Organic Chemistry 2 Quiz: Protecting Groups Alcohols Amines
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Protecting Groups Alcohols AminesQuestion 1 of 17

A student attempts to synthesize 5-methylhexane-1,5-diol by first treating 5-bromopentan-1-ol with magnesium turnings in diethyl ether, followed by the addition of acetone and an aqueous workup. The reaction yields very little of the desired product. What is the primary reason for this failure, which necessitates a protection strategy?

The magnesium preferentially coordinates with the hydroxyl oxygen, preventing the formation of the organometallic reagent.
The Grignard reagent, once formed, is intramolecularly quenched by the acidic proton of the alcohol group.
The Grignard reagent acts as a strong base, causing an E2 elimination to form pent-4-en-1-ol.
The ether solvent is deprotonated by the highly reactive Grignard reagent, rendering it inactive.
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Organic Chemistry 2 Quiz

Organic Chemistry 2 Quiz: Protecting Groups Alcohols Amines

Practice Protecting Groups Alcohols Amines in Organic Chemistry 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Protecting Groups Alcohols Amines, giving you a quick way to practice the rules, question types, and explanations that matter most for Organic Chemistry 2.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

A student attempts to synthesize 5-methylhexane-1,5-diol by first treating 5-bromopentan-1-ol with magnesium turnings in diethyl ether, followed by the addition of acetone and an aqueous workup. The reaction yields very little of the desired product. What is the primary reason for this failure, which necessitates a protection strategy?

  1. The magnesium preferentially coordinates with the hydroxyl oxygen, preventing the formation of the organometallic reagent.
  2. The Grignard reagent, once formed, is intramolecularly quenched by the acidic proton of the alcohol group. (correct answer)
  3. The Grignard reagent acts as a strong base, causing an E2 elimination to form pent-4-en-1-ol.
  4. The ether solvent is deprotonated by the highly reactive Grignard reagent, rendering it inactive.
Explanation: The primary issue is that Grignard reagents are strong bases. If a Grignard reagent is formed from 5-bromopentan-1-ol, its carbanionic carbon will immediately deprotonate the acidic hydroxyl group on the other end of the same molecule in an intramolecular acid-base reaction. This consumes the Grignard reagent, preventing it from reacting with acetone. To make this synthesis work, the alcohol must first be protected (e.g., as a silyl ether).

Question 2

A synthetic intermediate contains a primary alcohol protected as a tert-butyldimethylsilyl (TBDMS) ether and a primary amine protected as a tert-butoxycarbonyl (Boc) carbamate. The final step requires the selective removal of the TBDMS group while leaving the Boc group intact. Which reagent would be most effective for this selective deprotection?

  1. Trifluoroacetic acid (TFA) in CH₂Cl₂.
  2. Tetrabutylammonium fluoride (TBAF) in THF. (correct answer)
  3. H₂ gas with a Palladium on carbon (Pd/C) catalyst.
  4. Aqueous sodium hydroxide with heat.
Explanation: This is a problem of orthogonal deprotection. Silyl ethers are characteristically cleaved by fluoride ions, so TBAF is the ideal reagent. Boc groups are stable to fluoride but are labile to strong acid. Therefore, TFA (choice A) would remove the Boc group. H₂/Pd-C (choice C) is used for hydrogenolysis (e.g., cleaving Cbz or benzyl groups) and would not affect either group. NaOH (choice D) is not effective for cleaving TBDMS ethers under mild conditions.

Question 3

A synthetic intermediate contains a sterically hindered tertiary alcohol that must be protected. The subsequent reaction step involves treatment with a strong base. Which protecting group would be most difficult to install on this alcohol?

  1. Trimethylsilyl (TMS) group using TMS-Cl.
  2. tert-Butyldimethylsilyl (TBDMS) group using TBDMS-Cl. (correct answer)
  3. Tetrahydropyranyl (THP) group using dihydropyran.
  4. Acetyl (Ac) group using acetic anhydride.
Explanation: The installation of a protecting group is an SN2-like reaction at the silicon (for silyl ethers) or carbonyl carbon (for esters). The rate is highly sensitive to steric hindrance. The TBDMS group is significantly larger and more sterically demanding than TMS, THP, or acetyl groups. Attempting to attach the bulky TBDMS group to an already hindered tertiary alcohol would be extremely slow and inefficient.

