Chapter 17, Problem 17.67P

Interpretation Introduction

(a)

Interpretation:

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

Concept introduction:

Phosphorus tribromide (${\text{PBr}}_{\text{3}}$) can convert primary and secondary alcohols to alkyl halides via back-to-back ${\text{S}}_{\text{N}}\text{2}$ reactions. A Wittig reagent, also called a phosphonium ylide, is highly nucleophilic at the ${\text{C}}^{\text{-}}$ site. To generate a Wittig reagent, the alkyl halide is first treated with $\text{triphenylphosphine}$, $\text{P}{\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\right)}_{\text{3}}$, and the product of that reaction is treated with a very strong base such as $\text{butyllithium}$. Step 1 is an ${\text{S}}_{\text{N}}\text{2}$ reaction in which the nucleophile is $\text{P}{\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\right)}_{\text{3}}$ and the leaving group is the halide anion. The organic product possesses a positive charge on the P atom. Although $\text{P}{\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\right)}_{\text{3}}$ is uncharged, it is a good nucleophile because the P atom in the product can accommodate the positive charge rather well, due to both its large size and its modest electronegativity. Step 2 is deprotonation by the alkyllithium species. A very strong base such as $\text{butyllithium}$ is necessary because deprotonation occurs at a carbon atom. In a Wittig reaction, the $\text{C = O}$ bond of a ketone or aldehyde is converted to a $\text{C = C}$ bond. In Step 1, the highly nucleophilic C atom of the Wittig reagent attacks the electrophilic carbonyl C atom, yielding a betaine. Step 2 is a coordination step, in which a bond is formed between the negatively charged O atom and the positively charged P atom. This results in an oxaphosphetane that contains a four-membered ring, which, due to the strain, falls apart into the alkene and triphenylphosphine oxide in Step 3.

Answer to Problem 17.67P

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

Explanation of Solution

The given starting material and reagents are as shown below:

The reagent used for the first step of this reaction is Phosphorus tribromide (${\text{PBr}}_{\text{3}}$) which converts primary and secondary alcohols to alkyl halides via back-to-back ${\text{S}}_{\text{N}}\text{2}$ reactions. In Step 1, the O atom of the alcohol is electron rich and acts as the nucleophile, whereas the P atom of ${\text{PBr}}_{\text{3}}$ is electron poor and acts as the substrate. A ${\text{Br}}^{\text{-}}$ anion is the leaving group in this step, and is therefore liberated. In Step 2, the ${\text{Br}}^{\text{-}}$ anion generated in Step 1 acts as the nucleophile and the phosphorus-containing species acts as the substrate. The leaving group is ${\text{HOPBr}}_{\text{2}}$.

The above formed alkyl bromide further undergoes an ${\text{S}}_{\text{N}}\text{2}$ reaction in which the nucleophile is $\text{P}{\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\right)}_{\text{3}}$ and the leaving group is the bromide anion. The product of this reaction undergoes deprotonation by butyllithium species to form a Wittig reagent.

The above Wittig reagent undergoes Wittig reaction with acetone. In a Wittig reaction, the $\text{C = O}$ bond of acetone is converted to a $\text{C = C}$ bond. In Step 1, the highly nucleophilic C atom of the Wittig reagent attacks the electrophilic carbonyl C atom, yielding a betaine. Step 2 is a coordination step, in which a bond is formed between the negatively charged O atom and the positively charged P atom. This results in an oxaphosphetane that contains a four-membered ring, which, due to the strain, falls apart into the alkene and triphenylphosphine oxide in Step 3.

The overall mechanism and the major product of the given reaction is as shown below:

Conclusion

The product of the given reaction and complete mechanism is drawn on the basis of the given reagents.

Interpretation Introduction

(b)

Interpretation:

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

Concept introduction:

The organometallic compounds such as Grignard reagents (RMgX) and alkyllithium reagents behave both as strong bases and as strong nucleophiles. Both alkyllithium reagents and Grignard reagents can react rapidly with water in a substantially exothermic proton transfer reaction to produce an alkane and ${\text{HO}}^{\text{-}}$ i.e., alcohol. The Williamson ether synthesis can be used to synthesize either symmetric or unsymmetric ethers via an ${\text{S}}_{\text{N}}\text{2}$ reaction between an alkoxide anion and an alkyl halide (RX). The Williamson synthesis takes place under basic conditions.

Answer to Problem 17.67P

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

Explanation of Solution

The given starting material and reagents are as shown below:

In the first reaction, a new C-C bond is formed and a phenyl group is added to the carbonyl carbon. Grignard reagents can react rapidly with ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ in a substantially exothermic proton transfer reaction to produce an alkane and ${\text{HO}}^{\text{-}}$. Here, the Grignard reagent is phenylmagnesium bromide (${\text{C}}_6{\text{H}}_5\text{MgBr}$). An acid workup is done by ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ to protonate strongly basic ${\text{O}}^{\text{-}}$ generated in the first step to get uncharged alcohol.

Further, this alcohol undergoes deprotonation by NaH to form negatively charged alkoxide. This alkoxide acts as a strong nucleophile and attacks ethyl bromide via ${\text{S}}_{\text{N}}\text{2}$ mechanism to form the ether compound. This is Williamson ether synthesis.

The overall mechanism and the major product of the given reaction are as shown below:

Conclusion

The product of the given reaction and complete mechanism is drawn on the basis of the mechanism for Grignard reagent and Williamson ether synthesis.