Introduction: Understanding the basicity of aromatic amines is essential for B. Pharm students studying drug design, pharmacokinetics, and medicinal chemistry. Aromatic amines like aniline show reduced basicity compared to aliphatic amines due to resonance delocalization of the nitrogen lone pair into the benzene ring. Substituent effects (electron-donating or electron-withdrawing), position (ortho, meta, para), steric hindrance, solvent, and conjugation with carbonyls critically influence protonation, pKa, and biological ionization states. Mastery of these concepts helps predict drug ionization, absorption, and receptor interactions. This set of focused MCQs emphasizes mechanisms, pKa trends, and practical applications for pharmaceutical contexts. Now let’s test your knowledge with 50 MCQs on this topic.
Q1. Which factor primarily reduces the basicity of aniline compared to methylamine?
- Greater solvation of aniline in water
- Resonance delocalization of the nitrogen lone pair into the aromatic ring
- Stronger inductive donation from the benzene ring
- Higher steric hindrance around nitrogen
Correct Answer: Resonance delocalization of the nitrogen lone pair into the aromatic ring
Q2. Which substituent on the para position of aniline would most increase its basicity?
- –NO2 (nitro)
- –CF3 (trifluoromethyl)
- –OCH3 (methoxy)
- –Cl (chloro)
Correct Answer: –OCH3 (methoxy)
Q3. Why do electron-withdrawing groups decrease the basicity of aromatic amines?
- They increase solvation of the protonated form
- They decrease electron density on nitrogen via inductive or resonance effects
- They cause steric hindrance that prevents protonation
- They convert the amine to a quaternary ammonium
Correct Answer: They decrease electron density on nitrogen via inductive or resonance effects
Q4. Which position of a nitro group has the strongest effect in reducing aniline basicity?
- Ortho
- Meta
- Para
- All positions have equal effect
Correct Answer: Ortho
Q5. Which statement about halogen-substituted anilines is correct?
- Halogens always increase basicity by resonance donation
- Halogens decrease basicity due to the electron-withdrawing inductive effect despite resonance donation
- Halogens have no effect on basicity
- Halogens convert aniline into a strong base
Correct Answer: Halogens decrease basicity due to the electron-withdrawing inductive effect despite resonance donation
Q6. Which aromatic amine is more basic in water: aniline or p-anisidine (p-methoxyaniline)?
- Aniline
- p-Anisidine
- They have identical basicity
- Depends on temperature only
Correct Answer: p-Anisidine
Q7. How does acylation of aniline (forming anilide) affect basicity?
- Greatly increases basicity due to resonance stabilization
- Greatly decreases basicity because the nitrogen lone pair is delocalized into the carbonyl
- No change in basicity
- Converts the molecule into a strong base
Correct Answer: Greatly decreases basicity because the nitrogen lone pair is delocalized into the carbonyl
Q8. Among the following, which has the highest basicity in aqueous solution?
- Aniline (Ph-NH2)
- Methylamine (CH3NH2)
- Pyridine
- p-Nitroaniline
Correct Answer: Methylamine (CH3NH2)
Q9. Which heteroaromatic amine is typically more basic: aniline or pyridine?
- Aniline
- Pyridine
- They are equally basic
- Depends on substituents only
Correct Answer: Pyridine
Q10. What is the main reason pyridine is more basic than aniline?
- Pyridine nitrogen lone pair is in an sp2 orbital not conjugated with a benzene ring, making it more available for protonation
- Pyridine has greater resonance stabilization of its protonated form
- Aniline has no resonance effects
- Pyridine is aliphatic
Correct Answer: Pyridine nitrogen lone pair is in an sp2 orbital not conjugated with a benzene ring, making it more available for protonation
Q11. How does steric hindrance at the ortho position affect basicity of aromatic amines?
- Steric hindrance increases basicity by blocking resonance
- Steric hindrance decreases basicity by hindering solvation of the protonated form and sometimes twisting the ring to reduce resonance
- Steric hindrance has no effect
- Steric hindrance converts amine into an amide
Correct Answer: Steric hindrance decreases basicity by hindering solvation of the protonated form and sometimes twisting the ring to reduce resonance
Q12. Which aniline derivative would you expect to be the least basic?
