Protecting Groups: Amino protection – carbamates and amides MCQs With Answer

Introduction: This quiz set focuses on amino-protecting groups — specifically carbamates (Boc, Cbz, Fmoc, Alloc) and amide-type protections — as encountered in advanced organic and medicinal chemistry. Questions emphasize selection criteria, installation and removal methods, reaction compatibility, mechanisms, and orthogonality relevant to peptide synthesis and small-molecule drug design. Designed for M.Pharm students, the items probe practical decision-making (which protecting group to choose under given conditions), stability under acid/base/oxidative and reductive environments, and common side reactions. Mastery of these concepts enables correct protecting-group strategies to prevent undesired reactions, minimize racemization, and streamline multi-step syntheses in pharmaceutical research and development.

Q1. Which reagent is most commonly used to install a tert‑butoxycarbonyl (Boc) protecting group on an amine?

• Boc2O (di-tert-butyl dicarbonate)
• Fmoc-Cl (9‑fluorenylmethyloxycarbonyl chloride)
• Benzyl chloroformate (Cbz-Cl)
• Allyl chloroformate (Alloc-Cl)

Correct Answer: Boc2O (di-tert-butyl dicarbonate)

Q2. Which condition is typically used to remove a Boc protecting group from a secondary amine without affecting a Cbz group?

• TFA (trifluoroacetic acid) in dichloromethane
• Pd/C hydrogenolysis (H2, Pd/C)
• Piperidine in DMF
• TBAF (tetrabutylammonium fluoride)

Correct Answer: TFA (trifluoroacetic acid) in dichloromethane

Q3. The Fmoc protecting group is best characterized by which deprotection condition?

• Base-catalyzed cleavage (20% piperidine in DMF)
• Acid-catalyzed cleavage (TFA)
• Hydrogenolysis (H2, Pd/C)
• Reductive cleavage with LiAlH4

Correct Answer: Base-catalyzed cleavage (20% piperidine in DMF)

Q4. Which protecting group is most orthogonal to Fmoc in standard solid-phase peptide synthesis?

• Boc (tert‑butoxycarbonyl)
• Cbz (benzyloxycarbonyl)
• Trt (trityl on side-chain)
• Alloc (allyloxycarbonyl)

Correct Answer: Boc (tert‑butoxycarbonyl)

Q5. Which reagent cleaves the Cbz (benzyloxycarbonyl) protecting group selectively?

• Hydrogenation (H2, Pd/C)
• 20% piperidine in DMF
• TFA in DCM
• Oxidative cleavage with m‑CPBA

Correct Answer: Hydrogenation (H2, Pd/C)

Q6. Which statement correctly contrasts carbamates and amides as N‑protecting groups?

• Carbamates are generally more labile than amides under acidic or basic conditions.
• Amides are more labile than carbamates toward mild base (piperidine).
• Carbamates resist hydrogenolysis while amides are hydrogenolytically removed easily.
• Amides can be removed with dilute acids like Boc groups.

Correct Answer: Carbamates are generally more labile than amides under acidic or basic conditions.

Q7. Which method is commonly used to form carbamates from an amine in the laboratory?

• Reaction of amine with chloroformate (ROCOCl)
• Direct acylation with acid anhydride (RCO)2O
• Carboxylation with CO2 followed by reduction
• SN2 reaction with alkyl halide under basic conditions

Correct Answer: Reaction of amine with chloroformate (ROCOCl)

Q8. Alloc (allyloxycarbonyl) protecting groups on nitrogen are typically removed by which protocol?

• Pd(0)-catalyzed deprotection using Pd(PPh3)4 and a nucleophile (e.g., PhSiH3)
• Treatment with strong acid (HCl in EtOAc)
• Basic cleavage with 20% piperidine
• Hydrogenolysis with H2 and Pd/C

Correct Answer: Pd(0)-catalyzed deprotection using Pd(PPh3)4 and a nucleophile (e.g., PhSiH3)

Q9. During multi-step synthesis, which protecting group is preferred to avoid racemization of an adjacent stereocenter when coupling amino acids?

• Cbz is often preferred due to mild hydrogenolysis conditions that minimize racemization.
• Boc because TFA always prevents racemization during coupling.
• Fmoc because base-promoted deprotection accelerates racemization.
• Acetyl (N‑Ac) because it is very labile under coupling conditions.

Correct Answer: Cbz is often preferred due to mild hydrogenolysis conditions that minimize racemization.

Q10. Which side reaction is a known problem when removing Boc groups with strong acid in presence of acid‑sensitive functionalities?

