Organic Chemistry Quiz: Acid Base Concepts Pka Conjugates

What this quiz covers

This quiz focuses on Acid Base Concepts Pka Conjugates, giving you a quick way to practice the rules, question types, and explanations that matter most for Organic Chemistry.

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.

Question 1

1. A good leaving group is a strong base, and its conjugate acid has a high pKa.
2. A good leaving group is a weak base, and its conjugate acid has a low pKa. (correct answer)
3. A good leaving group is a strong base, and its conjugate acid has a low pKa.
4. There is no direct relationship between leaving group ability and the pKa of its conjugate acid.

Explanation: A good leaving group is one that is stable on its own after detaching from the substrate. Stable species are weak bases. The strength of a base is inversely related to the strength of its conjugate acid. Therefore, a weak base has a strong conjugate acid. A strong acid is characterized by a low pKa. So, a good leaving group is a weak base, and its conjugate acid has a low pKa. For example, I⁻ is an excellent leaving group; its conjugate acid, HI, is a very strong acid (pKa ≈ -10).

Question 2

1. $\text{CH}_3\text{CH}_2\text{NH}_2$ (ethylamine)
2. $\text{CF}_3\text{CH}_2\text{NH}_2$ (2,2,2-trifluoroethylamine)
3. $(\text{CH}_3\text{CH}_2)_2\text{NH}$ (diethylamine) (correct answer)
4. $(\text{CH}_3\text{CH}_2)_3\text{N}$ (triethylamine)

Explanation: In aqueous solution, diethylamine is the strongest base due to optimal balance of inductive donation from alkyl groups and hydrogen bonding stabilization of its conjugate acid. Choice A (primary amine) has fewer electron-donating groups. Choice B has electron-withdrawing CF₃ groups that destabilize the lone pair. Choice D (tertiary amine) cannot form hydrogen bonds to stabilize its conjugate acid in water, making it weaker than the secondary amine despite having more alkyl groups.

Question 3

1. Vigorous gas evolution occurs as 2,4-dinitrophenol acts as a strong acid toward bicarbonate (correct answer)
2. No reaction occurs because phenols do not react with bicarbonate under normal conditions
3. A slow equilibrium is established with approximately equal amounts of reactants and products
4. The phenol is completely deprotonated, but no gas evolution occurs due to carbonic acid stability

Explanation: When you encounter acid-base reactions, the key principle is that stronger acids will protonate the conjugate bases of weaker acids. To predict the outcome, compare the $pK_a$ values of the acids involved. Here, 2,4-dinitrophenol has a $pK_a$ of 4.0, making it significantly more acidic than carbonic acid ($pK_a$ = 6.4). Since the phenol is about 250 times stronger as an acid, it will readily donate its proton to bicarbonate ion. The reaction proceeds: 2,4-dinitrophenol + $\text{HCO}_3^-$ → 2,4-dinitrophenolate + $\text{H}_2\text{CO}_3$. The unstable carbonic acid immediately decomposes to $\text{CO}_2$ and $\text{H}_2\text{O}$, causing vigorous gas evolution. This confirms answer A is correct. Answer B is wrong because it makes the false generalization that phenols don't react with bicarbonate. While simple phenol ($pK_a$ ≈ 10) indeed doesn't react significantly with bicarbonate, electron-withdrawing groups like nitro groups dramatically increase acidity, making substituted phenols much more reactive. Answer C incorrectly suggests an equilibrium. With a $pK_a$ difference of 2.4 units, the equilibrium lies heavily toward products (about 99% complete), not equally balanced. Answer D correctly identifies complete deprotonation but wrongly claims no gas evolution. Carbonic acid is highly unstable and rapidly decomposes to produce $\text{CO}_2$ gas. Study tip: Remember that $pK_a$ differences greater than 2 units indicate essentially complete acid-base reactions. Always check for gas-producing decomposition reactions like carbonic acid breakdown.

