Organic Chemistry Quiz: Nucleophiles And Electrophiles Recognizing Reactivity

What this quiz covers

This quiz focuses on Nucleophiles And Electrophiles Recognizing Reactivity, 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. The nitrogen atom's lone pair is sterically blocked by the carbon atom, forcing attack through the carbon terminus.
2. Attack through nitrogen would form an isonitrile, which is a thermodynamically much less stable product than a nitrile.
3. The triple bond makes the nitrogen atom electron-deficient and thus a very poor nucleophilic site.
4. The carbon atom is less electronegative than nitrogen, so the highest occupied molecular orbital (HOMO) has a larger coefficient on carbon. (correct answer)

Explanation: When you encounter ambident nucleophiles (nucleophiles with multiple reactive sites), the key is understanding that nucleophilic attack occurs preferentially at the site with the highest electron density in the HOMO (highest occupied molecular orbital). The cyanide ion's preference for carbon attack stems from orbital coefficient distribution. In cyanide's HOMO, the molecular orbital has a larger coefficient on the carbon atom than on nitrogen. This occurs because carbon is less electronegative than nitrogen, meaning electrons are less tightly held by carbon. The larger orbital coefficient translates to higher electron density at carbon, making it the more nucleophilic site despite nitrogen having a formal lone pair. Let's examine why the other options miss the mark. Option A incorrectly suggests steric hindrance - the linear geometry of cyanide doesn't create significant steric blocking of nitrogen. Option B contains a grain of truth about isonitriles being less stable than nitriles, but this thermodynamic consideration doesn't explain the kinetic preference for carbon attack during the nucleophilic substitution reaction. Option C fundamentally misunderstands electronic structure - the triple bond doesn't make nitrogen electron-deficient; rather, it affects how electron density is distributed in the molecular orbitals. Remember this pattern: for ambident nucleophiles, nucleophilic strength correlates with orbital coefficients in the HOMO, not just formal charges or lone pairs. The most nucleophilic site is where electrons are most available for donation, which often occurs on the less electronegative atom due to orbital mixing effects.

Question 2

1. $CH_3CH_2OH$ (ethanol)
2. $CH_3CH_2O^-$ (ethoxide ion) (correct answer)
3. $H_2O$ (water)
4. $CH_3CH_2NH_3^+$ (ethylammonium ion)

Explanation: The correct answer is B. Nucleophilicity in protic solvents like water generally increases with increasing negative charge and decreasing solvation. Ethoxide ion (CH₃CH₂O⁻) carries a negative charge, making it highly nucleophilic. While it will be solvated by water, it retains significant nucleophilic character. Choice A (ethanol) is neutral and much less nucleophilic than its conjugate base. Choice C (water) is a weak nucleophile due to its neutral charge and small size leading to tight solvation. Choice D (ethylammonium ion) carries a positive charge, making it electrophilic rather than nucleophilic.

Question 3

1. The larger size and polarizability of sulfur makes hydrogen sulfide ion more nucleophilic than hydroxide ion in aprotic solvents (correct answer)
2. The higher electronegativity of oxygen makes hydroxide ion a stronger nucleophile than hydrogen sulfide ion in aprotic solvents
3. Both ions have similar nucleophilicity because they carry the same formal charge and have similar basicity in aprotic solvent systems
4. The shorter bond length of the O-H bond compared to S-H bond makes hydroxide ion more reactive toward electrophilic carbon centers

Explanation: When you encounter nucleophilicity questions, remember that the solvent environment dramatically affects how nucleophiles behave. In polar aprotic solvents like DMSO, the key factors are polarizability and size rather than basicity. In aprotic solvents, nucleophiles aren't heavily solvated by hydrogen bonding, so their intrinsic properties dominate. Sulfur is much larger than oxygen and significantly more polarizable, meaning its electron cloud can easily distort to form new bonds. This polarizability makes $HS^-$ a superior nucleophile compared to $HO^-$ in DMSO, even though hydroxide is more basic. The larger sulfur atom can better stabilize the transition state during the $S_N2$ reaction with $CH_3I$. Choice A correctly identifies that sulfur's size and polarizability make $HS^-$ more nucleophilic in aprotic solvents. Choice B incorrectly applies electronegativity logic—while oxygen is more electronegative, this actually makes it hold its electrons more tightly, reducing nucleophilicity in aprotic environments. Choice C wrongly assumes similar nucleophilicity based on charge and basicity, ignoring the crucial role of polarizability differences. Choice D confuses bond lengths with nucleophilicity; the O-H vs S-H bond comparison is irrelevant since we're comparing the anions' ability to attack electrophilic carbon. Study tip: Remember the nucleophilicity trend reverses between protic and aprotic solvents. In aprotic solvents, larger, more polarizable atoms down a group (like sulfur vs oxygen) are better nucleophiles, while in protic solvents, smaller, less solvated ions often win.

