Organic Chemistry Quiz: Sn2 Reactions Conditions Stereochemistry Rate

What this quiz covers

This quiz focuses on Sn2 Reactions Conditions Stereochemistry Rate, 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 reaction rate will be identical, and both reactions will produce the same enantiomer of the product in equal amounts
2. The reaction rate will be identical, and the products will be enantiomers of each other with opposite absolute configurations (correct answer)
3. The reaction rate will be different because the transition states have different energies, but the products will be enantiomers
4. The reaction rate will be identical, but the (S)-substrate will produce a racemic mixture while the (R)-substrate produces a single enantiomer

Explanation: In SN2 reactions, enantiomeric substrates react at identical rates because they form enantiomeric transition states that have identical energies. Both reactions proceed with complete inversion of configuration at the stereocenter. Starting with (R)-2-bromobutane produces (S)-2-azidobutane, while starting with (S)-2-bromobutane produces (R)-2-azidobutane. The products are enantiomers with opposite absolute configurations. Choice A is wrong because the products are enantiomers, not identical. Choice C is wrong because enantiomeric transition states have equal energies. Choice D is wrong because SN2 reactions proceed with complete inversion, not racemization.

Question 2

1. The rate increase follows the dielectric constant of the solvents, with higher dielectric constants stabilizing the charged nucleophile more effectively
2. Polar aprotic solvents accelerate SN2 by better solvating the partial positive charge that develops on the benzylic carbon during nucleophilic attack
3. The trend reflects the ability of solvents to stabilize the developing negative charge on the leaving bromide ion in the transition state
4. Protic solvents slow the reaction by hydrogen bonding to hydroxide, while aprotic solvents cannot hydrogen bond, leaving hydroxide more nucleophilic (correct answer)

Explanation: When analyzing SN2 reactions across different solvents, you need to focus on how solvents affect nucleophile reactivity. The key insight is understanding the difference between protic and aprotic solvents and their interactions with nucleophiles. In protic solvents like water and methanol, hydrogen bonding occurs between the solvent and the hydroxide nucleophile. This solvation "wraps" the nucleophile in a hydrogen-bonded shell, making it less available for attack on the electrophilic carbon. The nucleophile becomes less "naked" and therefore less reactive. Polar aprotic solvents like DMF and DMSO cannot form hydrogen bonds with anions because they lack hydrogen atoms bonded to highly electronegative atoms. This leaves the hydroxide ion much more nucleophilic - it's essentially "naked" and highly reactive. The dramatic rate increases (2400× and 7200×) reflect this enhanced nucleophilicity. Answer D correctly identifies this fundamental principle. Answer A is wrong because dielectric constant alone doesn't explain the dramatic differences - DMSO and water have similar dielectric constants but vastly different reaction rates. Answer B incorrectly focuses on stabilizing the electrophile rather than the nucleophile effect, which is the dominant factor in SN2 solvent effects. Answer C misidentifies the key interaction - it's the nucleophile solvation that matters most, not leaving group stabilization. Study tip: For SN2 solvent effects, remember "protic = slower nucleophile" due to hydrogen bonding, while "aprotic = faster nucleophile" because the nucleophile stays more reactive.

Question 3

1. The results show secondary isotope effects from hyperconjugation changes in the transition state, with deuterium substitution affecting the electron-donating ability of adjacent C-H bonds (correct answer)
2. The KIE demonstrates that C-H bond breaking occurs during the rate-determining step, with more C-H bonds involved giving larger isotope effects
3. The isotope effect arises from mass differences affecting the vibrational frequencies of C-C bonds that undergo stretching during nucleophilic attack
4. The KIE reflects changes in hybridization at the reaction center, with deuterium substitution altering the energy required for sp³ to sp² transition state geometry

