Organic Chemistry Quiz: Electrophilic Addition To Alkenes

What this quiz covers

This quiz focuses on Electrophilic Addition To Alkenes, 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. 1-chloro-2-methyl-2-deuteriopropane
2. 2-chloro-2-methyl-1-deuteriopropane (correct answer)
3. 1-chloro-2-(deuteriomethyl)propane
4. 2-chloro-1-methyl-1-deuteriopropane

Explanation: The reaction is an electrophilic addition where D⁺ is the electrophile. According to Markovnikov's rule, the electrophile (D⁺) adds to the carbon of the double bond that has more hydrogen atoms (the CH₂ group, C1). This forms the more stable tertiary carbocation at C2. The nucleophile, Cl⁻, then attacks the carbocation at C2. This results in the deuterium being on C1 and the chlorine on C2, which is 2-chloro-2-methyl-1-deuteriopropane.

Question 2

1. trans-1,2-dibromocyclohexane
2. cis-1,2-dibromocyclohexane
3. trans-1-bromo-2-methoxycyclohexane (correct answer)
4. cis-1-bromo-2-methoxycyclohexane

Explanation: This reaction involves the formation of a cyclic bromonium ion intermediate. Since methanol is used as a nucleophilic solvent and is present in large excess, it will attack the bromonium ion instead of the bromide ion (Br⁻). The attack occurs from the face opposite the bromine bridge (anti-addition) at one of the carbons of the former double bond. This ring-opening is an SN2-like process, resulting in a trans relationship between the bromine and methoxy groups.

Question 3

1. Competing anti-Markovnikov addition due to steric hindrance at the more substituted carbon
2. Syn-addition from the opposite face of the ring, giving the other possible stereoisomer
3. Direct addition to form the secondary carbocation without rearrangement to the tertiary carbocation (correct answer)
4. Formation of the tertiary carbocation followed by elimination and re-addition at the alternative position

Explanation: While most of the initially formed secondary carbocation rearranges to the more stable tertiary carbocation (major product), some reacts directly with bromide before rearrangement occurs, giving the minor product. Choice A incorrectly invokes anti-Markovnikov addition. Choice B misidentifies this as a stereochemical issue. Choice D describes an unrealistic elimination/re-addition mechanism.

Question 4

1. Oxymercuration avoids carbocation intermediates entirely, preventing rearrangements that occur in acid-catalyzed hydration (correct answer)
2. Oxymercuration proceeds through radical intermediates instead of carbocations, giving anti-Markovnikov selectivity
3. Oxymercuration uses a stronger electrophile than H⁺, allowing the reaction to proceed under milder conditions
4. Oxymercuration gives syn-addition while acid catalysis gives anti-addition, providing complementary stereochemistry

Explanation: Oxymercuration-demercuration proceeds through a mercurinium ion intermediate rather than a free carbocation, which prevents carbocation rearrangements that commonly occur in acid-catalyzed hydration. Both reactions give Markovnikov addition. Choice B incorrectly describes radical intermediates and wrong regioselectivity. Choice C doesn't address the rearrangement issue. Choice D incorrectly contrasts stereochemistry (both are typically syn).

Question 5

1. 3,3-dimethyl-2-butanol; direct Markovnikov addition without rearrangement due to steric hindrance
2. 2,3-dimethyl-2-butanol; carbocation rearrangement occurs via a 1,2-methyl shift to form a more stable tertiary carbocation (correct answer)
3. 3,3-dimethyl-1-butanol; anti-Markovnikov addition occurs due to the bulky tertiary carbon adjacent to the alkene
4. 2,3-dimethyl-1-butanol; carbocation rearrangement occurs but the hydroxyl group adds to the primary carbon

Explanation: Initial protonation at C1 (following Markovnikov's rule) forms a secondary carbocation at C2. This undergoes a 1,2-methyl shift to form the much more stable tertiary carbocation, giving 2,3-dimethyl-2-butanol as the major product. Choice A ignores the rearrangement. Choice C incorrectly invokes anti-Markovnikov addition. Choice D has the rearrangement but wrong regiochemistry for water addition.

Question 6

1. The same product distribution because temperature doesn't affect regioselectivity in electrophilic addition reactions
2. More 1-chloro-2-methylpentane because lower temperature favors the kinetically controlled anti-Markovnikov product
3. More 1-chloro-2-methylpentane because lower temperature reduces the extent of carbocation rearrangement (correct answer)
4. More 2-chloro-2-methylpentane because lower temperature increases the lifetime of the more stable tertiary carbocation

Explanation: The major product forms via rearrangement of the initially formed secondary carbocation to a more stable tertiary carbocation. At lower temperatures, this rearrangement is slower, allowing more of the secondary carbocation to react directly, giving more of the unrearranged product (1-chloro-2-methylpentane). Choice A is incorrect as temperature affects rearrangement rates. Choice B incorrectly invokes anti-Markovnikov selectivity. Choice D has the wrong temperature effect on rearrangement.

