- Which of the following alkynes will NOT produce a stereoisomeric mixture of alkenes (i.e., will yield a single, achiral alkene product) upon reaction with Na in liquid NH₃?
Answer: 3-Hexyne
The reaction of an internal alkyne with Na/NH₃ produces an (E)-alkene. For an unsymmetrical alkyne like 2-pentyne, 2-hexyne, or 3-heptyne, the product (E)-alkene will have different groups on each end of the double bond, and thus can exist as stereoisomers if other chiral centers are present or created. However, 3-hexyne is a symmetrical alkyne (ethyl group on both sides). Its reduction product, (E)-3-hexene, is symmetrical and achiral. It cannot form a mixture of stereoisomers because only one geometric isomer (E) is formed and it possesses a plane of symmetry.
- Predict the major organic product of the reaction of 4-octyne with deuterium gas (D₂) over Lindlar's catalyst.
Answer: (Z)-4,5-dideuteriooct-4-ene
Catalytic hydrogenation of an alkyne using Lindlar's catalyst occurs via a *syn*-addition mechanism, where both hydrogen (or deuterium) atoms are delivered to the same face of the alkyne from the catalyst surface. This results in the formation of a (Z)- or *cis*-alkene. Therefore, the reaction of 4-octyne with D₂ will yield (Z)-4,5-dideuteriooct-4-ene.
- The reaction of 1-hexyne with sodium metal in liquid ammonia (Na/NH₃) followed by an ammonium chloride (NH₄Cl) workup primarily yields 1-hexyne. Why does the expected reduction to 1-hexene not occur?
Answer: The strong basic conditions rapidly deprotonate the terminal alkyne, and the resulting acetylide anion is resistant to reduction by solvated electrons.
The terminal proton of a 1-alkyne is acidic (pKa ≈ 25) and is readily deprotonated by the strongly basic conditions of Na/NH₃ (which generates NaNH₂). The resulting negatively charged acetylide anion is electron-rich and repels the solvated electrons needed to initiate the reduction sequence. Thus, an acid-base reaction occurs instead of reduction. The NH₄Cl quench simply reprotonates the acetylide to regenerate the starting alkyne.
- Dissolving metal reduction of 3-hexyne fails to proceed when sodium metal is added to a solution of the alkyne in hexane. The reaction proceeds efficiently, however, upon addition of liquid ammonia. What is the essential role of the ammonia?
Answer: It acts as a Brønsted-Lowry acid to protonate the anionic intermediates.
The mechanism of dissolving metal reduction involves the formation of a radical anion and then a vinylic anion. Both of these intermediates must be protonated to yield the final alkene product. Hexane is an aprotic solvent and cannot provide these protons. Ammonia (NH₃) is a protic solvent that serves as the essential proton source to complete the reaction.
- Consider the reduction of 3-hexyne using Lindlar catalyst (Pd/CaCO₃, Pb(OAc)₂) in the presence of hydrogen gas. If this reaction is allowed to proceed under standard conditions, what is the major product and its stereochemical configuration?
Answer: (Z)-3-hexene with >95% stereoselectivity due to syn addition of hydrogen atoms
Lindlar catalyst provides a poisoned palladium surface that selectively reduces alkynes to alkenes without further reduction to alkanes. The syn addition mechanism occurs because both hydrogen atoms are delivered from the same face of the catalyst surface, resulting in (Z)-alkene formation with high stereoselectivity. Choice B is incorrect because Lindlar reduction involves syn addition, not anti addition. Choice C is wrong because this is not a radical mechanism but rather a heterogeneous catalytic process with high stereoselectivity. Choice D is incorrect because Lindlar catalyst is specifically designed to prevent over-reduction to alkanes.
- A chemist wishes to perform a dissolving metal reduction on 4-octyne but has run out of liquid ammonia. Which of the following solvent and proton source combinations could serve as a viable alternative to produce (E)-oct-4-ene?
Answer: Sodium metal in ethanol (CH₃CH₂OH)
Dissolving metal reductions require a metal (like Na or Li) that can donate an electron and a solvent that can serve as a proton source. Ethanol, like ammonia, has an acidic proton (on the OH group) that can protonate the anionic intermediates in the reduction mechanism. Water would react too violently with sodium metal. Hexane and diethyl ether are aprotic and cannot serve as the necessary proton source for the mechanism to proceed to completion.
