → (iii) Bromination: $Br_2$

(D)

(i) Acetylation: $CH_3COCl$ (or $Ac_2O$ )/pyridine → (ii) Bromination: $Br_2$ (in AcOH) → (iii) Hydrolysis of amide: $NaOH$ (hot), then acidify

Q5. When treated with $NaNO_2/HCl$ at $0$ – $5^\circ$ C to form diazonium salts (followed by warming to convert diazonium to phenol), which of the following anilines will form diazonium most readily (i.e., arrange in decreasing ease of diazotization / phenol yield): $p$ -nitroaniline, aniline, $p$ -methoxyaniline?

(A)

$p\text{-}MeO\text{-}C_6H_4\text{-}NH_2 > C_6H_5NH_2 > p\text{-}NO_2\text{-}C_6H_4\text{-}NH_2$

(B)

$p\text{-}NO_2\text{-}C_6H_4\text{-}NH_2 > C_6H_5NH_2 > p\text{-}MeO\text{-}C_6H_4\text{-}NH_2$

(C)

$C_6H_5NH_2 > p\text{-}NO_2\text{-}C_6H_4\text{-}NH_2 > p\text{-}MeO\text{-}C_6H_4\text{-}NH_2$

(D)

$C_6H_5NH_2 > p\text{-}MeO\text{-}C_6H_4\text{-}NH_2 > p\text{-}NO_2\text{-}C_6H_4\text{-}NH_2$

Q6. n-Butylamine ( $pK_b=3.25$ ) is in an aqueous solution at pH $7.50$ . What percentage of the amine is present in the protonated form $\,\mathrm{BH^+}\,$ ?

(A)

$99.50\%$

(B)

$99.94\%$

(C)

$0.056\%$

(D)

$3.30\%$

Q7. Arrange the following in order of basicity in aqueous solution (most basic first): $\mathrm{NH_3},\; \mathrm{CH_3NH_2},\; (CH_3)_2NH,\; (CH_3)_3N$ .

(A)

$(CH_3)_3N > (CH_3)_2NH > CH_3NH_2 > NH_3$

(B)

$(CH_3)_2NH > (CH_3)_3N > CH_3NH_2 > NH_3$

(C)

$CH_3NH_2 > (CH_3)_2NH > (CH_3)_3N > NH_3$

(D)

$(CH_3)_2NH > CH_3NH_2 > (CH_3)_3N > NH_3$

Q8. To prepare p-bromoaniline selectively from aniline ( $\mathrm{C_6H_5NH_2}$ ) with minimal polybromination or loss of the amino group, which reaction sequence is the correct laboratory procedure?

(A)

Protect the amino group by acetylation ( $\mathrm{C_6H_5NH_2}\xrightarrow{CH_3COCl/pyridine} \mathrm{C_6H_5NHCOCH_3}$ ), brominate the acetanilide (gives mainly $p$ -bromoacetanilide), then hydrolyse the amide to obtain $p$ -bromoaniline.

(B)

Directly treat aniline with $\mathrm{Br_2}$ in water to obtain $p$ -bromoaniline selectively.

(C)

Diazotize aniline and perform Sandmeyer with $\mathrm{CuBr}$ to obtain $p$ -bromoaniline.

(D)

Nitrate aniline to give nitroaniline, reduce the nitro group to amine, then brominate to get $p$ -bromoaniline.

Q9. Assertion (A): In the gas phase the basicity order of methylated amines is $(CH_3)_3N > (CH_3)_2NH > CH_3NH_2 > NH_3$ .
Reason (R): In aqueous solution the observed basicity order is $(CH_3)_2NH > CH_3NH_2 > (CH_3)_3N > NH_3$ because solvation stabilises smaller protonated species more effectively than bulky protonated ions.
Choose the correct option.

(A)

A is true, R is false.

(B)

A is false, R is true.

(C)

Both A and R are true, but R does not correctly explain A.

(D)

Both A and R are true and R correctly explains A.

Q10. Which of the following halonitrobenzenes will undergo nucleophilic aromatic substitution with ammonia most rapidly (addition–elimination SNAr mechanism) at room temperature? Compare: $1$ -fluoro- $2,4$ -dinitrobenzene, $1$ -chloro- $2,4$ -dinitrobenzene, $1$ -bromo- $2,4$ -dinitrobenzene.