Question 7

1. The transition state for protonation
2. The π-complex between the alkene and HBr
3. The final alkyl halide product
4. The carbocation intermediate (correct answer)

Explanation: In a two-step reaction mechanism like electrophilic addition, the species that exists between the two transition states is an intermediate. For the addition of HBr to an alkene, the first step is protonation to form a carbocation, and the second step is nucleophilic attack by bromide. The carbocation is a true chemical species that exists for a finite lifetime and corresponds to a local energy minimum on the reaction coordinate diagram.

Question 8

1. Concentrated H₂SO₄, heat
2. 1. Hg(OAc)₂, H₂O; 2. NaBH₄
   (correct answer)
3. Dilute H₂SO₄, room temperature
4. 1. BH₃·THF; 2. H₂O₂, NaOH

Explanation: When you see alkene hydration problems with side products like rearrangement and multiple regioisomers, you need to consider the mechanism and selectivity of different hydration methods. The key issue here is that 1-hexene is producing unwanted products through carbocation rearrangements and lack of regioselectivity. This tells you that acid-catalyzed hydration is problematic because it proceeds through carbocation intermediates that can rearrange and doesn't follow strict Markovnikov selectivity. Oxymercuration-demercuration (choice B) is the ideal solution. This two-step process involves mercuric acetate and water, followed by sodium borohydride reduction. It proceeds through a mercurinium ion intermediate rather than a carbocation, which prevents rearrangements entirely. Additionally, it follows strict Markovnikov regioselectivity, placing the OH group on the more substituted carbon to give exclusively 2-hexanol. Choice A (concentrated H₂SO₄, heat) would actually worsen the problem by promoting carbocation rearrangements and elimination reactions. Choice C (dilute H₂SO₄) still proceeds through carbocations, so rearrangements remain problematic even at room temperature. Choice D (hydroboration-oxidation) follows anti-Markovnikov selectivity, placing the OH on the less substituted carbon to give 1-hexanol, not 2-hexanol. Remember this pattern: when you need Markovnikov hydration without rearrangements, oxymercuration-demercuration is your go-to method. It's the "clean" alternative to acid-catalyzed hydration that avoids carbocation chemistry entirely.

Question 9

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

Explanation: When you encounter alkene hydrogenation questions, the key principle is that reaction rates decrease as the alkene becomes more substituted. This happens because bulky substituents around the double bond create steric hindrance, making it harder for the large palladium catalyst to approach and coordinate with the alkene. Let's analyze the substitution patterns: 1-butene is monosubstituted (one alkyl group on the double bond), 2-methyl-1-butene is disubstituted, (E)-2-butene is disubstituted, and 2-methyl-2-butene is trisubstituted. Generally, monosubstituted alkenes react fastest, followed by disubstituted, then trisubstituted alkenes react slowest. However, within the disubstituted alkenes, terminal alkenes (like 2-methyl-1-butene) experience more steric hindrance than internal alkenes (like (E)-2-butene) because the substituents are concentrated near one end of the molecule, creating a more crowded environment for the catalyst. Therefore, the correct order is: 1-butene > 2-methyl-1-butene > (E)-2-butene > 2-methyl-2-butene, which matches answer D. Answer A incorrectly places (E)-2-butene ahead of 2-methyl-1-butene, missing the steric difference between terminal and internal disubstituted alkenes. Answer B completely reverses the trend, suggesting more substituted alkenes react faster. Answer C also incorrectly ranks (E)-2-butene as fastest and places 1-butene ahead of 2-methyl-1-butene. Remember: for catalytic hydrogenation, less substitution means faster reaction due to reduced steric hindrance around the reactive double bond.

Question 10

1. 2-chloro-1-butanol; chlorine adds first to the more substituted carbon following Markovnikov's rule
2. 2-chloro-1-butanol; the chloronium ion opens at the less substituted carbon due to steric effects favoring attack there
3. 1-chloro-2-butanol; water acts as a nucleophile and attacks the less hindered carbon of the chloronium ion
4. 1-chloro-2-butanol; the chloronium ion opens at the more substituted carbon due to greater carbocation character (correct answer)