Explanation: When you encounter kinetic isotope effect (KIE) questions in SN2 reactions, focus on understanding how isotope substitution affects bond vibrations and hyperconjugation, not bond breaking during the reaction. The correct interpretation (A) recognizes that these are secondary isotope effects. In SN2 reactions, the C-H bonds adjacent to the reaction center don't break, but their vibrational frequencies change as the transition state forms. Deuterium makes C-D bonds vibrate more slowly than C-H bonds, affecting hyperconjugation - the overlap between C-H σ bonds and the developing p-orbital at the reaction center. When you have more deuterium atoms (CD₃CD₂Br vs CD₃CH₂Br), the hyperconjugative stabilization changes more dramatically, explaining why the KIE increases from 1.15 to 1.35. Choice B incorrectly assumes C-H bonds break during SN2 reactions - they don't. The rate-determining step involves only C-Br bond breaking and C-O bond formation. Choice C misidentifies the source as C-C bond stretching, but C-C bonds don't significantly stretch in SN2 mechanisms. Choice D incorrectly focuses on hybridization changes at the reaction center itself, but the isotope substitution is on adjacent carbons, not the reaction center. Remember that secondary isotope effects in SN2 reactions arise from hyperconjugation changes in bonds that remain intact throughout the reaction. The magnitude correlates with the number of deuterium atoms because more C-D bonds mean greater cumulative effects on the transition state stabilization.

Question 4

1. 4.8 × 10⁻³ M⁻¹s⁻¹, because only the alkyl halide concentration affects the rate in SN2 reactions
2. 7.2 × 10⁻³ M⁻¹s⁻¹, because the rate constant changes when nucleophile concentration increases significantly
3. 1.44 × 10⁻² M⁻¹s⁻¹, because both reactant concentrations affect the rate and multiply together in the rate expression (correct answer)
4. 2.4 × 10⁻³ M⁻¹s⁻¹, because changing concentrations does not affect the rate constant in second-order reactions

Explanation: SN2 reactions follow second-order kinetics: rate = k[RX][Nu⁻]. The rate depends on both the alkyl halide and nucleophile concentrations. When [RX] is doubled (×2) and [Nu⁻] is tripled (×3), the new rate = k(2[RX])(3[Nu⁻]) = 6 × k[RX][Nu⁻] = 6 × original rate = 6 × 2.4 × 10⁻³ = 1.44 × 10⁻² M⁻¹s⁻¹. Choice A incorrectly assumes first-order dependence on RX only. Choice B gives an incorrect calculation and wrong reasoning about rate constants. Choice D confuses rate with rate constant - the rate constant k remains the same, but the overall reaction rate changes with concentration.

Question 5

1. The rate will be faster in DMF because DMF is a polar aprotic solvent that solvates the Na⁺ cation but poorly solvates the CN⁻ anion, increasing its effective nucleophilicity. (correct answer)
2. The rate will be faster in ethanol because ethanol is a polar protic solvent whose hydrogen bonds stabilize the negatively charged transition state.
3. The rate will be faster in ethanol because its higher dielectric constant better supports the charge separation in the transition state.
4. The rates will be approximately equal because both DMF and ethanol are polar solvents capable of dissolving the ionic reactant.

Explanation: SN2 reactions with anionic nucleophiles are fastest in polar aprotic solvents like DMF. Polar protic solvents like ethanol have acidic protons that form a strong solvent shell around the anionic nucleophile (CN⁻) through hydrogen bonding. This solvation stabilizes the nucleophile, lowering its energy and increasing the activation energy of the reaction. DMF lacks these acidic protons, leaving the CN⁻ anion less solvated, higher in energy, and thus more reactive.

