- The reaction of 3,3-dimethyl-2-butanol with concentrated H₂SO₄ yields 2,3-dimethyl-2-butene as the major product. What would be the major product if 3,3-dimethyl-2-butanol were instead treated with POCl₃ in pyridine?
Answer: 3,3-dimethyl-1-butene
The reaction with H₂SO₄ proceeds via an E1 mechanism involving a carbocation intermediate, which allows for a 1,2-methyl shift to form a more stable tertiary carbocation, leading to the rearranged product 2,3-dimethyl-2-butene. In contrast, the reaction with phosphorus oxychloride (POCl₃) in pyridine is an E2 dehydration. The E2 mechanism is concerted and does not involve a carbocation intermediate, so rearrangements are not possible. Elimination occurs on the original carbon skeleton. The only β-hydrogens available for elimination are on the methyl group at C1, leading to the formation of 3,3-dimethyl-1-butene.
- Which first step best turns an -OH into a good leaving group for SN2 by I− on a primary carbon?
Answer: TsCl, pyridine
This question tests intermediate Organic Chemistry skills specifically related to alcohol substitution and elimination mechanisms. Alcohols often act as poor leaving groups; conversion to tosylates is common for SN2 reactions on primary carbons. In this scenario, TsCl with pyridine forms a good leaving group for subsequent iodide displacement. The correct answer is choice B because it accurately reflects the first step necessary for the SN2 mechanism. Choice A fails because it suggests deprotonation without activation, a common misconception about base roles. To help students avoid these errors, emphasize multi-step transformations and leaving group quality. Practice sequencing reactions with flowcharts and stereochemistry considerations.
- Why does the conversion of an alcohol to a p-toluenesulfonate (tosylate) make it a much better substrate for substitution reactions?
Answer: The tosylate anion (TsO⁻) is a very stable species due to resonance delocalization of the negative charge.
The effectiveness of a leaving group is determined by its stability once it has departed. Good leaving groups are weak bases. The hydroxide ion (-OH) is a strong base and therefore a poor leaving group. When an alcohol is converted to a tosylate, the leaving group becomes the tosylate anion (TsO⁻). This anion is the conjugate base of a very strong acid (p-toluenesulfonic acid). Its negative charge is extensively delocalized by resonance over the three oxygen atoms of the sulfonate group, making it a very stable and weak base. This stability is the primary reason it is an excellent leaving group. Choice A is the opposite of the correct reasoning. Choice C is incorrect; tosylation occurs at the oxygen and does not affect the stereocenter. Choice D describes an incorrect electronic/steric effect.
- The reaction of (R)-2-butanol with thionyl chloride (SOCl₂) in an inert solvent (e.g., ether) without pyridine is known to proceed via an Sₙi mechanism. What is the stereochemical outcome of this reaction?
Answer: Formation of (R)-2-chlorobutane (retention)
In the absence of a base like pyridine, the reaction of an alcohol with SOCl₂ proceeds through an Sₙi (substitution nucleophilic internal) mechanism. The alcohol first forms a chlorosulfite ester intermediate. This intermediate then collapses in a concerted step where the chloride is delivered from the same face as the departing SO₂ group. This internal delivery of the nucleophile results in overall retention of configuration at the stereocenter. Thus, (R)-2-butanol yields (R)-2-chlorobutane. If pyridine were present, it would intercept the intermediate and promote an Sₙ2 reaction, leading to inversion of configuration (Choice A).
- In converting a primary alcohol like 1-butanol to 1-bromobutane, PBr₃ is often preferred over concentrated HBr. What is the best chemical justification for this preference in a synthesis requiring high purity?
Answer: HBr's strong acidity can catalyze a competing E1 elimination to form butenes as side products.