Question 4

An intermediate has a primary alcohol protected as a tetrahydropyranyl (THP) ether and a phenol protected as a robust methyl ether. The synthetic goal is to deprotect the primary alcohol while leaving the phenol's methyl ether intact. Which set of deprotection conditions would best accomplish this selective transformation?

  1. BBr₃ in CH₂Cl₂, followed by H₂O.
  2. H₂, Pd/C in ethanol.
  3. Tetrabutylammonium fluoride (TBAF) in THF.
  4. Aqueous acetic acid (AcOH/H₂O) with gentle warming. (correct answer)
Explanation: A THP ether is a type of acetal and is readily cleaved under mild aqueous acid conditions. A methyl ether on a phenol is very stable and requires harsh conditions, like the strong Lewis acid BBr₃ (A), for cleavage. H₂/Pd-C (B) is for hydrogenolysis and would not affect either group. TBAF (C) is for cleaving silyl ethers. Therefore, mild aqueous acid is the best choice for selectively removing the THP group.

Question 5

A chemist wants to perform a Wittig reaction on the ketone in 4-hydroxycyclohexanone. They realize the alcohol must be protected first. Which of the following strategies for protecting the alcohol would fail because the protecting reagent would also react with the ketone?

  1. Using TBDMS-Cl and imidazole to form a silyl ether.
  2. Using dihydropyran and a catalytic amount of acid to form a THP ether.
  3. Using ethylene glycol and a catalytic amount of acid to form a cyclic acetal. (correct answer)
  4. Using benzyl bromide and NaH to form a benzyl ether.
Explanation: The Wittig reaction targets the ketone. Ethylene glycol with an acid catalyst is the standard reagent for protecting ketones and aldehydes as cyclic acetals. This reagent would react with the target ketone, protecting it and thus preventing the subsequent Wittig reaction. The other reagents (A, B, D) are standard methods for protecting alcohols that would not react with the ketone under the conditions used.

Question 6

In a synthesis starting from 4-nitrotoluene, the benzylic methyl group must be oxidized to a carboxylic acid using a strong oxidant like KMnO₄, and the nitro group must be reduced to an amine. Why is it synthetically advantageous to perform the oxidation before the reduction?

  1. The nitro group activates the benzylic position, making the oxidation reaction faster and higher yielding.
  2. The amino group is sensitive to oxidation and would likely be degraded by KMnO₄. (correct answer)
  3. The amino group is a strong deactivator, making the benzylic C-H bonds too strong to be oxidized.
  4. The reduction of the nitro group must be catalyzed by acid, which is incompatible with the basic oxidation conditions.
Explanation: The primary reason for this order of operations is chemoselectivity. Aromatic amino groups are very easily oxidized and would be destroyed by a powerful oxidizing agent like KMnO₄. The nitro group, however, is robust and stable under these conditions. Therefore, the oxidation must be done while the nitrogen is in the nitro oxidation state, which effectively serves as a protecting group for the future amine against oxidation. After oxidation, the nitro group can be safely reduced (e.g., with H₂, Pd/C).

Question 7

A chemist wishes to convert p-aminotoluene to p-aminobenzoic acid. This transformation requires oxidizing the methyl group to a carboxylic acid using KMnO₄. What is the best sequence of protection and oxidation steps?

    1. Oxidize with KMnO₄; 2. Protect amine with (Boc)₂O.
    1. Oxidize with KMnO₄; the amine does not require protection from this reagent.
    1. Protect amine with Boc group; 2. Oxidize with KMnO₄; 3. Deprotect with TFA.
    1. Protect amine with acetic anhydride; 2. Oxidize with KMnO₄; 3. Hydrolyze amide.
     (correct answer)
Explanation: When converting p-aminotoluene to p-aminobenzoic acid, you need to consider how the amine group will interact with your oxidizing conditions. Primary aromatic amines are basic and nucleophilic, making them reactive toward many reagents including strong oxidizers like KMnO₄. The best approach is option D: protect the amine with acetic anhydride, oxidize with KMnO₄, then hydrolyze the amide. Acetic anhydride converts the primary amine to an amide (acetamide), which is much less basic and nucleophilic than the free amine. This protection allows KMnO₄ to selectively oxidize the methyl group to a carboxylic acid without interfering with the nitrogen. The final hydrolysis step removes the acetyl protecting group, regenerating the free amine. Option A is backwards - you must protect before oxidizing, not after. Option B incorrectly assumes the amine won't interfere with KMnO₄. Primary amines can coordinate to metals and undergo unwanted side reactions under these harsh oxidizing conditions. Option C suggests using a Boc protecting group, but Boc groups are acid-labile and may not survive the strongly basic conditions typically used with KMnO₄ oxidations. Additionally, Boc protection is more commonly used for temporary protection during milder reactions. Remember this pattern: when planning multi-step syntheses involving reactive functional groups, always consider protection strategies. Amide protection (like acetylation) is particularly useful because amides are stable under most reaction conditions yet easily hydrolyzed when needed.

