Organic Chemistry 2 Quiz: Amines Basicity Trends And Salt Formation
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Amines Basicity Trends And Salt FormationQuestion 1 of 14

The pKa of the conjugate acid of triethylamine is approximately 10.7. An aqueous solution of triethylamine is buffered to a pH of 9.7. What is the approximate ratio of the concentration of neutral triethylamine to its protonated form, triethylammonium chloride, in this solution?

1:10
1:1
10:1
1:100
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Organic Chemistry 2 Quiz

Organic Chemistry 2 Quiz: Amines Basicity Trends And Salt Formation

Practice Amines Basicity Trends And Salt Formation in Organic Chemistry 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Amines Basicity Trends And Salt Formation, giving you a quick way to practice the rules, question types, and explanations that matter most for Organic Chemistry 2.

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Question 1

The pKa of the conjugate acid of triethylamine is approximately 10.7. An aqueous solution of triethylamine is buffered to a pH of 9.7. What is the approximate ratio of the concentration of neutral triethylamine to its protonated form, triethylammonium chloride, in this solution?

  1. 1:10 (correct answer)
  2. 1:1
  3. 10:1
  4. 1:100
Explanation: The relationship between pH, pKa, and the concentrations of an acid and its conjugate base is given by the Henderson-Hasselbalch equation: pH = pKa + log([Base]/[Acid]). Here, the 'Base' is triethylamine and the 'Acid' is its conjugate acid, triethylammonium. Plugging in the values: 9.7 = 10.7 + log([triethylamine]/[triethylammonium]). This simplifies to -1.0 = log([triethylamine]/[triethylammonium]). Taking the antilog of both sides gives 10⁻¹ = [triethylamine]/[triethylammonium], which is a ratio of 1:10.

Question 2

A chemist has a water-insoluble liquid amine and wants to convert it into a solid, water-soluble salt for purification by recrystallization. Which reagent would be most suitable for this transformation?

  1. Aqueous sodium hydroxide (NaOH)
  2. Anhydrous magnesium sulfate (MgSO₄)
  3. A solution of hydrochloric acid (HCl) in ether (correct answer)
  4. Liquid bromine (Br₂)
Explanation: To convert a basic amine into a salt, it must be reacted with an acid. The reaction of an amine with HCl produces an ammonium chloride salt. Using HCl dissolved in an organic solvent like ether is a common method to precipitate the ammonium salt from the solution in which the neutral amine is soluble, yielding a solid product. Aqueous NaOH (A) is a base and would not form a salt. MgSO₄ (B) is a drying agent. Liquid bromine (D) would perform electrophilic aromatic substitution if the amine were aromatic, not an acid-base reaction.

Question 3

The reaction of (R)-1-phenylethanamine with an excess of racemic (R,S)-mandelic acid is a key step in chiral resolution. What is the relationship between the ammonium carboxylate salts formed in this reaction?

  1. They are a racemic mixture of enantiomers.
  2. They are identical, achiral compounds.
  3. They are diastereomers with different physical properties. (correct answer)
  4. They are constitutional isomers with different connectivity.
Explanation: This reaction involves a single enantiomer of a base, (R)-amine, reacting with a racemic mixture of an acid, (R)-acid and (S)-acid. The two possible salt products are (R-amine, R-acid) and (R-amine, S-acid). These two products have a stereochemical relationship of (R,R) and (R,S). Since they are stereoisomers but not mirror images of each other, they are diastereomers. Diastereomers have different physical properties, such as solubility, which allows them to be separated by crystallization.

Question 4

The basicity of amines is often compared in aqueous solution. The observed order of basicity for methylamines in water is Me₂NH > MeNH₂ > Me₃N > NH₃. Why is trimethylamine (Me₃N) less basic than dimethylamine (Me₂NH) in water, despite having more electron-donating methyl groups?