- A student attempts to reduce 2-hexyne using catalytic hydrogenation. Instead of the expected alkene, the major product isolated is hexane. Which of the following reagent conditions was most likely used?
Answer: H₂ (excess), Pd/C
The formation of the fully saturated alkane (hexane) from an alkyne indicates complete reduction. Standard hydrogenation catalysts like Pd/C, Pt, or Ni are highly active and will reduce an alkyne all the way to an alkane, especially with excess H₂. Lindlar's catalyst is 'poisoned' to stop at the (Z)-alkene. Na/NH₃ produces the (E)-alkene. Hydroboration-oxidation would produce a ketone.
- When oct-2-en-6-yne is treated with one equivalent of H₂ gas and Lindlar's catalyst, which functional group is preferentially reduced and what is the geometry of the resulting product?
Answer: The alkyne is reduced, yielding (Z)-octa-2,6-diene.
This question tests your understanding of selective reduction reactions, specifically how Lindlar's catalyst behaves with different unsaturated functional groups. When you encounter problems involving multiple reducible groups, always consider the catalyst's selectivity and stereochemical outcomes.
Lindlar's catalyst (palladium on calcium carbonate with lead acetate and quinoline) is specifically designed for partial reduction of alkynes to alkenes. The catalyst is "poisoned" to prevent over-reduction, making it highly selective for triple bonds over double bonds. With one equivalent of H₂, the alkyne will be reduced preferentially because triple bonds are more reactive toward hydrogenation than double bonds.
The key stereochemical feature of Lindlar's catalyst is that it delivers hydrogen atoms to the same face of the triple bond (syn addition), invariably producing the (Z)-alkene geometry. Therefore, oct-2-en-6-yne becomes (Z)-octa-2,6-diene.
Choice A incorrectly suggests the alkene reduces preferentially due to sterics, but electronic factors (triple bond reactivity) dominate over steric considerations here. Choice B is wrong because one equivalent of H₂ with Lindlar's catalyst cannot reduce both functional groups - you'd need excess hydrogen and likely different conditions for complete reduction to octane. Choice C correctly identifies alkyne reduction but incorrectly predicts (E)-geometry, missing the crucial stereochemical outcome of syn addition.
Study tip: Remember "Lindlar = alkyne to Z-alkene." This catalyst is your go-to for converting terminal or internal alkynes to cis-alkenes without over-reducing to alkanes.
- If 3-hexyne is treated with excess H₂ gas over a standard palladium-on-carbon (Pd/C) catalyst, and the reaction is stopped when exactly one equivalent of H₂ has been consumed, what will be the composition of the product mixture?
Answer: A mixture containing 3-hexyne, (Z)-hex-3-ene, and hexane.
Unlike the poisoned Lindlar's catalyst, a standard Pd/C catalyst is highly active and not selective for the alkyne-to-alkene step. The hydrogenation of the alkyne produces the (Z)-alkene (syn-addition), but this alkene can be immediately hydrogenated further to the alkane. The catalyst does not stop at the alkene stage. Therefore, as soon as some alkene is formed, it begins to be converted to alkane. After one equivalent of H₂ is consumed, the mixture will contain unreacted alkyne, some intermediate alkene, and some fully reduced alkane.
- In the sodium-ammonia reduction of alkynes, the first electron transfer creates a radical anion. What structural feature makes internal alkynes more reactive toward this initial electron transfer compared to terminal alkynes?
Answer: Internal alkynes have lower-energy π* orbitals due to greater alkyl substitution, making them better electron acceptors
Internal alkynes are more electron-deficient due to alkyl substitution, which lowers the energy of their π* antibonding orbitals and makes them more susceptible to nucleophilic attack by solvated electrons. This electronic effect makes the initial electron transfer thermodynamically more favorable. Choice B is incorrect because while terminal alkynes can be deprotonated, this actually occurs after the reduction in most cases. Choice C is wrong because internal alkynes actually have more steric hindrance, not less. Choice D is incorrect because while hyperconjugation may provide some stabilization, the primary driving force is the electronic effect on the π* orbital energy.
- What would be the major product if an equimolar mixture of 2-butyne and (Z)-2-butene is subjected to one equivalent of H₂ with Lindlar's catalyst?
Answer: Primarily (Z)-2-butene, with most of the starting (Z)-2-butene remaining unreacted.