Question 4

1. $\text{HOOC-CH}_2\text{-COOH}$ (fully protonated form)
2. $\text{HOOC-CH}_2\text{-COO}^-$ (monoanion form) (correct answer)
3. $^-\text{OOC-CH}_2\text{-COO}^-$ (dianion form)
4. Equal mixture of $\text{HOOC-CH}_2\text{-COO}^-$ and $^-\text{OOC-CH}_2\text{-COO}^-$

Explanation: At pH 4.0, we compare to both p$K_a$ values. Since pH > p$K_a1$ (2.8), the first proton is mostly deprotonated. Since pH < p$K_a2$ (5.7), the second proton remains mostly protonated. Therefore, the monoanion $\text{HOOC-CH}_2\text{-COO}^-$ predominates. Choice A would predominate at pH < 2.8. Choice C would predominate at pH > 5.7. Choice D would occur only at pH ≈ 5.7 (near p$K_a2$).

Question 5

1. Equilibrium favors reactants; $K_{eq} < 1$
2. Cannot determine without initial concentrations
3. Reaction is at equilibrium with equal concentrations
4. Equilibrium favors products; $K_{eq} > 1$ (correct answer)

Explanation: When you encounter acid-base equilibrium problems, the key is understanding that equilibrium position depends on the relative strengths of the acids and bases involved. You can predict which direction is favored by comparing $pK_a$ and $pK_b$ values. To determine equilibrium position, first identify what's happening: acetic acid (weak acid) is donating a proton to methylamine (weak base), forming acetate ion and methylammonium ion. The equilibrium constant for this reaction is $K_{eq} = \frac{K_a(\text{acetic acid})}{K_w/K_b(\text{methylamine})} = \frac{K_a \times K_b}{K_w}$. Since $pK_a = 4.8$, then $K_a = 10^{-4.8} = 1.6 \times 10^{-5}$. Since $pK_b = 3.4$, then $K_b = 10^{-3.4} = 4.0 \times 10^{-4}$. Therefore: $K_{eq} = \frac{(1.6 \times 10^{-5})(4.0 \times 10^{-4})}{1.0 \times 10^{-14}} = 640$. Since $K_{eq} > 1$, equilibrium favors products. Answer D is correct because the large equilibrium constant indicates products are strongly favored. Answer A is wrong because $K_{eq} = 640$, which is much greater than 1. Answer B is incorrect because equilibrium position depends only on the inherent acid/base strengths ($pK_a$ and $pK_b$), not initial concentrations. Answer C is wrong because equal concentrations would require $K_{eq} = 1$, but we calculated $K_{eq} = 640$. Study tip: Remember that for acid-base reactions, when $K_a \times K_b > K_w$, the equilibrium favors products. Memorize that $K_w = 1.0 \times 10^{-14}$ at 25°C.

Question 6

1. The oxygen atom of the hydroxyl group acts as a Lewis acid, accepting an electron pair from the bromide ion.
2. The hydrogen atom of HBr acts as a Brønsted-Lowry acid, and the oxygen atom of tert-butanol acts as a Brønsted-Lowry base. (correct answer)
3. The bromide ion acts as a Brønsted-Lowry base, abstracting a proton from the hydroxyl group.
4. The tert-butanol acts as a Brønsted-Lowry acid, donating a proton to the bromide ion to form the tert-butoxide ion.

Explanation: In the first step of this reaction, the lone pair on the oxygen atom of the tert-butanol's hydroxyl group attacks the proton (H⁺) of HBr. This is a classic Brønsted-Lowry acid-base reaction. HBr donates a proton, so it is the Brønsted-Lowry acid. The oxygen atom of tert-butanol accepts the proton, so it is the Brønsted-Lowry base. This protonation converts the poor leaving group (-OH) into a good leaving group (-OH₂⁺).

