Reactions of imidazole MCQs With Answer

Reactions of imidazole MCQs With Answer

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Reactions of imidazole MCQs With Answer

Imidazole chemistry is central to medicinal chemistry and pharmaceutical synthesis because the five‑membered aromatic ring exhibits unique reactivity — nucleophilic/basic behavior at the pyridine‑like nitrogen, electrophilic substitution at C‑2/C‑5, N‑alkylation to form imidazolium salts and generation of N‑heterocyclic carbenes (NHCs) by deprotonation. B.Pharm students must master reaction mechanisms (Vilsmeier formylation, halogenation, metalation), tautomerism, acidity/basicity (pKa ≈ 7 for conjugate acid), ligand behavior in metalloenzymes, and implications for drug design and metabolism. This set of focused, application‑oriented MCQs will strengthen your understanding of imidazole reactions and pharmaceutical relevance. Now let’s test your knowledge with 50 MCQs on this topic.

Q1. Which statement correctly describes the basic structural features of imidazole?

  • Six-membered aromatic ring containing one nitrogen atom
  • Five-membered aromatic heterocycle with two nitrogen atoms at positions 1 and 3
  • Five-membered non-aromatic ring with two oxygen atoms
  • Linear chain compound with alternating double bonds and one nitrogen

Correct Answer: Five-membered aromatic heterocycle with two nitrogen atoms at positions 1 and 3

Q2. How many π electrons contribute to the aromaticity of imidazole?

  • 4 π electrons
  • 6 π electrons
  • 8 π electrons
  • 10 π electrons

Correct Answer: 6 π electrons

Q3. At which site is imidazole preferentially protonated (conjugate acid formation)?

  • At the carbon C‑2 position
  • At the pyridine‑like nitrogen (N3)
  • At the pyrrole‑like nitrogen (N1)
  • At the C‑5 position

Correct Answer: At the pyridine‑like nitrogen (N3)

Q4. The pKa of the conjugate acid of imidazole (pKaH) is approximately:

  • 1.0
  • 4.5
  • 7.0
  • 12.0

Correct Answer: 7.0

Q5. Which tautomeric process is characteristic of imidazole?

  • Ring opening to form an acyclic amide
  • Proton transfer between the two ring nitrogens (N1 and N3)
  • Migration of a double bond to form a six-membered ring
  • Interconversion between oxazole and imidazole

Correct Answer: Proton transfer between the two ring nitrogens (N1 and N3)

Q6. Electrophilic aromatic substitution on imidazole mainly occurs at which carbon?

  • C-4
  • C-3
  • C-2
  • C-1

Correct Answer: C-2

Q7. Under basic alkylation conditions (deprotonation followed by alkyl halide), which nitrogen is most commonly alkylated?

  • The pyrrole-like N1 after deprotonation
  • The pyridine-like N3 without deprotonation
  • The carbon C‑2 via direct alkylation
  • No alkylation occurs under basic conditions

Correct Answer: The pyrrole-like N1 after deprotonation

Q8. The Vilsmeier–Haack formylation of imidazole (DMF + POCl3) gives formylation primarily at which position?

  • N‑alkylated imidazole
  • C‑2 formylimidazole
  • C‑4 chlorinated imidazole
  • N‑oxide formation

Correct Answer: C‑2 formylimidazole

Q9. Deprotonation of an N,N′‑dialkylimidazolium salt at the C‑2 position yields which reactive species?

  • An imidazolium radical cation
  • An N‑heterocyclic carbene (NHC)
  • A diazonium salt
  • An imidazole N‑oxide

Correct Answer: An N‑heterocyclic carbene (NHC)

Q10. When imidazole acts as a ligand in metalloproteins (histidine side chain), which atom typically coordinates to metal ions?

  • The carbon C‑5 atom
  • The pyrrole-type nitrogen (N1)
  • The pyridine-type nitrogen (N3)
  • The protonated nitrogen

Correct Answer: The pyridine-type nitrogen (N3)

Q11. Bromination of imidazole using N‑bromosuccinimide (NBS) typically occurs at which site?