Question 6

1. The product will be 2,3-dimethylpent-2-ene, formed via a rapid E2 elimination pathway.
2. The major product will be (S)-3-azido-2,2,3-trimethylpentane, assuming an (R)-starting material was used.
3. A racemic mixture of the azido product will form via an SN1 mechanism, promoted by the polar solvent.
4. No reaction or a very slow elimination reaction will occur because SN2 attack is impossible at a sterically hindered tertiary carbon. (correct answer)

Explanation: When you encounter substitution reactions involving tertiary carbons, you need to carefully analyze both the substrate structure and the nucleophile to predict the mechanism and outcome. The substrate 3-bromo-2,2,3-trimethylpentane has a tertiary carbon (three methyl substituents) bonded to the bromine. Sodium azide ($\text{N}_3^-$) is a strong nucleophile that typically undergoes $S_N2$ reactions with primary and secondary substrates. However, $S_N2$ mechanisms require backside attack by the nucleophile, which becomes increasingly difficult as steric hindrance increases around the electrophilic carbon. At a tertiary carbon center, the extreme steric crowding from three bulky substituents effectively blocks the nucleophile's approach path, making $S_N2$ substitution virtually impossible. While DMSO is a polar aprotic solvent that enhances nucleophilicity, it cannot overcome the fundamental geometric constraints of the tertiary substrate. Option A is incorrect because E2 elimination requires a strong base, and azide is primarily nucleophilic rather than basic. Option B wrongly suggests an $S_N2$ inversion could occur at the hindered tertiary center. Option C incorrectly assumes an $S_N1$ mechanism would proceed readily, but tertiary alkyl halides need additional stabilization (like benzylic or allylic resonance) to form stable carbocations under typical $S_N1$ conditions. The correct answer is D because the severe steric hindrance prevents nucleophilic attack, resulting in no reaction or possibly very slow elimination if heated extensively. Study tip: Remember that tertiary carbons are "$S_N2$-dead" due to steric hindrance. Always check the substitution pattern before predicting substitution mechanisms.

Question 7

1. The reaction with potassium tert-butoxide proceeds via an SN1 mechanism, where the resulting primary carbocation rapidly eliminates a proton.
2. Potassium tert-butoxide is a much weaker base than potassium ethoxide, which thermodynamically favors the elimination pathway.
3. The tert-butanol solvent is less polar than ethanol, which fundamentally changes the mechanism from substitution to elimination.
4. Potassium tert-butoxide is a sterically hindered, strong base that acts as a poor nucleophile, favoring E2 elimination over SN2 substitution. (correct answer)

Explanation: When you encounter competing reaction pathways between substitution and elimination, focus on how the base/nucleophile's properties determine the outcome. Both reactions involve the same substrate (1-bromopropane), so the reagent difference drives the selectivity. Potassium tert-butoxide is both a strong base and sterically bulky. The tert-butyl group creates significant steric hindrance around the oxygen atom, making it difficult for KOtBu to approach the electrophilic carbon for backside attack in an SN2 mechanism. However, its strong basicity allows it to easily abstract a proton from the β-carbon, promoting E2 elimination to form propene. In contrast, potassium ethoxide is less hindered and can readily approach the carbon for SN2 substitution, forming ethyl propyl ether. Choice A is incorrect because primary alkyl halides don't undergo SN1 reactions—primary carbocations are too unstable. Choice B mischaracterizes the basicity: tert-butoxide is actually a stronger base than ethoxide, not weaker. The stronger base favors elimination kinetically, not thermodynamically. Choice C incorrectly emphasizes solvent polarity as the determining factor. While solvent can influence reaction rates, the key difference here is the steric and basic properties of the reagent, not the solvent polarity change. Remember this pattern: bulky, strong bases favor elimination (E2) over substitution (SN2) due to steric hindrance preventing nucleophilic attack while maintaining the ability to abstract protons. When analyzing base/nucleophile selectivity, always consider both steric accessibility and basicity strength.

Question 8

1. It is a racemic mixture of (R)- and (S)-2-butanol, both containing oxygen-18.
2. It consists exclusively of (S)-2-butanol containing oxygen-18.
3. It consists exclusively of (R)-2-butanol containing oxygen-18. (correct answer)
4. It consists exclusively of (R)-2-butanol containing normal oxygen-16.