Question 7

1. Water acts as a Lewis acid and coordinates more strongly to the larger iodide ion, deactivating it.
2. Polar protic solvents increase the electronegativity of fluorine relative to iodine, which decreases the basicity of the fluoride ion.
3. In water, iodide becomes a stronger base due to the leveling effect, which makes all bases appear equally strong.
4. The protic solvent molecules form a tight salvation shell via hydrogen bonding around the small, charge-dense fluoride ion, making it less available to act as a base. (correct answer)

Explanation: When comparing basicity across different solvents, you need to consider how solvent interactions affect the availability of lone pairs for proton acceptance. The dramatic reversal in basicity trends between polar aprotic and polar protic solvents reveals the crucial role of solvation. In polar aprotic solvents like DMSO, fluoride's small size and high charge density make it an excellent base because there's no hydrogen bonding to interfere with its lone pairs. Iodide, being larger and less charge-dense, is naturally a weaker base. However, polar protic solvents like water completely flip this relationship through differential solvation effects. Water molecules form extensive hydrogen bonds with the small, highly charged fluoride ion, creating a tight solvation shell that essentially "wraps up" the fluoride and makes its lone pairs much less accessible for acting as a base. The larger, more diffuse iodide ion experiences weaker solvation interactions, leaving it more available to accept protons. This explains why option D is correct. Option A incorrectly describes water as a Lewis acid coordinating to iodide - water acts as a hydrogen bond donor, not a Lewis acid. Option B misunderstands electronegativity as a variable property when it's actually constant for each element. Option C incorrectly invokes the leveling effect, which applies to very strong acids or bases being "leveled" to the solvent's strength, not the relative comparison described here. Remember: small, highly charged ions experience the strongest solvation effects in protic solvents, which can dramatically reduce their reactivity compared to larger, more diffuse ions.

Question 8

1. Inductive electron-withdrawal by fluorine atoms stabilizing the alkoxide conjugate base
2. Distance between the electron-withdrawing groups and the oxygen bearing the negative charge
3. Solvation differences of the various alkoxide ions in protic solvents like water
4. Resonance stabilization of the alkoxide conjugate bases through π-orbital overlap (correct answer)

Explanation: Resonance stabilization is least important because simple alkoxide ions cannot participate in significant resonance delocalization—the negative charge is localized on oxygen with no adjacent π-system. Choices A and B are crucial factors: fluorine substitution provides strong inductive stabilization, and proximity matters ($\text{CF}_3\text{CH}_2\text{O}^-$ vs $\text{CH}_3\text{CF}_2\text{O}^-$). Choice C is also relevant as different substitution patterns affect hydrogen bonding and solvation of the conjugate bases.

Question 9

1. 9.5
2. 12.5
3. 15.5 (correct answer)
4. 12.8

Explanation: The equilibrium constant is related to the pKa values by the equation: Keq = 10^(pKa_product_acid - pKa_reactant_acid). Here, Keq = 0.001 = 10⁻³. The reactant acid is HA and the product acid is HB. So, -3 = pKa(HB) - pKa(HA). We are given pKa(HB) = 12.5. Therefore, -3 = 12.5 - pKa(HA). Solving for pKa(HA) gives pKa(HA) = 12.5 + 3 = 15.5. A common mistake is to subtract 3 from 12.5, which would result from reversing the acids in the equation.

Question 10

1. Both acids react equally with ammonia since the p$K_a$ difference is small (~1 unit)
2. Formic acid reacts preferentially, but significant amounts of both ammonium salts form
3. Formic acid reacts almost exclusively, with minimal acetate formation until formic acid is consumed (correct answer)
4. The order of reaction depends on the concentration of water and cannot be predicted

Explanation: The equilibrium constants favor formic acid reaction by a factor of 10¹ (since Δp$K_a$ = 1.0). In a competitive situation with limited base, the stronger acid (formic) will react almost exclusively first. Only when formic acid is largely consumed will significant amounts of acetate begin to form. Choice A underestimates the significance of a 1 p$K_a$ unit difference. Choice B suggests more simultaneous reaction than expected. Choice D is incorrect because the relative reaction preference depends on intrinsic acidity, not water concentration.