• Cleavage of other acid-labile protecting groups and tertiary carbocation formation leading to alkylation
• Hydrogenolytic removal of benzyl groups
• Base-catalyzed β‑elimination on saturated substrates
• Oxidative cleavage of sulfides to sulfoxides

Correct Answer: Cleavage of other acid-labile protecting groups and tertiary carbocation formation leading to alkylation

Q11. Which reagent sequence installs an Fmoc group on an amine most efficiently?

• Fmoc‑Cl with a weak base like Na2CO3 or NaHCO3 in aqueous/organic biphasic medium
• Treatment with Boc2O and triethylamine
• Reaction with benzyl chloroformate and then hydrogenation
• Acylation with acetic anhydride followed by oxidation

Correct Answer: Fmoc‑Cl with a weak base like Na2CO3 or NaHCO3 in aqueous/organic biphasic medium

Q12. For selective deprotection of Cbz in presence of Boc and Fmoc, which condition is most appropriate?

• Hydrogenolysis (H2, Pd/C) — removes Cbz selectively
• TFA treatment — removes Cbz selectively
• 20% piperidine — removes Cbz selectively
• Strong base like NaOH — removes Cbz selectively

Correct Answer: Hydrogenolysis (H2, Pd/C) — removes Cbz selectively

Q13. What factor most strongly influences the lability of a carbamate protecting group under acidic conditions?

• Stability of the leaving group after protonation (e.g., tert‑butyl cation stability)
• The size of the amide N–H bond
• Presence of heavy metals in the reaction mixture
• Optical rotation of the substrate

Correct Answer: Stability of the leaving group after protonation (e.g., tert‑butyl cation stability)

Q14. Which protecting group would you choose for an amine if the synthetic route requires strong acid but must avoid hydrogenation?

• Boc, because it is removed by acid and does not require hydrogenation
• Cbz, because it is acid‑stable and removed by hydrogenation
• Fmoc, because it is removed by hydrogenation
• Benzyl (Bn), because it is base-labile

Correct Answer: Boc, because it is removed by acid and does not require hydrogenation

Q15. Which of the following is a common method to form amides as protecting groups on nitrogen?

• Acylation with acyl chlorides or activated esters (e.g., RCOCl or RCOOSu)
• Treatment with diazomethane
• Oxidation with Dess‑Martin periodinane
• Nitration followed by reduction

Correct Answer: Acylation with acyl chlorides or activated esters (e.g., RCOCl or RCOOSu)

Q16. Which protecting group is most appropriate when a temporary, base‑labile protection of the amine is required during SPPS (solid phase peptide synthesis)?

• Fmoc, because it is base-labile and commonly used in SPPS
• Boc, because it is base-labile and removed by piperidine
• Cbz, because it is base-labile
• Trt, because it is base-labile

Correct Answer: Fmoc, because it is base-labile and commonly used in SPPS

Q17. Which cleavage condition would remove both Cbz and benzyl ether groups simultaneously?

• Hydrogenolysis (H2, Pd/C)
• TFA in DCM
• 20% piperidine in DMF
• Oxidative conditions with KMnO4

Correct Answer: Hydrogenolysis (H2, Pd/C)

Q18. In the Curtius rearrangement pathway used to introduce carbamates, what intermediate rearranges to the isocyanate?

• Acyl azide formed from acyl chloride and sodium azide
• Carbamate directly formed from alcohol and CO2
• Amide formed by direct acylation of amine
• Nitrene generated by diazotization of an amine

Correct Answer: Acyl azide formed from acyl chloride and sodium azide

Q19. Which protecting group would be least compatible with strong nucleophiles like hydride (LiAlH4) during later steps?

• Cbz and benzyl-derived carbamates because they are hydrogenolytically and reductively labile
• Fmoc because it is inert to hydride reductions
• Boc because it is fully resistant to hydride donors
• Simple amides (N‑acyl) because they are reduced only under very mild conditions

Correct Answer: Cbz and benzyl-derived carbamates because they are hydrogenolytically and reductively labile

Q20. When planning orthogonal protection, which pair is the most orthogonal (i.e., each deprotected under distinct, nonoverlapping conditions)?

• Fmoc (base‑labile) and Cbz (hydrogenolysis) — highly orthogonal
• Fmoc and Boc — both removed under the same conditions, not orthogonal
• Cbz and benzyl ether — orthogonal because one is acid-labile and the other base-labile
• Boc and t‑butyl ester — orthogonal because both require hydrogenolysis

Correct Answer: Fmoc (base‑labile) and Cbz (hydrogenolysis) — highly orthogonal

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