Explanation: When you encounter a substitution reaction with a chiral substrate, the key is identifying whether the mechanism proceeds via S_N1 or S_N2, as this determines both stereochemistry and which nucleophile attacks. Starting with (S)-2-bromobutane and hydroxide ion, this reaction follows an S_N2 mechanism. Secondary alkyl halides with strong nucleophiles like hydroxide typically undergo S_N2 substitution. In S_N2 reactions, the nucleophile attacks the carbon from the backside (opposite to the leaving group), causing inversion of configuration at the chiral center. Since you start with (S)-2-bromobutane, inversion gives you (R)-2-butanol exclusively. The isotopic labeling with oxygen-18 in Na¹⁸OH allows you to track the nucleophile. Since hydroxide is the attacking nucleophile, the oxygen-18 becomes incorporated into the product alcohol. Answer A is incorrect because S_N2 reactions don't produce racemic mixtures - they give complete inversion, not a 50:50 mixture of stereoisomers. Answer B is wrong because it suggests retention of configuration, which would occur in an S_N1 mechanism (where you'd actually get racemization). Answer D incorrectly suggests the product contains normal oxygen-16, but since the hydroxide nucleophile contains oxygen-18, this isotope must appear in the product. Remember this pattern: S_N2 reactions always involve inversion of stereochemistry and direct incorporation of the nucleophile. When you see isotopic labeling experiments, they're designed to trace exactly which species becomes incorporated into the product, helping confirm the mechanism.

Question 9

1. In NaNO₂, the harder oxygen atom of the nitrite ion preferentially attacks the hard carbon center of the alkyl halide.
2. The reaction with AgNO₂ proceeds via an SN1 mechanism, and the resulting carbocation is preferentially attacked by the nitrogen atom.
3. In AgNO₂, the Ag-O bond has significant covalent character, making the nitrogen's lone pair the more available nucleophilic site for attack. (correct answer)
4. The solvents typically used for each reaction are different, and this solvent effect is solely responsible for the change in selectivity.

Explanation: When you encounter questions about nucleophilic substitution reactions involving ambidentate nucleophiles (nucleophiles with two potential attacking sites), focus on how the metal cation affects nucleophilicity and bond character. The nitrite ion ($NO_2^-$) is ambidentate, meaning it can attack through either nitrogen or oxygen. The key difference lies in how silver versus sodium affects the nucleophile's behavior. In AgNO₂, silver forms a bond with oxygen that has significant covalent character due to silver's intermediate electronegativity and polarizability. This covalent Ag-O interaction effectively "ties up" the oxygen atom, making nitrogen's lone pair the primary nucleophilic site available to attack the carbon center, yielding 1-nitrobutane (C-N bond formation). Option A incorrectly applies hard-soft acid-base theory backwards - if this were the controlling factor, we'd expect the same selectivity regardless of the metal cation. Option B is wrong because 1-iodobutane is a primary alkyl halide that reacts via SN2, not SN1 mechanisms, and doesn't form stable carbocations. Option D oversimplifies the phenomenon - while solvents can influence reactions, the fundamental difference stems from the metal cation's effect on nucleophile availability, not solvent choice alone. The correct answer is C because the covalent character of the Ag-O bond makes nitrogen the dominant nucleophilic site, while in sodium nitrite, the more ionic Na-O interaction leaves oxygen more available for nucleophilic attack. Study tip: Remember that metal cations can dramatically alter nucleophile selectivity by affecting bond character and electron availability - always consider how the counterion influences the nucleophile's reactive sites.

Question 10

1. Iodide is a vastly superior nucleophile to chloride, making the forward reaction rate effectively irreversible under these conditions.
2. Sodium chloride (NaCl) is insoluble in acetone and precipitates, driving the equilibrium to the right via Le Châtelier's principle. (correct answer)
3. The C-I bond is much weaker than the C-Cl bond, making the product thermodynamically more stable and favored at equilibrium.
4. Acetone is a polar aprotic solvent that specifically solvates the iodide ion, increasing its reactivity and pushing the reaction forward.