Question 11

1. The benzene ring withdraws electron density through inductive effects, stabilizing the phenoxide anion
2. Resonance delocalization of the negative charge in phenoxide ion stabilizes the conjugate base (correct answer)
3. The sp² hybridization of carbon atoms in benzene makes them more electronegative than sp³ carbons
4. Aromatic compounds are always more acidic than aliphatic compounds due to planarity effects

Explanation: Phenol's enhanced acidity is primarily due to resonance stabilization of the phenoxide ion, where the negative charge can be delocalized into the benzene ring through multiple resonance structures. Choice A is incorrect because inductive withdrawal by benzene is minimal. Choice C confuses hybridization effects, which are more relevant for carbon acidity. Choice D is an overgeneralization that is not always true and doesn't explain the specific mechanism.

Question 12

1. The triple bond creates significant ring strain that is relieved upon deprotonation.
2. The conjugate base of the alkyne (an acetylide anion) is stabilized by resonance, while the alkane's conjugate base is not.
3. The carbon atom of the alkyne C-H bond is sp-hybridized, which has more s-character than an sp³-hybridized carbon, making it more electronegative. (correct answer)
4. The inductive effect of the alkyl group attached to the alkyne withdraws electron density and stabilizes the negative charge of the conjugate base.

Explanation: The acidity of a C-H bond is related to the stability of the carbanion formed upon deprotonation. The key difference is the hybridization of the carbon atom bearing the negative charge. An alkyne carbon is sp-hybridized (50% s-character), while an alkane carbon is sp³-hybridized (25% s-character). Since s-orbitals are closer to the nucleus than p-orbitals, electrons in an sp-orbital are held more tightly. This means an sp-hybridized carbon is more electronegative and can better stabilize a negative charge, making the corresponding C-H bond more acidic.

Question 13

1. BF₃ is a Brønsted-Lowry acid and NH₃ is a Brønsted-Lowry base.
2. BF₃ is a Lewis acid and NH₃ is a Lewis base. (correct answer)
3. BF₃ is a Lewis base and NH₃ is a Lewis acid.
4. BF₃ is an oxidizing agent and NH₃ is a reducing agent.

Explanation: This reaction does not involve the transfer of a proton, so the Brønsted-Lowry definition does not apply. Instead, it involves the donation and acceptance of an electron pair. Ammonia (NH₃) has a lone pair of electrons on the nitrogen atom, which it can donate. An electron-pair donor is a Lewis base. Boron trifluoride (BF₃) has an electron-deficient boron atom (an incomplete octet), which can accept an electron pair. An electron-pair acceptor is a Lewis acid. The lone pair from NH₃ is donated to the empty p-orbital of boron to form a new covalent bond.

Question 14

1. Ethanol, CH₃CH₂OH
2. Ethane, CH₃CH₃
3. 2,2,2-Trifluoroethanol, CF₃CH₂OH
4. Acetic acid, CH₃COOH (correct answer)

Explanation: When evaluating acid strength, you need to consider the stability of the conjugate base that forms after the acid donates a proton. The more stable the conjugate base, the stronger the original acid, because a stable conjugate base makes the deprotonation reaction more favorable. Acetic acid (D) has the most stable conjugate base because when it loses a proton, it forms the acetate ion (CH₃COO⁻). This conjugate base is stabilized by resonance - the negative charge can be delocalized between both oxygen atoms in the carboxyl group. This electron delocalization significantly lowers the energy of the conjugate base, making acetic acid a relatively strong acid among these choices. 2,2,2-Trifluoroethanol (C) does form a somewhat stable conjugate base due to the electron-withdrawing fluorine atoms, which help stabilize the negative charge through inductive effects. However, this stabilization is less effective than resonance stabilization. Ethanol (A) forms an ethoxide ion (CH₃CH₂O⁻) when deprotonated. This conjugate base has no special stabilization - the negative charge is localized on oxygen with no resonance or significant inductive stabilization, making ethanol a much weaker acid. Ethane (B) is not even considered an acid under normal conditions. Its conjugate base would be a carbanion (CH₃CH₂⁻), which is extremely unstable and highly basic. Remember: when comparing acid strength, look for structural features that stabilize the conjugate base - resonance is typically the strongest stabilizing factor, followed by inductive effects from electronegative atoms.

