Question

Outline a synthesis of cis-\beta-bromostyrene from cinnamic acid \((\mathrm{PhCH}=\mathrm{CHCOOH})\). Which isomer of the acid is required, cis or trans.

Short answer

To synthesize cis-β-bromostyrene from cinnamic acid, the cis isomer of cinnamic acid is required. First, perform the Hell-Volhard-Zelinsky (HVZ) reaction on the cis-cinnamic acid to introduce a bromine atom at the α-position. Then, perform an elimination reaction on the α-bromocarboxylic acid obtained in the HVZ reaction using a strong base like NaOH. This will result in the formation of the desired product, cis-β-bromostyrene.

Step-by-step solution

Hell-Volhard-Zelinsky (HVZ) reaction

First, we need to perform the HVZ reaction on the cinnamic acid to introduce a bromine atom at the α-position. In this reaction, halogenation of the carboxylic acid takes place in the presence of phosphorus tribromide (PBr3) or thionyl bromide (SBr2). The reaction proceeds via the formation of an acid bromide, an acyl halide, followed by α-bromination resulting in an α-bromoacyl halide which is then hydrolyzed back to an α-bromocarboxylic acid. The general reaction can be summarized as follows: \[ \mathrm{R-COOH} \xrightarrow {PBr_3} \mathrm{R-COBr} \xrightarrow {Br_2} \mathrm{R-C(O)CH(Br)R'} \xrightarrow {H_2O} \mathrm{R-C(O)CH(Br)R'-COOH} \] For our case, the reaction can be represented as: \[ \mathrm{PhCH=CHCOOH} \xrightarrow {PBr_3} \mathrm{PhCH=CHCOBr} \xrightarrow {Br_2} \mathrm{PhCH(Br)CHCOBr} \xrightarrow {H_2O} \mathrm{PhCH(Br)CHCOOH} \]

Analyze the isomers and stereochemistry

Now, let's determine which isomer of cinnamic acid is required, cis or trans. In the final product (cis-β-bromostyrene), the bromine atom and the phenyl group are on the same side of the alkene. Since the HVZ reaction doesn't change the stereochemistry of the double bond, the cinnamic acid isomer used must also have the phenyl group and the carboxylic acid group on the same side of the double bond. Thus, the cis isomer of cinnamic acid is required for this synthesis.

Perform elimination reaction to form cis-β-bromostyrene

Finally, we will perform an elimination reaction on the α-bromocarboxylic acid obtained in step 1 to form the desired product, cis-β-bromostyrene. The reaction involves treatment with a strong base, such as sodium or potassium hydroxide (NaOH or KOH), which will facilitate the removal of the α-bromo group and the carboxyl group, resulting in the formation of cis-β-bromostyrene and a molecule of water. The reaction can be represented as: \[ \mathrm{PhCH(Br)CHCOOH} \xrightarrow {NaOH} \mathrm{PhCH(Br)=CH_2 + H_2O} \] Now, we have successfully synthesized cis-β-bromostyrene from the cis isomer of cinnamic acid.

Key concepts

Hell-Volhard-Zelinsky Reaction

Understanding the Hell-Volhard-Zelinsky (HVZ) reaction is essential for students aiming to synthesize halogenated compounds from carboxylic acids. This chemical reaction introduces a halogen at the alpha position of a carboxylic acid. The mechanism typically includes the use of a phosphorus trihalide (like phosphorus tribromide, PBr₃) or thionyl halide as a reagent.

The HVZ reaction unfolds in three primary steps:

• Formation of an acyl halide from the carboxylic acid by reacting with PBr₃
• Alpha halogenation of the acyl halide with bromine (Br₂)
• Hydrolysis of the alpha-bromoacyl halide to form the alpha-bromocarboxylic acid

Students must grasp that the double bond's configuration remains unaffected during this reaction. Therefore, to synthesize cis-β-bromostyrene, the starting material must be cis-cinnamic acid to retain the desired stereochemistry.

Cinnamic Acid Isomerism

Discerning between different isomers of cinnamic acid is pivotal in the synthesis of targeted compounds since the isomerism influences the ultimate stereochemistry of the product. Cinnamic acid, known as phenylacrylic acid, can exist in two geometric isomers: cis (Z) and trans (E).

The cis isomer features the phenyl and carboxylic groups on the same side of the double bond, which is crucial when synthesizing cis-β-bromostyrene to maintain the positions of the substituents relative to the double bond. In contrast, the trans isomer has these groups on opposing sides, which would lead to the undesired trans product. Students must recognize the necessity of choosing the correct isomer of cinnamic acid at the beginning of the synthesis process, as this directly correlates to the formation of the correct positional isomer of \(\beta\)-bromostyrene.

Elimination Reaction

The final stage of synthesizing cis-β-bromostyrene involves an elimination reaction, which allows the transformation from an alpha-bromocarboxylic acid to an alkene. This reaction type is characterized by the removal of two atoms or groups from a molecule, resulting in the formation of a double bond.

For this synthesis, treatment with a strong base, such as sodium hydroxide (NaOH), induces the elimination of a bromine atom and a hydroxyl group from the carboxylic acid moiety. This process forms a molecule of water and the desired cis-β-bromostyrene. Students must acknowledge the necessity of a strong base and understand the elimination mechanism, where anti-periplanar geometry is often required for optimal removal of the leaving groups, to successfully execute this reaction.