  • N‑bromination only
  • C‑2 bromination
  • C‑4 bromination exclusively
  • Ring cleavage

Correct Answer: C‑2 bromination

Q12. Oxidation of imidazole can give an N‑oxide. Which nitrogen is most likely to form the N‑oxide?

  • The protonated nitrogen only
  • The pyridine-like nitrogen (N3)
  • The pyrrole-like nitrogen (N1)
  • The carbon C‑2

Correct Answer: The pyridine-like nitrogen (N3)

Q13. Catalytic hydrogenation of imidazole under mild conditions primarily reduces which bonds to give dihydro derivatives?

  • The N–C single bonds
  • The C=C double bonds in the ring
  • The aromaticity is preserved; no reduction occurs
  • The N–H bond only

Correct Answer: The C=C double bonds in the ring

Q14. Which classical multicomponent synthesis builds the imidazole ring from a 1,2‑dicarbonyl compound, an aldehyde and ammonia or amine?

  • Buchwald–Hartwig synthesis
  • Debus–Radziszewski imidazole synthesis
  • Mannich reaction
  • Friedel–Crafts acylation

Correct Answer: Debus–Radziszewski imidazole synthesis

Q15. Why is imidazole commonly used as a biological buffer in biochemistry?

  • Because it is non-polar and insoluble in water
  • Because its conjugate acid has a pKa close to physiological pH (~7)
  • Because it forms permanent covalent bonds with proteins
  • Because it is a strong acid

Correct Answer: Because its conjugate acid has a pKa close to physiological pH (~7)

Q16. Which reagent is commonly used to generate a 2‑lithioimidazole for further C‑2 functionalization?

  • Sodium borohydride (NaBH4)
  • Butyllithium (n‑BuLi) or LDA at low temperature
  • Sodium hydroxide at room temperature
  • Hydrochloric acid

Correct Answer: Butyllithium (n‑BuLi) or LDA at low temperature

Q17. N‑Alkylation of imidazole with an alkyl halide under neutral conditions typically gives which product type?

  • C‑alkylated imidazole only
  • N‑alkylated imidazolium salt if complete alkylation occurs
  • Ring opened acyclic products
  • No reaction occurs with alkyl halides

Correct Answer: N‑alkylated imidazolium salt if complete alkylation occurs

Q18. Which base is typically strong enough to deprotonate an imidazolium salt to form an NHC?

  • Sodium chloride (NaCl)
  • Sodium hydride (NaH) or potassium tert‑butoxide (KOtBu)
  • Water
  • Acetic acid

Correct Answer: Sodium hydride (NaH) or potassium tert‑butoxide (KOtBu)

Q19. Which electrophilic reagent pair is often used to perform formylation at C‑2 of imidazole (Vilsmeier reagent)?

  • DMF + POCl3
  • NaBH4 + MeOH
  • H2/Pd
  • NBS + light

Correct Answer: DMF + POCl3

Q20. Which position on imidazole is most activated toward electrophilic acylation?

  • C‑2 position
  • N‑3 position only
  • C‑4 position exclusively
  • C‑5 but not C‑2

Correct Answer: C‑2 position

Q21. Nitration of imidazole is challenging because:

  • Imidazole is too electron rich and polymerizes under nitration conditions
  • The ring is inert to electrophiles under all conditions
  • Strong oxidizing acidic conditions can oxidize or destroy the heterocycle
  • Nitric acid immediately converts imidazole into benzene

Correct Answer: Strong oxidizing acidic conditions can oxidize or destroy the heterocycle

Q22. Imidazole‑containing antifungal drugs (like ketoconazole) act by:

  • Blocking cell wall synthesis by binding peptidoglycan
  • Complexing to fungal cytochrome P450 haem iron and inhibiting 14α‑demethylase
  • Inhibiting DNA polymerase directly
  • Acting as protonophores to collapse membrane potential

Correct Answer: Complexing to fungal cytochrome P450 haem iron and inhibiting 14α‑demethylase

Q23. Which analytical technique is best to confirm N‑alkylation of imidazole (appearance of imidazolium)?