  1. The inductive effect of three methyl groups is electron-withdrawing, destabilizing the conjugate acid.
  2. Trimethylamine is a weaker base due to resonance delocalization of its lone pair.
  3. The conjugate acid of trimethylamine is sterically hindered, leading to poor solvation by water molecules, which destabilizes it. (correct answer)
  4. The nitrogen in trimethylamine is sp² hybridized, which decreases its basicity compared to the sp³ nitrogen in dimethylamine.
Explanation: In the gas phase, basicity increases with alkylation (Me₃N > Me₂NH > MeNH₂) due to induction. However, in aqueous solution, the stability of the protonated ammonium ion is also dependent on solvation via hydrogen bonding. The conjugate acid of dimethylamine (Me₂NH₂⁺) has two N-H protons and can be well-solvated. The conjugate acid of trimethylamine (Me₃NH⁺) has only one N-H proton and is sterically crowded by the three methyl groups, hindering access by water molecules. This poor solvation destabilizes the trimethylammonium ion relative to the dimethylammonium ion, making trimethylamine a weaker base in water.

Question 5

A student attempts to synthesize an amide by reacting ethylamine with acetyl chloride. To neutralize the HCl byproduct, the student adds one equivalent of triethylamine (pKaH ≈ 10.7) to the reaction mixture. However, a significant portion of the acetyl chloride reacts with the triethylamine instead of the desired ethylamine (pKaH ≈ 10.6). What is the primary reason for this undesired side reaction?

  1. Triethylamine is a stronger base than ethylamine, so it reacts faster with the acid byproduct.
  2. Triethylamine is significantly more nucleophilic than ethylamine due to greater steric hindrance.
  3. The basicities and nucleophilicities of ethylamine and triethylamine are comparable, leading to a competitive reaction with the electrophile. (correct answer)
  4. Ethylamine is too sterically hindered to react efficiently with acetyl chloride, allowing the triethylamine to react instead.
Explanation: Both ethylamine and triethylamine are nucleophiles that can react with the electrophilic acetyl chloride. Their basicities are very similar (pKaH ~10.6 vs 10.7), and their nucleophilicities are also comparable. Although triethylamine is more sterically hindered, it is still a potent nucleophile. Because it is present in the reaction mixture, it competes with ethylamine for the acetyl chloride, leading to the formation of an undesired acylammonium salt. This illustrates that a non-nucleophilic base (like pyridine or a hindered base like DIPEA) should have been used.

Question 6

An industrial process requires the separation of a mixture containing aniline, benzoic acid, and toluene, all dissolved in diethyl ether. The first step involves extraction with a 1 M aqueous HCl solution. After this step, which compound(s) will have predominantly moved into the aqueous layer?

  1. Aniline, as anilinium chloride (correct answer)
  2. Benzoic acid, as sodium benzoate
  3. Toluene, due to its polarity
  4. Both aniline and benzoic acid
Explanation: Amines are basic and react with strong acids like HCl to form water-soluble ammonium salts. Aniline will react with HCl to form anilinium chloride, which is ionic and will partition into the aqueous layer. Benzoic acid is an acid and will not react with HCl. Toluene is a neutral, nonpolar hydrocarbon. Both benzoic acid and toluene will remain in the diethyl ether layer during the acidic extraction.

Question 7

An amine-containing drug has a conjugate acid pKa of 8.0. For the drug to passively diffuse across a cell membrane, it must be in its neutral, uncharged form. In which physiological environment will the concentration of the neutral form of the drug be highest?

  1. The stomach, with a pH of approximately 2.0
  2. The upper small intestine, with a pH of approximately 6.0
  3. The blood plasma, with a pH of approximately 7.4
  4. The lower small intestine, with a pH of approximately 8.5 (correct answer)
Explanation: The equilibrium is between the neutral amine (B) and its protonated, charged form (BH⁺). The Henderson-Hasselbalch equation is pH = pKa + log([B]/[BH⁺]). To maximize the concentration of the neutral form [B], the ratio [B]/[BH⁺] must be maximized. This occurs when the pH is greater than the pKa. At pH 8.5, which is above the pKa of 8.0, the equilibrium will favor the neutral, uncharged form. In all other environments listed (pH 2.0, 6.0, 7.4), the pH is below the pKa, meaning the protonated, charged form (BH⁺) will be the dominant species.

Question 8

Which of the following amines is expected to be the LEAST basic?

  1. Aniline
  2. Diphenylamine (correct answer)
  3. Cyclohexylamine
  4. N,N-Dimethylaniline
Explanation: Basicity in amines is reduced by delocalization of the nitrogen lone pair. Cyclohexylamine is a standard alkylamine with a localized lone pair, making it the most basic. Aniline has one phenyl group, which delocalizes the lone pair via resonance, significantly reducing basicity. N,N-Dimethylaniline is slightly more basic than aniline due to the inductive effect of the methyl groups. Diphenylamine has two phenyl groups, allowing the nitrogen lone pair to delocalize over two aromatic rings. This extensive delocalization makes the lone pair much less available for protonation, rendering diphenylamine the least basic of the choices.