Question 15

1. An electron-withdrawing halogen substituent on the carbon α to the carboxyl group (correct answer)
2. An electron-donating alkyl group adjacent to the carboxyl group, such as in propionic acid
3. A benzene ring directly attached to the carboxyl group, as in benzoic acid
4. Multiple carboxyl groups allowing for intramolecular hydrogen bonding effects

Explanation: When you encounter a pKa value for a carboxylic acid, you're being tested on how structure affects acidity. Remember that lower pKa values indicate stronger acids, and carboxylic acids become more acidic when their conjugate bases are stabilized by electron-withdrawing effects. A typical carboxylic acid like acetic acid has a pKa around 4.8. The given pKa of 3.2 is significantly lower, meaning this unknown acid is much stronger than a simple carboxylic acid. This increased acidity must result from structural features that stabilize the carboxylate anion after deprotonation. Option A is correct because electron-withdrawing halogens on the α-carbon create a strong inductive effect that pulls electron density away from the carboxylate group. This stabilizes the conjugate base and dramatically increases acidity. Chloroacetic acid (ClCH₂COOH), for example, has a pKa of 2.9, which matches our target value perfectly. Option B is wrong because electron-donating alkyl groups actually decrease acidity through inductive donation, making the pKa higher than acetic acid, not lower. Option C is incorrect because benzoic acid has a pKa of 4.2 - lower than aliphatic acids due to resonance stabilization, but still too high to match 3.2. Option D is wrong because while multiple carboxyl groups can affect acidity, the pKa of 3.2 is more characteristic of a monocarboxylic acid with strong electron-withdrawing substituents. Study tip: When comparing carboxylic acid acidity, remember that electron-withdrawing groups (especially halogens) near the carboxyl group cause the most dramatic pKa decreases through inductive effects.

Question 16

1. CH₄ (Methane) (correct answer)
2. H₂S (Hydrogen sulfide)
3. H₂O (Water)
4. NH₃ (Ammonia)

Explanation: When comparing acidity across different compounds, you need to consider how readily each molecule can donate a proton (H⁺). The key factors are the stability of the conjugate base that forms after losing H⁺ and the bond strength of the H-X bond being broken. Methane (CH₄) has the least acidic character because carbon and hydrogen have very similar electronegativities, creating a strong, nonpolar C-H bond. When methane would theoretically lose H⁺, it would form CH₃⁻ (methide ion), which is an extremely unstable, highly basic species. This makes proton loss from methane essentially impossible under normal conditions. Looking at why the other options are more acidic: Option B (H₂S) is moderately acidic because sulfur is larger and less electronegative than oxygen, making the H-S bond weaker and the resulting HS⁻ ion more stable than OH⁻. Option C (H₂O) shows weak acidity since oxygen's high electronegativity stabilizes the OH⁻ conjugate base somewhat. Option D (NH₃) exhibits very weak acidity, but nitrogen's lone pair can stabilize the NH₂⁻ conjugate base better than carbon can stabilize CH₃⁻. The order from most to least acidic is: H₂S > H₂O > NH₃ > CH₄. Study tip: Remember the periodic trend for acidity of binary hydrides: acidity increases going down a group (bond strength decreases) and across a period (electronegativity increases). Hydrocarbons like methane are essentially non-acidic in aqueous solutions.

Question 17

1. Sodium ethoxide (NaOCH₂CH₃), the conjugate base of ethanol (pKa ≈ 16)
2. Sodium hydroxide (NaOH), the conjugate base of water (pKa ≈ 15.7)
3. Sodium amide (NaNH₂), the conjugate base of ammonia (pKa ≈ 38) (correct answer)
4. Sodium acetate (NaOCOCH₃), the conjugate base of acetic acid (pKa ≈ 4.8)

Explanation: For a base to deprotonate an acid, the conjugate acid of the base must be weaker (have a higher pKa) than the acid being deprotonated. The reaction is: Propyne + Base⁻ ⇌ Propynide⁻ + Base-H. The equilibrium favors the products if the pKa of Base-H is greater than the pKa of propyne (25). Of the options, only ammonia (pKa ≈ 38) is a weaker acid than propyne. Therefore, its conjugate base, sodium amide (NaNH₂), is strong enough.

