Organic Chemistry Quiz: E Z And Alkene Stereochemistry

What this quiz covers

This quiz focuses on E Z And Alkene Stereochemistry, 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. Two different alkenes including the original compound
2. Three different alkenes including the original compound
3. Four different alkenes including the original compound (correct answer)
4. Six different alkenes including the original compound

Explanation: Each double bond can exist in either E or Z configuration independently. With two double bonds, there are 2² = 4 possible combinations: (3Z,5Z), (3Z,5E), (3E,5Z), and (3E,5E). The original compound (3Z,5E) is one of these four possibilities. Choice A only accounts for one double bond. Choice B undercounts the combinations. Choice D incorrectly applies 3! = 6, which would be relevant for three independent stereocenters, not two.

Question 2

1. The student failed to recognize that the reaction proceeds through a carbocation intermediate that leads to racemization
2. The student failed to account for the syn stereochemistry of the addition reaction
3. The student should have predicted Markovnikov addition instead of anti-Markovnikov addition under these conditions
4. The student correctly identified the regiochemistry but failed to recognize that no stereocenters are formed in the product (correct answer)

Explanation: 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.

Question 3

1. The alkene has Z configuration because the highest priority groups are on the same side of the double bond (correct answer)
2. The alkene has E configuration because the highest priority groups are on the same side of the double bond
3. The alkene has Z configuration because the lowest priority groups are on opposite sides of the double bond
4. The configuration cannot be determined without knowing the exact spatial arrangement of all substituents

Explanation: 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.

Question 4

1. (E)-pent-2-ene
2. (Z)-pent-2-ene (correct answer)
3. Pentane
4. Pent-1-ene

Explanation: 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.

Question 5

1. The severe steric hindrance between substituents on adjacent carbons.
2. The high bond dissociation energy of the carbon-carbon sigma bond.
3. The electrostatic repulsion between the sp² hybridized carbons.
4. The requirement to break the pi (π) bond during the rotational transition state. (correct answer)

Explanation: 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.

Question 6

1. H₂, Pd/C
2. H₂, Lindlar's catalyst
3. 1. BH₃·THF; 2. H₂O₂, NaOH
4. Na, NH₃ (l) (correct answer)

Explanation: 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.

Question 7

1. Propanal (correct answer)
2. Butan-2-one
3. Ethanal
4. Acetone (Propan-2-one)

Explanation: 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.

Question 8

1. Hyperconjugation is completely absent in the trans-configured ring system.
2. The molecule is forced into a planar geometry, creating extreme angle strain.
3. A trans double bond in a six-membered ring induces unfavorable aromatic character.
4. The p-orbitals of the pi bond are twisted, leading to poor orbital overlap and high strain. (correct answer)

Explanation: 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.

Question 9

1. The higher stability of the carbocation leading to the (E)-isomer.
2. The kinetic preference for abstracting a proton leading to the Hofmann product.
3. The lower energetic barrier of the transition state leading to the less sterically hindered alkene. (correct answer)
4. The requirement for syn-periplanar geometry in the transition state.

Explanation: This reaction is an E2 elimination, which proceeds through a single concerted transition state, not a carbocation intermediate (eliminating A). The major products are the Zaitsev products (pent-2-enes), not the Hofmann product (pent-1-ene), eliminating B. The E2 mechanism requires an anti-periplanar geometry, not syn-periplanar (eliminating D). The preference for the (E)-isomer over the (Z)-isomer arises because both are possible via anti-periplanar eliminations, but the transition state leading to the (E)-alkene is lower in energy. This is because the bulky alkyl groups (methyl and ethyl) are oriented away from each other in the E-transition state, minimizing steric strain. The transition state leading to the Z-alkene has these groups in a higher-energy gauche interaction.

Question 10

1. They are identical (a meso compound).
2. They are enantiomers.
3. They are diastereomers. (correct answer)
4. They are the same achiral molecule.

Explanation: 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).

Question 11

1. These are enantiomers with identical physical properties except for optical rotation
2. These are diastereomers with different physical properties such as boiling points and dipole moments (correct answer)
3. These are constitutional isomers with significantly different chemical reactivity patterns
4. These are conformational isomers that interconvert rapidly at room temperature

Explanation: 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).

Question 12

1. E/Z isomers are possible because the molecule contains a trisubstituted alkene with different substituents
2. E/Z isomers are not possible because the molecule is too small to exhibit stereoisomerism
3. E/Z isomers are possible, but they would have identical physical properties and cannot be distinguished
4. E/Z isomers are not possible because one carbon of the double bond bears two identical methyl groups (correct answer)

Explanation: 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.

Question 13

1. The student incorrectly identified the highest priority groups and should have compared Cl with H instead of with the ethyl group (correct answer)
2. The student placed the chlorine on the wrong carbon of the double bond
3. The student correctly identified the highest priority groups but confused E and Z notation definitions
4. The student drew the correct stereoisomer but used incorrect nomenclature for the base chain numbering

Explanation: 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).

Question 14

1. $-CH_2CH_2Br$ has higher priority because Br has higher atomic mass than Cl
2. $-CH_2CH_2Cl$ has higher priority because Cl is more electronegative than Br
3. $-CH_2CH_2Br$ has higher priority because Br has higher atomic number than Cl (correct answer)
4. The two groups have equal priority because the first point of difference is at the same position

Explanation: 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.

Question 15

1. isopropyl (–CH(CH₃)₂) > n-propyl (–CH₂CH₂CH₃) (correct answer)
2. ethyl (–CH₂CH₃) > vinyl (–CH=CH₂)
3. tert-butyl (–C(CH₃)₃) > sec-butyl (–CH(CH₃)CH₂CH₃)
4. hydroxyl (–OH) > formyl (–CHO)

Explanation: 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.

Question 16

1. 1,2-dichloroethene
2. but-2-ene
3. 1-bromo-2-chloroethene (correct answer)
4. 2-methylbut-2-ene

Explanation: 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).

Question 17

1. (E)-4-methylhex-2-ene
2. (Z)-4-methylhex-3-ene (correct answer)
3. (E)-4-methylhex-3-ene
4. (Z)-4-methylhex-2-ene

Explanation: 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.

Question 18

1. The (E)-isomer has a higher boiling point due to its greater thermodynamic stability.
2. The (Z)-isomer has a higher boiling point because its molecular dipole moments do not cancel. (correct answer)
3. Both isomers have identical boiling points and dipole moments because they are stereoisomers.
4. The (E)-isomer has a measurable molecular dipole moment, while the (Z)-isomer does not.

Explanation: 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.

Question 19

1. E/Z notation can be applied to chiral centers, whereas cis/trans cannot.
2. E/Z notation is based on the relative molecular weight of substituents, which is more precise.
3. E/Z notation can unambiguously describe stereochemistry in tri- and tetrasubstituted alkenes. (correct answer)
4. E/Z notation only applies to cyclic alkenes, where cis/trans is often confusing.

Explanation: 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.

Question 20

1. 1-bromo-2-methylpropene (correct answer)
2. 3-methylpent-2-ene
3. 1,2-dideuterioethene
4. penta-1,3-diene

Explanation: 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).