Organic Chemistry Quiz: Addition To Alkynes And Partial Reductions

What this quiz covers

This quiz focuses on Addition To Alkynes And Partial Reductions, 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. H₂ (1 equivalent), Lindlar's catalyst
2. Na, NH₃(l)
3. H₂ (excess), Pd/C (correct answer)
4. BH₃·THF, followed by H₂O₂, NaOH

Explanation: 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.

Question 2

1. Bicyclo[4.2.0]oct-7-ene
2. 1,2-dimethylcyclohex-1-ene
3. Cyclooctyne (correct answer)
4. Octa-1,7-diyne

Explanation: 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.

Question 3

1. 1. H₂, Lindlar's catalyst ; 2. mCPBA
2. 1. Na, NH₃(l) ; 2. mCPBA
   (correct answer)
3. 1. H₂, Lindlar's catalyst ; 2. OsO₄, NMO
4. 1. Na, NH₃(l) ; 2. OsO₄, NMO

Explanation: The target is a trans-epoxide. Epoxidation with a peroxyacid like mCPBA is a syn-addition of an oxygen atom. To obtain a trans product from a syn-addition, the starting alkene must have trans or (E) stereochemistry. The conversion of 2-pentyne to (E)-pent-2-ene is achieved with a dissolving metal reduction (Na, NH₃(l)). Subsequent reaction with mCPBA yields the desired trans-epoxide.

Question 4

1. Lindlar involves concerted syn-addition while dissolving metal involves stepwise anti-elimination of hydrogen atoms
2. Lindlar occurs through surface-bound intermediates while dissolving metal involves solution-phase radical anions and vinyl anions (correct answer)
3. Lindlar uses heterogeneous catalysis with orbital control while dissolving metal uses homogeneous conditions with charge control
4. Lindlar reduction is kinetically controlled while dissolving metal reduction is thermodynamically controlled, favoring the more stable (E)-isomer

Explanation: 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.

Question 5

1. Internal alkynes have lower-energy π* orbitals due to greater alkyl substitution, making them better electron acceptors (correct answer)
2. Terminal alkynes are deprotonated by the basic conditions before electron transfer can occur
3. Internal alkynes have less steric hindrance around the triple bond, allowing better orbital overlap with the solvated electron
4. The radical anion intermediate from internal alkynes is stabilized by hyperconjugation from adjacent alkyl groups

Explanation: 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.

Question 6

1. (Z)-3-hexene with >95% stereoselectivity due to syn addition of hydrogen atoms (correct answer)
2. (E)-3-hexene with >95% stereoselectivity due to anti addition of hydrogen atoms
3. A 1:1 mixture of (E)- and (Z)-3-hexene due to non-selective radical mechanism
4. Hexane as the major product due to over-reduction under these mild conditions

Explanation: 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.

Question 7

1. Sodium metal in H₂O
2. Sodium metal in ethanol (CH₃CH₂OH) (correct answer)
3. Sodium metal in hexane
4. Sodium metal in diethyl ether

Explanation: 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.

Question 8

1. It acts as a Brønsted-Lowry acid to protonate the anionic intermediates. (correct answer)
2. It acts as a catalyst to increase the rate of electron transfer from sodium.
3. It coordinates to the alkyne and directs the anti-addition stereochemistry.
4. It is a polar aprotic solvent required to dissolve the sodium metal.

Explanation: 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.

Question 9

1. Almost exclusively (Z)-hex-3-ene.
2. Almost exclusively (E)-hex-3-ene.
3. A mixture containing 3-hexyne, (Z)-hex-3-ene, and hexane. (correct answer)
4. A mixture of (Z)-hex-3-ene and (E)-hex-3-ene.

Explanation: 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.

Question 10

1. The vinyl carbocation formed after first addition is stabilized by hyperconjugation from adjacent methyl groups
2. The vinyl carbocation intermediate is stabilized by resonance with the π-system of the remaining double bond
3. Both additions follow identical mechanisms because vinyl carbocations have similar stability to alkyl carbocations
4. The electron-withdrawing bromine substituent increases the electrophilicity of the double bond toward the second HBr addition (correct answer)

Explanation: 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.

Question 11

1. Identical products from both reductions because ozonolysis cleaves C=C bonds regardless of stereochemistry
2. Different products because the (Z)- and (E)-alkenes have different regioselectivity in ozonolysis reactions
3. Different products because Lindlar reduction produces a mixture while Na/NH₃ gives pure stereoisomer
4. Identical products from both reductions because ozonolysis of internal alkenes gives the same carbonyl fragments (correct answer)

Explanation: 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.

