Organic Chemistry Quiz: Conjugated Dienes And Allylic Stability

What this quiz covers

This quiz focuses on Conjugated Dienes And Allylic Stability, giving you a quick way to practice the rules, question types, and explanations that matter most for Organic Chemistry.

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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. P_kinetic is the product with the higher overall energy, and it is formed via the transition state with the higher activation energy from the intermediate.
2. P_thermo is the product with the lower overall energy, and it is formed via the transition state with the lower activation energy from the intermediate.
3. P_kinetic is the product formed via the transition state with the lower activation energy from the intermediate, regardless of its final thermodynamic stability. (correct answer)
4. P_thermo is always formed in greater yield than P_kinetic, because it is the most stable product and its formation is irreversible under all conditions.

Explanation: Kinetic control favors the product that is formed the fastest, which corresponds to the reaction pathway with the lowest activation energy (Ea). The kinetic product is not necessarily the most stable product. Thermodynamic control favors the most stable (lowest energy) product, which corresponds to the pathway with the most negative Gibbs free energy change (ΔG). This is favored at higher temperatures where the reaction is reversible, allowing equilibrium to be established. Statement C correctly defines the kinetic product as the one formed through the lowest energy barrier from the common intermediate, irrespective of its final stability compared to the thermodynamic product.

Question 2

1. 3-bromo-1-butene (1,2-addition product) (correct answer)
2. (E)-1-bromo-2-butene (1,4-addition product)
3. 4-bromo-1-butene (anti-Markovnikov product)
4. 1-bromo-1-butene (vinylic halide)

Explanation: At low temperatures (-80 °C), the reaction is under kinetic control, meaning the product that forms fastest will be the major product. The mechanism involves protonation of the diene to form a resonance-stabilized allylic carbocation. The nucleophile (Br-) can then attack at either C2 or C4. Attack at C2 (1,2-addition) has a lower activation energy due to the proximity effect—the bromide ion is closer to C2 in the intermediate—and is therefore faster. This leads to 3-bromo-1-butene as the kinetic product. The 1,4-addition product, (E)-1-bromo-2-butene, is more stable (thermodynamic product) but forms more slowly.

Question 3

1. The product is achiral and does not have any stereocenters.
2. A single enantiomer, (R)-3-bromocyclopentene, is formed selectively.
3. A racemic mixture of (R)- and (S)-3-bromocyclopentene is formed. (correct answer)
4. A pair of diastereomers, cis- and trans-3-bromocyclopentene, is formed.

Explanation: The reaction is an allylic bromination that proceeds through a radical mechanism. The starting material, cyclopentene, is achiral. The mechanism involves the abstraction of an allylic hydrogen to form a cyclopentenyl radical. This radical intermediate is planar (or very nearly planar) at the radical center due to sp2 hybridization. The bromine atom can then add to this planar intermediate from either the top face or the bottom face with equal probability. This non-selective addition creates a new stereocenter at C3. Since attack from both faces is equally likely, both the (R) and (S) enantiomers are produced in equal amounts, resulting in a racemic mixture.

Question 4

1. C1: sp2, C2: sp2, C3: sp3
2. All three carbon atoms are sp2 hybridized. (correct answer)
3. C1: sp3, C2: sp2, C3: sp3
4. The carbon atoms rapidly interconvert between sp2 and sp3 hybridization.

Explanation: For resonance to occur, the p-orbitals on adjacent atoms must be aligned. In the allyl anion, the negative charge (a lone pair) on one terminal carbon is delocalized over all three carbon atoms. This delocalization requires a p-orbital on that carbon. Therefore, all three carbon atoms must be sp2 hybridized to allow for a continuous system of parallel p-orbitals through which the pi electrons and the lone pair are delocalized. Looking at a single resonance structure (e.g., with a lone pair on C1) might incorrectly suggest sp3 hybridization for C1, but the reality of the resonance hybrid requires all three carbons to be sp2.

Question 5

1. The sp2-sp2 single bond in a conjugated system is shorter and stronger than a typical sp3-sp3 single bond.
2. There is a greater degree of hyperconjugation between the pi bonds and adjacent C-H sigma bonds in the conjugated system.
3. The parallel alignment of four adjacent p-orbitals allows for the delocalization of pi electrons over the entire system, lowering the overall energy. (correct answer)
4. Conjugated dienes adopt a rigid s-trans conformation that minimizes torsional strain compared to the more flexible isolated dienes.

Explanation: The primary source of stability in conjugated systems is the delocalization of pi electrons. In 1,3-butadiene, the four p-orbitals on the sp2-hybridized carbons overlap to form a single, extended pi system. This allows the four pi electrons to be spread out over all four atoms, which is a lower energy arrangement than having them confined to two isolated double bonds. While the sp2-sp2 bond is indeed shorter and stronger (A), this is a consequence of the delocalization, not the fundamental cause. Hyperconjugation (B) and conformational preferences (D) are real effects but are secondary to the primary stabilization from pi electron delocalization.

Question 6

1. Because one resonance structure is significantly more stable and is the only form that actually exists.
2. Because resonance structures are theoretical constructs; the true molecule is a single, static hybrid structure. (correct answer)
3. Because the activation energy for the interconversion is too high at room temperature.
4. Because equilibrium describes the interconversion of different molecules, not different electronic representations of the same molecule.

Explanation: A common misconception is that resonance involves molecules flipping back and forth between different forms. This is incorrect. Resonance structures are not real, distinct molecules. They are Lewis structure representations used to describe the delocalization of electrons within a single, real molecule. The actual molecule, the resonance hybrid, has a structure that is a weighted average of all its valid resonance contributors. It does not interconvert between them; it exists as the hybrid at all times. Therefore, the concept of equilibrium, which applies to the interconversion of distinct chemical species, is not applicable.

