Chapter 12, Problem 12.37P

Interpretation Introduction

(a)

Interpretation:

The complete, detailed mechanism and the major organic product for the given reaction are to be drawn.

Concept introduction:

When an alkene reacts with molecular bromine $\left({\text{Br}}_{\text{2}}\right)$ in carbon tetrachloride ${\text{(CCl}}_{\text{4}}\text{)}$, a mixture of dihaloalkane is produced. Molecular bromine undergoes anti addition across a carbon-carbon double bond. To account for the stereochemistry of the reaction, the mechanism must proceed through a halonium ion intermediate, hence anti addition takes place. Even though ${\text{CCl}}_{\text{4}}$ is much more abundant than molecular bromine, ${\text{CCl}}_{\text{4}}$ is a non-nucleophilic solvent.

Answer to Problem 12.37P

The complete, detailed mechanism and the major organic product for the given reaction are shown below:

Explanation of Solution

The starting material for the given reaction is

When alkene is treated with molecular bromine $\left({\text{Br}}_{\text{2}}\right)$ in $\text{tetrachloromethane}$ (${\text{CCl}}_{\text{4}}$), also called carbon tetrachloride, a racemic mixture of alkene is produced. Anti-addition of ${\text{Br}}_{\text{2}}$ takes place. The mechanism proceeds through a bromonium ion intermediate, which is produced in Step 1. The curved arrow from the $\text{π}$ bond to Br represents the flow of electrons from an electron-rich site to an electron-poor site. Also, a lone pair of electrons on Br forms a bond back to one of the alkene C atoms to avoid breaking the C atom’s octet resulting in a three-membered ring. In this case, the ring consists of two C atoms and one Br atom. Simultaneously, the weak $\text{Br-Br}$ bond breaks and one of the atoms leaves as ${\text{Br}}^{\text{-}}$.

In step 2, the ${\text{Br}}^{\text{-}}$ ion produced in Step 1 acts as the nucleophile, and the positively charged Br atom in the ring becomes the leaving group. The nucleophile in an ${\text{S}}_{\text{N}}\text{2}$ reaction must attack from the side opposite to the leaving group. This results in the formation of a mixture of enantiomers. The complete, detailed mechanism and the major organic product for the given reaction are shown below:

Conclusion

The complete, detailed mechanism and the major organic product for the given reaction is drawn by electrophilic addition of ${\text{Br}}_{\text{2}}$ molecule to $\text{C=C}$ bond followed by ${\text{S}}_{\text{N}}\text{2}$ reaction.

Interpretation Introduction

(b)

Interpretation:

The complete, detailed mechanism and the major organic product for the given reaction is to be drawn.

Concept introduction:

Molecular halogen undergoes anti addition across a $\text{C=C}$ double bond. A halohydrin is produced when an alkene is treated with molecular halogen in water. In the formation of a halohydrin from an alkene, the OH group and the halogen atom gets added anti to each other. In the first step, halogen molecule adds to an alkene. It is electrophilic addition that produces the halonium ion intermediate. The specific isomer that is formed is dictated by Step 2 of the mechanism— namely, nucleophilic attack by water. ${\text{H}}_{\text{2}}\text{O}$ attacks the more substituted carbon. In Step 3, a proton transfer produces the uncharged halohydrin. Regiochemistry becomes an issue with halohydrin formation if the alkene reactant is unsymmetrical.

Answer to Problem 12.37P

The complete, detailed mechanism and the major product for the given reaction is as shown below:

Explanation of Solution

The starting material for given reaction is

In the first step, ${\text{Br}}_{\text{2}}$ adds to the $\text{C=C}$ bond of the given alkene. It is electrophilic addition that produces the bromonium ion intermediate. The specific isomer that is formed is dictated by Step 2 of the mechanism— namely, nucleophilic attack by water. In Step 3, a proton transfer produces uncharged bromohydrin. Water attacks the more substituted carbon. With distinct $\text{C=C}$ atoms, two possible constitutional isomers can be produced, more substituted carbon ends up bonded to the OH group, and the other carbon atom ends up bonded to the Br atom.

The complete, detailed mechanism and the major product for the given reaction is as shown below:

Conclusion

The detailed mechanism for the given reaction is described by explaining the appropriate stereochemistry with the formation of bromonium ion intermediate.

Interpretation Introduction

(c)

Interpretation:

The complete, detailed mechanism and the major organic product for the given reaction is to be drawn.

Concept introduction:

When an alkene reacts with molecular chlorine $\left({\text{Cl}}_{\text{2}}\right)$ in carbon tetrachloride ${\text{(CCl}}_{\text{4}}\text{)}$, a mixture of dihaloalkane is produced. Molecular chlorine undergoes anti addition across a carbon-carbon double bond. To account for the stereochemistry of the reaction, the mechanism must proceed through a halonium ion intermediate, hence anti addition takes place. Even though ${\text{CCl}}_{\text{4}}$ is much more abundant than molecular chlorine, ${\text{CCl}}_{\text{4}}$ is a non-nucleophilic solvent. An excess ${\text{Cl}}_{\text{2}}$ yields a tetrachloride compound.