Question 8

A synthetic plan requires an alcohol to be protected. The intermediate must survive treatment with n-butyllithium and then be deprotected under non-acidic, non-basic, and non-reducing conditions. Which protecting group is most suitable?

  1. A tetrahydropyranyl (THP) ether.
  2. An acetate ester.
  3. A tert-butyldimethylsilyl (TBDMS) ether. (correct answer)
  4. A tosylate ester.
Explanation: The protecting group must be stable to a strong base/nucleophile (n-BuLi) and be removable under neutral conditions. A TBDMS ether is stable to n-BuLi and is removed by a fluoride source (TBAF), which is non-acidic/basic/reducing. A THP ether (A) is stable to n-BuLi but requires acid for removal. An acetate ester (B) and a tosylate ester (D) would both be attacked by n-butyllithium.

Question 9

A synthesis requires the selective reduction of an ester to a primary alcohol in a molecule that also contains a secondary alcohol. The chosen reducing agent is lithium aluminum hydride (LiAlH₄). Which protecting group for the secondary alcohol would be most suitable for this transformation?

  1. An acetate group, formed with acetic anhydride.
  2. A tosylate group, formed with tosyl chloride.
  3. A tert-butyldimethylsilyl (TBDMS) ether group. (correct answer)
  4. No protecting group is needed as LiAlH₄ is selective for esters.
Explanation: LiAlH₄ is a very powerful, non-selective reducing agent that will react with both esters and acidic protons from alcohols. A TBDMS ether is stable to the basic, nucleophilic conditions of a LiAlH₄ reduction. An acetate is an ester and would also be reduced. A tosylate can be reduced or displaced by hydride. Leaving the alcohol unprotected would cause it to be deprotonated by LiAlH₄, consuming the reagent and potentially causing a failed reaction.

Question 10

A reaction to protect an alcohol with TBDMS-Cl proceeds very slowly. The addition of a catalytic amount of imidazole significantly increases the reaction rate. What is the primary role of imidazole in this reaction?

  1. To fully deprotonate the alcohol, forming a highly nucleophilic alkoxide species.
  2. To act as a shuttle, forming a reactive silyl-imidazolium intermediate and also neutralizing the HCl byproduct. (correct answer)
  3. To function as a phase-transfer catalyst, increasing the solubility of the reagents.
  4. To prevent the acidic HCl byproduct from isomerizing the alcohol's stereocenter.
Explanation: Imidazole plays a dual role. It is a nucleophilic catalyst that reacts with TBDMS-Cl to form a highly reactive silyl-imidazolium ion, which is then more readily attacked by the alcohol. Secondly, and just as importantly, it is a base that neutralizes the HCl generated as a byproduct. This prevents the buildup of acid which would establish an unfavorable equilibrium and slow the reaction down (Le Châtelier's principle). Imidazole is not a strong enough base to fully deprotonate an alcohol (A).

Question 11

A chemist attempts to synthesize p-bromoaniline by treating aniline directly with one equivalent of Br₂. The major product isolated is 2,4,6-tribromoaniline. What modification to the synthesis would best achieve the desired mono-bromination at the para position?

  1. Protect the amine as an acetanilide before bromination, followed by amide hydrolysis. (correct answer)
  2. Add a strong acid like H₂SO₄ to the reaction mixture to protonate the aniline.
  3. Run the reaction at a very low temperature (-78 °C) to control the reaction rate.
  4. Use a milder brominating agent, such as N-bromosuccinimide (NBS), instead of Br₂.
Explanation: The amino group of aniline is a powerful activating group, leading to uncontrollable polybromination. Converting the amine to an acetanilide (-NHCOCH₃) moderates its activating effect and adds steric bulk, favoring mono-bromination at the para position. The amide can then be hydrolyzed back to the amine. Adding strong acid (B) would form the anilinium ion, which is a meta-director. Low temperature (C) or a milder reagent (D) are generally insufficient to prevent over-reaction due to the extremely high reactivity of the aniline ring.