- p-Toluidine (para-methyl aniline)
- m-Nitroaniline (meta-nitro aniline)
- p-Anisidine (para-methoxy aniline)
- o-Methylaniline
Correct Answer: m-Nitroaniline (meta-nitro aniline)
Q13. Protonation of aniline occurs primarily at which atom?
- Ring carbon
- Nitrogen
- Hydrogen of the ring
- Oxygen
Correct Answer: Nitrogen
Q14. How does hydrogen bonding intramolecularly (e.g., o-hydroxyaniline) affect basicity?
- Increases basicity by stabilizing the neutral form
- Decreases basicity because intramolecular H-bonding can reduce availability of the lone pair for protonation
- No effect on basicity
- Converts the amine to a stronger base
Correct Answer: Decreases basicity because intramolecular H-bonding can reduce availability of the lone pair for protonation
Q15. Which quantitative measure is commonly used to compare basicity of aromatic amines?
- Log P
- pKa of the conjugate acid
- Boiling point
- Melting point
Correct Answer: pKa of the conjugate acid
Q16. For drug absorption, why is knowing the pKa of aromatic amines important?
- It determines melting point
- It predicts the degree of ionization at physiological pH, affecting solubility and membrane permeability
- It affects color of the drug
- It only matters for solid-state stability
Correct Answer: It predicts the degree of ionization at physiological pH, affecting solubility and membrane permeability
Q17. Which effect describes the ability of a substituent to donate or withdraw electron density through sigma bonds?
- Resonance effect
- Inductive effect
- Hyperconjugation effect
- Hydrogen bonding effect
Correct Answer: Inductive effect
Q18. Which resonance form explains decreased basicity of aniline?
- Nitrogen lone pair localized on nitrogen exclusively
- Delocalization of nitrogen lone pair into the aromatic ring forming a delocalized positive charge upon protonation
- Complete transfer of electron density to solvent
- Formation of a carbonyl at nitrogen
Correct Answer: Delocalization of nitrogen lone pair into the aromatic ring forming a delocalized positive charge upon protonation
Q19. In gas phase basicity comparisons, which factor is less important compared to solution phase?
- Intrinsic proton affinity
- Solvation and hydrogen bonding
- Electronic structure
- Bare ion stability
Correct Answer: Solvation and hydrogen bonding
Q20. Which compound is expected to have the highest pKa for its conjugate acid (i.e., most basic) in water?
- Aniline
- N,N-dimethylaniline
- Nitroaniline
- Pyridinium ion from pyridine
Correct Answer: N,N-dimethylaniline
Q21. How does conjugation of an aromatic amine with a carbonyl (e.g., acetanilide) alter basicity?
- Increases basicity by hyperconjugation
- Decreases basicity because the lone pair is delocalized into the carbonyl
- No change in basicity
- Converts it into a stronger base than aliphatic amines
Correct Answer: Decreases basicity because the lone pair is delocalized into the carbonyl
Q22. What is the expected effect of para-NH2 substituent on aniline basicity compared to aniline itself?
- Decrease basicity due to strong electron withdrawal
- Increase basicity due to electron donation via resonance
- No effect on basicity
- Convert to a non-basic compound
Correct Answer: Increase basicity due to electron donation via resonance
Q23. Which technique helps experimentally determine pKa of aromatic amines?
- IR spectroscopy
- UV-visible spectrophotometric titration
- Thin-layer chromatography
- X-ray crystallography
Correct Answer: UV-visible spectrophotometric titration
Q24. Which of the following increases the gas-phase basicity of an aromatic amine relative to solution?
- Stronger solvent stabilization of the protonated form
- Absence of solvation leading to intrinsic proton affinity dominating
- Increased hydrogen bonding in water
- Lower dielectric constant
Correct Answer: Absence of solvation leading to intrinsic proton affinity dominating
Q25. What is the effect of methyl substitution on nitrogen (forming N-methylaniline) compared to aniline?
- Usually increases basicity slightly due to electron-donating alkyl group
- Greatly decreases basicity because alkyl groups withdraw electrons
- No change in basicity
- Makes it non-basic
Correct Answer: Usually increases basicity slightly due to electron-donating alkyl group
Q26. Which position of a substituent will not significantly affect basicity through resonance?