Question 4

1. The electrophilic carbon's accessibility decreases with the bulky base, while the acidic β-hydrogens become more accessible, shifting reactivity toward elimination (correct answer)
2. The electrophilic carbon becomes more reactive due to increased steric hindrance from the bulkier base, favoring nucleophilic substitution over elimination
3. The change in solvent from methanol to tert-butanol increases the electrophilicity of the carbon-chlorine bond, promoting elimination reactions
4. Potassium tert-butoxide is a weaker nucleophile but stronger base than sodium methoxide, making the β-hydrogens more electrophilic than the α-carbon

Explanation: When you encounter competing substitution versus elimination reactions, the key factor is understanding how steric hindrance affects the accessibility of different reaction sites on the substrate. In this case, 2-chlorobutane is a secondary alkyl halide that can undergo both $S_N2$ substitution (attack at the carbon bearing chlorine) and $E2$ elimination (removal of a β-hydrogen). With small nucleophiles like methoxide ($CH_3O^-$), the carbon center is accessible for backside attack, favoring substitution. However, when you use the much bulkier tert-butoxide base, steric hindrance blocks access to the carbon center for substitution, while the β-hydrogens remain accessible for elimination. This shifts the reaction pathway from substitution to elimination. Choice A correctly identifies this principle: the bulky base reduces accessibility to the electrophilic carbon while β-hydrogens remain accessible, favoring elimination. Choice B incorrectly suggests that steric hindrance makes the carbon more reactive for substitution—it actually blocks substitution. Choice C wrongly attributes the change to solvent effects on the C-Cl bond electrophilicity, when sterics is the primary factor. Choice D contains a terminology error by calling β-hydrogens "electrophilic"—they're actually acidic and removed by the base. Remember this pattern: small nucleophiles favor substitution with secondary substrates, while bulky bases favor elimination due to steric accessibility differences. When you see bulky reagents in mechanism problems, always consider how sterics affects site accessibility.

Question 5

1. Trimethylamine acts as an electrophile because nitrogen is more electronegative than carbon, while boron trifluoride acts as a nucleophile due to its empty p orbital
2. Both molecules act as nucleophiles because they both contain lone pairs of electrons that can be donated to form the coordinate bond
3. Trimethylamine acts as a nucleophile by donating its lone pair, while boron trifluoride acts as an electrophile by accepting electrons into its vacant p orbital (correct answer)
4. Boron trifluoride acts as a nucleophile because boron is less electronegative than fluorine, while trimethylamine acts as an electrophile due to steric crowding

Explanation: The correct answer is C. In this Lewis acid-base reaction, trimethylamine has a lone pair on nitrogen, making it electron-rich and nucleophilic (Lewis base). BF₃ has an empty p orbital on boron, making it electron-deficient and electrophilic (Lewis acid). The coordinate bond forms when the nucleophilic nitrogen donates its lone pair to the electrophilic boron. Choice A incorrectly reverses the roles and misapplies electronegativity. Choice B is wrong because BF₃ doesn't have available lone pairs - it's electron-deficient. Choice D completely reverses the correct assignments and incorrectly invokes steric effects and electronegativity comparisons.

Question 6

1. The bromide ion (Br⁻), because it is a better nucleophile than water due to its negative charge.
2. Water (H₂O), because it is the solvent and present in a much higher concentration than the bromide ion. (correct answer)
3. The alkene, because its pi bond attacks the bromonium ion to form a more stable carbocation.
4. The bromide ion (Br⁻), because it is less sterically hindered than a water molecule.

Explanation: In halohydrin formation, although bromide ion is generated, water is the solvent and is present in a vast excess. Due to this large concentration difference, the statistical probability of a water molecule attacking the bromonium ion is much higher than that of a bromide ion. Therefore, water acts as the effective nucleophile, leading to the formation of the bromohydrin after a final deprotonation step.

