E/Z and Alkene Stereochemistry - Organic Chemistry 1

1. A two-step synthesis is performed starting from propyne. Step 1 involves treatment with NaNH₂ followed by ethyl iodide. Step 2 involves treatment of the intermediate product with H₂ gas and Lindlar's catalyst. What is the major product of this sequence?
   Answer: (Z)-pent-2-ene Step 1 is an alkylation of a terminal alkyne. The strong base NaNH₂ deprotonates propyne to form a propynide anion. This anion then acts as a nucleophile, attacking ethyl iodide in an SN2 reaction to form pent-2-yne. Step 2 is the reduction of an internal alkyne. H₂ with Lindlar's catalyst is a poisoned catalyst system that performs a syn-addition of hydrogen across the triple bond, stopping at the alkene stage. Syn-addition to an alkyne always produces the (Z)- or cis-alkene. Therefore, the final major product is (Z)-pent-2-ene.
2. When assigning CIP priorities to substituents on a double bond, which of the following pairs is correctly ranked from higher priority to lower priority?
   Answer: isopropyl (–CH(CH₃)₂) > n-propyl (–CH₂CH₂CH₃) Priorities are determined by comparing atoms at the first point of difference. A) Isopropyl: The first carbon is attached to (C, C, H). n-Propyl: The first carbon is attached to (C, H, H). Comparing the lists, C vs H at the second position gives isopropyl higher priority. This is correct. B) Vinyl: The first carbon is treated as being attached to (C, C, H) due to the double bond. Ethyl: The first carbon is attached to (C, H, H). Vinyl has higher priority. C) tert-butyl: The first carbon is attached to (C, C, C). sec-butyl: The first carbon is attached to (C, C, H). tert-butyl has higher priority. D) Formyl: The first atom is carbon, C(O, O, H). Hydroxyl: The first atom is oxygen, O(H). Comparing the first atoms, O (Z=8) > C (Z=6), so hydroxyl has higher priority.
3. Treatment of (3R,4R)-3-bromo-4-methylhexane with a strong, non-bulky base such as sodium ethoxide (NaOEt) results in an E2 elimination. What is the major alkene product of this reaction?
   Answer: (Z)-4-methylhex-3-ene E2 elimination requires an anti-periplanar arrangement between the leaving group (Br) and the β-hydrogen being removed. According to Zaitsev's rule, the major product forms by removing a hydrogen from the more substituted β-carbon (C4). In the (3R,4R) substrate, when the molecule adopts the required anti-periplanar conformation with H on C4 and Br on C3, the two ethyl substituents (one from each carbon) end up on the same side of the forming double bond. This stereospecific geometry leads to formation of the (Z)-4-methylhex-3-ene as the major product, despite the (E)-isomer being thermodynamically more stable.
4. An alkene has the structure $CH_3CH_2CH=CHCH_2CH_3$. A student claims this alkene exhibits E/Z isomerism and draws both stereoisomers. What is the error in the student's reasoning?
   Answer: The alkene cannot exhibit E/Z isomerism because both carbons of the double bond have identical substituents For E/Z isomerism to occur, each carbon of the double bond must have two different substituents. In 3-hexene ($CH_3CH_2CH=CHCH_2CH_3$), each carbon has an H and an ethyl group, so the two sides of the molecule are identical. This makes E/Z isomerism impossible. Choice A incorrectly assumes isomerism exists. Choice C confuses stability with the fundamental requirement for different substituents. Choice D incorrectly relates E/Z isomerism to conjugation, which is not required.
5. A student draws the structure of (E)-3-chloropent-2-ene and places the chlorine and ethyl group on opposite sides of the double bond. However, their instructor marks this as incorrect. What is the most likely error in the student's structure?