Answer to Problem 12.37P

The complete, detailed mechanism and the major organic product for the given reaction are shown below:

Explanation of Solution

The starting material for given reaction is

When alkene is treated with molecular chlorine in $\text{tetrachloromethane}$ (${\text{CCl}}_{\text{4}}$), also called carbon tetrachloride, a racemic mixture of alkene is produced. Anti-addition of ${\text{Cl}}_{\text{2}}$ takes place. The mechanism proceeds through a chloronium ion intermediate, which is produced in Step 1. The curved arrow from the $\text{π}$ bond to Cl represents the flow of electrons from an electron-rich site to an electron-poor site. Also, a lone pair of electrons on Cl forms a bond back to one of the alkene C atoms to avoid breaking the C atom’s octet, resulting in a three-membered ring. In this case, the ring consists of two C atoms and one Cl atom. Simultaneously, the weak $\text{Cl-Cl}$ bond breaks, and one of the atoms leaves as ${\text{Cl}}^{\text{-}}$.

In step 2, the ${\text{Cl}}^{\text{-}}$ ion produced in Step 1 acts as the nucleophile, and the positively charged Cl atom in the ring becomes the leaving group. The nucleophile in an ${\text{S}}_{\text{N}}\text{2}$ reaction must attack from the side opposite to the leaving group. This results in the formation of a mixture of enantiomers. An excess addition of ${\text{Cl}}_{\text{2}}$ is followed in the same way as one equivalent ${\text{Cl}}_{\text{2}}$ which yields a tetrachloride compound. The mixture of configurations is produced at each of the two new stereocenters. This gives three stereoisomers. The complete, detailed mechanism and the major organic product for the given reaction are shown below:

Conclusion

The complete, detailed mechanism and the major organic product for the given reaction are drawn by electrophilic addition of ${\text{Cl}}_{\text{2}}$ molecule to $\text{C=C}$ bond followed by ${\text{S}}_{\text{N}}\text{2}$ reaction.

Interpretation Introduction

(d)

Interpretation:

The complete, detailed mechanism and the major organic product for the given reaction is to be drawn.

Concept introduction:

When an alkyne reacts with molecular bromine $\left({\text{Br}}_{\text{2}}\right)$ in carbon tetrachloride ${\text{(CCl}}_{\text{4}}\text{)}$, a mixture of both E and Z isomers of dihaloalkene is produced. ${\text{Br}}_{\text{2}}$ adds to an alkyne via both syn and anti addition. Even though ${\text{CCl}}_{\text{4}}$ is much more abundant than molecular bromine, ${\text{CCl}}_{\text{4}}$ is a non-nucleophilic solvent. An excess ${\text{Br}}_{\text{2}}$ yields a tetrabrominated compound.

Answer to Problem 12.37P

The complete, detailed mechanism and the major organic product for the given reaction are shown below:

Explanation of Solution

The starting material for the given reaction is

When alkyne is treated with molecular bromine $\left({\text{Br}}_{\text{2}}\right)$ in $\text{tetrachloromethane}$ (${\text{CCl}}_{\text{4}}$), also called carbon tetrachloride, both E and Z isomers of alkene are produced. Syn and anti addition of ${\text{Br}}_{\text{2}}$ takes place. The mechanism proceeds through a bromonium ion intermediate which is produced in Step 1. The curved arrow from the $\text{C}\equiv \text{C}$ bond to Br represents the flow of electrons from an electron-rich site to an electron-poor site. Also, a lone pair of electrons on Br forms a bond back to one of the alkene C atoms to avoid breaking the C atom’s octet resulting in a three-membered ring. In this case, the ring consists of two C atoms and one Br atom. Simultaneously, the weak $\text{Br-Br}$ bond breaks, and one of the atoms leaves as ${\text{Br}}^{\text{-}}$.

In step 2, the ${\text{Br}}^{\text{-}}$ ion produced in Step 1 acts as the nucleophile, and the positively charged Br atom in the ring becomes the leaving group. The nucleophile attacks on more substituted carbon. The reaction proceeds two times to give achiral tetrabrominated alkene compound. The complete, detailed mechanism and the major organic product for the given reaction are shown below:

Conclusion

The complete, detailed mechanism and the major organic product for the given reaction are drawn by electrophilic addition of ${\text{Br}}_{\text{2}}$ molecule to $\text{C}\equiv \text{C}$ bond followed by ${\text{S}}_{\text{N}}\text{2}$ reaction.