Question 7

1. The tert-butyl group is more electron-donating, making tert-butoxide a much stronger base that can only perform elimination.
2. Ethoxide is a softer nucleophile than tert-butoxide, which favors attack at the carbon center over proton abstraction.
3. The significant steric hindrance of the bulky tert-butyl group prevents it from acting as an effective nucleophile, so it primarily functions as a base. (correct answer)
4. The reaction with ethoxide is reversible and thermodynamically controlled, while the reaction with tert-butoxide is irreversible and kinetically controlled.

Explanation: Nucleophilicity is a kinetic property that is highly sensitive to steric bulk. The large tert-butyl group on tert-butoxide sterically hinders its approach to the electrophilic carbon required for SN2 substitution. However, this bulk does not prevent it from abstracting a small, accessible proton from the substrate, allowing it to function effectively as a strong base for E2 elimination. Ethoxide is less hindered and can act as both a strong nucleophile and a strong base.

Question 8

1. The alkyl halide acts as a nucleophile toward the methanol electrophile, forming a carbocation intermediate that determines the final product distribution
2. The carbocation intermediate acts as an electrophile toward methanol, while the original halide acts as a nucleophile in a competing reaction pathway
3. Methanol acts as a nucleophile toward the carbocation electrophile formed after ionization of the alkyl halide in the rate-determining step (correct answer)
4. The halide leaving group acts as a nucleophile toward the carbocation, while methanol serves as an electrophilic proton source for the elimination pathway

Explanation: The correct answer is C. In the SN1 mechanism, the alkyl halide first ionizes to form a carbocation (electrophile) and halide ion in the rate-determining step. The methanol then acts as a nucleophile, attacking the electron-deficient carbocation. Choice A incorrectly reverses the nucleophile-electrophile roles. Choice B incorrectly suggests the halide ion acts as a nucleophile toward the carbocation, when it actually leaves as a leaving group. Choice D incorrectly describes the halide as a nucleophile toward the carbocation and mischaracterizes methanol's role in elimination.

Question 9

1. The positive charge is localized primarily on C2, making it the most electrophilic site for nucleophilic attack by bromide ion
2. The carbocation rearranges rapidly between primary and secondary forms, creating multiple electrophilic sites through dynamic equilibrium processes
3. The π electrons in the double bond system act as nucleophiles, while the sp² hybridized carbons act as electrophiles throughout the molecule
4. Resonance delocalizes the positive charge between C1 and C3, creating two electrophilic sites of equal reactivity toward nucleophiles (correct answer)

Explanation: When you encounter resonance structures and nucleophilic attack patterns, focus on how electron delocalization affects reactivity sites. Resonance structures show where electrons and charges can be distributed in a molecule. The allyl cation demonstrates classic allylic resonance. The two structures $CH_2=CH-CH_2^+ \leftrightarrow CH_2^+-CH=CH_2$ aren't separate molecules rapidly interconverting—they're different ways to draw the same species. The actual structure is a hybrid where the positive charge is shared equally between C1 and C3, while C2 remains neutral. This delocalization stabilizes the cation and creates two equivalent electrophilic sites where nucleophiles like $Br^-$ can attack, explaining why products form at both positions. Option A incorrectly places the charge on C2. In the resonance hybrid, C2 actually bears no positive charge—it's the bridge carbon in the three-carbon π system. Option B misunderstands resonance as a dynamic equilibrium between different carbocations, when it's actually about charge delocalization in a single species. Option C confuses the roles of nucleophiles and electrophiles—the π electrons don't act as nucleophiles here, and all carbons aren't electrophilic. Only C1 and C3 bear positive charge. Option D correctly identifies that resonance delocalizes the positive charge between the terminal carbons, making both electrophilic and equally reactive toward nucleophiles. Remember: resonance structures with equal stability contribute equally to the hybrid. When you see symmetrical allylic or benzylic systems, expect charge delocalization to create multiple reactive sites.

Question 10

1. Acetone is a nonpolar solvent, which better solubilizes the nonpolar 1-iodopropane.
2. Ethanol is a protic solvent that forms a strong hydrogen-bond cage around the cyanide nucleophile, deactivating it. (correct answer)
3. Acetone actively participates in the reaction mechanism as a catalyst, lowering the activation energy.
4. Ethanol is a weak acid and protonates the cyanide ion, converting it to HCN, a much weaker nucleophile.

