Reactions at the Allylic Position: Videos & Practice Problems

Allylic compounds exhibit unique reactivity due to the stability of allylic carbocations, which are stabilized by resonance, making them more reactive in SN1 and E1 reactions compared to alkyl counterparts. Allylic anions are more acidic than alkyl anions, leading to increased elimination products. In radical reactions, allylic halogenation occurs when halogens replace allylic protons, facilitated by radical initiators. Additionally, allylic oxidation selectively transforms allylic alcohols into carbonyls using manganese(IV) oxide, preserving the alkene double bond. Understanding these reactions is crucial for mastering organic synthesis and reaction mechanisms.

Reactions at the Allylic Position Concept 1

Reactions at the allylic position involve the carbon atom adjacent to a double bond, known as the allylic carbon. In a molecular structure featuring double-bonded carbons, these carbons are referred to as the vinylic carbons, while the allylic position is the carbon next to them. Understanding the behavior of allylic compounds is crucial, as they exhibit distinct reactivity compared to alkenes.

In the context of nucleophilic substitution and elimination reactions, allylic compounds demonstrate faster reaction rates than their alkyl counterparts in S_N1 and E₁ reactions. Specifically, allylic halides react more rapidly than alkyl halides in S_N2 and E₂ reactions. This increased reactivity can be attributed to the stability provided by resonance in allylic systems.

When considering radical reactions, substitution occurs preferentially at the allylic position, highlighting the unique reactivity of these compounds. Additionally, allylic anions are notable for their rapid rearrangement due to resonance stabilization, which allows for the delocalization of negative charge across the molecule.

Overall, the allylic position plays a significant role in determining the pathways and mechanisms of various chemical reactions, making it an essential concept in organic chemistry.

Reactions at the Allylic Position Concept 2

In the study of organic chemistry, understanding the mechanisms of SN1 and E1 reactions is crucial, particularly when examining allylic compounds. These reactions are characterized by a two-step process where the first step is the rate-determining step. A key aspect to note is that allylic carbocations are formed more rapidly than their similarly substituted alkyl counterparts due to their enhanced stability.

Alkyl carbocations gain stability primarily through hyperconjugation, which involves the interaction of the empty p-orbital of the carbocation with adjacent C-H bonds. However, allylic carbocations benefit from an additional stabilizing factor: resonance. This resonance stabilization occurs because the positive charge on the carbocation can be delocalized over adjacent double bonds, leading to a more stable structure.

To illustrate this concept, consider the formation of carbocations from tertiary alkyl halides. When a tertiary alkyl halide loses a leaving group, such as a chloride ion, it forms a tertiary carbocation. The relative rate of this reaction is assigned a value of 1. In contrast, when a tertiary allylic halide undergoes the same process, it forms a tertiary allylic carbocation, which has a relative rate of approximately 162. This significant difference highlights that allylic carbocations are formed much more readily than typical tertiary carbocations, underscoring the powerful stabilizing effect of resonance compared to hyperconjugation.

In summary, when dealing with allylic positions, it is essential to recognize their propensity to undergo SN1 and E1 reactions more efficiently due to the resonance stabilization of the resulting carbocations, making them 162 times more reactive than standard tertiary carbocations.

Reactions at the Allylic Position Example 1

In the context of nucleophilic substitution reactions, particularly under S_N1 conditions, the reactivity of alkyl halides is influenced by their structure. To determine the relative reaction rates of various alkyl halides, it is essential to classify them based on their carbon connectivity and the presence of resonance stabilization.

First, consider the classification of the alkyl halides in question. The first compound features a carbon atom bonded to a bromine atom that is also adjacent to a double-bonded carbon, making it a tertiary allylic halide. This structure is particularly reactive due to its ability to stabilize the resulting carbocation through resonance, as well as its tertiary nature, which provides additional stability.

The second compound is a secondary allylic halide, where the carbon bonded to bromine is secondary and also next to a double bond. While it is less substituted than the tertiary allylic halide, it still benefits from resonance stabilization, making it more reactive than a regular secondary alkyl halide.

