Organic Chemistry 2 Quiz: Amine Alkylation And Reductive Amination
14 questions · exam conditions
0:00
Amine Alkylation And Reductive AminationQuestion 1 of 14

While often a problematic side reaction, exhaustive alkylation can be a deliberate synthetic goal. Treatment of ammonia with a large excess of ethyl iodide will primarily produce which of the following?

Tetraethylammonium iodide
Ethylamine
A mixture of diethylamine and triethylamine
Ethene
← Back to quizzes

Organic Chemistry 2 Quiz

Organic Chemistry 2 Quiz: Amine Alkylation And Reductive Amination

Practice Amine Alkylation And Reductive Amination 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 Amine Alkylation And Reductive Amination, giving you a quick way to practice the rules, question types, and explanations that matter most for Organic Chemistry 2.

How to use this quiz

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.

All questions

Question 1

While often a problematic side reaction, exhaustive alkylation can be a deliberate synthetic goal. Treatment of ammonia with a large excess of ethyl iodide will primarily produce which of the following?

  1. Tetraethylammonium iodide (correct answer)
  2. Ethylamine
  3. A mixture of diethylamine and triethylamine
  4. Ethene
Explanation: The correct answer is A. When the alkylating agent (ethyl iodide) is present in large excess relative to the amine (ammonia), the reaction is driven to the highest possible state of alkylation. Ammonia is alkylated to ethylamine. Ethylamine is more nucleophilic and is rapidly alkylated to diethylamine. Diethylamine is further alkylated to triethylamine. Finally, the tertiary amine is alkylated to form the quaternary ammonium salt, tetraethylammonium iodide. Since the salt is no longer nucleophilic, the reaction stops there. B would be the major product if ammonia were used in large excess. C represents intermediates. D would be a minor elimination side product.

Question 2

A chemist aims to synthesize N-ethylpropan-2-amine from propan-2-amine. Which statement best evaluates the two common synthetic approaches: direct alkylation with ethyl bromide versus reductive amination with acetaldehyde?

  1. Direct alkylation is superior because it is a simple, one-step Sₙ2 reaction, while reductive amination is a more complex, multi-reagent process.
  2. Both methods are equally effective, but direct alkylation is often preferred due to the lower cost of reagents compared to sodium cyanoborohydride.
  3. Reductive amination with acetaldehyde would be superior because direct alkylation would likely result in significant over-alkylation to a tertiary amine and a quaternary salt. (correct answer)
  4. Reductive amination with acetone would be the most efficient method, providing the target compound in the highest yield and purity.
Explanation: The correct answer is C. Direct alkylation of a primary amine (propan-2-amine) with an alkyl halide (ethyl bromide) is notoriously difficult to control. The product, a secondary amine (N-ethylpropan-2-amine), is generally more nucleophilic than the starting primary amine, leading to a second alkylation to form a tertiary amine, and even a third to form a quaternary ammonium salt. Reductive amination, which involves forming an imine with acetaldehyde followed by reduction, is a highly controlled method that avoids this over-alkylation problem. Distractor A is incorrect because it ignores the severe practical limitation of direct alkylation. Distractor B is incorrect because the methods are not equally effective; reductive amination gives much better control and yield of the desired product. Distractor D incorrectly identifies the carbonyl compound; reacting propan-2-amine with acetone would add an isopropyl group, not an ethyl group.

Question 3

Reductive amination is typically carried out in a mildly acidic buffer (e.g., pH 4-6). Why is performing the reaction in a strongly basic solution (e.g., pH > 10) likely to fail or proceed very slowly?

  1. The high concentration of hydroxide ions would deprotonate the starting amine, making it a poorer nucleophile.
  2. The reducing agent, such as NaBH₃CN, is unstable and rapidly decomposes under basic conditions.
  3. The formation of the imine/iminium ion intermediate is acid-catalyzed, specifically at the dehydration step of the carbinolamine. (correct answer)
  4. The carbonyl group is converted to an unreactive enolate under strongly basic conditions, preventing nucleophilic attack by the amine.
Explanation: The correct answer is C. The mechanism of imine formation involves two key stages: 1) nucleophilic attack of the amine on the carbonyl to form a carbinolamine intermediate, and 2) dehydration of the carbinolamine to form the imine. The second step, the elimination of water, requires protonation of the hydroxyl group to turn it into a good leaving group (H₂O). This step is acid-catalyzed and is very slow without an acid catalyst. In a strongly basic solution, there is no acid source to facilitate this critical dehydration step. Distractor D is a plausible misconception, but amine addition is generally faster than enolate formation for most simple amines/carbonyls. Distractor A is incorrect; deprotonation does not occur and even if it did, the anion would be more nucleophilic. Distractor B is incorrect; borohydride reagents are generally more stable in basic solution.