   Answer: The student incorrectly identified the highest priority groups and should have compared Cl with H instead of with the ethyl group When determining E/Z stereochemistry, you must correctly identify the highest priority groups on each carbon of the double bond using Cahn-Ingold-Prelog rules, where higher atomic number takes priority. For (E)-3-chloropent-2-ene, the double bond is between carbons 2 and 3. On carbon 2, you compare the attached groups: hydrogen (atomic number 1) versus a methyl group (carbon, atomic number 6). The methyl group has higher priority. On carbon 3, you compare chlorine (atomic number 17) versus an ethyl group (carbon, atomic number 6). Chlorine has higher priority. In E configuration, the highest priority groups on each carbon are on opposite sides of the double bond. So the methyl group (C2) and chlorine (C3) should be on opposite sides. The student likely saw chlorine and ethyl on opposite sides and thought this was correct, but they should have been comparing the chlorine with the methyl group on the other carbon. Answer A correctly identifies this error - the student failed to properly identify which groups to compare for priority determination. Answer B is wrong because chlorine belongs on carbon 3 in this molecule. Answer C is incorrect because the student didn't confuse E/Z definitions; they confused which groups to compare. Answer D is wrong because the issue isn't nomenclature or chain numbering, but stereochemistry assignment. Remember: Always identify the highest priority group on each carbon of the double bond first, then determine whether those two groups are on the same side (Z) or opposite sides (E).
6. While (E)-cyclooctene is a stable, isolable molecule, (E)-cyclohexene is extremely unstable and has only been observed transiently at low temperatures. What is the primary source of instability in (E)-cyclohexene?
   Answer: The p-orbitals of the pi bond are twisted, leading to poor orbital overlap and high strain. When analyzing the stability of cyclic alkenes, you need to consider how the ring size affects the geometry of the double bond. In normal alkenes, the carbons involved in the double bond are sp² hybridized and prefer a planar arrangement with the p-orbitals perfectly aligned for optimal π bond overlap. The key issue with (E)-cyclohexene is geometric constraint. A trans double bond naturally wants the substituents on opposite sides of the double bond, but in a six-membered ring, this creates severe strain. The ring forces the p-orbitals of the π bond to twist away from their optimal parallel alignment, dramatically reducing orbital overlap. This poor overlap weakens the π bond significantly and destabilizes the entire molecule, making answer D correct. Answer A is wrong because hyperconjugation isn't the primary stability factor here – the issue is direct π bond orbital overlap. Answer B incorrectly suggests the molecule becomes planar; actually, the ring puckering and orbital twisting are what create the instability, not planarity. Answer C is nonsensical because trans double bonds don't induce aromaticity – aromatic character requires a conjugated cyclic system with 4n+2 π electrons, which isn't present here. In contrast, (E)-cyclooctene is stable because the larger eight-membered ring has enough flexibility to accommodate the trans geometry without severely twisting the π bond orbitals. Remember: when evaluating cyclic alkene stability, always consider whether the ring size can accommodate the preferred geometry of the double bond without forcing unfavorable orbital orientations.
7. The energy barrier for rotation around the carbon-carbon double bond in ethene is approximately 264 kJ/mol, which is high enough to allow for the isolation of stable E and Z isomers in substituted alkenes. What is the primary reason for this high barrier to rotation?
   Answer: The requirement to break the pi (π) bond during the rotational transition state. When you encounter questions about rotation barriers in alkenes, focus on the structural differences between single and double bonds. Double bonds consist of both a sigma (σ) and a pi (π) bond, and this π bond is crucial to understanding rotational restrictions. The high energy barrier exists because rotating around a C=C double bond requires breaking the π bond during the transition state. The π bond forms from the sideways overlap of p orbitals that are perpendicular to the molecular plane. When rotation occurs, these p orbitals must become perpendicular to each other, completely disrupting their overlap and effectively breaking the π bond. Since π bonds have significant bond energy, breaking this interaction requires substantial energy input—hence the 264 kJ/mol barrier. Let's examine why the other options miss the mark. Choice A incorrectly focuses on steric hindrance, but the question specifically mentions ethene, which has only hydrogen substituents that create minimal steric clash. Choice B confuses the issue by referencing σ bond dissociation energy, but the σ bond remains intact during rotation—only the π bond is disrupted. Choice C suggests electrostatic repulsion between sp² carbons, but this doesn't explain why rotation specifically is restricted, as these carbons maintain their hybridization and electron density throughout rotation. Remember this key principle: π bonds are "fragile" compared to σ bonds because of their sideways orbital overlap. Whenever you see questions about restricted rotation in alkenes, immediately think about π bond disruption rather than steric effects or σ bond strength.
8. An organic chemistry student needs to distinguish between (E)-2-bromo-2-butene and (Z)-2-bromo-2-butene using physical properties. Which statement best describes the relationship between these isomers and their properties?