The third compound is a regular secondary alkyl halide, which does not have the resonance stabilization that the allylic halides possess. Finally, the fourth compound is a tertiary alkyl halide, which, while stable due to its tertiary nature, lacks the resonance stabilization that the allylic structures provide.

In terms of reactivity, the order from fastest to slowest reaction rates under S_N1 conditions is as follows: the tertiary allylic halide (1) is the fastest due to resonance and substitution, followed by the secondary allylic halide (2), then the tertiary alkyl halide (4), and finally the regular secondary alkyl halide (3). Thus, the correct arrangement of these alkyl halides in decreasing order of reaction rates is 1, 2, 4, and 3.

This analysis aligns with the understanding that resonance stabilization significantly enhances the reactivity of certain alkyl halides in nucleophilic substitution reactions, particularly under S_N1 conditions.

Which of the following alcohols will undergo acid-catalyzed dehydration at the fastest rate?

• A.
• B.
• C.
• D.

Answer: (C)

What are the substitution products of the following reaction?

• A.
• B.
• C.
• D.

Answer: (D)

Reactions at the Allylic Position Concept 3

In organic chemistry, the acidity of protons in allylic positions is significantly greater than that of protons in alkyl groups due to the stability of the resulting conjugate bases. Allylic protons typically have a pKa of around 44, while alkyl protons have a much higher pKa of approximately 50, indicating that they are much less acidic. This difference arises because when an allylic proton is removed by a base, the resulting allylic carbanion can stabilize itself through resonance. The negative charge on the carbon can delocalize, allowing for the formation of a double bond and enhancing stability.

For example, when a strong base interacts with an alkyl halide, it may remove a proton from an allylic position, leading to the formation of a double bond and the expulsion of a leaving group, such as bromide (Br^-). This reaction pathway, known as E2 elimination, is favored in the presence of allylic protons due to their increased acidity and the resonance stabilization of the resulting alkene product. The alkene formed is further stabilized by resonance, creating a conjugated system that enhances its stability.

In contrast, when dealing with a primary alkyl halide, the reaction may proceed via an SN2 mechanism instead of E2, as the steric hindrance prevents the elimination pathway. In this case, the strong base acts as a nucleophile, attacking the carbon and displacing the leaving group without forming a double bond.

Overall, the enhanced acidity of allylic protons and the resonance stabilization of the resulting products make E2 eliminations more favorable in reactions involving allylic positions compared to those involving alkyl protons. This understanding is crucial for predicting the outcomes of reactions in organic synthesis.

Reactions at the Allylic Position Example 2

In the Fong reaction, we start with a secondary alkyl halide, where the halogen is attached to the alpha carbon, and adjacent to it are the beta carbons. In this case, the beta carbons each have two hydrogens. The reaction utilizes ethoxide ion as a strong base in an ethanol solvent, which is crucial for the elimination mechanism.

Given the structure, the elimination process follows the E2 mechanism due to the presence of a secondary alkyl halide and a strong base. The key step involves the abstraction of a hydrogen atom from the allylic position, which is more acidic (with a pKa around 44) compared to the typical alkyl hydrogens (with a pKa around 50). The ethoxide ion deprotonates one of these allylic hydrogens, leading to the formation of a double bond as the bond between the alpha carbon and the halogen (chlorine in this case) breaks, resulting in the expulsion of the halogen.

The major elimination product is an alkene, characterized by the formation of a double bond between the alpha carbon and the beta carbon from which the hydrogen was removed. The final structure retains the double bond and the two methyl groups branching off the remaining carbon atoms. This highlights the importance of recognizing the acidity of the protons involved and the role of the strong base in facilitating the E2 elimination process.

In summary, the mechanism involves the removal of an allylic hydrogen, leading to the formation of a double bond and the generation of the major elimination product, an alkene, through the E2 pathway.

Which of the following compounds will undergo E2 elimination at the fastest rate?