- Ortho
- Meta
- Para
- All positions affect resonance equally
Correct Answer: Meta
Q27. For pKa prediction, which of the following descriptors is most relevant for aromatic amines?
- Log P
- Electron donating/withdrawing capacity (Hammett sigma values)
- Boiling point
- Color
Correct Answer: Electron donating/withdrawing capacity (Hammett sigma values)
Q28. Which aniline derivative will show intramolecular hydrogen bonding that reduces basicity?
- p-Aminoanisole (para-methoxy aniline)
- o-Aminophenol (ortho-aminophenol)
- m-Toluidine (meta-methyl aniline)
- p-Chloroaniline
Correct Answer: o-Aminophenol (ortho-aminophenol)
Q29. Which statement best summarizes the effect of resonance on aromatic amine basicity?
- Resonance always makes the amine a stronger base
- Resonance delocalization of the lone pair into the ring decreases availability for protonation, lowering basicity
- Resonance has no effect on basicity
- Resonance converts the amine into an acid
Correct Answer: Resonance delocalization of the lone pair into the ring decreases availability for protonation, lowering basicity
Q30. Which factor can make an ortho-substituted aniline more basic than para-substituted analogue despite similar electronics?
- Increased conjugation with the ring
- Steric hindrance preventing resonance delocalization of the lone pair
- Higher molecular weight
- Greater solubility in organic solvents
Correct Answer: Steric hindrance preventing resonance delocalization of the lone pair
Q31. Which experimental observation indicates reduced basicity of an aromatic amine?
- Higher pKa of its conjugate acid
- Lower pKa of its conjugate acid
- Higher melting point
- Greater UV absorbance
Correct Answer: Lower pKa of its conjugate acid
Q32. How does solvent polarity generally affect the basicity of aromatic amines?
- Polar solvents stabilize the protonated form and can increase observed basicity (higher pKa of conjugate acid)
- Solvent polarity has no effect
- Nonpolar solvents always increase basicity
- Polar solvents always decrease basicity by solvating the neutral amine
Correct Answer: Polar solvents stabilize the protonated form and can increase observed basicity (higher pKa of conjugate acid)
Q33. Which is more basic in water: anilinium ion or protonated pyridinium ion?
- Anilinium ion
- Protonated pyridinium ion
- They have identical basicity
- Depends only on temperature
Correct Answer: Protonated pyridinium ion
Q34. What is the effect of multiple electron-withdrawing substituents (e.g., di-nitro) on aromatic amine basicity?
- They significantly increase basicity
- They significantly decrease basicity, often making the amine essentially non-basic
- No effect unless the substituents are ortho
- They convert amine to a stronger base than aliphatic amines
Correct Answer: They significantly decrease basicity, often making the amine essentially non-basic
Q35. Which pharmacological property is directly influenced by the basicity of an aromatic amine drug?
- Color of the tablet
- Degree of ionization at physiological pH affecting absorption and distribution
- Drug’s molecular weight
- Melting point of the active pharmaceutical ingredient
Correct Answer: Degree of ionization at physiological pH affecting absorption and distribution
Q36. Which of the following is a weaker base than aniline?
- Aniline
- p-Phenylenediamine (para-diamine)
- p-Nitroaniline
- o-Toluidine
Correct Answer: p-Nitroaniline
Q37. In terms of resonance, why do para electron-donating substituents often increase aniline basicity?
- They withdraw electrons from nitrogen
- They push electron density into the ring, increasing overall electron density on nitrogen, making the lone pair more available
- They cause steric hindrance that blocks protonation
- They lower molecular polarity
Correct Answer: They push electron density into the ring, increasing overall electron density on nitrogen, making the lone pair more available
Q38. Which of these aromatic amines will be mostly protonated at physiological pH (7.4) given a conjugate acid pKa ~9.4?
- No protonation (less than 1%)
- About 50% protonated
- Mostly protonated (>90%)
- Exactly 0% protonated
Correct Answer: Mostly protonated (>90%)
Q39. Which measurement can indicate the site of protonation in multifunctional aromatic amines?