  • IR showing strong carbonyl peak at 1700 cm⁻¹
  • 1H NMR showing disappearance of N‑H signal and appearance of new alkyl signals
  • Thin layer chromatography only
  • UV‑Vis showing large bathochromic shift

Correct Answer: 1H NMR showing disappearance of N‑H signal and appearance of new alkyl signals

Q24. Generation of a stable N‑heterocyclic carbene (NHC) from an imidazolium salt typically requires deprotonation at which site?

  • The N‑alkyl side chain
  • The C‑2 position (between the two nitrogens)
  • The C‑4 methyl substituent
  • The ring nitrogen N1

Correct Answer: The C‑2 position (between the two nitrogens)

Q25. Which substituent on imidazole will most strongly activate the ring toward electrophilic substitution?

  • A strongly electron-withdrawing nitro group at C‑4
  • An N‑alkyl substituent increasing electron density
  • A carbonyl group directly attached to C‑2
  • A halogen substituent at C‑5

Correct Answer: An N‑alkyl substituent increasing electron density

Q26. Reaction of imidazole with chloroform and strong base (Carbene insertion conditions) is most likely to give which outcome?

  • Formation of trichloromethyl imidazole without further change
  • Carbene insertion into N–H giving N‑CCl3 substituted imidazole or decomposition
  • Direct oxidation to imidazole N‑oxide
  • No reaction under any conditions

Correct Answer: Carbene insertion into N–H giving N‑CCl3 substituted imidazole or decomposition

Q27. Which condition favors C‑alkylation of imidazole over N‑alkylation?

  • Using hard nucleophiles in protic solvent
  • Using strong base and polar aprotic solvent to generate C‑metalated species
  • Carrying out reaction at very low temperature without base
  • Using only dilute acid

Correct Answer: Using strong base and polar aprotic solvent to generate C‑metalated species

Q28. In an electrophilic aromatic substitution mechanism on imidazole, the rate‑determining intermediate is:

  • A free radical located at C‑4
  • A protonated imidazole cation radical
  • A σ‑complex (Wheland intermediate) stabilized by adjacent nitrogens
  • A neutral carbocation off the ring

Correct Answer: A σ‑complex (Wheland intermediate) stabilized by adjacent nitrogens

Q29. Which reagent is commonly used to brominate activated aromatic heterocycles selectively at electron-rich carbons?

  • N‑Bromosuccinimide (NBS)
  • Sodium hypochlorite (NaOCl)
  • Sodium azide (NaN3)
  • Hydrochloric acid

Correct Answer: N‑Bromosuccinimide (NBS)

Q30. Why is directed metalation often used to functionalize imidazole at C‑2?

  • Because C‑2 cannot undergo electrophilic substitution
  • Metalation gives a carbon nucleophile (e.g., 2‑lithioimidazole) that reacts with electrophiles selectively
  • Metalation destroys the aromaticity making reactions faster
  • Metalation only occurs at N‑positions, not carbon

Correct Answer: Metalation gives a carbon nucleophile (e.g., 2‑lithioimidazole) that reacts with electrophiles selectively

Q31. Which product results from N‑oxidation of imidazole followed by hydrolysis under mild conditions?

  • Imidazole is reduced to ammonia
  • Formation of imidazole N‑oxide which can be further transformed to 2‑oxo derivatives
  • Direct formation of 2‑chloroimidazole
  • Polymeric tar without defined structure

Correct Answer: Formation of imidazole N‑oxide which can be further transformed to 2‑oxo derivatives

Q32. Which type of reaction is used to convert an imidazole into its corresponding imidazolium salt?