Question 12

A molecule contains both a cis-1,2-diol and a carboxylic acid. The goal is to reduce the carboxylic acid to a primary alcohol using LiAlH₄ without affecting the diol. What is the most efficient strategy to achieve this selective reduction?

  1. Protect the diol as a cyclic acetal using acetone and an acid catalyst before the reduction step. (correct answer)
  2. Protect each hydroxyl group of the diol individually with a TBDMS ether before the reduction step.
  3. Use NaBH₄ instead of LiAlH₄, as it will selectively reduce the carboxylic acid.
  4. No protection is needed; the acidic protons of the diol will not interfere with the reduction.
Explanation: LiAlH₄ will be quenched by the acidic protons of both the carboxylic acid and the diol. Therefore, the diol must be protected. The most efficient way to protect a 1,2-diol is to form a cyclic acetal (an acetonide in this case) in a single step with acetone. Protecting each alcohol individually (B) is less efficient. NaBH₄ (C) is not strong enough to reduce a carboxylic acid. No protection (D) would result in a failed reaction due to consumption of the LiAlH₄ by the acidic protons.

Question 13

A synthetic goal is to nitrate aniline to produce primarily meta-nitroaniline. Direct nitration is unsuitable as it yields ortho/para products. Which strategy would best achieve the desired meta selectivity?

  1. Protect the amine as an acetanilide using acetic anhydride prior to nitration.
  2. Use a milder nitrating agent like dilute nitric acid at 0 °C to favor kinetic meta-product.
  3. Protect the amine with a bulky Boc group to sterically block the ortho/para positions.
  4. Perform the nitration in the presence of a strong acid like H₂SO₄ to form the anilinium ion. (correct answer)
Explanation: When approaching aromatic substitution problems, you need to consider how substituents already on the benzene ring direct incoming groups. Aniline's amino group is a strong ortho/para director, which explains why direct nitration fails to give the desired meta product. The key insight is that you can change the directing properties by altering the electronic nature of the substituent. When aniline is treated with strong acid, the amino group becomes protonated to form the anilinium ion (−NH3+-NH_3^+). This positively charged group is now a strong meta-directing, electron-withdrawing substituent rather than an electron-donating one. The anilinium ion deactivates the ortho and para positions through its strong electron-withdrawing inductive effect, making the meta position most favorable for electrophilic attack. This is exactly why option D achieves the desired meta selectivity. Option A is incorrect because acetanilide still contains an electron-donating amide group that directs ortho/para, though less strongly than the free amine. Option B misunderstands the fundamental issue—changing reaction conditions doesn't alter the inherent directing effects of substituents; aniline will always direct ortho/para regardless of temperature or reagent concentration. Option C is flawed because steric hindrance from a Boc group might reduce ortho substitution but wouldn't favor meta over para positions, and the nitrogen would still be electron-donating. Remember this pattern: when you need to reverse the directing properties of an amine, consider protonating it with acid to convert the electron-donating −NH2-NH_2 into the electron-withdrawing −NH3+-NH_3^+.

Question 14

A student attempts to synthesize 4-aminoacetophenone via a Friedel-Crafts acylation of aniline using acetyl chloride and AlCl₃. The reaction fails to produce the desired product. What is the fundamental reason for this failure?

  1. The amino group's lone pair acts as a Lewis base, coordinating with the AlCl₃ catalyst and deactivating the ring. (correct answer)
  2. The amino group is a deactivator, which prevents the Friedel-Crafts reaction from occurring.
  3. The reaction produces an inseparable mixture of ortho and para isomers, leading to a low yield of the desired product.
  4. The acetyl chloride reacts preferentially with the solvent rather than with the aniline.
Explanation: When approaching Friedel-Crafts reactions with substituted benzenes, you need to consider how existing substituents interact with the Lewis acid catalyst, not just their electronic effects on the ring. The correct answer is A because aniline's amino group has a lone pair of electrons that acts as a Lewis base. This lone pair coordinates directly with the AlCl3AlCl_3 catalyst, forming a stable complex that removes the catalyst from solution. Without free AlCl3AlCl_3 available to activate the acetyl chloride, the Friedel-Crafts mechanism cannot proceed. Additionally, this coordination creates a positively charged nitrogen, which converts the normally activating amino group into a strongly deactivating ammonium-like substituent. Option B is incorrect because while the coordinated amino group becomes deactivating, this isn't the fundamental reason for failure—it's the catalyst deactivation that prevents the reaction entirely. Option C misses the point; the reaction doesn't produce any significant amount of product to create selectivity issues because the catalyst is deactivated from the start. Option D is wrong because acetyl chloride doesn't preferentially react with typical Friedel-Crafts solvents like dichloromethane or nitrobenzene under normal conditions. Remember this key principle: before analyzing electronic effects in Friedel-Crafts reactions, first check if the substrate has lone pairs that can coordinate with the Lewis acid catalyst. Amines, alcohols, and other Lewis basic functional groups will typically shut down Friedel-Crafts reactions by sequestering the catalyst, regardless of their activating or deactivating effects on the aromatic ring.