• A. I
• B. II
• C. Both will proceed at equal rates.

Answer: (B)

Reactions at the Allylic Position Concept 4

In the study of radical reactions involving allylic compounds, it is essential to understand the mechanisms of halogen addition to double bonds. Initially, halogens such as bromine or chlorine react with alkenes to form a bridged ion intermediate, known as a bromonium ion or chloronium ion, respectively. This process begins with the alkene reacting with a dihalogen (X₂), resulting in a dark solution. The bridged ion intermediate is formed when one halogen atom attaches to the double bond, creating a three-membered ring structure. Subsequently, the second halogen atom attacks one of the bridge carbons, leading to the formation of vicinal dihalides, where the two halogens are added in an anti or trans configuration to neighboring carbons.

When a radical initiator, such as heat, UV light, or peroxides, is introduced, the reaction pathway shifts from simple halogenation to allylic halogenation. In this scenario, the alkene reacts with X₂ or N-bromosuccinimide (NBS) under radical conditions. The presence of these initiators allows for the substitution of an allylic proton with a halogen atom. For instance, in the case of NBS, the allylic proton is replaced by a bromine atom, resulting in the formation of an allylic halide. The remaining halogen combines with the liberated proton to yield HX as a byproduct.

It is crucial to differentiate between these two types of reactions: halogenation of alkenes occurs without radical initiators, while allylic halogenation involves the replacement of allylic protons in the presence of radical initiators. Understanding these mechanisms is vital for predicting the outcomes of reactions involving allylic compounds and their derivatives.

Reactions at the Allylic Position Example 3

In the context of allylic bromination, the reaction involving an alkene and N-bromosuccinimide (NBS) in carbon tetrachloride serves as a classic example of how to generate allylic radicals. The process begins with the abstraction of an allylic proton, which can occur at two positions adjacent to the double bond, leading to the formation of two possible allylic radicals. These radicals are resonance-stabilized, meaning they can delocalize their unpaired electron across the structure, resulting in multiple resonance forms.

For the given alkene, the two allylic positions can generate four distinct resonance-stabilized radicals. Each of these radicals can then react with bromine, leading to different brominated products. The key to determining the major product lies in the stability of the resulting alkene. The most substituted alkene, which is typically the most stable due to hyperconjugation and the inductive effect, will be favored in this reaction.

In this case, the major product is identified as the trisubstituted alkene, where the double bond is flanked by three carbon groups rather than hydrogen atoms. This structure is more stable than the others, making it the predominant product of the reaction. When drawing the final structures, it is important to indicate the stereochemistry of the bromine attachment, as the positions can lead to chiral centers. However, the focus remains on the stability of the product, which is the most substituted alkene formed from the reaction.

In summary, when working with NBS for allylic bromination, remember to consider the formation of allylic radicals, their resonance stabilization, and the stability of the resulting alkenes to predict the major product accurately. The major product will always be the one that results in the most substituted alkene, reflecting the principles of stability in organic chemistry.

Draw the structures of the compounds A and B for the following synthesis.

In this synthesis problem, we begin with compound A, which contains a double bond and an allylic hydrogen. The reaction involves N-bromosuccinimide (NBS) in carbon tetrachloride, which facilitates the substitution of the allylic hydrogen with a bromine atom. This step is crucial as it highlights the role of NBS in bromination, specifically targeting the allylic position due to the presence of the double bond.

After the formation of compound B, which now includes the bromine substitution, the next step is an elimination reaction, specifically an E2 mechanism. This reaction is driven by a strong base, the ethoxide ion, in an ethanol solvent, and is further promoted by the application of heat. The heat indicates that elimination is favored, leading to the formation of a dialkene product.

To visualize the structures, compound A can be represented with a double bond and a methyl group branching off a ring. Upon reacting with NBS, the bromine replaces the allylic hydrogen, resulting in compound B. The E2 elimination then occurs at the beta position, where the base removes a hydrogen atom, allowing the formation of a new double bond while expelling the bromine atom. This process follows Zaitsev's rule, as the elimination occurs at the less substituted carbon, leading to the final product with two double bonds.