   Answer: These are diastereomers with different physical properties such as boiling points and dipole moments E/Z isomers are diastereomers (stereoisomers that are not mirror images). Diastereomers have different physical properties including boiling points, melting points, and dipole moments, making them distinguishable by physical methods. Choice A incorrectly classifies them as enantiomers. Choice C incorrectly calls them constitutional isomers (they have the same connectivity). Choice D incorrectly suggests they are conformers that interconvert (E/Z isomers are stable and do not interconvert under normal conditions).
9. Which of the following alkenes cannot exhibit E/Z isomerism?
   Answer: 1-bromo-2-methylpropene For an alkene to exhibit E/Z isomerism, each carbon atom of the double bond must be attached to two different groups. Let's analyze the options: A) 1-bromo-2-methylpropene: The structure is Br-CH=C(CH₃)₂. The second carbon of the double bond is attached to two identical methyl groups, so E/Z isomerism is not possible. B) 3-methylpent-2-ene: The structure is CH₃-CH=C(CH₃)-CH₂CH₃. C2 is bonded to H and CH₃. C3 is bonded to CH₃ and CH₂CH₃ (two different groups). E/Z isomerism is possible. C) 1,2-dideuterioethene: D-CH=CH-D. Each carbon has H and D, which are different isotopes, so E/Z isomerism is possible. D) Penta-1,3-diene: CH₂=CH-CH=CH-CH₃. The C3=C4 double bond can exhibit E/Z isomerism (C3 has H and vinyl group, C4 has H and methyl).
10. Consider the separate reactions of (E)-but-2-ene and (Z)-but-2-ene with bromine (Br₂) in an inert solvent. What is the stereochemical relationship between the major product formed from the (E)-isomer and the major product formed from the (Z)-isomer?
    Answer: They are diastereomers. This question probes the stereospecificity of halogen addition to alkenes. The addition of Br₂ across a double bond occurs via a three-membered bromonium ion intermediate, resulting in anti-addition of the two bromine atoms. - (E)-but-2-ene is an achiral (meso-like) starting material. Anti-addition of Br₂ leads to the formation of a single product, the meso compound (2R,3S)-2,3-dibromobutane. - (Z)-but-2-ene is also an achiral starting material. Anti-addition of Br₂ leads to the formation of a racemic mixture of two enantiomers: (2R,3R)- and (2S,3S)-2,3-dibromobutane. The relationship between the meso product from the (E)-isomer and either of the enantiomeric products from the (Z)-isomer is that they are diastereomers (stereoisomers that are not mirror images).
11. Why is the E/Z notation system considered superior to the cis/trans system for naming stereoisomeric alkenes?
    Answer: E/Z notation can unambiguously describe stereochemistry in tri- and tetrasubstituted alkenes. The primary advantage of the E/Z system is its universality. The cis/trans system works well for disubstituted alkenes, especially when two substituents are identical (e.g., a hydrogen on each carbon), but it becomes ambiguous or unusable for alkenes with three or four different substituents. The E/Z system, based on the rigorous and sequential Cahn-Ingold-Prelog (CIP) priority rules, provides an unambiguous descriptor for any alkene that can exhibit stereoisomerism, regardless of the substitution pattern. E/Z is not used for chiral centers (R/S is used), is based on atomic number not weight, and is not restricted to cyclic alkenes.
12. A researcher synthesizes 2-methylbut-2-ene and wants to determine if it can exist as E/Z stereoisomers. After examining the structure $CH_3C(CH_3)=CHCH_3$, what should be the researcher's conclusion and reasoning?
    Answer: E/Z isomers are not possible because one carbon of the double bond bears two identical methyl groups When determining if an alkene can exhibit E/Z stereoisomerism, you need to examine what substituents are attached to each carbon of the double bond. E/Z isomers exist when each carbon of the double bond has two different groups attached, allowing for distinct spatial arrangements. Looking at 2-methylbut-2-ene ($CH_3C(CH_3)=CHCH_3$), let's identify what's attached to each carbon. The left carbon of the double bond has two methyl groups ($CH_3$) attached to it, while the right carbon has one hydrogen and one methyl group. Since the left carbon bears two identical methyl groups, there's no way to arrange them differently in space—they're the same regardless of orientation. For E/Z isomerism to occur, you need four different groups total (two different groups on each carbon), or at minimum, each carbon must have two different substituents. Here, the presence of two identical methyl groups on one carbon eliminates any possibility of stereoisomerism. Choice A incorrectly suggests that having a trisubstituted alkene with different substituents is sufficient—but it ignores the critical requirement that each carbon needs different substituents. Choice B wrongly attributes the absence of isomerism to molecular size rather than the actual structural requirement. Choice C assumes E/Z isomers could exist but be indistinguishable, which misunderstands that if two identical groups are on one carbon, no isomers form at all. Study tip: Always check both carbons of a double bond systematically. If either carbon has two identical substituents, E/Z isomerism is impossible—this is the quickest way to eliminate stereoisomerism in alkenes.
