Chapter 10, Problem 10.3P

Interpretation Introduction

(a)

Interpretation:

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

Concept introduction:

A way to convert an alcohol into an alkyl halide is to use phosphorus tribromide, ${\text{PBr}}_{\text{3}}$, or phosphorus trichloride, ${\text{PCl}}_{\text{3}}$. The reaction yields no elimination products. The conversion is stereospecific with an inversion of configuration at the alcohol carbon atom. No carbocation rearrangement occurs in the conversion.

In the first step of the reaction, the oxygen atom of an alcohol is electron-rich and acts as a nucleophile, whereas the phosphorus atom in ${\text{PBr}}_{\text{3}}$ is electron-deficient and acts as the substrate. This reaction generates bromide ion which is a good leaving group. In step 2, the halide ion generated in the previous step acts as a nucleophile and the phosphorus-containing species acts as the substrate. The leaving group is ${\text{HOPBr}}_{\text{2}}{\text{or HOPCl}}_{\text{2}}$. Thus, overall two back-to-back ${\text{S}}_{\text{N}}\text{2}$ reactions occur which results in the inversion of configuration at the chiral carbon atom. The ${\text{S}}_{\text{N}}\text{2}$ reactions often make use of a strong nucleophile (electron-rich), thus, ${\text{S}}_{\text{N}}\text{2}$ reactions occur most rapidly at the carbon atom which is ${\text{sp}}^3$ hybridized. The reactions are less likely on a ${\text{sp}}^2$ hybridized carbon atom, because the electron density on an ${\text{sp}}^2$ hybridized carbon atom is high, which would repel the electron-rich nucleophile.

Answer to Problem 10.3P

The complete and detailed mechanism for the given reaction is:

The product formed in given reaction and the overall reaction is:

Explanation of Solution

The given reaction is:

Deuterium is an isotope of hydrogen. Its chemical behavior is identical to hydrogen but serves as a label in reactions. $\text{-OH}$ is not a good leaving group. The reagent used in the reaction is ${\text{PCl}}_{\text{3}}$. Reactions of alcohols with ${\text{PCl}}_{\text{3}}$ converts them into alkyl chlorides.

As a first step, the oxygen atom of an alcohol is electron-rich and acts as a nucleophile, whereas the phosphorus atom in ${\text{PCl}}_{\text{3}}$ is electron-deficient and acts as the substrate. The lone pairs on oxygen attack the electron-deficient phosphorus atom and converts the $\text{-OH}$ into a good leaving group, ${\text{ HO}}^{+}{\text{PCl}}_{\text{2}}$. The first step is an ${\text{S}}_{\text{N}}\text{2}$ reaction and is shown as:

Note that, the stereochemical configuration of the carbon atom does not change at the end of this step.

In the second step, the chloride ion generated in the previous step acts as a nucleophile and the phosphorus-containing species acts as the substrate. The chloride ion attacks the carbon atom to which ${\text{ HO}}^{+}{\text{PCl}}_{\text{2}}$ is attached and a new bond between carbon and chlorine is formed and leaving group, ${\text{ HO}}^{+}{\text{PCl}}_{\text{2}}$, departs as a neutral ${\text{ HOPCl}}_{\text{2}}$ molecule. This is the second ${\text{S}}_{\text{N}}\text{2}$ reaction. As the nucleophilic attack is on the carbon atom, its configuration is reversed. The second step is shown as:

Note that, though bromine is a good leaving group, but, since it is attached to the ${\text{sp}}^2$ hybridized carbon atom, it will be unaffected and does not participate in the reaction.

Conclusion

When an alcohol is treated with ${\text{PCl}}_{\text{3}}$_, the alcoholic hydroxyl group ($\text{-OH}$) of a substrate is replaced by chlorine atom with inversion of configuration in an ${\text{S}}_{\text{N}}\text{2}$ reaction.

Interpretation Introduction

(b)

Interpretation:

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

Concept introduction:

A way to convert an alcohol into an alkyl halide is to use phosphorus tribromide, ${\text{PBr}}_{\text{3}}$, or phosphorus trichloride, ${\text{PCl}}_{\text{3}}$. The reaction yields no elimination products and only substitution products. The conversion is stereospecific with an inversion of configuration at the alcohol carbon atom. No carbocation rearrangement occurs in the conversion.

In the first step of the reaction, the oxygen atom of an alcohol is electron-rich and acts as a nucleophile, whereas the phosphorus atom in ${\text{PBr}}_{\text{3}}$ is electron-deficient and acts as the substrate. This reaction generates bromide ion which is a good leaving group. In step 2, the halide ion generated in the previous step acts as a nucleophile and the phosphorus-containing species acts as the substrate. The leaving group is ${\text{HOPBr}}_{\text{2}}{\text{or HOPCl}}_{\text{2}}$. Thus, overall two back-to-back ${\text{S}}_{\text{N}}\text{2}$ reactions occur which results in the inversion of configuration at the chiral carbon atom. The ${\text{S}}_{\text{N}}\text{2}$ reactions often make use of a strong nucleophile (electron-rich), thus, ${\text{S}}_{\text{N}}\text{2}$ reactions occur most rapidly at the carbon atom which is ${\text{sp}}^3$ hybridized. The reactions are less likely on a ${\text{sp}}^2$ hybridized carbon atom, because the electron density on an ${\text{sp}}^2$ hybridized carbon atom is high, which would repel the electron-rich nucleophile.

Answer to Problem 10.3P

The complete and detailed mechanism for the given reaction is:

The product formed in given reaction and the overall reaction is:

Explanation of Solution

The given reaction is:

In the substrate above, $\text{-OH}$ is not a good leaving group. The reagent used in the reaction is ${\text{PBr}}_{\text{3}}$. Reactions of alcohols with ${\text{PBr}}_{\text{3}}$ converts them into alkyl bromides.

As a first step, the oxygen atom of an alcohol is electron-rich and acts as a nucleophile, whereas the phosphorus atom in ${\text{PBr}}_{\text{3}}$ is electron-deficient and acts as the substrate. The lone pairs on oxygen attack the electron-deficient phosphorus atom and converts the $\text{-OH}$ into a good leaving group ${\text{ HO}}^{+}{\text{PBr}}_{\text{2}}$. The first step is an ${\text{S}}_{\text{N}}\text{2}$ reaction and is shown as:

Note that, the stereochemical configuration of the carbon atom does not change at the end of this step as the carbon atom is not involved in this step.

In the second step, the bromide ion generated in the previous step acts as a nucleophile and the phosphorus containing species acts as the substrate. The bromide ion attacks the carbon atom to which ${\text{ HO}}^{+}{\text{PBr}}_{\text{2}}$ is attached and a new bond between carbon and bromine is formed and leaving group, ${\text{ HO}}^{+}{\text{PBr}}_{\text{2}}$, departs as a neutral ${\text{ HOPBr}}_{\text{2}}$ molecule. This is the second ${\text{S}}_{\text{N}}\text{2}$ reaction. As the nucleophilic attack is on the carbon atom, its configuration is reversed as the nucleophile attacks the carbon from the backside of the plane. The second step is shown as:

Conclusion

When an alcohol is treated with ${\text{PBr}}_{\text{3}}$_, the alcoholic hydroxyl group ($\text{-OH}$) of a substrate is replaced by bromine atom with inversion of configuration in an ${\text{S}}_{\text{N}}\text{2}$ reaction.