Question 9

How does the hybridization of the nitrogen atom's lone pair orbital affect the basicity of an amine?

  1. Basicity increases as the s-character of the orbital increases, as s-orbitals are more diffuse.
  2. Basicity decreases as the s-character of the orbital increases, as electrons in s-orbitals are held more tightly to the nucleus. (correct answer)
  3. Hybridization has no significant effect on basicity; only inductive and resonance effects are important.
  4. Basicity is maximized with sp hybridization because the linear geometry minimizes steric hindrance.
Explanation: The s-character of a hybrid orbital affects how tightly the electrons in that orbital are held by the nucleus. An s-orbital is closer to the nucleus than a p-orbital. Therefore, as the percent s-character increases (sp > sp² > sp³), the lone pair electrons are held more tightly and are less available for donation to a proton. This results in decreased basicity. The general trend is sp³ amines (alkylamines, ~25% s-character) are more basic than sp² amines (imines, pyridine, ~33% s-character), which are more basic than sp nitriles (~50% s-character).

Question 10

During an E2 elimination reaction (the Hofmann elimination), a tetraalkylammonium salt is used as the substrate. Why is the trialkylamine group, -NR₃⁺, an effective leaving group in this reaction?

  1. The -NR₃⁺ group is a strong base, which facilitates its own departure.
  2. The -NR₃⁺ group departs as a stable, neutral trialkylamine molecule, making it a good leaving group. (correct answer)
  3. The positive charge on the nitrogen atom inductively weakens the adjacent C-H bonds, promoting elimination.
  4. The bulky -NR₃⁺ group creates steric strain that is relieved upon its elimination from the substrate.
Explanation: The quality of a leaving group is determined by its stability after it has departed with the electron pair from the bond. Good leaving groups are weak bases. In the Hofmann elimination, the positively charged -NR₃⁺ group departs as a neutral, stable trialkylamine molecule (NR₃). Because the departing species is a stable, uncharged molecule, it is an excellent leaving group. Strong bases (A) are poor leaving groups. While inductive effects (C) and sterics (D) are relevant to the reaction's regioselectivity and rate, the fundamental reason it's a good leaving group is the stability of the neutral amine that is formed.

Question 11

An equimolar mixture of aniline (pKaH ≈ 4.6) and pyridine (pKaH ≈ 5.2) is treated with 0.5 equivalents of a strong acid (HCl). Which statement best describes the resulting mixture at equilibrium?

  1. Aniline will be protonated preferentially, forming anilinium chloride, because it is an aromatic amine.
  2. Pyridine will be protonated preferentially, forming pyridinium chloride, because it is the stronger base. (correct answer)
  3. Both amines will be protonated equally, resulting in a 1:1 mixture of anilinium and pyridinium salts.
  4. Neither amine will be significantly protonated because the amount of acid is insufficient.
Explanation: When a limited amount of acid is added to a mixture of bases, the stronger base will be protonated preferentially. Basicity is determined by the pKa of the conjugate acid (pKaH); a higher pKaH indicates a stronger base. Since pyridine (pKaH ≈ 5.2) has a higher pKaH than aniline (pKaH ≈ 4.6), it is the stronger base. Therefore, the 0.5 equivalents of HCl will react preferentially with pyridine to form pyridinium chloride.

Question 12

Which of the following acid-base reactions is predicted to have an equilibrium constant (Keq) significantly greater than 1?