13. A student attempts to assign E/Z configuration to the alkene 2-bromo-3-methylpent-2-ene. After applying the Cahn-Ingold-Prelog priority rules, they identify the highest priority groups on each carbon of the double bond. On C2, the highest priority group is Br, and on C3, the highest priority group is the ethyl group ($-CH_2CH_3$). If these two highest priority groups are on the same side of the double bond, what is the correct stereochemical designation?
    Answer: The alkene has Z configuration because the highest priority groups are on the same side of the double bond When the highest priority groups on each carbon of a double bond are on the same side, the alkene has Z configuration (from German 'zusammen' meaning together). Choice B incorrectly assigns E configuration to same-side arrangement. Choice C mentions correct Z assignment but gives wrong reasoning about lowest priority groups. Choice D is incorrect because the spatial arrangement of highest priority groups is sufficient to determine configuration.
14. When applying Cahn-Ingold-Prelog rules to assign E/Z configuration to an alkene, a student encounters the partial structure $-CH_2CH_2Br$ on one side and $-CH_2CH_2Cl$ on the other side of a double bond carbon. Both groups also have hydrogen substituents. How should the student determine the priority order for these specific substituents?
    Answer: $-CH_2CH_2Br$ has higher priority because Br has higher atomic number than Cl CIP priority rules are based on atomic number, not atomic mass or electronegativity. We compare atoms at each position moving outward from the point of attachment. Both groups have C-C-halogen, so the first point of difference is at the halogen. Br (atomic number 35) has higher priority than Cl (atomic number 17). Choice A incorrectly uses atomic mass. Choice B incorrectly uses electronegativity. Choice D incorrectly suggests equal priority when there is a clear difference in atomic numbers.
15. Which of the following reaction conditions would convert hept-3-yne into (E)-hept-3-ene with the highest yield and selectivity?
    Answer: Na, NH₃ (l) When you encounter alkyne reduction problems, focus on the stereochemistry of the products. Different reduction methods give different stereochemical outcomes due to their distinct mechanisms. The dissolving metal reduction using sodium in liquid ammonia (choice D) is the only method that produces the (E)-alkene with high selectivity. This reaction proceeds through a radical anion mechanism where electrons are added sequentially to the triple bond. The intermediate radical anion is protonated to form a vinyl radical, which then receives another electron and proton. The trans geometry is favored because the bulky alkyl groups adopt positions that minimize steric hindrance, consistently producing the (E)-alkene. Choice A (H₂, Pd/C) would completely reduce the alkyne to an alkane, giving heptane instead of the desired alkene. Choice B (H₂, Lindlar's catalyst) does stop at the alkene stage, but it produces the (Z)-alkene through syn addition. The poisoned palladium catalyst forces both hydrogens to add from the same face of the triple bond, placing the alkyl groups on the same side. Choice C (hydroboration-oxidation) is used for terminal alkynes and wouldn't work effectively on this internal alkyne. Even if it did react, it wouldn't provide the stereoselectivity needed for the (E)-product. Remember this pattern: dissolving metal reduction (Na/NH₃) gives (E)-alkenes, while Lindlar reduction gives (Z)-alkenes. When you see stereochemistry specified in alkyne reduction problems, immediately consider which method produces the desired geometry.
16. An internal alkyne is subjected to dissolving metal reduction using Na in liquid NH₃. Which of the following statements best describes the product and the mechanism?
    Answer: An (E)-alkene is formed via anti-addition of two hydrogen atoms. Dissolving metal reduction (Na in liquid NH₃) of an internal alkyne is a stereoselective reaction that produces an (E)-alkene. The mechanism involves a radical anion intermediate, and the stereochemical outcome is determined by the preference of the bulky alkyl groups to be on opposite sides in the key vinyl radical intermediate to minimize steric repulsion. This leads to the anti-addition of two hydrogen atoms across the triple bond, resulting in the more stable (E)- or trans-alkene.
17. Which of the following molecules can exist as E/Z isomers but cannot be unambiguously named using cis/trans notation?