In summary, the synthesis involves the bromination of an allylic hydrogen in compound A to form compound B, followed by an E2 elimination reaction that results in a dialkene product. Understanding the mechanisms and the role of reagents like NBS and ethoxide is essential for predicting the outcomes of such reactions.

Reactions at the Allylic Position Concept 5

Allylic anions are a significant topic in organic chemistry, particularly due to their enhanced stability compared to alkyl anions, primarily attributed to resonance effects. In an allylic anion, the negative charge on the carbon can delocalize through resonance, allowing for the formation of multiple resonance structures. This delocalization stabilizes the anion, making it more favorable for reactions.

When considering allylic anions formed from Grignard reagents, two isomeric anions can exist in equilibrium. For instance, if we analyze two Grignard reagents, one can be a trisubstituted alkene while the other is disubstituted. The stability of these alkenes is crucial, as it follows Zaitsev's rule, which states that the more substituted alkene is generally more stable. In this context, the trisubstituted alkene (with three groups attached to the double-bonded carbons) is favored over the disubstituted alkene (with only two groups). This preference is illustrated by a major arrow indicating the predominant formation of the trisubstituted product and a minor arrow for the disubstituted product.

When these isomers react with an electrophile, they yield distinct products. The Grignard reagent, upon interacting with an alkyl halide, facilitates the nucleophilic attack where the carbon atom adjacent to the magnesium halide bond (MgX) undergoes substitution. This results in the attachment of an R group to the carbon that was previously bonded to the halide. Thus, the outcome of the reaction is influenced by the stability of the alkene formed during the rearrangement process.

In summary, understanding allylic anion rearrangements involves recognizing the importance of resonance stabilization, the implications of Zaitsev's rule on alkene stability, and the role of Grignard reagents in facilitating nucleophilic substitutions. This knowledge is essential for predicting the products of reactions involving allylic anions and their corresponding electrophiles.

Reactions at the Allylic Position Example 4

In this reaction, a Grignard reagent interacts with formaldehyde, which is the simplest aldehyde. A Grignard reagent is characterized as an ionic species, where the carbon atom carries a negative charge due to its higher electronegativity compared to magnesium. This results in a structure where the carbon is negatively charged, while the magnesium-bromide component is positively charged.

Allylic anions, which are formed during this reaction, are resonance-stabilized. This means that the structure of the allylic anion can be represented in two ways, showcasing its ability to resonate. The resonance forms involve the movement of electrons, leading to the formation of a double bond and the shifting of bonds within the molecule.

When the Grignard reagent reacts with formaldehyde, the negatively charged carbon from the Grignard reagent attacks the carbonyl carbon of formaldehyde. This nucleophilic addition results in the breaking of the double bond between the carbon and oxygen, leading to the formation of an alkoxide intermediate. The resulting structure features a carbon connected to a negatively charged oxygen atom (–O⁻), which will eventually be protonated to yield a neutral alcohol.

The final products of this reaction are two distinct alcohols, which can be represented as R–CH₂–OH. Among these products, one is typically more stable due to being more substituted, making it the major product. This stability arises from the nature of alkenes, where more substituted alkenes are generally favored due to their lower energy states.

In summary, when working with Grignard reagents and formaldehyde, it is essential to recognize the ionic nature of the Grignard reagent, illustrate the resonance forms of the allylic anion, and perform the nucleophilic addition to derive the final alcohol products. Understanding these concepts is crucial for predicting the outcomes of reactions involving Grignard reagents.

Reactions at the Allylic Position Concept 6

Allylic oxidation is a selective oxidation process that targets the hydroxyl (OH) group on an allylic carbon, while preserving the integrity of the alkene double bond. This reaction typically employs manganese(IV) oxide as the oxidizing agent, dissolved in methylene chloride as the solvent. The primary outcome of this reaction is the conversion of an allylic alcohol into an allylic carbonyl compound, specifically a ketone.