The primary reason for preferring PBr₃ is to avoid the strongly acidic conditions of concentrated HBr. While the reaction of a primary alcohol with HBr proceeds via an Sₙ2 mechanism, the presence of a strong acid catalyst (H⁺) and heat can promote a competing dehydration (E1 or E2) pathway, leading to the formation of butene impurities. The mechanism with PBr₃ does not involve free protons or carbocations, thus minimizing elimination side reactions and ensuring a cleaner conversion to the alkyl bromide. Choice D is incorrect because primary alcohols react with HBr via an Sₙ2, not Sₙ1, mechanism, so rearrangements are not a concern for 1-butanol itself, but elimination is.
- Which alcohol class reacts fastest by SN1 with HCl (ZnCl2) to form an alkyl chloride?
Answer: Tertiary alcohol
This question tests intermediate Organic Chemistry skills specifically related to alcohol substitution and elimination mechanisms. Alcohols often act as poor leaving groups; reactivity varies by class in SN1 conditions. In this scenario, tertiary alcohols form stable carbocations fastest with HCl/ZnCl2. The correct answer is choice C because it accurately reflects the fastest-reacting class. Choice A fails because it suggests primary alcohols, a common error ignoring carbocation stability. To help students avoid these errors, emphasize alkyl group effects and rate orders. Practice ranking reactivities with structural examples.
- Which step makes an alcohol a better leaving group for E1 dehydration in acid?
Answer: Protonation to form H2O leaving group
This question tests intermediate Organic Chemistry skills specifically related to alcohol substitution and elimination mechanisms. Alcohols often act as poor leaving groups; acid protonation converts them to water. In this scenario, protonation facilitates carbocation formation for E1 dehydration. The correct answer is choice B because it accurately reflects the activation step. Choice A fails because it suggests base-catalyzed processes, a common error confusing E1 and E2. To help students avoid these errors, emphasize acid's role in leaving group improvement and mechanism initiation. Practice drawing stepwise mechanisms with protonation emphasis.
- What is the major product when 5-bromo-1-pentanol is treated with sodium hydride (NaH)?
Answer: Tetrahydropyran
This is an intramolecular Williamson ether synthesis. Sodium hydride (NaH) is a strong, non-nucleophilic base that deprotonates the alcohol to form a sodium alkoxide. The resulting alkoxide ion is a potent nucleophile. It attacks the electrophilic carbon bearing the bromine atom at the other end of the molecule in an intramolecular Sₙ2 reaction. This cyclization reaction forms a six-membered ring containing an oxygen atom, which is named tetrahydropyran. This 6-exo-tet cyclization is favorable according to Baldwin's rules. Choice B would require substitution with hydroxide. Choice C is an E2 product, which is kinetically slower than the intramolecular Sₙ2 reaction in this case. Choice D is an incorrect ring size.
- Which reagent best converts 2-butanol to 2-chlorobutane mainly via SN1 (racemization possible)?
Answer: HCl, ZnCl2
This question tests intermediate Organic Chemistry skills specifically related to alcohol substitution and elimination mechanisms. Alcohols often act as poor leaving groups; certain reagents promote SN1 for secondary substrates. In this scenario, HCl with ZnCl2 generates carbocations leading to racemized chlorides. The correct answer is choice B because it accurately reflects the conditions necessary for the SN1 mechanism. Choice A fails because it suggests SN2-favoring conditions, a common error mixing activation methods. To help students avoid these errors, emphasize catalyst roles in mechanism shifts and stereochemical outcomes. Practice distinguishing SN1/SN2 with examples and condition analyses.
- (R)-2-pentanol is treated with tosyl chloride (TsCl) in pyridine, followed by reaction with sodium iodide in acetone. What is the major product of this two-step sequence?
Answer: (S)-2-iodopentane
The first step is the conversion of the alcohol to a tosylate. This reaction occurs at the oxygen atom and does not break the C-O bond, so the configuration at the chiral center is retained. The product is (R)-2-pentyl tosylate. The second step is an Sₙ2 reaction. Iodide ion is a good nucleophile, and tosylate is an excellent leaving group. The Sₙ2 mechanism proceeds with a backside attack, causing an inversion of configuration at the chiral center. Therefore, the (R) stereocenter is inverted to (S), yielding (S)-2-iodopentane. Choice B (retention) and C (racemization) are incorrect stereochemical outcomes. Choice D is an elimination product, which is a minor pathway in the presence of a good nucleophile like iodide in a polar aprotic solvent.
