Question

The interconversion of oxidation states of metal(s) observed in Wacker process is(are)

• A. \(\mathrm{Pd(II)/Pd(0)}\)
• B. \(\mathrm{Cu(II)/Cu(I)}\)
• C. \(\mathrm{Pd(IV)/Pd(II)}\)
• D. \(\mathrm{Cu(II)/Cu(0)}\)

Hint:

Trace the Wacker catalytic cycle: palladium activates and releases the alkene product, then copper reoxidises palladium so oxygen can finish the loop.

Answer 1

The Correct Option is A, B

Step 1: Recall the overall Wacker process.
The Wacker process converts ethylene into acetaldehyde using a \(\mathrm{PdCl_2}\)/\(\mathrm{CuCl_2}\) catalyst system in water, with \(\mathrm{O_2}\) as the final oxidant.
\[ \mathrm{CH_2=CH_2 + \tfrac{1}{2}O_2 \xrightarrow{PdCl_2,\ CuCl_2} CH_3CHO} \]

Step 2: Follow the palladium cycle.
Ethylene coordinates to \(\mathrm{Pd(II)}\), and after attack by water and beta-hydride steps, acetaldehyde is released and the palladium is reduced from \(\mathrm{Pd(II)}\) to \(\mathrm{Pd(0)}\). So the palladium interconversion is \(\mathrm{Pd(II)/Pd(0)}\), matching option (A).

Step 3: Follow the copper cycle that reoxidises palladium.
\(\mathrm{Pd(0)}\) on its own would stop the catalytic cycle after one turnover. \(\mathrm{Cu(II)}\) chloride reoxidises \(\mathrm{Pd(0)}\) back to \(\mathrm{Pd(II)}\), and in doing so \(\mathrm{Cu(II)}\) itself is reduced to \(\mathrm{Cu(I)}\). So the copper interconversion is \(\mathrm{Cu(II)/Cu(I)}\), matching option (B). The \(\mathrm{Cu(I)}\) formed is then reoxidised back to \(\mathrm{Cu(II)}\) by \(\mathrm{O_2}\), closing the catalytic loop.

Step 4: Rule out options (C) and (D).
\(\mathrm{Pd(IV)/Pd(II)}\) chemistry appears in some C-H activation catalytic cycles, but not in the classic Wacker mechanism, which never forms \(\mathrm{Pd(IV)}\); this rules out (C). Copper never gets reduced all the way to metallic \(\mathrm{Cu(0)}\) in this cycle, it only shuttles between \(\mathrm{Cu(II)}\) and \(\mathrm{Cu(I)}\); this rules out (D).

Final Answer:
The observed interconversions are \(\mathrm{Pd(II)/Pd(0)}\) and \(\mathrm{Cu(II)/Cu(I)}\). \[ \boxed{\text{A, B}} \]