Explanation: Acetone is a polar aprotic solvent, while ethanol is a polar protic solvent. Polar protic solvents like ethanol have acidic protons (O-H) that can form strong hydrogen bonds with anionic nucleophiles like CN⁻. This 'solvation shell' or 'cage' stabilizes the nucleophile and sterically hinders its attack on the electrophile, slowing the reaction. In contrast, polar aprotic solvents like acetone solvate the cation (Na⁺) but not the anion, leaving the 'naked' nucleophile highly reactive.

Question 11

1. It acts as a Brønsted-Lowry base, accepting a proton from BH₃.
2. It acts as a Lewis base (nucleophile), donating electron density from its pi bond to the boron atom. (correct answer)
3. It acts as a Lewis acid (electrophile), accepting a hydride from the boron atom.
4. It acts as a radical initiator, homolytically cleaving the B-H bond.

Explanation: The boron atom in BH₃ has an empty p-orbital and is electron-deficient, making it a Lewis acid (electrophile). The alkene's pi bond is a region of high electron density, capable of donating electrons. Therefore, the alkene acts as a Lewis base (nucleophile), attacking the electrophilic boron atom. This is the key interaction in the first step of hydroboration.

Question 12

1. (CH₃)₃N > CH₃NH₂ > NH₃
2. CH₃NH₂ > (CH₃)₃N > NH₃
3. NH₃ > CH₃NH₂ > (CH₃)₃N
4. CH₃NH₂ > NH₃ > (CH₃)₃N (correct answer)

Explanation: When evaluating nucleophilicity of nitrogen compounds, you need to consider two competing factors: electron density on nitrogen and steric hindrance around the nucleophilic center. All three compounds have a lone pair on nitrogen, making them nucleophiles. However, alkyl groups affect nucleophilicity in two ways. First, they're electron-donating through inductive effects, increasing electron density on nitrogen and enhancing nucleophilicity. Second, they create steric bulk that can hinder the nucleophile's approach to electrophilic centers. Methylamine (CH₃NH₂) is the strongest nucleophile here because it gains significant electron density from one methyl group without excessive steric hindrance. Ammonia (NH₃) ranks second—while it lacks the electron-donating methyl groups, it's completely unhindered sterically. Trimethylamine ((CH₃)₃N) is the weakest nucleophile despite having three electron-donating methyl groups because the steric crowding around nitrogen severely impedes its ability to attack electrophiles. Looking at the choices: A) incorrectly ranks trimethylamine as most nucleophilic, overestimating the electronic effect while ignoring sterics. B) correctly identifies methylamine as strongest but wrongly places trimethylamine above ammonia. C) suggests ammonia is most nucleophilic, ignoring the beneficial electronic effects of alkyl substitution entirely. Only D) correctly recognizes that methylamine balances electronic enhancement with manageable sterics, making it superior to both unsubstituted ammonia and over-substituted trimethylamine. Study tip: In nucleophilicity problems, moderate substitution often wins—enough electron donation to help, but not so much bulk that it hurts.

Question 13

1. Methoxide, because its negative charge makes it a much stronger nucleophile than neutral methanol. (correct answer)
2. Methanol, because as the solvent, its high concentration outweighs its lower intrinsic nucleophilicity.
3. Bromide, because it is generated as a leaving group and can re-attack the carbocation intermediate.
4. Both methanol and methoxide will act as nucleophiles with roughly equal effectiveness.

Explanation: When analyzing nucleophile competition in organic reactions, you need to consider both the intrinsic nucleophilicity of each species and their relative concentrations. This question tests your understanding of how charge dramatically affects nucleophilic strength. Methoxide ion ($\text{OCH}_3^-$) is an exceptionally strong nucleophile due to its negative charge, which makes it electron-rich and highly reactive toward electrophiles like 1-bromobutane. Even though methoxide is present in only small amounts compared to methanol, its vastly superior nucleophilicity means it will dominate the reaction. Charged nucleophiles are orders of magnitude more reactive than their neutral counterparts. Looking at the incorrect options: Option B incorrectly assumes that concentration alone determines nucleophile effectiveness. While methanol is present in large excess, its neutral charge makes it a weak nucleophile that cannot compete effectively with the charged methoxide. Option C misunderstands the reaction mechanism—this is an $S_N2$ reaction (primary alkyl halide with strong nucleophile), so no carbocation intermediate forms. Bromide ion, even if present, would be a poor nucleophile in the protic solvent methanol. Option D fails to recognize the enormous difference in nucleophilic strength between charged and neutral species. The answer is A because charge trumps concentration when the nucleophilicity difference is this dramatic. Study tip: Remember that in nucleophile competition, intrinsic reactivity usually beats concentration unless the amounts are extremely disproportionate. Negatively charged nucleophiles almost always outcompete their neutral analogs, even when present in much smaller quantities.