(A)

Cl $>$ F $>$ Br

(B)

F $>$ Cl $>$ Br

(C)

Br $>$ Cl $>$ F

(D)

F $>$ Br $>$ Cl

Q11. A sample of methylamine, $\mathrm{CH_3NH_2}$ , has $pK_b=3.35$ . At $25^\circ\mathrm{C}$ what percentage of methylamine molecules are present as the protonated form $\mathrm{CH_3NH_3^+}$ in a solution at $pH=9.00$ ? (Use $pK_a=14.00-pK_b$ and the Henderson–Hasselbalch relation $pH=pK_a+\log\frac{[B]}{[BH^+]}$ .)

(A)

$2.2\%$

(B)

$50.0\%$

(C)

$97.8\%$

(D)

$99.9\%$

Q12. Which of the following amines is the most basic in aqueous solution at $25^\circ\mathrm{C}$ ? (Consider resonance, inductive and solvation effects.)

(A)

Aniline, $\mathrm{C_6H_5NH_2}$

(B)

Benzylamine, $\mathrm{C_6H_5CH_2NH_2}$

(C)

p-Nitroaniline, $\mathrm{p\!-\!NO_2C_6H_4NH_2}$

(D)

p-Anisidine (p-methoxyaniline), $\mathrm{p\!-\!CH_3OC_6H_4NH_2}$

Q13. To synthesize $p$ -bromoaniline ( $p$ - $\mathrm{C_6H_4NH_2}$ ) from aniline ( $\mathrm{C_6H_5NH_2}$ ) while avoiding polybromination, which sequence of reagents applied in the order shown is the correct choice?

(A)

(i) $\mathrm{Br_2/Fe}$ (direct bromination of aniline), (ii) aqueous $\mathrm{NaOH}$ work-up, (iii) acidification

(B)

(i) $\mathrm{HNO_3/H_2SO_4}$ (nitration of aniline), (ii) $\mathrm{Br_2/Fe}$ (bromination), (iii) $\mathrm{Sn/HCl}$ (reduction of nitro to amino)

(C)

(i) $\mathrm{NaNO_2/HCl}$ (diazotization), (ii) $\mathrm{CuBr}$ (Sandmeyer to bromobenzene), (iii) nitration then reduction to introduce $-NH_2$

(D)

(i) Acetylation with $(\mathrm{CH_3CO})_2\mathrm{O}$ to give acetanilide, (ii) $\mathrm{Br_2/H_2O}$ to give $p$ -bromoacetanilide, (iii) hydrolysis with $\mathrm{NaOH}$ (then acidify) to yield $p$ -bromoaniline

Q14. For the three amines methylamine $\mathrm{CH_3NH_2}$ , dimethylamine $\mathrm{(CH_3)_2NH}$ and trimethylamine $\mathrm{(CH_3)_3N}$ , which option gives the correct order of basicity (I) in the gas phase and (II) in aqueous solution at $25^\circ\mathrm{C}$ ? (Consider proton affinity in the gas phase and solvation/steric effects in water.)

(A)

(I) $\mathrm{(CH_3)_3N}>\mathrm{(CH_3)_2NH}>\mathrm{CH_3NH_2}$ ; (II) $\mathrm{(CH_3)_2NH}>\mathrm{CH_3NH_2}>\mathrm{(CH_3)_3N}$

(B)

(I) $\mathrm{CH_3NH_2}>\mathrm{(CH_3)_2NH}>\mathrm{(CH_3)_3N}$ ; (II) $\mathrm{(CH_3)_3N}>\mathrm{(CH_3)_2NH}>\mathrm{CH_3NH_2}$

(C)

(I) $\mathrm{(CH_3)_2NH}>\mathrm{(CH_3)_3N}>\mathrm{CH_3NH_2}$ ; (II) $\mathrm{CH_3NH_2}>\mathrm{(CH_3)_2NH}>\mathrm{(CH_3)_3N}$

(D)

(I) $\mathrm{(CH_3)_3N}>\mathrm{CH_3NH_2}>\mathrm{(CH_3)_2NH}$ ; (II) $\mathrm{(CH_3)_2NH}>\mathrm{(CH_3)_3N}>\mathrm{CH_3NH_2}$

Q15. When benzylamine $\mathrm{C_6H_5CH_2NH_2}$ is treated with $\mathrm{NaNO_2/HCl}$ at $0\!-\!5^\circ\mathrm{C}$ and the reaction mixture is then warmed, the major stable organic product obtained is:

(A)

Benzyl diazonium chloride, $\mathrm{C_6H_5CH_2N_2^+Cl^-}$

(B)

Azobenzene, $\mathrm{C_6H_5NH\!-\!NHC_6H_5}$

(C)

Benzyl alcohol, $\mathrm{C_6H_5CH_2OH}$

(D)

Phenol, $\mathrm{C_6H_5OH}$

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