Explanation: When you encounter questions about the Finkelstein reaction, remember that understanding why equilibrium reactions go to completion requires thinking about what removes products from the reaction mixture. The Finkelstein reaction works because of a clever solvent choice that exploits solubility differences. When 1-chlorobutane reacts with NaI in acetone, the reaction produces 1-iodobutane and NaCl. The key insight is that acetone is a polar aprotic solvent that dissolves organic halides and NaI well, but NaCl is essentially insoluble in acetone and precipitates out as a white solid. According to Le Châtelier's principle, removing NaCl from solution shifts the equilibrium toward products, driving the reaction to completion. Let's examine why the other options miss the mark. Choice A incorrectly focuses on nucleophile strength - while iodide is indeed a better nucleophile than chloride, this alone wouldn't drive an equilibrium reaction to completion since the reverse reaction would still occur. Choice C gets the thermodynamics backward - the weaker C-I bond actually makes the product less thermodynamically stable, not more. Choice D mentions acetone's ability to solvate ions, but this affects both forward and reverse reactions similarly and doesn't explain why the equilibrium shifts. Study tip: For halide exchange reactions, always consider solubility! The Finkelstein reaction is a classic example of using precipitation to drive equilibrium. Remember that NaCl and NaBr are poorly soluble in acetone, while NaI dissolves well - this solubility pattern determines which direction these reactions can proceed.

Question 11

1. The ground-state stability of the alkyl halide determines the rate, with less stable primary halides reacting faster than more stable secondary halides.
2. Transition state energy increases with steric hindrance: primary (1°) < secondary (2°) < heavily β-branched primary (neopentyl). (correct answer)
3. The reaction rate follows the stability of the partial positive charge in the transition state, which is greater for a secondary carbon than a primary one.
4. Neopentyl bromide is slow because it first rearranges to a more stable tertiary carbocation, which is a slow, rate-limiting step.

Explanation: When analyzing SN2 reaction rates, you need to focus on how steric hindrance affects the transition state energy. SN2 reactions proceed through a single concerted step where the nucleophile attacks the carbon while the leaving group departs simultaneously. The correct answer is B because steric hindrance directly determines how easily the nucleophile can approach the electrophilic carbon. In 1-bromopropane (primary), there's minimal steric crowding around the reaction center, allowing easy nucleophilic attack. 2-bromopropane (secondary) has more steric hindrance from the additional methyl group, raising the transition state energy and slowing the reaction. Neopentyl bromide, though technically primary, has three bulky methyl groups on the β-carbon that severely block nucleophilic approach, making it the slowest. Option A incorrectly focuses on ground-state stability rather than transition state effects. SN2 rates aren't determined by starting material stability but by the energy barrier to reach the transition state. Option C confuses SN2 with SN1 mechanisms. SN2 reactions don't involve partial positive charges or carbocation-like character in the transition state—the mechanism is concerted with simultaneous bond breaking and forming. Option D also incorrectly invokes SN1 thinking. Neopentyl bromide doesn't rearrange through carbocation intermediates in SN2 conditions; it's simply too sterically hindered for the concerted mechanism to proceed efficiently. Remember: SN2 rates depend on sterics, not electronics. The more crowded the reaction center, the higher the activation energy and the slower the reaction.

Question 12

1. In the polar protic solvent ethanol, SH⁻ is a better nucleophile than OH⁻ because sulfur is larger and more polarizable than oxygen. (correct answer)
2. NaOH is a much stronger base than NaSH, causing it to favor a competing E2 elimination reaction over SN2 substitution.
3. The C-S bond is thermodynamically more stable than the C-O bond, which results in a faster rate of formation.
4. The hydroxide ion (OH⁻) is smaller than the hydrosulfide ion (SH⁻), and its smaller size leads to greater steric hindrance in the transition state.

Explanation: Nucleophilicity trends are key here. When comparing nucleophiles from the same group in the periodic table (like O and S) in a polar protic solvent (like ethanol), the larger atom is the better nucleophile. This is because the larger atom (S) is more polarizable, meaning its electron cloud can be more easily distorted to form a bond in the transition state. Additionally, the smaller, more charge-dense OH⁻ ion is more heavily solvated by ethanol via hydrogen bonding, which stabilizes it and makes it less reactive.