Question 4

The Eschweiler-Clarke reaction uses formic acid (HCOOH) and formaldehyde (CH₂O) to convert a primary or secondary amine to a tertiary N,N-dimethylamine. This reaction is mechanistically a type of reductive amination. What is the specific role of formic acid in this transformation?

  1. It acts as a Brønsted-Lowry acid to catalyze the initial imine formation from formaldehyde and the amine.
  2. It serves as the in situ source of a hydride ion to reduce the iminium ion intermediate. (correct answer)
  3. It acts as a dehydrating agent to drive the equilibrium toward the iminium ion.
  4. It reversibly protects the amine nitrogen, allowing for controlled methylation by formaldehyde.
Explanation: The correct answer is B. In the Eschweiler-Clarke reaction, the amine first reacts with formaldehyde to form an iminium ion. Formic acid then serves as the reducing agent. Specifically, a hydride is transferred from the formate ion to the electrophilic carbon of the iminium ion, reducing it to the amine and producing CO₂. While formic acid is an acid (A), its unique and critical role in this named reaction is as the hydride donor, making it a special case of reductive amination where the reductant is not a metal hydride.

Question 5

To synthesize the primary amine benzylamine from benzaldehyde, a student proposes two methods: (I) reductive amination with excess ammonia and NaBH₃CN, and (II) direct alkylation of excess ammonia with benzyl bromide. Why is method I strongly preferred?

  1. Method II is not feasible because ammonia is not nucleophilic enough to displace bromide.
  2. Method I is a one-pot reaction, whereas Method II requires isolation of benzyl bromide, which is unstable.
  3. Method II generates significant amounts of di- and tribenzylamine byproducts even with excess ammonia. (correct answer)
  4. Method I avoids the use of toxic cyanide-containing reagents, making it safer than Method II.
Explanation: The correct answer is C. While using a large excess of ammonia in direct alkylation (Method II) can favor the primary amine product, it is very difficult to completely prevent over-alkylation. The product, benzylamine, is more nucleophilic than ammonia, so it will compete for the benzyl bromide, leading to dibenzylamine and tribenzylamine byproducts, which are difficult to separate. Reductive amination (Method I) is a much more controlled reaction that cleanly forms the C-N bond without the possibility of over-alkylation, yielding a much purer product. Distractor A is false; ammonia is a good nucleophile. Distractor B misrepresents the situation; both starting materials are common and stable, but Method I is simply cleaner. Distractor D has the logic reversed; Method I uses NaBH₃CN, a cyanide derivative, while Method II does not.

Question 6

A reaction is performed between aniline and an excess of methyl iodide, yielding product P. A separate reaction is performed between aniline, excess formaldehyde, and NaBH₃CN, yielding product Q. Which spectroscopic feature would most reliably distinguish ionic product P from neutral product Q?

  1. The IR spectrum of P will show a strong N-H stretch around 3400 cm⁻¹, which is absent in Q.
  2. The mass spectrum of Q will show a molecular ion peak, while P is a non-volatile salt that will not show a molecular ion under standard EI conditions.
  3. The ¹³C NMR spectrum of P will show one fewer aromatic signal than Q due to higher symmetry.
  4. The ¹H NMR spectrum of P will show a singlet for the methyl protons integrating to 9H, while Q will show a singlet integrating to 6H. (correct answer)
Explanation: The correct answer is D. Product P is the result of exhaustive methylation (a form of direct alkylation), which converts aniline to the quaternary ammonium salt, N,N,N-trimethylanilinium iodide. Product Q is the result of exhaustive reductive amination, which converts aniline to the tertiary amine, N,N-dimethylaniline. In the ¹H NMR, the three equivalent methyl groups of P will appear as a single peak integrating to 9 protons. The two equivalent methyl groups of Q will appear as a single peak integrating to 6 protons. This difference in integration is a definitive way to distinguish them. Distractor A is incorrect as neither product has an N-H bond. B is often true, but NMR integration provides more quantitative and reliable structural information. C is incorrect as the aromatic ring has the same C₂ᵥ symmetry in both, leading to the same number of aromatic signals (4).

Question 7

A student attempts to synthesize diethylamine by reacting one equivalent of ethylamine with one equivalent of ethyl bromide. Analysis of the product mixture reveals significant amounts of ethylamine, diethylamine, triethylamine, and tetraethylammonium bromide. What is the primary reason for this complex mixture?

