Question

HBr reacts with \(\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{OCH}_{3}\) under anhydrous conditions at room temperature to give: (a) \(\mathrm{CH}_{3} \mathrm{CHO}\) and \(\mathrm{CH}_{3} \mathrm{Br}\) (b) \(\mathrm{BrCH}_{2} \mathrm{CHO}\) and \(\mathrm{CH}_{3} \mathrm{OH}\) (c) \(\mathrm{BrCH}_{2}^{2}-\mathrm{CH}_{2}-\mathrm{OCH}_{3}\) (d) \(\mathrm{H}_{3} \mathrm{C}-\mathrm{CHBr}-\mathrm{OCH}_{3}\)

Short answer

The correct product is (d) \(\text{H}_3\text{C}-\text{CHBr}-\text{OCH}_3\).

Step-by-step solution

Identify the Reaction Type

The reaction involves the addition of HBr to an alkene, \(\text{CH}_2=\text{CH}-\text{OCH}_3\). This is an example of an electrophilic addition reaction.

Determine Markovnikov's Rule

According to Markovnikov's Rule, during the addition of HX (where X is a halogen like Br) to an alkene, the hydrogen (H) will add to the carbon with more hydrogen atoms already attached, while the halogen will add to the carbon with fewer hydrogen atoms.

Apply Markovnikov's Rule

In \(\text{CH}_2=\text{CH-OCH}_3\), the H from HBr will add to the terminal \(\text{CH}_2\) group because it has more hydrogen atoms, and \(\text{Br}\) will add to the \(\text{CH}\) connected to \(\text{OCH}_3\).

Determine the Product Structure

The addition of HBr to \(\text{CH}_2=\text{CH-OCH}_3\) results in the following structure: \(\text{CH}_3-\text{CHBr}-\text{OCH}_3\), where \(\text{Br}\) is attached to the \(\text{CH}\) with the ether (\(\text{OCH}_3\)).

Identify the Correct Product Option

The correct product matches option (d): \(\text{H}_3\text{C}-\text{CHBr}-\text{OCH}_3\).

Key concepts

Markovnikov's Rule

In organic chemistry, understanding how a molecule reacts often comes down to knowing where atoms prefer to add in a reaction. Markovnikov's Rule is a key concept that helps predict this behavior in electrophilic addition reactions, particularly with alkenes. According to Markovnikov's Rule, when a hydrohalic acid, such as HBr, approaches an alkene, the hydrogen atom attaches to the carbon with more hydrogens, and the halogen attaches to the carbon with fewer hydrogens.

This can be simplified into the idea that 'the rich get richer,' meaning the carbon atom already richer in hydrogen gains another one. Applying this rule helps ensure accurate predictions of the major product formed during these reactions. This rule was employed in the exercise to deduce that Br would attach itself to the carbon linked with the \(-\text{OCH}_3\) group, while H would bond with the terminal \(\text{CH}_2\).

Alkene Reaction

Alkenes are hydrocarbons that contain carbon-carbon double bonds, represented as \(\text{C=C}\). These structures serve as common active sites in organic molecules because of their high reactivity compared to single bonds. The double bond is made up of one sigma and one pi bond, with the pi bond being susceptible to attack by electrophiles.

In the context of electrophilic addition reactions, such as the reaction of \(\text{CH}_2=\text{CH}-\text{OCH}_3\) with HBr, the \(\text{C=C}\) bond acts as an attractant for the incoming electrophile, which is the hydrogen ion (H) in this case.

• The hydrogen approaches the double bond, attacking the area of higher electron density provided by the pi electrons.
• This leads to the formation of a carbocation intermediate, facilitating subsequent addition steps in the mechanism.

By understanding these facets, students can better visualize the steps leading to product formation.

Organic Chemistry Reaction Mechanism

Organic chemistry revolves around understanding reaction mechanisms, which describe the step-by-step transformations that lead from reactants to products. For an electrophilic addition mechanism, such as with HBr and an alkene, several key steps occur:

• Formation of carbocation: The pi electrons of the alkene attack the hydrogen of HBr, leading to the cleavage of the H-Br bond and forming a transient carbocation.
• Halide Ion Attack: The bromide ion (Br-), a strong nucleophile, quickly attacks the positively charged carbon of the carbocation, forming a stable alkyl halide product.

Envisioning the mechanistic path gives insight into how and why the reaction proceeds in a particular manner. Mastery of such mechanisms helps in predicting the outcome of similar organic reactions, enhancing both theoretical and practical problem-solving abilities. Through practice, these concepts are naturally integrated into the chemistry toolkit of a student.