This question tests your understanding of selective hydrogenation and catalyst reactivity. When you encounter problems involving different unsaturated compounds with limited reagents, you need to consider which functional group reacts preferentially.
Lindlar's catalyst (Pd/CaCO₃/PbO) is specifically designed for partial hydrogenation of alkynes to alkenes, and it shows much higher reactivity toward triple bonds than double bonds. With only one equivalent of H₂ available for two different substrates, the catalyst will preferentially reduce the more reactive alkyne (2-butyne) to form (Z)-2-butene, while leaving the existing (Z)-2-butene largely untouched due to its much slower reaction rate with this catalyst.
Since you start with equimolar amounts and the alkyne consumes most of the available hydrogen, you'll end up with primarily (Z)-2-butene: the original amount plus the newly formed amount from alkyne reduction.
Answer A is incorrect because Lindlar's catalyst doesn't readily reduce alkenes to alkanes - that would require a different catalyst like Pd/C. Answer B is wrong because with limited H₂, you won't have significant amounts of all three compounds; the alkyne will be preferentially consumed. Answer C is incorrect because it reverses the reactivity - alkynes are much more reactive than alkenes toward Lindlar's catalyst, so the alkyne won't remain unreacted while forming butane.
Remember: Lindlar's catalyst shows strong selectivity for alkynes over alkenes. When hydrogen is limiting, always consider which functional group reacts fastest with your specific catalyst.
- The purpose of including quinoline or lead(II) acetate in Lindlar's catalyst is to decrease the catalyst's activity. What is the primary undesired side reaction that this 'poisoning' prevents?
Answer: Further reduction of the intermediate (Z)-alkene to an alkane.
Lindlar's catalyst is designed for the partial reduction of an alkyne to a (Z)-alkene. Standard catalysts like Pd/C are active enough to also reduce the newly formed alkene to an alkane. The catalyst 'poison' (e.g., quinoline) deactivates the catalyst just enough so that it reacts rapidly with the alkyne but very slowly with the alkene product, allowing the (Z)-alkene to be isolated before it is over-reduced.
- When 1-pentyne is treated with two equivalents of HBr in the absence of peroxides, the major product formed follows Markovnikov's rule. What structural feature of the intermediate explains why the second addition also follows Markovnikov selectivity?
Answer: The electron-withdrawing bromine substituent increases the electrophilicity of the double bond toward the second HBr addition
After the first Markovnikov addition of HBr to 1-pentyne, the resulting vinyl bromide has an electron-withdrawing bromine substituent that makes the alkene more electrophilic. This increased electrophilicity promotes the second addition, and Markovnikov selectivity is maintained because the more substituted carbocation is still more stable. Choice A is incorrect because the vinyl carbocation doesn't benefit significantly from hyperconjugation. Choice B is wrong because there's no remaining π-system for resonance after the first addition creates a vinyl bromide. Choice C is incorrect because vinyl carbocations are actually less stable than alkyl carbocations, but the reaction still proceeds due to the electron-withdrawing effect.
- An alkyne substrate undergoes partial reduction under two different conditions: Condition A gives 85% (Z)-alkene and 15% (E)-alkene, while Condition B gives 95% (E)-alkene and 5% (Z)-alkene. Based on these selectivities, which reduction methods were most likely used?
Answer: Condition A: Lindlar catalyst with partial catalyst poisoning; Condition B: Na/NH₃ under standard conditions
Lindlar catalyst typically gives >95% (Z)-selectivity, but if the catalyst is partially poisoned or reaction conditions are suboptimal, selectivity can decrease to ~85% (Z). Na/NH₃ under standard conditions reliably gives >95% (E)-selectivity through the trans-diaxial elimination mechanism. Choice A is backwards regarding the expected selectivities. Choice B is incorrect because extended reaction time with Na/NH₃ doesn't change stereoselectivity. Choice D is wrong because H₂/Pt would likely give over-reduction to alkane rather than controlled partial reduction.
- A synthetic chemist needs to convert 1-octyne to (Z)-1-octene with >90% stereochemical purity. However, when using standard Lindlar conditions (Pd/CaCO₃, H₂, quinoline), only 75% (Z)-selectivity is achieved. What modification would most likely improve the stereoselectivity?