  1. The reaction proceeds via an E2 mechanism, which inherently favors multiple products over a single substitution product.
  2. The product, diethylamine, acts as a base to deprotonate the starting ethylamine, creating multiple nucleophiles.
  3. The product, diethylamine, is a stronger nucleophile than the starting material, ethylamine, leading to competitive subsequent alkylations. (correct answer)
  4. The ethyl bromide undergoes a competing Sₙ1 reaction, which leads to carbocation rearrangements and multiple products.
Explanation: The correct answer is C. This is the classic problem of direct amine alkylation. The initial Sₙ2 reaction produces diethylamine. However, alkyl groups are electron-donating, making the nitrogen in diethylamine more electron-rich and thus more nucleophilic (and basic) than the nitrogen in the starting ethylamine. This causes the diethylamine product to compete effectively with the remaining ethylamine for the ethyl bromide, leading to triethylamine. Triethylamine is also nucleophilic and reacts further to form the tetraethylammonium salt. Distractor A is incorrect; while some E2 elimination to form ethene might occur, it does not explain the polyalkylation. Distractor B is an incorrect acid-base equilibrium; ethylamine is not acidic enough to be deprotonated by diethylamine. Distractor D is incorrect as primary alkyl halides like ethyl bromide react exclusively via the Sₙ2 mechanism.

Question 8

A student attempts to synthesize N-benzylaniline by adding benzylamine and NaBH₄ to a flask containing benzoyl chloride. The reaction fails to produce the desired secondary amine. What is the most likely reason for this failure?

  1. NaBH₄ is too weak of a reducing agent to reduce the imine formed from benzylamine and benzoyl chloride.
  2. Benzoyl chloride is immediately reduced to benzyl alcohol by NaBH₄ before it can react with the amine.
  3. The highly reactive benzoyl chloride undergoes rapid nucleophilic acyl substitution with benzylamine to form a stable amide. (correct answer)
  4. The reaction requires a strong acid catalyst to form the necessary imine intermediate, which was not added.
Explanation: The correct answer is C. The concept of reductive amination applies to aldehydes and ketones. Acid chlorides, like benzoyl chloride, are much more reactive electrophiles at the carbonyl carbon. When mixed with an amine, they will undergo a very rapid and essentially irreversible nucleophilic acyl substitution to form an amide (N-benzylbenzamide in this case). Amides are very stable functional groups and are not reducible by NaBH₄. Distractor A is incorrect because an imine does not form from an acid chloride. Distractor D is incorrect for the same reason. Distractor B is incorrect because the reaction of the amine with the acid chloride (acylation) is typically much faster than hydride reduction of the acid chloride.

Question 9

Which of the following synthetic sequences is the most efficient and selective laboratory method to prepare pure N,N-dimethylaniline from aniline?

  1. Treatment of aniline with two equivalents of methyl iodide in the presence of a non-nucleophilic base.
  2. Treatment of aniline with excess formaldehyde and sodium cyanoborohydride (NaBH₃CN). (correct answer)
  3. Treatment of aniline with two equivalents of methanol and a strong acid catalyst like H₂SO₄.
  4. Treatment of aniline with acetyl chloride, followed by reduction with LiAlH₄, and then repeating this sequence.
Explanation: The correct answer is B. This describes a double reductive amination, which is a highly effective way to convert a primary amine to a dimethylated tertiary amine. The reaction proceeds through the formation and reduction of an imine, followed by formation and reduction of a second iminium ion. This is a clean, one-pot procedure. Distractor A describes direct alkylation. While it can work, it's very difficult to stop cleanly at the tertiary amine stage and often produces significant amounts of the quaternary ammonium salt (N,N,N-trimethylanilinium iodide). Distractor C describes conditions that do not typically form C-N bonds. Distractor D is an unnecessarily long, multi-step process for achieving the same transformation.

Question 10

Which set of reagents is most appropriate for the selective, high-yield conversion of cyclohexanone and methylamine to N-methylcyclohexylamine?

    1. CH₃NH₂; 2. LiAlH₄; 3. H₂O
  2. CH₃NH₂, NaBH₃CN, mild acid (e.g., CH₃COOH) (correct answer)
  3. CH₃NH₂, NaBH₄, methanol
    1. CH₃MgBr; 2. H₂O; 3. H₂SO₄, heat
Explanation: The correct answer is B. This transformation is a reductive amination. It requires the formation of an iminium ion intermediate from cyclohexanone and methylamine, which is then reduced. Sodium cyanoborohydride (NaBH₃CN) or sodium triacetoxyborohydride are ideal reagents because they are mild hydride donors that selectively reduce the protonated iminium ion much faster than they reduce the starting ketone. Mild acid catalyzes imine formation. Distractor A is incorrect because LiAlH₄ is a very strong reducing agent that would rapidly reduce cyclohexanone to cyclohexanol before it could react with methylamine. Distractor C is plausible, but NaBH₄ is less selective than NaBH₃CN and can reduce a significant amount of the starting ketone, lowering the yield. Distractor D describes a Grignard reaction, which would form 1-methylcyclohexanol, followed by dehydration, which is an entirely different transformation.

Question 11

Anisaldehyde (4-methoxybenzaldehyde) is treated with triethylamine (Et₃N) and sodium triacetoxyborohydride (NaBH(OAc)₃) in an acidic buffer. What is the expected major outcome?