- A solution of tert-butanol in aqueous sulfuric acid reacts to form both substitution and elimination products. If the temperature of the reaction is significantly increased from 0 °C to 50 °C, what change in the product distribution is expected?
Answer: The proportion of 2-methylpropene will increase relative to the substitution products.
Both Sₙ1 and E1 reactions proceed through a common carbocation intermediate. Elimination reactions are generally more favored by higher temperatures than substitution reactions. This is because elimination reactions typically have a larger positive entropy change (ΔS) since the number of moles of products is greater than the moles of reactants. According to the Gibbs free energy equation (ΔG = ΔH - TΔS), the -TΔS term becomes more negative at higher T, making ΔG more favorable for elimination. Thus, increasing the temperature increases the E1/Sₙ1 ratio. Choice C is incorrect as tertiary substrates do not undergo Sₙ2 reactions. Choice D is incorrect as the tert-butyl carbocation is already the most stable C4 carbocation and will not rearrange.
- Which reagent best converts 1-propanol to 1-bromopropane via SN2 under mild conditions?
Answer: PBr3
This question tests intermediate Organic Chemistry skills specifically related to alcohol substitution and elimination mechanisms. Alcohols often act as poor leaving groups; mild reagents are preferred for primary alcohols to avoid side reactions. In this scenario, the choice of reagent such as PBr3 enables SN2 substitution under mild conditions without rearrangement. The correct answer is choice B because it accurately reflects the conditions necessary for the SN2 mechanism to occur efficiently. Choice A fails because it suggests harsher conditions that may lead to elimination, which is a common error when confusing primary and secondary alcohol reactivities. To help students avoid these errors, emphasize the importance of selecting reagents based on substrate type and understanding temperature effects on mechanism selection. Practice mapping reagents to mechanisms using flowcharts and example reactions.
- A student attempts to synthesize tert-butyl methyl ether by adding sodium methoxide (NaOCH₃) to tert-butyl bromide. This method fails to produce a significant amount of the ether. What is the major product and the reason for this outcome?
Answer: 2-methylpropene, because methoxide acts as a base promoting E2 elimination.
This is a classic example of the limitations of the Williamson ether synthesis. The reaction involves a strong nucleophile/strong base (methoxide) and a tertiary alkyl halide. Due to the significant steric hindrance at the tertiary carbon, the Sₙ2 pathway is blocked. Instead, the methoxide acts as a base and removes a β-proton from one of the methyl groups. This initiates an E2 elimination reaction, where the bromide ion is ejected, forming the alkene 2-methylpropene as the major product. Sₙ1/E1 could also occur, but with a strong base like methoxide, E2 is dominant.
- How does polar protic solvent affect SN2 on a primary alkyl halide made from an alcohol?
Answer: Slows SN2 by solvating nucleophile
This question tests intermediate Organic Chemistry skills specifically related to alcohol substitution and elimination mechanisms. Alcohols often act as poor leaving groups; derived halides react under solvent-influenced conditions. In this scenario, polar protic solvents solvate nucleophiles, reducing SN2 rates on primary halides. The correct answer is choice B because it accurately reflects the solvent's inhibitory effect. Choice A fails because it suggests acceleration, a common error confusing protic and aprotic effects. To help students avoid these errors, emphasize solvation and nucleophilicity trends. Practice solvent classification with rate impact examples.
- Which reagent is commonly used to convert an alcohol to a mesylate (good leaving group) without changing C stereochemistry?
