Amines are a high-yield topic in Class 12 Chemistry because they connect fundamental concepts of basicity, resonance, and equilibrium with important reaction patterns (acylation, diazotization, Hofmann elimination, Gabriel synthesis). Board and competitive exams frequently test both qualitative reasoning (direction of reactions, effect of substituents on basicity) and quantitative calculations (pH/pKb/pKa relationships), so mastering this chapter improves accuracy across multiple question types.
15
Minutes
10
Questions
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Marking
Q1. Calculate the pH of a solution of methylamine () at , given . (Assume .)
Q2. Two amines X and Y are each in the same aqueous solution at . Given and , calculate the ratio of their unprotonated (free base) concentrations at . (Use and Henderson–Hasselbalch.)
Q3. Assertion (A): In the gas phase the basicity order is .
Reason (R): In the gas phase there is no solvation, so inductive and hyperconjugation effects dominate and the protonated tertiary amine is stabilized more than protonated secondary or primary amines.
Both (A) and (R) are false.
(A) is true but (R) is false.
Both (A) and (R) are true but (R) does not correctly explain (A).
Both (A) and (R) are true and (R) correctly explains (A).
Q4. Which of the following reagent sequences will give p-nitroaniline (-) from aniline () with maximum para-selectivity and minimal oxidation/over-nitration?
Direct nitration: aniline mixture of poly-nitrated/oxidised products.
Acetylation nitration hydrolysis: aniline acetanilide -nitroacetanilide -nitroaniline.
Diazotization then Sandmeyer-type replacement: aniline diazonium introduce .
Protonate then nitrate: aniline anilinium chloride then neutralise.
Q5. Which of the following amines cannot be prepared by Gabriel phthalimide synthesis starting from the corresponding halide under standard SN2 conditions?
aniline ()
butylamine ()
benzylamine ()
1-aminopropane ()
Q6. Given the of anilinium ion () is at , calculate the of aniline in water (assume ).
Q7. Arrange the following compounds in order of decreasing basicity in aqueous solution: -toluidine (-CHCHNH), aniline (PhNH), o-nitroaniline and m-nitroaniline. Which order is correct?
-toluidine ;>; aniline ;>; m-nitroaniline ;>; o-nitroaniline
-toluidine ;>; m-nitroaniline ;>; aniline ;>; o-nitroaniline
aniline ;>; -toluidine ;>; m-nitroaniline ;>; o-nitroaniline
-toluidine ;>; aniline ;>; o-nitroaniline ;>; m-nitroaniline
Q8. On exhaustive methylation of 2-aminobutane () followed by treatment with and heat (Hofmann elimination), the major alkene formed will be:
trans-2-butene (, trans)
cis-2-butene (, cis)
a mixture of cis- and trans-2-butene
1-butene ()
Q9. Aniline gives 2,4,6-tribromoaniline on treatment with /HO, whereas acetanilide gives predominantly -bromoacetanilide. Which of the following explanations best accounts for this difference?
Acetylation makes the ring strongly deactivated and directs bromination mainly to the meta position, so para substitution is expected to be minimal.
Acetylation converts into which delocalizes the lone pair into the carbonyl, reducing electron donation to the ring and, combined with steric hindrance at ortho positions, makes bromination less vigorous and mainly para-selective, giving -bromoacetanilide.
Acetylation delocalizes the lone pair only by effect, and this resonance withdrawal alone directs bromination exclusively to para position (steric effects are not responsible).
Acetylation protonates the nitrogen under reaction conditions, blocking ortho positions and forcing para substitution.
Q10. Which of the following aromatic amines has the lowest basicity in aqueous solution? (Consider resonance and possible intramolecular hydrogen bonding.)
aniline (PhNH)
p-aminophenol (-HO-CH-NH)
o-aminophenol (-HO-CH-NH)
m-aminophenol (-HO-CH-NH)