  1. N,N-Diethyl-4-methoxybenzylamine is formed via displacement.
  2. 4-Methoxybenzyl alcohol is formed from slow reduction of the aldehyde.
  3. The triethylamine is dealkylated to form diethylamine and other products.
  4. No reaction occurs and the starting aldehyde and amine are recovered. (correct answer)
Explanation: The correct answer is D. Reductive amination requires the formation of an imine or enamine intermediate. This process requires an amine with at least one N-H bond (i.e., a primary or secondary amine). Triethylamine is a tertiary amine; it has no N-H protons. Therefore, it cannot form a carbinolamine that can dehydrate to an iminium ion. Without the formation of the highly electrophilic iminium ion intermediate, the mild reducing agent NaBH(OAc)₃ will not reduce the aldehyde at an appreciable rate. Thus, no reaction occurs. Distractors A and C describe reactions that do not occur under these conditions. Distractor B is incorrect because the reduction of an aldehyde by NaBH(OAc)₃ is very slow in the absence of an iminium intermediate.

Question 12

Glutaraldehyde (pentane-1,5-dial) is treated with a large excess of dimethylamine ((CH₃)₂NH) and NaBH(OAc)₃. What is the expected major product?

  1. 3,3-Bis(dimethylaminomethyl)propan-1-ol
  2. 1,5-Bis(dimethylamino)pentane
  3. N¹,N¹,N⁵,N⁵-Tetramethylpentane-1,5-diamine (correct answer)
  4. Pentane-1,5-diol
Explanation: The correct answer is C. This reaction involves the reductive amination of both aldehyde functional groups in glutaraldehyde (OHC-(CH₂)₃-CHO). Dimethylamine is a secondary amine that reacts with aldehydes to form iminium ions, which are then reduced to tertiary amines. Since there are two aldehyde groups and excess dimethylamine, both ends react to give (CH₃)₂N-CH₂-(CH₂)₃-CH₂-N(CH₃)₂. The systematic IUPAC name is N¹,N¹,N⁵,N⁵-tetramethylpentane-1,5-diamine. Answer B describes the same structure but uses a less systematic naming convention. Answer A represents an incorrect structural isomer. Answer D would result from simple reduction of the dialdehyde without the amine.

Question 13

Which statement accurately describes a key difference between NaBH₄ and NaBH₃CN that makes the latter superior for most reductive aminations?

  1. NaBH₃CN is a much stronger reducing agent, ensuring the reaction goes to completion rapidly.
  2. The electron-withdrawing cyanide group deactivates NaBH₃CN, making it selective for the highly electrophilic iminium ion over a ketone/aldehyde. (correct answer)
  3. NaBH₄ is insoluble in the organic solvents typically used for reductive amination, whereas NaBH₃CN is highly soluble.
  4. NaBH₃CN contains a nitrogen atom which helps coordinate the reagent to the iminium ion intermediate, increasing the reaction rate.
Explanation: The correct answer is B. The key to a successful reductive amination is the selectivity of the reducing agent. It must reduce the iminium ion intermediate much faster than it reduces the starting carbonyl compound. The electron-withdrawing cyanide group in NaBH₃CN reduces the hydridic character of the B-H bonds, making it a much milder and more selective reducing agent than NaBH₄. This deactivation means it is slow to react with a moderately electrophilic carbonyl group but reacts rapidly with the highly electrophilic, positively charged iminium ion. Distractor A is incorrect; NaBH₃CN is a weaker reducing agent. Distractor C is incorrect; both have comparable solubility profiles in relevant solvents like methanol. Distractor D provides a false mechanistic explanation; the selectivity is electronic, not based on coordination through the cyanide's nitrogen.

Question 14

A student plans to synthesize N-isopropylcyclohexylamine by reacting cyclohexylamine with 2-bromopropane in ethanol. What is the most likely major organic product derived from 2-bromopropane?

  1. Propene (correct answer)
  2. Propan-2-ol
  3. The desired N-isopropyl group attached to the amine
  4. Diisopropyl ether
Explanation: The correct answer is A. This reaction involves a secondary alkyl halide (2-bromopropane) and a primary amine (cyclohexylamine). Amines are not only nucleophiles but also moderately strong bases. When a base reacts with a secondary alkyl halide, the E2 elimination pathway strongly competes with, and often dominates, the Sₙ2 substitution pathway. Therefore, the major reaction will be the dehydrohalogenation of 2-bromopropane to form propene. Distractor C represents the desired Sₙ2 product, which would be a minor product at best. Distractor B (propan-2-ol) would be the product of an Sₙ1 reaction with the ethanol solvent, which is less likely than E2. Distractor D would come from a Williamson ether synthesis type reaction with ethoxide, not the amine.