Answer: MsCl, Et3N
This question tests intermediate Organic Chemistry skills specifically related to alcohol substitution and elimination mechanisms. Alcohols often act as poor leaving groups; mesylates provide good alternatives without stereochemical change. In this scenario, MsCl with Et3N activates alcohols retaining carbon configuration. The correct answer is choice A because it accurately reflects the reagent for stereochemistry-preserving conversion. Choice C fails because it suggests elimination conditions, a common misconception in activation versus reaction. To help students avoid these errors, emphasize sulfonate formation and stereochemical retention. Practice comparing sulfonates with halides in mechanisms.
- Which base is most likely to give E2 over SN2 with a secondary alkyl tosylate from an alcohol?
Answer: t-BuO−
This question tests intermediate Organic Chemistry skills specifically related to alcohol substitution and elimination mechanisms. Alcohols often act as poor leaving groups; tosylates allow base-promoted eliminations. In this scenario, bulky bases like t-BuO- favor E2 over SN2 on secondary substrates. The correct answer is choice B because it accurately reflects the base promoting elimination. Choice A fails because it suggests good nucleophiles for substitution, a common misconception in base selection. To help students avoid these errors, emphasize steric effects and mechanism competition. Practice predicting products with base type comparisons. - A chemist aims to convert (S)-butan-2-ol into (R)-2-bromobutane with the highest possible yield and stereochemical purity. Which reagent or sequence is the most appropriate choice?
Answer: PBr₃ in pyridine
The desired transformation is a substitution with inversion of stereochemistry (S to R). This requires an Sₙ2 reaction. Choice B, phosphorus tribromide (PBr₃), reacts with primary and secondary alcohols via an Sₙ2-type mechanism, leading to a clean inversion of configuration. Choice A, HBr, would react with the secondary alcohol via an Sₙ1 mechanism, forming a carbocation intermediate and leading to significant racemization, not a clean inversion. Choice C is an incorrect sequence; deprotonation followed by reaction with bromine would not lead to the desired product. Choice D involves elimination followed by addition, which would also lead to a racemic product.
- Predict the major product of the reaction of 1-methylcyclohexanol with concentrated aqueous HBr.
Answer: 1-bromo-1-methylcyclohexane
This reaction involves a tertiary alcohol and HBr, proceeding via an SN1 mechanism. The hydroxyl group is protonated by HBr, water departs to form a stable tertiary carbocation at C1, and bromide ion attacks to give 1-bromo-1-methylcyclohexane. While some E1 elimination to form 1-methylcyclohexene occurs as a competing reaction, SN1 substitution predominates with HBr due to the strong nucleophilicity of bromide ion. Choice C would require rearrangement, which is unnecessary since the tertiary carbocation is already stable. Choice D would require an impossible rearrangement.
- Which condition most favors SN2 when converting a primary alcohol-derived tosylate to a bromide?
Answer: Polar aprotic solvent, strong nucleophile
This question tests intermediate Organic Chemistry skills specifically related to alcohol substitution and elimination mechanisms. Alcohols often act as poor leaving groups; tosylates facilitate SN2 with optimized conditions. In this scenario, polar aprotic solvents enhance strong nucleophile activity for bromide formation. The correct answer is choice B because it accurately reflects the conditions necessary for the SN2 mechanism. Choice A fails because it suggests protic solvents that hinder SN2, a common misconception about solvent effects. To help students avoid these errors, emphasize aprotic solvent advantages and nucleophile strength. Practice classifying conditions with mechanism maps and rate comparisons.
- Which condition would most likely shift 2-butanol dehydration toward substitution instead of elimination?
Answer: Lower temperature, high [Br−]
This question tests intermediate Organic Chemistry skills specifically related to alcohol substitution and elimination mechanisms. Alcohols often act as poor leaving groups; conditions can shift dehydration outcomes. In this scenario, lower temperatures and high nucleophile concentrations favor substitution over elimination. The correct answer is choice A because it accurately reflects the shift toward substitution. Choice B fails because it suggests elimination-favoring heat, a common misconception in competition control. To help students avoid these errors, emphasize temperature and concentration effects on pathways. Practice predicting ratios with variable condition examples.