Question 13

1. The new rate will be 9 times the original rate.
2. The new rate will be 5 times the original rate.
3. The new rate will be 3 times the original rate.
4. The new rate will be 6 times the original rate. (correct answer)

Explanation: When you encounter SN2 reaction kinetics problems, remember that SN2 reactions follow second-order kinetics, meaning the rate depends on the concentrations of both reactants. The rate law for the SN2 reaction between methyl iodide (CH₃I) and hydroxide ion (OH⁻) is: $\text{Rate} = k[\text{CH₃I}][\text{OH⁻}]$ To find how the rate changes, you need to substitute the new concentrations into this rate law. If the initial concentration of methyl iodide is doubled (2×) and the initial concentration of sodium hydroxide is tripled (3×), the new rate becomes: $\text{New Rate} = k[2 \times \text{CH₃I}][3 \times \text{OH⁻}] = k \times 2 \times 3 \times [\text{CH₃I}][\text{OH⁻}] = 6 \times \text{Original Rate}$ Therefore, answer D is correct - the new rate will be 6 times the original rate. Answer A (9 times) incorrectly assumes you square one of the concentration changes (3² = 9) or multiply the changes incorrectly. Answer B (5 times) mistakenly adds the multiplication factors (2 + 3 = 5) instead of multiplying them. Answer C (3 times) only accounts for the change in hydroxide concentration while ignoring the change in methyl iodide concentration. For SN2 kinetics problems, always remember that both reactant concentrations matter equally in the rate law. When concentrations change, multiply all the changes together to find the overall rate change.

Question 14

1. The product is (S)-2-azidopentane; the rate-determining step is the bimolecular collision between the azide ion and the alkyl halide. (correct answer)
2. The product is (R)-2-azidopentane; the rate-determining step is the formation of a carbocation intermediate after the bromide leaves.
3. The product is a racemic mixture of (R)- and (S)-2-azidopentane; the rate-determining step is the bimolecular collision between reactants.
4. The product is (S)-2-azidopentane; the rate-determining step is the dissociation of the sodium azide salt in acetone.

Explanation: The reaction conditions (secondary alkyl halide, strong nucleophile, polar aprotic solvent) are ideal for an SN2 reaction. The SN2 mechanism proceeds via backside attack, which results in inversion of configuration at the stereocenter. Thus, (R)-2-bromopentane yields (S)-2-azidopentane. The SN2 reaction is a concerted, one-step process, so the rate-determining step is the bimolecular collision of the two reactants.

Question 15

1. The ethyl bromide reaction has a higher activation energy than ethyl iodide, and this difference becomes more pronounced at higher temperatures
2. Both reactions have similar activation energies, but the pre-exponential factor is different due to the different leaving group sizes
3. The ethyl bromide reaction has a higher activation energy, but the temperature effect partially compensates, making the rate difference smaller at 45°C
4. The activation energy difference cannot be determined without additional temperature data points for the ethyl bromide reaction (correct answer)

Explanation: To determine activation energy differences, we need rate constants at different temperatures for both substrates. We have temperature-dependent data for ethyl iodide (showing the typical increase with temperature), but only one temperature point for ethyl bromide. While we can see that ethyl bromide reacts slower than ethyl iodide at 45°C (2.8 × 10⁻² vs 1.1 × 10⁻¹), we cannot calculate its activation energy or compare activation energies without knowing how the ethyl bromide rate changes with temperature. Choice A assumes higher Ea but lacks data support. Choice B incorrectly assumes similar Ea values. Choice C makes assumptions about temperature effects without sufficient data.

