Using alkenes and any other necessary inorganic reagents, write reactions showing the formation of (a) \(\mathrm{CH}_{3} \mathrm{CHBrCH}_{2} \mathrm{Br}\) (b) \(\mathrm{CH}_{3} \mathrm{CHBrCH}_{3}\)

In organic chemistry, vicinal dibromide synthesis is an essential reaction that illustrates the addition of bromine atoms to adjacent carbon atoms in an alkene. The term 'vicinal' refers to neighboring atoms. To understand how we derive compounds like \(\mathrm{CH_3CHBrCH_2Br}\), let's consider an alkene such as \(\mathrm{CH_2=CHCH_3}\). When this alkene reacts with molecular bromine (\(\mathrm{Br_2}\)), each carbon atom of the double bond forms a new bond with a bromine atom, thus transforming the double bond into a vicinal dibromide.

This reaction is stereospecific, resulting in trans-addition where the bromine atoms are added on opposite sides of the former double bond, maintaining the maximum distance between them to minimize steric strain. The product, being an addition across the double bond, no longer contains a double bond, resulting in a more saturated, dihalogenated molecule.

A hydrohalogenation reaction involves the addition of hydrohalic acids, such as hydrobromic acid (\(\mathrm{HBr}\)), to alkenes. To form compounds like \(\mathrm{CH_3CHBrCH_3}\), we begin with an alkene such as \(\mathrm{CH_3CH=CH_2}\). The addition of \(\mathrm{HBr}\) to this molecule will lead to the attachment of hydrogen (\(\mathrm{H}\)) to one carbon of the double bond and bromine (\(\mathrm{Br}\)) to the other.

During the reaction, the component atoms of the hydrohalic acid are not added randomly; rather, the reaction follows a specific rule, known as Markovnikov's rule, which dictates the placement of the hydrohalic acid's hydrogen and halogen atoms. This rule and its implications are so significant that they merit their own section of discussion.

Markovnikov's rule is a guiding principle in organic chemistry, which predicts the outcome of certain types of addition reactions, particularly hydrohalogenation. According to this rule, when a hydrogen halide like \(\mathrm{HBr}\) is added to an unsymmetrical alkene, the hydrogen atom (\(\mathrm{H}\)) will bond with the less substituted carbon atom—the carbon with more hydrogen atoms already attached—while the halide (\(\mathrm{Br}\)) will bond with the more substituted carbon atom.

This rule is a result of the stability of the intermediate carbocation that forms during the reaction; more substituted carbocations are more stable due to hyperconjugation and inductive effects. Therefore, in the formation of \(\mathrm{CH_3CHBrCH_3}\), we see that the bromine atom bonds with the carbon atom in the middle of the compound, which is the more substituted than the end carbon atoms.

When dealing with a reaction that involves a symmetrical double bond addition, the molecule in question has a double bond with identical groups on both sides. This symmetry allows for a particular type of addition where the added groups can attach to either side of the double bond without creating different products. For instance, when bromine (\(\mathrm{Br_2}\)) is added to a symmetrical alkene such as ethylene (\(\mathrm{CH_2=CH_2}\)), it results in 1,2-dibromoethane regardless of which carbon the bromine atoms attach to because both carbons are equivalent.

In our exercise, the starting alkene, \(\mathrm{CH_2=CHCH_3}\), is symmetrical in terms of its reaction with \(\mathrm{Br_2}\) despite having different groups attached to the carbons. This stability allows for the creation of \(\mathrm{CH_3CHBrCH_2Br}\) through a straightforward reaction where either carbon of the double bond can form a bond with a bromine atom, leading to the same product without concern for regioselectivity as seen in unsymmetrical alkenes.