Question 14

1. Propanal (CH₃-CH₂-C*HO) (correct answer)
2. Propane (CH₃-C*H₂-CH₃)
3. Propyne (CH₃-C*≡CH)
4. Propene (CH₃-C*H=CH₂)

Explanation: When evaluating nucleophilic susceptibility, you need to consider the electronic environment around the carbon atom. Nucleophiles are electron-rich species that seek electron-deficient (electrophilic) carbon centers. The key factor here is the presence of electron-withdrawing groups that create partial positive charge on carbon. In propanal (A), the carbonyl oxygen is highly electronegative and pulls electron density away from the carbon through both inductive and resonance effects. This creates a significant partial positive charge ($\delta^+$) on the carbonyl carbon, making it highly electrophilic and susceptible to nucleophilic attack. The carbonyl carbon is also sp²-hybridized, which is more electronegative than sp³, further enhancing its electrophilicity. Option B (propane) features a saturated carbon surrounded only by other carbons and hydrogens—no electron-withdrawing groups are present, making this carbon essentially neutral and unreactive toward nucleophiles. Option C (propyne) has an sp-hybridized carbon that, while more electronegative than sp² or sp³, lacks electron-withdrawing substituents and is involved in a stable triple bond. Option D (propene) contains an sp²-hybridized carbon, but it's part of an electron-rich alkene system where the π-electrons actually make the carbon more electron-rich, not electron-poor. Study tip: Look for carbons adjacent to highly electronegative atoms (especially oxygen in carbonyls) or electron-withdrawing groups. Carbonyl carbons are classic electrophilic centers and frequently appear in nucleophilic addition reactions throughout organic chemistry.

Question 15

1. Sodium azide (NaN₃)
2. Sodium hydrosulfide (NaSH)
3. Water (H₂O) (correct answer)
4. Sodium iodide (NaI)

Explanation: SN2 reactions require good nucleophiles. Nucleophilicity is enhanced by a negative charge and decreased by high electronegativity and solvation in protic solvents. Azide (N₃⁻), hydrosulfide (SH⁻), and iodide (I⁻) are all strong to excellent nucleophiles because they are anionic. Water (H₂O) is a neutral molecule and a very weak nucleophile. While it can participate in substitution reactions (typically SN1 solvolysis), it is a very poor choice for promoting a fast SN2 reaction.

Question 16

1. HO⁻ is a stronger base and a stronger nucleophile than HS⁻.
2. HO⁻ is a stronger base, but HS⁻ is a stronger nucleophile. (correct answer)
3. HS⁻ is a stronger base and a stronger nucleophile than HO⁻.
4. HS⁻ is a stronger base, but HO⁻ is a stronger nucleophile.

Explanation: Basicity is determined by pKa. A stronger base has a weaker conjugate acid. Since H₂O (pKa=15.7) is a much weaker acid than H₂S (pKa=7.0), the conjugate base HO⁻ is a much stronger base than HS⁻. Nucleophilicity in a polar protic solvent for atoms in the same group is determined by polarizability. Sulfur is larger and more polarizable than oxygen, and HS⁻ is less strongly solvated than HO⁻. Both factors make HS⁻ the stronger nucleophile in a protic solvent. This is a classic example of nucleophilicity and basicity trends not being parallel.

Question 17

1. The sp²-hybridized oxygen in phenoxide is more electronegative than the sp³-hybridized oxygen in cyclohexoxide.
2. The negative charge on the phenoxide oxygen is delocalized into the aromatic ring by resonance, making it less available for attack. (correct answer)
3. The cyclohexane ring is more sterically hindering than the flat phenyl ring, which concentrates the charge on the oxygen.
4. The phenyl group is strongly electron-withdrawing by induction, pulling electron density away from the nucleophilic oxygen atom.