Question 18

1. 10:1 (correct answer)
2. 1:1
3. 1:10
4. 1:100

Explanation: When you encounter pH and pKa problems involving weak acids and their conjugate bases, you're working with buffer systems and the Henderson-Hasselbalch equation. This equation relates pH, pKa, and the ratio of conjugate base to weak acid: $pH = pKa + \log\left(\frac{[A^-]}{[HA]}\right)$ To find the ratio of acetate to acetic acid, substitute the given values: $5.76 = 4.76 + \log\left(\frac{[CH_3COO^-]}{[CH_3COOH]}\right)$ Solving for the log term: $\log\left(\frac{[CH_3COO^-]}{[CH_3COOH]}\right) = 5.76 - 4.76 = 1.00$ Taking the antilog: $\frac{[CH_3COO^-]}{[CH_3COOH]} = 10^1 = 10$ This gives us a 10:1 ratio of acetate to acetic acid, confirming answer A. Answer B (1:1) would occur when pH equals pKa (4.76), since the log term would be zero. Answer C (1:10) represents the inverse ratio—this would happen if the pH were 3.76 (one unit below the pKa). Answer D (1:100) would occur at pH 2.76, two units below the pKa. Remember this pattern: when pH is above pKa, the conjugate base predominates; when pH is below pKa, the weak acid predominates. Each pH unit difference from pKa represents a 10-fold change in the ratio. Master the Henderson-Hasselbalch equation—it's essential for buffer calculations throughout organic chemistry.

Question 19

1. The sp² hybridization of the carbon atoms increases the electronegativity and stabilizes the resulting carbanion.
2. The resulting conjugate base, the cyclopentadienyl anion, is stabilized by extensive resonance across all five carbon atoms.
3. The resulting conjugate base, the cyclopentadienyl anion, satisfies Hückel's rules for aromaticity, granting it extraordinary stability. (correct answer)
4. The five-membered ring structure experiences significant angle strain, which is relieved upon deprotonation.

Explanation: Upon deprotonation, cyclopentadiene forms the cyclopentadienyl anion. This anion has a continuous ring of p-orbitals, is planar, and contains 6 π-electrons (4 from the double bonds and 2 from the lone pair). This satisfies Hückel's rules (4n+2 π-electrons, where n=1), making the anion aromatic. This aromatic stabilization is a very powerful stabilizing effect, far greater than typical resonance (choice B) or hybridization effects (choice A), and is the primary reason for cyclopentadiene's unusual acidity.

Question 20

1. An arrow from the H of NaH to the H of the methanol -OH group, and an arrow from the O-H bond to the oxygen atom.
2. An arrow from the lone pair of the hydride ion (H⁻) to the H of the methanol -OH group, and an arrow from the O-H bond to the oxygen atom. (correct answer)
3. An arrow from the methanol oxygen's lone pair to the Na⁺ ion, and an arrow from the H⁻ to the methyl group carbon.
4. An arrow from the H⁻ to the oxygen atom of methanol, and an arrow from the O-H bond to the hydrogen atom.

Explanation: Curved arrows show the movement of electron pairs. In this reaction, the hydride ion (H⁻ from NaH) acts as the base. Its lone pair of electrons forms a new bond with the acidic proton of the methanol's hydroxyl group. This is shown by an arrow starting at the H⁻ lone pair and pointing to the H of the -OH group. Simultaneously, the O-H bond must break, with its electrons moving onto the oxygen atom to form the methoxide anion. This is shown by an arrow starting at the middle of the O-H bond and pointing to the oxygen atom.