Question 16

1. All relative rates will increase proportionally because tert-butoxide is a stronger base than methoxide, making it more nucleophilic
2. The relative rate differences will become more pronounced, with (CH₃)₃CBr showing virtually no reaction and CH₃CHBrCH₃ reacting even slower relative to CH₃CH₂Br (correct answer)
3. The relative rates will remain approximately the same because the steric hindrance affects all substrates equally in SN2 reactions
4. The primary bromide will react slower while the secondary and tertiary bromides will react faster due to the increased basicity favoring more substituted carbons

Explanation: Tert-butoxide ion is much more sterically hindered than methoxide ion. In SN2 reactions, steric hindrance in the nucleophile compounds the steric problems already present with more substituted alkyl halides. The bulky tert-butoxide will have even more difficulty approaching the backside of secondary and tertiary carbons, making the rate differences more extreme. Primary alkyl halides will still react fastest, but secondary will be much slower relative to primary, and tertiary will be essentially unreactive. Choice A is wrong because increased basicity doesn't overcome steric hindrance in SN2. Choice C is wrong because steric effects multiply when both nucleophile and substrate are hindered. Choice D incorrectly suggests that basicity favors more substituted substrates, but SN2 reactions depend on sterics, not substrate stability.

Question 17

1. The ratio will shift to approximately 60% bromide and 40% chloride because thiophenoxide is a softer nucleophile that discriminates less between leaving groups (correct answer)
2. The ratio will remain essentially the same (85:15) because the leaving group ability difference between Br⁻ and Cl⁻ is independent of nucleophile identity
3. The ratio will become more extreme, approximately 95% bromide and 5% chloride, because sulfur nucleophiles are more sensitive to leaving group quality
4. The ratio will reverse to favor chloride because thiophenoxide forms stronger bonds with chlorine than bromine in the transition state

Explanation: According to hard-soft acid-base theory, azide (N₃⁻) is a harder nucleophile that shows greater discrimination between hard (Cl⁻) and soft (Br⁻) leaving groups. Thiophenoxide is a softer nucleophile that discriminates less between leaving groups of different hardness. While Br⁻ is still a better leaving group than Cl⁻, the difference in reaction rates will be smaller with the softer nucleophile, resulting in a less extreme ratio. Choice B is wrong because nucleophile-leaving group matching does affect selectivity. Choice C is incorrect because softer nucleophiles are less discriminating, not more. Choice D is wrong because it ignores the fundamental principle that Br⁻ is a better leaving group than Cl⁻.

Question 18

1. The methoxide reaction proceeds via pure SN2 with complete inversion, while the iodide reaction has competing SN1 character due to the trace water stabilizing carbocation intermediates
2. The larger size of iodide ion compared to methoxide leads to some frontside attack in addition to the normal backside attack, producing both inversion and retention products
3. Both reactions are SN2, but the iodide reaction undergoes some double displacement due to the reversible nature of iodide as both nucleophile and leaving group (correct answer)
4. The trace water promotes competing E2 elimination that produces an alkene, which then undergoes addition reactions to generate the minor stereoisomeric product

Explanation: When analyzing stereochemical outcomes in substitution reactions, you need to consider not just the mechanism but also the reversibility of the process and the nature of the nucleophile/leaving group. The methoxide reaction gives a single stereoisomeric product because it follows a straightforward SN2 mechanism with complete inversion of configuration. Methoxide is a strong nucleophile and poor leaving group, so the reaction proceeds in only one direction. However, the iodide reaction produces two products because iodide is unique—it's both an excellent nucleophile and an excellent leaving group. This creates a reversible SN2 process. The initial SN2 attack by iodide inverts the stereochemistry, but then the newly formed alkyl iodide can undergo another SN2 reaction with another iodide ion, inverting the configuration back to the original. This double displacement mechanism explains the 9:1 ratio: most product retains the inverted configuration from the first SN2, but some undergoes the second inversion. Option A incorrectly suggests SN1 character, but secondary alkyl halides typically don't form stable carbocations, especially without strong ionizing conditions. Option B misunderstands SN2 stereochemistry—the mechanism always involves backside attack regardless of nucleophile size. Option D incorrectly invokes elimination followed by addition, which wouldn't produce the observed stereoisomeric distribution. Remember: when iodide is both the nucleophile and leaving group, consider the possibility of reversible reactions leading to stereochemical scrambling. This is a unique feature of halide exchange reactions.