    Answer: 1-bromo-2-chloroethene Cis/trans notation is typically used when each carbon of the double bond has a hydrogen and one non-hydrogen substituent, or when two of the substituents (one on each carbon) are identical. E/Z notation is a more general system based on CIP priorities. A) 1,2-dichloroethene has identical chloro groups, so cis/trans is unambiguous. B) but-2-ene has identical methyl groups, so cis/trans is unambiguous. C) 1-bromo-2-chloroethene has four different substituents (H, Br, H, Cl). There are no identical groups to reference for cis/trans, making it ambiguous. However, priorities can be assigned (Br>H and Cl>H), so E/Z isomers exist. D) 2-methylbut-2-ene has two identical methyl groups on one of its sp² carbons, so it cannot have any form of stereoisomerism (neither E/Z nor cis/trans).
18. A student is asked to predict the major product when (Z)-2-butene reacts with HBr in the presence of peroxides. They correctly identify that anti-Markovnikov addition will occur, but then claim the product will be optically active. What is the flaw in their reasoning?
    Answer: The student correctly identified the regiochemistry but failed to recognize that no stereocenters are formed in the product When alkenes react with HBr in the presence of peroxides, you need to consider both regiochemistry (where the atoms add) and stereochemistry (what happens at potential stereocenters). The peroxides trigger a radical mechanism that gives anti-Markovnikov addition, so the student correctly predicted that the Br will add to the less substituted carbon. The key insight is analyzing the product structure. When (Z)-2-butene undergoes anti-Markovnikov addition with HBr, the bromine adds to C-1 (the terminal carbon) and hydrogen adds to C-2. This produces 1-bromobutane: CH₃-CH₂-CH₂-CH₂Br. Looking at this structure, you'll notice that none of the carbons are bonded to four different groups - there are no stereocenters formed. Without stereocenters, the molecule cannot be optically active. Answer choice A is incorrect because this reaction proceeds through radical intermediates, not carbocations. The radical mechanism is what causes the anti-Markovnikov selectivity. Choice B is wrong because the stereochemistry of addition is irrelevant here - no stereocenters are created regardless of whether addition is syn or anti. Choice C is incorrect because the student was right about anti-Markovnikov addition; peroxides definitely reverse the normal regiochemistry through the radical pathway. The correct answer is D because the student properly identified anti-Markovnikov regiochemistry but failed to recognize that 1-bromobutane has no stereocenters. Study tip: Always draw out the complete product structure and check each carbon for four different substituents before predicting optical activity. Regiochemistry and stereochemistry are separate considerations that both need evaluation.
19. Comparing (Z)-1,2-dichloroethene and (E)-1,2-dichloroethene, which statement accurately predicts a difference in their physical properties?
    Answer: The (Z)-isomer has a higher boiling point because its molecular dipole moments do not cancel. Physical properties like boiling point depend on intermolecular forces. In (Z)-1,2-dichloroethene, the two polar C-Cl bonds are on the same side of the double bond, creating a net molecular dipole moment. In the (E)-isomer, the two C-Cl bond dipoles are on opposite sides and cancel each other out, resulting in a zero or near-zero molecular dipole moment. The Z-isomer, being polar, experiences stronger dipole-dipole interactions between molecules, which requires more energy to overcome. Consequently, the (Z)-isomer has a higher boiling point. Stability (E is more stable than Z) does not directly determine boiling point.
20. An unknown alkene with the molecular formula C₆H₁₂ exists as a pair of E/Z isomers. When this alkene undergoes ozonolysis (1. O₃; 2. DMS), it yields only a single carbonyl compound as the product. What is the identity of this single product?
    Answer: Propanal Ozonolysis cleaves a double bond and replaces it with two carbonyl groups. The fact that only a single product is formed implies that the starting alkene is symmetrical. The formula is C₆H₁₂. A symmetrical C₆ alkene would be formed from two C₃ units. Thus, the alkene must be hex-3-ene (CH₃CH₂CH=CHCH₂CH₃). Cleaving the double bond in hex-3-ene would produce two molecules of propanal (CH₃CH₂CHO). Hex-3-ene has one substituent (H) and one ethyl group on each carbon of the double bond, so it can and does exist as E/Z isomers. The other options are incorrect because they would arise from different alkenes: butan-2-one from 2,3-dimethylhex-3-ene, ethanal from but-2-ene, and acetone from 2,3-dimethylbut-2-ene.

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