  • Oxidation with KMnO4
  • N‑alkylation with an alkyl halide
  • Reduction with LiAlH4
  • Photochemical rearrangement

Correct Answer: N‑alkylation with an alkyl halide

Q33. In medicinal chemistry, why is the imidazole ring often used as a bioisostere for other heterocycles?

  • Because it is completely inert biologically
  • Because its pKa, hydrogen bonding and aromaticity can mimic other functional groups and tune binding and solubility
  • Because it always improves metabolic stability
  • Because it prevents any protein binding

Correct Answer: Because its pKa, hydrogen bonding and aromaticity can mimic other functional groups and tune binding and solubility

Q34. Which reaction condition is most likely to favor N‑oxide formation over C‑electrophilic substitution?

  • Mild base in non‑polar solvent
  • Strong oxidant like m‑CPBA under controlled conditions
  • Hydrogenation using Pd/C
  • Alkylation with methyl iodide

Correct Answer: Strong oxidant like m‑CPBA under controlled conditions

Q35. Which of the following best explains why electrophilic substitution at C‑2 is favored over C‑5 in unsubstituted imidazole?

  • C‑2 position is sterically more hindered
  • The σ‑complex at C‑2 is better stabilized by adjacent nitrogen lone pairs and resonance
  • C‑5 cannot form a σ‑complex
  • Electrophiles cannot approach C‑5 due to positive charge on N1

Correct Answer: The σ‑complex at C‑2 is better stabilized by adjacent nitrogen lone pairs and resonance

Q36. Which transformation converts an imidazole into an N‑protected derivative useful for selective C‑functionalization?

  • N‑oxidation to make it more nucleophilic
  • N‑sulfonylation or N‑acylation to protect the nitrogen
  • Hydrogenation to saturate the ring
  • Direct chlorination at C‑2

Correct Answer: N‑sulfonylation or N‑acylation to protect the nitrogen

Q37. The C‑2 hydrogen of imidazole is relatively acidic because:

  • Its removal leads to aromatic disruption
  • Deprotonation yields a carbanion stabilized by adjacent nitrogens and resonance
  • The C‑2 hydrogen is actually the least acidic in the molecule
  • It forms a stable free radical upon removal

Correct Answer: Deprotonation yields a carbanion stabilized by adjacent nitrogens and resonance

Q38. Which reagent set is suitable to selectively introduce an acyl group at C‑2 of imidazole?

  • Acyl chloride with Lewis acid after protecting N if necessary
  • Hydrogen peroxide only
  • Sodium hydroxide and ethanol
  • Photochemical chlorination

Correct Answer: Acyl chloride with Lewis acid after protecting N if necessary

Q39. Which metabolism pathway commonly affects imidazole‑containing drugs in vivo?

  • Complete aromatic ring cleavage only
  • CYP‑mediated N‑dealkylation or N‑oxidation
  • Spontaneous dimerization in blood
  • No metabolism occurs for imidazole drugs

Correct Answer: CYP‑mediated N‑dealkylation or N‑oxidation

Q40. In cross‑coupling chemistry, a 2‑halogenated imidazole (e.g., 2‑bromoimidazole) can be used as a substrate for which reaction to install aryl groups at C‑2?

  • Mitsunobu reaction
  • Pd‑catalyzed Suzuki or Negishi coupling
  • Hydride reduction
  • Williamson ether synthesis

Correct Answer: Pd‑catalyzed Suzuki or Negishi coupling

Q41. What is the primary reason imidazole rings can stabilize carbocations formed during electrophilic attack?

  • Imidazole is electron-withdrawing and destabilizes carbocations
  • Lone pairs on nitrogens provide resonance stabilization to the σ‑complex
  • Carbocations are not formed in these reactions
  • The ring fragments before any stabilization occurs

Correct Answer: Lone pairs on nitrogens provide resonance stabilization to the σ‑complex

Q42. Which protecting group strategy is commonly used when performing C‑2 lithiation to avoid N‑metalation?

