What is the product of the following Suzuki coupling reaction sequence? Include correct regio/stereochemistry. (6pts)a) Draw the

Answer

Suzuki Coupling Reaction Explanation

Suzuki Coupling Reaction: Detailed Explanation

What is Suzuki Coupling?

The Suzuki-Miyaura coupling reaction is a palladium-catalyzed cross-coupling reaction that forms carbon–carbon (C–C) bonds between an organoboron compound and an organohalide.

It typically uses a palladium catalyst such as Pd(PPh3)4 and a base like NaOEt.

Mechanism Overview

  1. Oxidative Addition: Pd(0) inserts into the carbon–halogen bond of the halide to form a Pd(II) complex.
  2. Transmetalation: The organoboron transfers its organic group to the palladium center with the help of a base.
  3. Reductive Elimination: The two organic groups couple to form the final product and regenerate Pd(0).

Reaction Context

In the provided reaction, a vinyl boronic ester couples with a vinyl halide under the influence of Pd(PPh3)4 and NaOEt.

The reaction leads to the formation of a conjugated diene structure through a vinyl–vinyl coupling.

Product Analysis

The final product is typically a 1,3-diene, often with E,E-configuration if no steric hindrance or electronic bias dictates otherwise.

This structure results from the coupling of two vinyl groups into a conjugated system, such as:

CH2=CH–CH=CH–R (general diene framework)

Conclusion

  • Reaction type: Suzuki-Miyaura coupling
  • Key reagents: Vinyl boronic ester, Vinyl halide, Pd(PPh₃)₄, NaOEt
  • Product: Conjugated diene (E,E-isomer in most cases)
  • Steps involved: Oxidative addition → Transmetalation → Reductive elimination

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