Question 12

1. Condition A: Na/NH₃; Condition B: Lindlar catalyst with careful temperature control
2. Condition A: Lindlar catalyst; Condition B: Na/NH₃ with extended reaction time
3. Condition A: Lindlar catalyst with partial catalyst poisoning; Condition B: Na/NH₃ under standard conditions (correct answer)
4. Condition A: H₂/Pt with limited hydrogen pressure; Condition B: dissolving metal reduction in liquid ammonia

Explanation: 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.

Question 13

1. Steric hindrance favors protonation at the less substituted carbon to minimize 1,3-diaxial interactions
2. Electronic effects favor protonation at the more substituted carbon due to better orbital overlap
3. Kinetic factors favor protonation at the less substituted carbon due to reduced steric crowding around the reaction site (correct answer)
4. Thermodynamic stability favors protonation at the more substituted carbon to form the more stable (E)-alkene product

Explanation: In the Na/NH₃ reduction, the vinyl anion intermediate is protonated under kinetic control. The less substituted carbon of the vinyl anion is more accessible to the proton source (ammonia), leading to preferential protonation at that site and ultimately forming the (E)-alkene. Choice A incorrectly applies cyclohexane conformational concepts. Choice B is wrong because electronic effects would actually favor the opposite regioselectivity. Choice D is incorrect because while the (E)-product is thermodynamically favored, the regioselectivity is determined by kinetic factors during the protonation step, not the final alkene stability.

Question 14

1. Increase the hydrogen pressure to accelerate the syn-addition process and minimize side reactions
2. Lower the reaction temperature and increase the amount of poison (quinoline) to enhance catalyst selectivity (correct answer)
3. Switch to a different solvent system with lower dielectric constant to favor the kinetic product
4. Add a Lewis acid cocatalyst to increase the electrophilicity of the alkyne substrate

Explanation: 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.

Question 15

1. First: (Z)-4-octene; Second: (E)-4-octene; Third: octane (correct answer)
2. First: octane; Second: (Z)-4-octene; Third: (E)-4-octene by isomerization
3. First: equal mixture of (Z)- and (E)-4-octene; Second: octane by over-reduction
4. First: (Z)-4-octene; Second: octane; Third: (E)-4-octene through thermodynamic equilibration

Explanation: 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.

Question 16

1. 2-Pentyne
2. 2-Hexyne
3. 3-Hexyne (correct answer)
4. 3-Heptyne

Explanation: 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.

Question 17

1. 1. Na, NH₃(l) ; 2. Br₂ in CCl₄
2. 1. H₂, Lindlar's catalyst ; 2. Br₂ in CCl₄
   (correct answer)
3. 1. Na, NH₃(l) ; 2. HBr (2 equivalents)
4. 1. H₂, Pd/C ; 2. Br₂ in CCl₄

Explanation: To form meso-2,3-dibromobutane, a meso compound, one needs a stereospecific sequence. The addition of Br₂ to an alkene is an anti-addition. To get a meso product from an anti-addition, the starting alkene must be the (Z)-isomer. Step 1 must therefore be the reduction of 2-butyne to (Z)-2-butene, which is accomplished using H₂ and Lindlar's catalyst. Step 2 is the anti-addition of Br₂, which converts the (Z)-alkene into the desired meso product.

Question 18

1. The terminal alkyne rearranges to an internal alkyne, which is unreactive under these conditions.
2. The strong basic conditions rapidly deprotonate the terminal alkyne, and the resulting acetylide anion is resistant to reduction by solvated electrons. (correct answer)
3. Liquid ammonia is not a sufficient proton source to complete the reduction mechanism for terminal alkynes.
4. The reaction produces a stable sodium-alkyne complex that is only reversed upon addition of a strong acid like NH₄Cl.

Explanation: 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.

Question 19

1. (Z)-4,5-dideuteriooct-4-ene (correct answer)
2. (E)-4,5-dideuteriooct-4-ene
3. A mixture of (Z)- and (E)-4,5-dideuteriooct-4-ene
4. 4,4,5,5-tetradeuteriooctane

Explanation: 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.

Question 20

1. The alkene is reduced, because it is less sterically hindered.
2. Both functional groups are reduced simultaneously, yielding octane.
3. The alkyne is reduced, yielding (E)-octa-2,6-diene.
4. The alkyne is reduced, yielding (Z)-octa-2,6-diene. (correct answer)

Explanation: 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.