Question 15

A complex molecule contains a primary alcohol protected as a TMS ether, a secondary alcohol protected as a TBDMS ether, and an amine protected as a Cbz carbamate. To selectively remove the Cbz group, which reagent would be unsuitable because it would also cleave one or more of the other protecting groups?

  1. H₂ with Pd/C catalyst in methanol.
  2. Transfer hydrogenation using ammonium formate.
  3. Na in liquid ammonia (Birch reduction conditions).
  4. HBr in acetic acid. (correct answer)
Explanation: This question tests your understanding of protecting group selectivity and deprotection mechanisms. When faced with multiple protecting groups, you need to analyze which deprotection conditions will affect each group present. The Cbz (benzyloxycarbonyl) protecting group is designed to be removed under specific conditions that cleave the benzyl-oxygen bond. Let's examine each option: Option D (HBr in acetic acid) is unsuitable because strong acid conditions will cleave multiple protecting groups. HBr is sufficiently acidic to protonate and cleave both TMS ethers (which are very acid-labile) and TBDMS ethers (also acid-sensitive, though more stable than TMS). This makes it non-selective - you'd lose all three protecting groups simultaneously. Option A (H₂ with Pd/C) works through catalytic hydrogenation, specifically targeting the benzyl C-O bond in the Cbz group. The TMS and TBDMS ethers contain Si-O bonds that aren't affected by these hydrogenation conditions, making this selective. Option B (transfer hydrogenation with ammonium formate) operates similarly to option A, using a different hydrogen source but the same mechanism - it selectively reduces the benzyl group without affecting silicon-based protecting groups. Option C (Na in liquid ammonia) performs Birch reduction, which can cleave benzyl groups through electron transfer mechanisms while leaving the silicon ethers intact. Remember this key principle: acid-based deprotection methods tend to be less selective because many protecting groups are acid-labile. When you need selectivity, look for mechanism-specific approaches like hydrogenation that target specific bond types rather than broad reactivity patterns.

Question 16

A student removes a TBDMS protecting group from an alcohol using a standard procedure with dilute aqueous acid (H₃O⁺). The starting molecule also contains a carbon-carbon double bond. An unexpected side product is formed in addition to the desired alcohol. What is the most likely structure of this side product?

  1. An epoxide, formed via reaction with adventitious oxygen.
  2. A diol, formed by dihydroxylation of the double bond.
  3. An alcohol, formed by the acid-catalyzed hydration of the double bond. (correct answer)
  4. A cyclopropane, formed by rearrangement of the carbocation intermediate.
Explanation: The deprotection conditions involve aqueous acid. These are the exact conditions required for the acid-catalyzed hydration of an alkene (a Markovnikov addition of water across the double bond). This represents a common problem in synthesis where the conditions required to deprotect one functional group are not compatible with another functional group in the molecule. Using a non-acidic deprotection method, like TBAF, would have avoided this side reaction.

Question 17

The N-terminus of an amino acid is protected with a Boc group, and its C-terminus is an ethyl ester. Which reagent is ideal for selectively removing the Boc group without hydrolyzing the ester?

  1. Concentrated aqueous HCl at reflux.
  2. Sodium hydroxide in ethanol/water.
  3. Trifluoroacetic acid (TFA) in dichloromethane. (correct answer)
  4. Lithium aluminum hydride (LiAlH₄) in THF.
Explanation: The Boc group is designed to be removed under strong acid conditions that do not involve water, thus avoiding ester hydrolysis. TFA in an anhydrous solvent like DCM is the standard method. Concentrated aqueous HCl (A) would also hydrolyze the ester. Sodium hydroxide (B) would saponify the ester, and the Boc group is stable to base. LiAlH₄ (D) is a reducing agent and would reduce both functional groups.