Explanation: When you encounter halohydrin formation reactions, you're dealing with the addition of a halogen and water across a double bond through a three-membered halonium ion intermediate. The key to predicting regioselectivity lies in understanding how this cyclic intermediate opens. In this reaction, (E)-2-butene first forms a chloronium ion when $Cl_2$ approaches the alkene. Water then acts as a nucleophile and attacks one of the carbons in this three-membered ring. The critical insight is that chloronium ions have partial carbocation character at both carbons, but more so at the more substituted position due to better stabilization of positive charge by alkyl groups. Water preferentially attacks the more substituted carbon (C2) because it has greater carbocation character, leading to 1-chloro-2-butanol as the major product. This makes answer choice D correct. Answer A incorrectly suggests chlorine adds first to the more substituted carbon following Markovnikov's rule, but halohydrin formation doesn't follow simple Markovnikov addition patterns. Answer B gives the wrong product (2-chloro-1-butanol) and incorrectly attributes regioselectivity to steric effects favoring the less substituted carbon. Answer C predicts the correct mechanism (nucleophilic attack at less hindered carbon) but arrives at the wrong product name. Remember that in halohydrin formation, the nucleophile (water) attacks the more substituted carbon of the halonium ion due to greater carbocation stability, not sterics. This often trips up students who try to apply simple Markovnikov rules.

Question 11

1. The bulky borane approaches from the less hindered face, and subsequent oxidation proceeds with retention of configuration (correct answer)
2. The borane adds in a Markovnikov fashion, placing boron on the more substituted carbon with inversion of configuration
3. The borane approaches from the more hindered face to minimize steric interactions, followed by syn-oxidation
4. The borane adds anti-Markovnikov with the bulky BH₂ group oriented away from the methyl substituent, maintaining syn-stereochemistry throughout

Explanation: Hydroboration-oxidation involves syn-addition of BH₃ from the less sterically hindered face of the alkene, followed by oxidation with retention of stereochemistry. The anti-Markovnikov regioselectivity places OH on the less substituted carbon. Choice B incorrectly describes Markovnikov addition. Choice C has the wrong face selectivity. Choice D correctly identifies anti-Markovnikov but incorrectly explains the steric approach.

Question 12

1. 4-chloro-4-methyl-1-pentene (correct answer)
2. 2-chloro-4-methyl-1,4-pentadiene
3. 5-chloro-2-methyl-1-pentene
4. 1,4-dichloro-2-methylpentane

Explanation: The starting material has two double bonds. The double bond between C1 and C2 is monosubstituted, while the double bond between C4 and C5 is disubstituted. Electrophilic addition occurs preferentially at the more substituted double bond because it leads to a more stable carbocation intermediate. Protonation of the C4-C5 double bond at C5 (the carbon with more hydrogens) forms a stable tertiary carbocation at C4. The other double bond would only form a secondary carbocation. Chloride ion then attacks the tertiary carbocation at C4, yielding 4-chloro-4-methyl-1-pentene as the major product.

Question 13

1. 1. H₂, Lindlar's catalyst; 2. HBr, ROOR
2. 1. Na, NH₃ (l); 2. 1. BH₃·THF, 2. H₂O₂, NaOH
3. 1. H₂SO₄, H₂O, HgSO₄; 2. NaBH₄, CH₃OH
   (correct answer)
4. 1. (sia)₂BH; 2. H₂O₂, NaOH; 3. H₂/Pd

Explanation: This is a two-step transformation. First, 1-pentyne needs to be converted to a functional group that can be reduced to the desired alcohol. Markovnikov hydration of the terminal alkyne 1-pentyne (using H₂SO₄, H₂O, HgSO₄) produces the methyl ketone 2-pentanone. In the second step, the ketone is selectively reduced to a secondary alcohol using a mild reducing agent like sodium borohydride (NaBH₄). This sequence correctly places the oxygen functionality at C2 and then converts it to the alcohol.

Question 14

1. 1-bromo-1-phenylpropane (correct answer)
2. 2-bromo-1-phenylpropane
3. (E)-1-bromo-1-phenylpropene
4. 1-bromo-3-phenylpropane

Explanation: In this electrophilic addition, the proton (H⁺) can add to either C1 or C2 of the double bond. If H⁺ adds to C2, a secondary carbocation forms at C1. This carbocation is also benzylic, meaning it is stabilized by resonance with the phenyl ring. If H⁺ adds to C1, a non-benzylic secondary carbocation forms at C2. The benzylic carbocation is significantly more stable and forms preferentially. Subsequent attack by Br⁻ at the benzylic position (C1) yields 1-bromo-1-phenylpropane as the major product.

Question 15

1. 3,3-dimethyl-1-butene + H₂SO₄/H₂O
2. 2,3-dimethyl-2-butene + 1. BH₃·THF / 2. H₂O₂, NaOH
3. 3,3-dimethyl-1-butene + 1. Hg(OAc)₂, H₂O / 2. NaBH₄ (correct answer)
4. 2,3-dimethyl-1-butene + H₂SO₄/H₂O

Explanation: The target molecule, 3,3-dimethyl-2-butanol, is the Markovnikov hydration product of 3,3-dimethyl-1-butene without rearrangement. Oxymercuration-demercuration (1. Hg(OAc)₂, H₂O / 2. NaBH₄) is the ideal method for this because it follows Markovnikov's rule but proceeds through a bridged mercurinium ion intermediate, which prevents the carbocation rearrangement that would occur under standard acid-catalyzed hydration conditions (Choice A). Choices B and D start with different alkenes and would lead to different products.