Answer: Lower the reaction temperature and increase the amount of poison (quinoline) to enhance catalyst selectivity
Lower temperature reduces the rate of over-reduction and side reactions, while additional quinoline (catalyst poison) further deactivates the palladium surface to prevent complete reduction to alkane and improve (Z)-selectivity. This combination enhances the selectivity of the desired partial reduction. Choice A is incorrect because higher H₂ pressure typically leads to over-reduction and lower selectivity. Choice C is wrong because solvent polarity doesn't significantly affect the stereochemical outcome of heterogeneous catalysis. Choice D is incorrect because Lewis acids would likely interfere with the catalyst and potentially lead to alternative reaction pathways.
- When 4-octyne is treated with excess hydrogen gas over a palladium catalyst that has been partially poisoned with lead acetate, three different products can be isolated depending on reaction time. If the reaction is monitored by GC-MS, what would be the expected order of product appearance over time?
Answer: First: (Z)-4-octene; Second: (E)-4-octene; Third: octane
With partially poisoned palladium (Lindlar-like conditions), the initial reduction gives predominantly (Z)-4-octene due to syn-addition. As reaction time increases, some of the (Z)-alkene undergoes isomerization to the more stable (E)-isomer through catalyst-mediated hydrogen transfer. Finally, with prolonged exposure, over-reduction to octane occurs despite the catalyst poisoning. Choice B is incorrect because over-reduction to alkane doesn't occur first. Choice C is wrong because initial selectivity favors the (Z)-isomer. Choice D incorrectly suggests that (E)-alkene forms after the alkane, which is not the typical progression.
- A student observes that Lindlar reduction of 3-hexyne gives exclusively (Z)-3-hexene, while dissolving metal reduction gives exclusively (E)-3-hexene. What fundamental difference in mechanism accounts for this complementary stereoselectivity?
Answer: Lindlar occurs through surface-bound intermediates while dissolving metal involves solution-phase radical anions and vinyl anions
The key difference is that Lindlar reduction occurs on a catalyst surface where both hydrogen atoms are delivered from the same face (giving syn-addition and Z-product), while dissolving metal reduction proceeds through discrete solution-phase intermediates (radical anion, then vinyl anion) that are protonated in a way that ultimately leads to the E-product. Choice A is incorrect because dissolving metal reduction doesn't involve elimination. Choice C uses imprecise terminology. Choice D is incorrect because both reactions are essentially irreversible under their respective conditions; the selectivity arises from mechanistic differences, not thermodynamic vs kinetic control.
- A researcher compares the reduction of 2-butyne using (1) Lindlar catalyst and (2) sodium in liquid ammonia. Both reactions go to completion. If the products are then subjected to ozonolysis followed by reductive workup, what would be observed?
Answer: Identical products from both reductions because ozonolysis of internal alkenes gives the same carbonyl fragments
Both Lindlar catalyst and Na/NH₃ reduce 2-butyne to 2-butene with different stereochemistry: Lindlar gives (Z)-2-butene while Na/NH₃ gives (E)-2-butene. However, ozonolysis cleaves the C=C bond completely and produces the same carbonyl fragments (two equivalents of acetaldehyde) regardless of the alkene's stereochemistry. Choice A is close but imprecise in reasoning. Choice B is incorrect because stereochemistry doesn't affect regioselectivity in ozonolysis of symmetric alkenes. Choice C is wrong because both methods give high stereoselectivity, not mixtures.
- What is the expected major product when (R)-3-methyl-1-pentyne is treated with sodium metal in liquid ammonia?
Answer: (R, E)-3-methylpent-1-ene
The reaction is the reduction of a terminal alkyne to a trans (E) alkene. The reaction occurs at the C1-C2 triple bond and does not involve the chiral center at C3. Therefore, the stereochemistry at the existing chiral center is retained. The product will be (R, E)-3-methylpent-1-ene.
- A compound with the molecular formula C₈H₁₂ is treated with excess H₂ over Lindlar's catalyst to yield (Z)-cyclooctene. What is the structure of the starting compound?
Answer: Cyclooctyne
The reaction is a partial reduction that produces a (Z)-alkene, which is characteristic of the reaction of an alkyne with H₂/Lindlar's catalyst. The product is (Z)-cyclooctene, an eight-membered ring with a cis double bond. Therefore, the starting material must be the corresponding alkyne, which is cyclooctyne. The molecular formula for cyclooctyne is C₈H₁₂, which matches the information given.