  1. Aniline (pKaH ≈ 4.6) + H₂O ⇌ Anilinium ion + OH⁻
  2. Ammonia (pKaH ≈ 9.2) + Sodium acetate ⇌ Ammonium ion + Acetic acid (pKa ≈ 4.8)
  3. Pyridinium ion (pKa ≈ 5.2) + NaHCO₃ ⇌ Pyridine + H₂CO₃ (pKa ≈ 6.4)
  4. Triethylammonium ion (pKa ≈ 10.7) + NaOH ⇌ Triethylamine + H₂O (pKa ≈ 15.7) (correct answer)
Explanation: When predicting acid-base equilibria, you need to remember that reactions favor the formation of weaker acids and bases. The equilibrium constant depends on the difference in pKa values between the acid on the reactant side and the acid on the product side. For a reaction to have Keq >> 1, the acid on the reactant side must be significantly stronger (lower pKa) than the acid formed on the product side. You can estimate Keq using: Keq=10(pKaproduct−pKareactant)K_{eq} = 10^{(pKa_{product} - pKa_{reactant})} Option D shows the triethylammonium ion (pKa ≈ 10.7) reacting with hydroxide to form water (pKa ≈ 15.7). Since Keq=10(15.7−10.7)=105K_{eq} = 10^{(15.7-10.7)} = 10^5, this reaction strongly favors products. Option A involves aniline acting as a base against water. Since aniline's conjugate acid has pKa ≈ 4.6, aniline is a very weak base, making this reaction unfavorable (Keq << 1). Option B has ammonia (pKaH ≈ 9.2) competing with acetate for protons, but since acetic acid (pKa ≈ 4.8) is much stronger than the ammonium ion, the equilibrium favors reactants. Option C shows pyridinium ion (pKa ≈ 5.2) transferring a proton to bicarbonate, forming carbonic acid (pKa ≈ 6.4). Since the reactant acid is stronger than the product acid, this favors products, but the difference is small (Keq ≈ 16). Study tip: Always compare pKa values of the acids on both sides. Large positive differences (product pKa - reactant pKa) mean strongly favorable equilibria.

Question 13

Why is the nitrogen atom in acetamide (CH₃CONH₂) significantly less basic than the nitrogen atom in ethylamine (CH₃CH₂NH₂)?

  1. The carbonyl group has a strong inductive electron-withdrawing effect that destabilizes the conjugate acid.
  2. The lone pair on the amide nitrogen is delocalized by resonance onto the adjacent carbonyl oxygen, making it less available. (correct answer)
  3. The nitrogen in acetamide is sp hybridized, while the nitrogen in ethylamine is sp³, making acetamide less basic.
  4. Acetamide can form intermolecular hydrogen bonds, which reduces the availability of the nitrogen lone pair.
Explanation: The primary reason for the very low basicity of amides is resonance. The lone pair of electrons on the nitrogen atom is delocalized through resonance with the adjacent carbonyl group, creating a resonance structure with a C=N double bond and a negative charge on the oxygen. This delocalization makes the lone pair much less available to accept a proton. While the inductive effect of the carbonyl (A) also plays a role, the resonance effect is dominant. The nitrogen in an amide is sp² hybridized, not sp (C). Hydrogen bonding (D) affects physical properties but is not the primary reason for the difference in basicity.

Question 14

An unknown compound C₅H₁₁N is insoluble in water but dissolves readily in 1 M aqueous HCl. This observation suggests that the unknown compound is most likely what type of amine?

  1. A quaternary ammonium salt
  2. An amide
  3. A nitrile
  4. A primary, secondary, or tertiary amine (correct answer)
Explanation: When you encounter a question about compound solubility behavior with acids, you're dealing with the basicity and protonation of nitrogen-containing functional groups. The key insight is understanding how different nitrogen functionalities respond to acidic conditions. The compound C₅H₁₁N is water-insoluble but becomes soluble in 1 M HCl because it can be protonated by the acid. Primary, secondary, and tertiary amines all have lone pairs of electrons on nitrogen that can accept protons from HCl, forming water-soluble ammonium salts (R₃NH⁺Cl⁻). This protonation creates ionic character, dramatically increasing water solubility. Option A is incorrect because quaternary ammonium salts (R₄N⁺) already carry a positive charge and are typically water-soluble without needing acid treatment. They also cannot be protonated further since the nitrogen has no lone pair. Option B is wrong because amides have their nitrogen lone pair delocalized into the carbonyl group through resonance, making them much less basic than amines. They don't readily protonate in dilute HCl and wouldn't show this dramatic solubility change. Option C is incorrect because nitriles (R-C≡N) are very weakly basic due to the sp hybridization of the nitrogen. They require much stronger acids than 1 M HCl for protonation and wouldn't exhibit this solubility behavior. Remember this pattern: when an organic compound becomes soluble in dilute acid, look for basic functional groups that can be protonated. Amines are among the most common basic organic compounds you'll encounter.