Question 16

1. 1-methylcyclohexan-1-ol
2. cis-2-methylcyclohexan-1-ol
3. trans-2-methylcyclohexan-1-ol (correct answer)
4. 2-methylcyclohexanone

Explanation: Hydroboration-oxidation results in the anti-Markovnikov, syn-addition of H and OH across the double bond. For 1-methylcyclohexene, the boron (part of BH₃) adds to the less substituted carbon (C2), and the hydrogen adds to the more substituted carbon (C1). This addition occurs from the same face of the double bond (syn-addition). The subsequent oxidation replaces boron with an OH group with retention of stereochemistry. This results in the methyl group at C1 and the hydroxyl group at C2 having a trans relationship.

Question 17

1. The reaction proceeds through a more stable secondary carbocation intermediate rather than a primary carbocation. (correct answer)
2. The chloride ion attacks the internal carbon because it is less sterically hindered than the terminal carbon.
3. The reaction is under thermodynamic control, and 2-chlorobutane is the most stable constitutional isomer.
4. The hydrogen atom is the electrophile and adds to the more substituted carbon of the double bond.

Explanation: This reaction is an electrophilic addition following Markovnikov's rule. The rule is an empirical observation based on the underlying mechanism. The rate-determining step is the protonation of the alkene to form a carbocation. Protonating C1 of 1-butene generates a secondary carbocation at C2, while protonating C2 would generate a primary carbocation at C1. Since the secondary carbocation is significantly more stable, it forms preferentially, leading to the major product, 2-chlorobutane, after attack by Cl⁻. Therefore, intermediate stability dictates the regioselectivity.

Question 18

1. I > II > III (correct answer)
2. III > II > I
3. II > I > III
4. I > III > II

Explanation: The rate-determining step in the electrophilic addition of HBr to an alkene is the formation of the carbocation intermediate. The stability of this carbocation determines the activation energy and thus the reaction rate. 2-Methyl-2-butene (I) forms a tertiary carbocation, which is the most stable. 1-Butene (II) forms a secondary carbocation, which is less stable than tertiary. Ethene (III) forms a primary carbocation, which is the least stable. Therefore, the order of reactivity is I > II > III, following the order of carbocation stability.

Question 19

1. The bromine radical is a strong electrophile that adds to the most substituted carbon.
2. The key propagation step involves addition of a bromine radical to form the most stable possible carbon radical. (correct answer)
3. The reaction intermediate is a carbocation, and the primary carbocation is more stable under radical conditions.
4. Steric hindrance prevents the bulky bromine radical from approaching the more substituted carbon atom.

Explanation: When you encounter questions about radical addition reactions, focus on the mechanism and which intermediates are most stable. Radical addition of HBr to alkenes follows a different regioselectivity pattern than ionic addition because radicals have different stability preferences than carbocations. The key to anti-Markovnikov addition lies in the propagation steps. After the initial hydrogen radical abstracts from HBr, a bromine radical is generated. This bromine radical then adds to the alkene carbon that will produce the most stable carbon radical intermediate. Since tertiary carbon radicals are more stable than secondary, which are more stable than primary, the bromine radical adds to the less substituted carbon, leaving the radical on the more substituted carbon. This creates the most stable possible carbon radical intermediate, making option B correct. Option A is incorrect because bromine radical acts as a radical species, not an electrophile, and it adds to form the most stable radical intermediate, not based on electrophilic attraction. Option C misidentifies the intermediate—radical reactions form carbon radicals, not carbocations, and primary carbocations are actually less stable than tertiary ones. Option D incorrectly suggests steric hindrance as the primary factor, when thermodynamic stability of the radical intermediate is what drives regioselectivity. Remember that radical stability follows the same order as carbocation stability (3° > 2° > 1°), but the mechanism is entirely different. When you see radical addition questions, immediately think about which carbon radical intermediate would be most stable.

Question 20

1. tert-butanol
2. 1-methoxy-2-methylpropane
3. 2-methoxy-2-methylpropane (correct answer)
4. 2-methyl-1-propanol

Explanation: This reaction is an acid-catalyzed addition of an alcohol across a double bond, mechanistically analogous to acid-catalyzed hydration. The alkene is protonated by the acid to form the most stable carbocation, which is the tertiary tert-butyl cation. Methanol, acting as a nucleophile, attacks the carbocation. A final deprotonation step (by methanol or HSO₄⁻) yields the ether product, 2-methoxy-2-methylpropane, also known as methyl tert-butyl ether (MTBE).