In this transformation, the hydroxyl group on the allylic carbon is oxidized to form a carbonyl group, effectively changing the functional group without affecting the adjacent double bond. This selective oxidation is crucial when the goal is to modify only the allylic alcohol, distinguishing it from other types of alcohol oxidation that may involve different oxidizing agents.

To summarize, when performing allylic oxidation, the use of manganese(IV) oxide in methylene chloride is essential for achieving the desired conversion from an allylic alcohol to an allylic carbonyl, specifically a ketone, while maintaining the alkene structure intact.

Reactions at the Allylic Position Example 5

In the context of organic chemistry, the oxidation reaction involving manganese(IV) oxide in a methylene chloride solvent is a specific process known as allylic oxidation. This reaction selectively targets allylic alcohols, which are alcohols where the hydroxyl group (-OH) is attached to a carbon adjacent to a double bond. In this scenario, the double-bonded carbons are referred to as vinylic carbons, while the carbon connected to them is the allylic carbon.

During the reaction, the allylic alcohol undergoes oxidation, resulting in the transformation of the hydroxyl group into a carbonyl group (C=O). It is important to note that only the allylic alcohol is affected by this oxidation; any hydroxyl groups that are not in the allylic position remain unchanged. Therefore, in the final product, the allylic alcohol is converted to a carbonyl compound, while the other hydroxyl group remains intact. Additionally, the alkene double bond is preserved throughout the reaction.

To summarize, the key outcome of this allylic oxidation is the selective conversion of the allylic hydroxyl group to a carbonyl group, while maintaining the integrity of other functional groups and the alkene double bond in the molecule.

Draw the structures of products A and B for the given reactions.

• A.
• B.
• C.
• D.

Answer: (D)

Here’s what students ask on this topic:

What is the allylic position in organic chemistry?

The allylic position in organic chemistry refers to the carbon atom adjacent to a double bond. This position is significant because it exhibits unique reactivity due to the stability provided by resonance. For example, in a molecule with a double bond between two carbons (C=C), the carbon atoms directly attached to these double-bonded carbons are in the allylic position. This position is more reactive in certain reactions, such as SN1, E1, and radical reactions, compared to non-allylic positions.

Why are allylic carbocations more stable than alkyl carbocations?

Allylic carbocations are more stable than alkyl carbocations due to resonance stabilization. In an allylic carbocation, the positive charge can be delocalized over multiple atoms through resonance structures. This delocalization distributes the charge more evenly, reducing the energy of the carbocation and increasing its stability. In contrast, alkyl carbocations are primarily stabilized by hyperconjugation, which is less effective than resonance.

How do allylic anions differ from alkyl anions in terms of acidity?

Allylic anions are more acidic than alkyl anions because the conjugate base of an allylic anion is resonance-stabilized. When a proton is removed from the allylic position, the resulting negative charge can be delocalized over multiple atoms through resonance. This stabilization lowers the pKa of the allylic proton, making it more acidic compared to an alkyl proton, which lacks such resonance stabilization.

What is allylic halogenation and how does it occur?

Allylic halogenation is a reaction where a halogen replaces an allylic hydrogen atom. This process is facilitated by radical initiators such as heat, UV light, or peroxides. For example, using N-bromosuccinimide (NBS) in the presence of UV light can selectively brominate the allylic position. The reaction involves the formation of a radical intermediate, which allows the halogen to replace the allylic hydrogen, resulting in an allylic halide.

What is the role of manganese(IV) oxide in allylic oxidation?

Manganese(IV) oxide (MnO₂) is used as an oxidizing agent in allylic oxidation to selectively oxidize allylic alcohols to carbonyl compounds. This reaction occurs without affecting the alkene double bond. For instance, an allylic alcohol can be converted to an allylic ketone or aldehyde using MnO₂ in a methylene chloride solvent. This selective oxidation is useful in organic synthesis for transforming specific functional groups while preserving others.