- NMR chemical shift changes upon titration
- Melting point measurement
- Thin-layer chromatography Rf values
- Mass spectrometry alone
Correct Answer: NMR chemical shift changes upon titration
Q40. Why are tertiary aliphatic amines sometimes less basic than secondary ones in aqueous solution?
- Tertiary amines have less electron-donating groups
- Steric hindrance reduces solvation of the protonated form, lowering observed basicity
- Tertiary amines always form stable amides
- Secondary amines are aromatic
Correct Answer: Steric hindrance reduces solvation of the protonated form, lowering observed basicity
Q41. Which concept explains how substituent effects on aromatic amines can be correlated quantitatively?
- Hückel theory
- Hammett equation and sigma constants
- VSEPR theory
- Raoult’s law
Correct Answer: Hammett equation and sigma constants
Q42. Which amine is expected to have the highest percent ionization at pH 7.4: aniline (pKa 4.6) or piperidine (pKa ~11.1)?
- Aniline
- Piperidine
- Both equally ionized
- Neither ionized
Correct Answer: Piperidine
Q43. How does para-carboxyl group (–COOH) on an aromatic ring affect aniline basicity?
- Increases basicity due to resonance donation
- Decreases basicity due to strong electron-withdrawing resonance and inductive effects
- No effect on basicity
- Converts amine to a tertiary amine
Correct Answer: Decreases basicity due to strong electron-withdrawing resonance and inductive effects
Q44. Which structural change most lowers basicity of an aromatic amine used in drug molecules?
- Introducing alkyl groups on nitrogen
- Acylation of the amino group to form amide
- Adding electron-donating alkoxy substituents
- Converting ring to cyclohexane
Correct Answer: Acylation of the amino group to form amide
Q45. Which aromatic amine will be least soluble in water at neutral pH due to lack of protonation?
- p-Nitroaniline (pKa ~1.0)
- Aniline (pKa ~4.6)
- N,N-Dimethylaniline (pKa ~5.1)
- Pyridine (pKa ~5.25)
Correct Answer: p-Nitroaniline (pKa ~1.0)
Q46. What is the effect of resonance stabilization of the conjugate acid on observed basicity?
- Resonance stabilization of the conjugate acid increases basicity (higher pKa)
- Resonance stabilization of the conjugate acid decreases basicity
- It has no effect
- It always converts base to non-basic compound
Correct Answer: Resonance stabilization of the conjugate acid increases basicity (higher pKa)
Q47. Which technique reliably distinguishes between protonation at nitrogen versus protonation on ring sites?
- UV-visible spectroscopy without titration
- NMR titration and chemical shift analysis
- Melting point comparison
- Thin-layer chromatography
Correct Answer: NMR titration and chemical shift analysis
Q48. In medicinal chemistry, why might one intentionally reduce basicity of an aromatic amine in a lead compound?
- To increase basicity of metabolite
- To alter pKa to reduce off-target binding, improve permeability, or modify metabolic stability
- To ensure the drug is fully ionized at all pH
- To increase water solubility only
Correct Answer: To alter pKa to reduce off-target binding, improve permeability, or modify metabolic stability
Q49. Which of the following increases the nucleophilicity of an aromatic amine but may not increase its basicity proportionally?
- Resonance delocalization into the ring
- Steric hindrance blocking the lone pair
- Electron-donating substituents increasing electron density at nitrogen
- Conversion to amide
Correct Answer: Electron-donating substituents increasing electron density at nitrogen
Q50. Which observation suggests that a substituted aniline is a very weak base in aqueous media?
- It has a conjugate acid pKa around 9–10
- It remains largely unprotonated at pH 1
- It has a conjugate acid pKa < 2 and shows negligible protonation at physiological pH
- It forms salts readily with weak acids
Correct Answer: It has a conjugate acid pKa < 2 and shows negligible protonation at physiological pH
I am a Registered Pharmacist under the Pharmacy Act, 1948, and the founder of PharmacyFreak.com. I hold a Bachelor of Pharmacy degree from Rungta College of Pharmaceutical Science and Research. With a strong academic foundation and practical knowledge, I am committed to providing accurate, easy-to-understand content to support pharmacy students and professionals. My aim is to make complex pharmaceutical concepts accessible and useful for real-world application.
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