  • No protection is needed; proceed directly
  • N‑silylation or N‑acylation to block the nitrogen
  • Oxidation to N‑oxide before lithiation
  • Alkylation with benzyl halide only

Correct Answer: N‑silylation or N‑acylation to block the nitrogen

Q43. In designing imidazole‑based drugs, which property of the ring can be tuned to improve membrane permeability?

  • Increasing the number of free NH groups
  • Alkylation of N and introduction of lipophilic substituents to modulate pKa and lipophilicity
  • Always adding charged groups to the ring
  • Removing aromaticity completely

Correct Answer: Alkylation of N and introduction of lipophilic substituents to modulate pKa and lipophilicity

Q44. Which of the following is true about imidazole N‑protonation versus N‑alkylation?

  • Protonation and alkylation always occur at the same nitrogen
  • Protonation occurs at the basic pyridine‑like nitrogen, while alkylation often requires deprotonation and can occur at the other nitrogen
  • Alkylation only occurs after oxidation
  • Protonation prevents any subsequent alkylation

Correct Answer: Protonation occurs at the basic pyridine‑like nitrogen, while alkylation often requires deprotonation and can occur at the other nitrogen

Q45. Which experimental observation indicates successful formation of a 2‑substituted imidazole via electrophilic substitution?

  • Complete disappearance of the aromatic proton signals in NMR
  • Appearance of a new substituent signal and downfield shift of nearby aromatic proton(s) in 1H NMR
  • No change in mass spectrometry data
  • Loss of all nitrogen signals in IR

Correct Answer: Appearance of a new substituent signal and downfield shift of nearby aromatic proton(s) in 1H NMR

Q46. Which heterocyclic carbene derived from imidazolium salts is widely used as a ligand in organometallic catalysis?

  • Free imidazole without deprotonation
  • Imidazol-2-ylidene (an NHC derived from C‑2 deprotonation)
  • Imidazole N‑oxide radical
  • Isoimidazole anion

Correct Answer: Imidazol-2-ylidene (an NHC derived from C‑2 deprotonation)

Q47. Which of the following statements about imidazole ring halogenation is correct?

  • Halogenation cannot occur on imidazole
  • Electrophilic halogenation tends to occur at C‑2 and C‑5, with C‑2 often favored
  • Halogenation gives exclusive N‑halogenation under all conditions
  • Halogenation always destroys aromaticity irreversibly

Correct Answer: Electrophilic halogenation tends to occur at C‑2 and C‑5, with C‑2 often favored

Q48. During synthesis of imidazole derivatives, why is selective N‑protection important?

  • To make the molecule colorless
  • To prevent unwanted N‑alkylation or N‑metalation and allow selective C‑functionalization
  • Protection is never necessary for imidazole chemistry
  • To increase the acidity of the ring

Correct Answer: To prevent unwanted N‑alkylation or N‑metalation and allow selective C‑functionalization

Q49. Which feature of the imidazole ring enables it to act as an acid–base catalyst in enzyme active sites?

  • Its inability to hydrogen bond
  • The presence of a protonatable nitrogen with pKa near physiological pH enabling both proton donation and acceptance
  • Its high molecular weight
  • Its covalent attachment to the active site metal only

Correct Answer: The presence of a protonatable nitrogen with pKa near physiological pH enabling both proton donation and acceptance

Q50. Which laboratory strategy is best to introduce an electrophile selectively at C‑5 of an imidazole when C‑2 is blocked?

  • Use strong oxidants to remove C‑2 first
  • Block C‑2 (e.g., via substitution or protection) and then perform electrophilic substitution which will direct to C‑5
  • Perform direct N‑alkylation to force C‑5 substitution
  • Heating in water without reagents will selectively functionalize C‑5

Correct Answer: Block C‑2 (e.g., via substitution or protection) and then perform electrophilic substitution which will direct to C‑5

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