Question

Alkenes are a highly reactive group of compounds as they undergo a wide variety of reactions like hydrogenation, ozonolysis, halogenation, epoxidation and halohydrin formation. Many of the reactions are regioselective or stereoselective reactions. When bromine is added to a solution of 1 -hexene in methanol, the product formed is (a) \(\mathrm{BrCH}_{2}-\mathrm{CHBr}-\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\) (b) \(\mathrm{H}_{3} \mathrm{C}-\mathrm{CHBr}-\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\) (c) a mixture of \(\mathrm{BrCH}_{2} \mathrm{CHBrCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\) and \(\mathrm{BrCH}_{2}-\mathrm{CH}\left(\mathrm{OCH}_{3}\right)-\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\) (d) \(\mathrm{H}_{3} \mathrm{COCH}_{2}-\mathrm{CH}_{2}-\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\)

Short answer

(a) $\mathrm{BrCH}_{2}-\mathrm{CHBr}-\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}$ (b) $\mathrm{H}_{3} \mathrm{C}-\mathrm{CHBr}-\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}$ (c) a mixture of $\mathrm{BrCH}_{2} \mathrm{CHBrCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}$ and $\mathrm{BrCH}_{2}-\mathrm{CH}\left(\mathrm{OCH}_{3}\right)-\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}$ (d) $\mathrm{H}_{3} \mathrm{COCH}_{2}-\mathrm{CH}_{2}-\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}$ Answer: (c)

Step-by-step solution

Identify the reaction

1-hexene is reacting with bromine (Br2) in methanol, so this reaction is a halogenation reaction. Specifically, it is an electrophilic addition reaction, where the alkene reacts with an electrophilic halogen species (in this case, bromine) to add both atoms of the halogen onto the double bond.

Reaction mechanism

The mechanism of halogenation of alkenes involves several key steps: 1. The alkene double bond acts as a nucleophile, attacking bromine and forming a bromonium ion (a three-membered cyclic intermediate) with one bromine atom still attached. 2. The solvent methanol will open the bromonium ion by acting as a nucleophile, attacking the more substituted carbon and forming an intermediate (Bridgehead Bromonium Methoxide Intermediate). 3. A second nucleophile, a bromide ion, will displace the methoxide ion to give the final product.

Identify the product

Now, let's analyze the given options to determine which corresponds to the correct product: (a) $\mathrm{BrCH}_{2}-\mathrm{CHBr}-\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}$ - This option represents a dibromoalkane, which would not occur in this reaction due to the presence of methanol to give a bromohydrin product instead. (b) $\mathrm{H}_{3} \mathrm{C}-\mathrm{CHBr}-\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}$ - This option represents a monobrominated product where bromine has been added to the less substituted carbon of the alkene. This would be the correct product if the bromide ion opened the bromonium ion, but as described in the mechanism, methanol first opens the bromonium ion to form an intermediate. (c) a mixture of $\mathrm{BrCH}_{2} \mathrm{CHBrCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}$ and $\mathrm{BrCH}_{2}-\mathrm{CH}\left(\mathrm{OCH}_{3}\right)-\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}$ - This option represents a mixture of a dibromo compound and a bromohydrin compound where the methoxy group has been added to a primary carbon. However, this is not the correct product since methanol would attack the more substituted carbon instead. (d) $\mathrm{H}_{3} \mathrm{COCH}_{2}-\mathrm{CH}_{2}-\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}$ - This option represents a product where methanol has simply replaced the double bond, which does not occur in this reaction because it involves bromine addition. The correct answer is therefore (c), as it represents the mixture of products that form after methanol first attacks the bromonium ion followed by the bromide ion displacement.

Key concepts

Halogenation

Halogenation is a significant type of reaction in organic chemistry that involves the addition of halogen atoms to a molecule. In the context of alkenes, halogenation generally refers to the addition of halogens like bromine (Br₂) or chlorine (Cl₂) to the carbon-carbon double bond.

This reaction is an example of an electrophilic addition reaction, where the alkene double bond acts as a nucleophile to attack an electrophilic halogen.

• The double bond is rich in electrons and capable of attacking the halogen molecule, which in this case is bromine.
• The mechanism starts when the alkene approaches the halogen, causing polarization of the halogen molecule and leading to the formation of a halogen ion.
• Ultimately, the halogen adds across the double bond, forming a vicinal dihalide if no other nucleophile is present.

This phase of halogenation is foundational for understanding how more complex products, such as bromohydrins, are formed.

Bromonium Ion

The bromonium ion is a crucial intermediate in the halogenation of alkenes, especially when bromine is added. When an alkene reacts with bromine, this intermediate is formed as a three-membered cyclic ion.

A bromonium ion is typically stabilized by its cyclic structure, representing a bridging bromine atom that forms covalent bonds with both carbons of the original double bond.

• In the course of forming a bromonium ion, the alkene attacks the bromine molecule, resulting in the formation of this temporary intermediate.
• Inside the bromonium ion, bromine bears a positive charge while simultaneously providing a pathway for the reaction to proceed through ring-opening or further nucleophilic attack.
• This step is essential because it sets up the stage for additional reactions to occur, such as nucleophilic attacks by solvents like methanol, which lead to subsequent products like bromohydrins.

The existence of bromonium ions is a testament to the dynamic nature of chemical bonding, significantly impacting the reactivity and outcome of halogenation reactions.

Electrophilic Addition

Electrophilic addition is a fundamental type of reaction mechanism seen in the reactivity of alkenes and other unsaturated compounds. In this process, the electron-rich double bond of an alkene reacts with an electrophile, an electron-deficient species, to form a new molecule where the double bond is converted to a single bond.

The process of electrophilic addition is central to many transformations in organic chemistry due to its versatility.

• Initially, the double bond acts as a nucleophile and attacks the electrophile, often a halogen or other activator, leading to the formation of an intermediate.
• This addition reaction can vary based on the nature of the electrophile and the conditions of the reaction, adding different groups to the molecule.
• In halogenation, it allows for the addition of halogens across a double bond, forming new, sometimes cyclic, intermediates like bromonium ions.
• The subsequent steps might include further reactions, such as nucleophilic attacks or rearrangements, making electrophilic addition a crucial interaction in synthetic pathways.

Understanding electrophilic addition is vital for predicting and controlling the reaction pathways to generate desired products.

Bromohydrin Formation

Bromohydrin formation is an outcome of the reaction between an alkene, such as 1-hexene, and bromine in a protic solvent like methanol. This reaction is notable for producing a compound that contains both a bromine atom and a hydroxyl group (alcohol group) added across the alkene double bond.

In bromohydrin formation:

• The process starts with the electrophilic addition of bromine across the alkene's double bond to form a bromonium ion intermediate.
• Following the formation of the bromonium ion, the methanol, a nucleophilic solvent, attacks the more substituted carbon of the cyclic intermediate, resulting in ring-opening.
• This attack by methanol leads to the substitution of one of the bromine atoms with a methoxy group, forming a univalent molecule known as a bromohydrin.
• The reaction ultimately produces a mixture of products due to the potential diversity in the attack sites and intermediates formed within the solution.

Bromohydrin formation showcases the interplay between different reactants and solvents in altering the outcome of alkene reactions, making it integral to complex organic synthesis.