Question 7

1. The carbocation formed by protonating isoprene at C1. (correct answer)
2. The carbocation formed by loss of chloride from 3-chloro-1-butene.
3. The carbocation formed by protonating 1,3-butadiene at C1.
4. The carbocation formed by loss of bromide from 1-bromo-1-methylcyclohexene.

Explanation: We need to identify the carbocation that meets two criteria: allylic (the positive charge is on a carbon adjacent to a C=C double bond) and tertiary (the positively charged carbon is bonded to three other carbon atoms). (A) Protonating isoprene (2-methyl-1,3-butadiene) at C1 places the positive charge at C2. C2 is bonded to C1, C3, and the methyl group, making it tertiary. It is also adjacent to the C3=C4 double bond, making it allylic. This fits both criteria. (B) This forms a secondary allylic carbocation. (C) This forms a secondary allylic carbocation. (D) This would form a vinylic carbocation, which is very unstable and not allylic.

Question 8

1. 1,5-hexadiene < (Z,Z)-2,4-hexadiene < (E,E)-2,4-hexadiene
2. (E,E)-2,4-hexadiene < (Z,Z)-2,4-hexadiene < 1,5-hexadiene (correct answer)
3. (Z,Z)-2,4-hexadiene < (E,E)-2,4-hexadiene < 1,5-hexadiene
4. 1,5-hexadiene < (E,E)-2,4-hexadiene < (Z,Z)-2,4-hexadiene

Explanation: Heat of hydrogenation is inversely proportional to alkene stability. We need to rank the dienes from most stable to least stable. 1,5-hexadiene is an isolated diene. Both 2,4-hexadiene isomers are conjugated dienes. Conjugated dienes are more stable than isolated dienes. Therefore, 1,5-hexadiene is the least stable and will have the most exothermic heat of hydrogenation. Between the two conjugated isomers, trans (E) double bonds are generally more stable than cis (Z) double bonds due to reduced steric strain. Therefore, (E,E)-2,4-hexadiene is more stable than (Z,Z)-2,4-hexadiene. The stability order is: (E,E)-2,4-hexadiene > (Z,Z)-2,4-hexadiene > 1,5-hexadiene. The order of increasing heat of hydrogenation is the reverse of the stability order: (E,E)-2,4-hexadiene < (Z,Z)-2,4-hexadiene < 1,5-hexadiene.

Question 9

1. 1-chloro-2-butene, because it forms a primary allylic carbocation that is resonance stabilized.
2. 3-chloro-1-butene, because steric hindrance is minimized at the primary carbon.
3. 1-chlorobutane, because it lacks pi bonds that could electronically hinder the departure of the leaving group.
4. 3-chloro-2-methyl-1-butene, because it forms a tertiary allylic carbocation that has extensive resonance stabilization. (correct answer)

Explanation: The rate of an SN1 reaction is determined by the stability of the carbocation intermediate formed in the rate-determining step. We need to compare the stability of the carbocations formed from each starting material. (A) 1-chloro-2-butene forms a primary allylic carbocation. (B) 3-chloro-1-butene forms a secondary allylic carbocation. (C) 1-chlorobutane forms an unstable primary carbocation. (D) 3-chloro-2-methyl-1-butene forms a tertiary allylic carbocation. The order of carbocation stability is tertiary allylic > secondary allylic > primary allylic > primary. Therefore, 3-chloro-2-methyl-1-butene will react the fastest because it forms the most stable carbocation intermediate.

Question 10

1. The 1,2-addition product is a less substituted alkene and therefore has lower steric hindrance in its transition state.
2. The secondary carbocation character at C2 of the allylic intermediate is significantly more stable than the primary character at C4.
3. After protonation at C1, the bromide ion is statistically closer to C2 than to C4, leading to a lower activation energy for attack at C2. (correct answer)
4. The 1,2-addition product is thermodynamically more stable, and according to the Hammond postulate, its transition state is also lower in energy.

Explanation: Both the 1,2- and 1,4-addition products form from the same resonance-stabilized allylic carbocation intermediate. Therefore, the relative stability of the contributors to the intermediate (B) does not explain the difference in rates of product formation from that intermediate. The 1,4-product is actually more stable (D is incorrect). The kinetic preference for the 1,2-product is explained by the proximity effect (C). When the proton adds to C1, the bromide counter-ion is located closer to C2 than to C4. This means that attack at C2 requires less movement and has a smaller entropic barrier, resulting in a lower activation energy for the 1,2-addition pathway.

Question 11

1. 3-chloro-3-methyl-1-butene
2. 1-chloro-3-methyl-2-butene (correct answer)
3. 4-chloro-2-methyl-2-butene
4. 3-chloro-2-methyl-1-butene

Explanation: Under thermodynamic control (higher temperature), the most stable product is favored. The mechanism starts with protonation of the diene to form the most stable carbocation. Protonation of 2-methyl-1,3-butadiene at C1 gives a tertiary allylic carbocation at C2, which is resonance-stabilized with a primary carbocation at C4. This is the most stable intermediate. The chloride ion can attack at C2 (1,2-addition) or C4 (1,4-addition). Attack at C2 gives 3-chloro-2-methyl-1-butene (D). Attack at C4 gives 1-chloro-3-methyl-2-butene (B). To determine the thermodynamic product, we compare the stability of these two alkenes. Product (B) is a trisubstituted alkene, while product (D) is a disubstituted alkene. Trisubstituted alkenes are more thermodynamically stable than disubstituted alkenes. Therefore, 1-chloro-3-methyl-2-butene is the major product under thermodynamic control.