Explanation: The key difference is resonance. In the phenoxide ion, the lone pair on the oxygen atom (and thus the negative charge) is delocalized over the entire aromatic ring through resonance. This stabilization makes the charge less concentrated on the oxygen and less available to act as a nucleophile. In the cyclohexoxide ion, the negative charge is localized entirely on the oxygen atom, making it a more potent nucleophile.

Question 18

1. CH₃O⁻
2. CH₃SH
3. CH₃OH
4. CH₃S⁻ (correct answer)

Explanation: When evaluating nucleophile strength in polar, protic solvents, you need to consider how solvation affects the nucleophile's ability to donate electrons. In protic solvents like methanol, hydrogen bonding significantly influences nucleophilic behavior. The key principle is that smaller, more electronegative atoms become weaker nucleophiles in protic solvents due to extensive hydrogen bonding that stabilizes and "ties up" the nucleophile. Conversely, larger atoms with lower electronegativity are less tightly solvated and remain more nucleophilic. CH₃S⁻ (option D) is the strongest nucleophile because sulfur is larger and less electronegative than oxygen. The negative charge on sulfur experiences less hydrogen bonding with methanol molecules, leaving it more available for nucleophilic attack. Option A (CH₃O⁻) is incorrect because the smaller, more electronegative oxygen atom becomes heavily solvated through hydrogen bonding, dramatically reducing its nucleophilicity despite being negatively charged. Option B (CH₃SH) is wrong because it's neutral - the sulfur still has its proton attached, making it much less electron-rich than the deprotonated thiolate ion. Option C (CH₃OH) is incorrect for the same reason as B, plus oxygen is less nucleophilic than sulfur even when both are neutral. Study tip: Remember the periodic trend reversal in protic solvents - going down a group increases nucleophile strength (opposite of basicity trends) because larger atoms resist solvation. This is a classic MCAT-style concept that frequently appears on organic chemistry exams.

Question 19

1. H₃O⁺
2. CH₃OH
3. BF₃ (correct answer)
4. CH₃COOH

Explanation: An electrophile is a Lewis acid, an electron-pair acceptor. A Brønsted-Lowry acid is a proton (H⁺) donor. Boron trifluoride (BF₃) has an incomplete octet on the boron atom, making it a potent electron-pair acceptor (Lewis acid/electrophile). However, it has no protons to donate, so it cannot be a Brønsted-Lowry acid. H₃O⁺, CH₃OH, and CH₃COOH all have acidic protons and can act as Brønsted-Lowry acids.

Question 20

1. The methyl group in CH₃S⁻ is electron-donating by induction, increasing the electron density on the sulfur atom. (correct answer)
2. The methyl group in CH₃S⁻ sterically hinders the sulfur atom, increasing its reactivity.
3. HS⁻ is a smaller ion and is more heavily solvated in most solvents, which deactivates it.
4. HS⁻ is a stronger base than CH₃S⁻, and stronger bases are always weaker nucleophiles.

Explanation: When comparing nucleophilicity, you need to consider how electron density and solvation effects influence a nucleophile's ability to attack an electrophile. Nucleophilicity generally increases with greater electron density on the attacking atom and decreases when the nucleophile is heavily solvated. The methyl group in CH₃S⁻ acts as an electron-donating group through inductive effects. Alkyl groups like methyl are electron-releasing because they're less electronegative than hydrogen, pushing electron density toward the sulfur atom. This increased electron density makes the sulfur more nucleophilic and better able to attack the electrophilic carbon in methyl iodide. Answer A correctly identifies this key factor. Answer B incorrectly suggests steric hindrance increases reactivity. Steric hindrance actually decreases nucleophilicity by making it harder for the nucleophile to approach the electrophile. The methyl group doesn't create significant steric problems here anyway. Answer C contains a grain of truth—smaller ions can be more solvated—but this isn't the primary factor explaining the reactivity difference between these two sulfur nucleophiles of similar size and charge. Answer D states a false general rule. While there's often an inverse relationship between basicity and nucleophilicity for different elements, this doesn't apply universally, especially when comparing similar species where other factors like electron donation dominate. Study tip: Remember that electron-donating groups (like alkyl groups) increase nucleophilicity by increasing electron density on the nucleophilic atom. This inductive effect is a key factor in predicting relative nucleophile strength.