Question 19

1. The hydrosulfide nucleophile preferentially attacks the carbon at position 1 due to lower steric hindrance compared to position 3.
2. The C-Cl bond is stronger than the C-Br bond, making the carbon attached to bromine a more reactive electrophilic site.
3. The reaction proceeds via an SN1 mechanism, and the carbocation forms at the site that allows for better hyperconjugation.
4. Bromide is a better leaving group than chloride because its conjugate acid, HBr, is stronger than HCl. (correct answer)

Explanation: This question tests your understanding of nucleophilic substitution reactions and leaving group ability. When a molecule has two different halides, you need to predict which one will be displaced based on their relative leaving group abilities. The key principle here is that better leaving groups make carbons more electrophilic and reactive toward nucleophiles. Leaving group ability correlates with the stability of the departing anion, which relates to the strength of its conjugate acid. Since HBr (pKa ≈ -9) is a much stronger acid than HCl (pKa ≈ -7), bromide ion is more stable than chloride ion, making bromide the superior leaving group. Therefore, the hydrosulfide nucleophile preferentially attacks the carbon bearing bromine, displacing Br⁻ and forming 3-bromopropanethiol. Answer A incorrectly focuses on sterics, but both carbons have similar steric environments in this linear molecule. Answer B gets the bond strength relationship backwards - while C-Cl bonds are indeed stronger than C-Br bonds, this makes bromine easier to displace, not harder. The stronger bond would make chlorine less likely to leave. Answer C incorrectly invokes an SN1 mechanism, but primary alkyl halides typically react via SN2, and neither carbon would form a stable carbocation anyway. Remember this pattern: when comparing halide leaving groups, the order is I⁻ > Br⁻ > Cl⁻ > F⁻, following the strength of their conjugate acids. In mixed halide substrates, nucleophiles will preferentially displace the better leaving group.

Question 20

1. KCN is a weak nucleophile and will not displace the chloride from a secondary alkyl halide at a reasonable rate.
2. The SN2 reaction will cause an inversion of stereochemistry, leading to the formation of (R)-2-cyanobutane. (correct answer)
3. The secondary alkyl halide will primarily undergo an E2 elimination reaction because KCN is a strong, sterically hindered base.
4. The polar aprotic solvent DMSO will promote an SN1 mechanism, leading to a racemic mixture of (R)- and (S)-2-cyanobutane.

Explanation: When you encounter a substitution reaction involving stereochemistry, you need to consider both the mechanism and its stereochemical consequences. This question tests your understanding of how SN2 reactions affect chiral centers. The proposed synthesis involves KCN attacking (S)-2-chlorobutane. Since this is a secondary alkyl halide with a good leaving group (Cl⁻) and a strong nucleophile (CN⁻) in a polar aprotic solvent (DMSO), the reaction will proceed via an SN2 mechanism. The critical point is that SN2 reactions always proceed with inversion of configuration at the chiral center. When the cyanide ion attacks the backside of the carbon bearing the chlorine, it flips the stereochemistry from S to R configuration. Therefore, the student would obtain (R)-2-cyanobutane, not the desired (S)-2-cyanobutane, making answer B correct. Answer A is incorrect because KCN is actually a strong nucleophile that readily participates in SN2 reactions with secondary alkyl halides. Answer C mischaracterizes KCN—while it can act as a base, it's not particularly sterically hindered, and under these conditions, substitution dominates over elimination. Answer D incorrectly suggests an SN1 mechanism; secondary alkyl halides typically don't form stable enough carbocations for SN1 reactions, especially with strong nucleophiles present. Remember this key principle: SN2 reactions always invert stereochemistry. When planning syntheses involving chiral molecules, you must account for whether the mechanism will retain or invert the configuration at each stereogenic center.