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The CORRECT distribution of the products in the following reaction is

Octahedral CH3C(O)Mn(CO)4(13CO) with the 13CO cis to the acetyl group, heated (decarbonylation)

  • A.CH3Mn(CO)5 with no 13CO (25 percent) plus CH3Mn(CO)5 with 13CO cis to the methyl (75 percent)
  • B.CH3Mn(CO)5 with no 13CO (20 percent) plus CH3Mn(CO)5 with 13CO (80 percent)
  • C.Three CH3Mn(CO)5 products: no 13CO (25 percent), 13CO in one position (25 percent) and 13CO in another position (50 percent)
  • D.Three CH3Mn(CO)5 products: no 13CO (20 percent), 13CO in one position (20 percent) and 13CO in another position (60 percent)

Answer: C

The distribution is (C): 25% unlabelled CH3Mn(CO)5, 25% with the CH3 trans to 13CO and 50% with the CH3 cis to 13CO.

Decarbonylation is the reverse of migratory insertion: a CO cis to the acetyl group dissociates, and the methyl then migrates from the acyl carbon onto the vacated site, while the acyl CO stays where the acetyl group was. The four CO ligands cis to the acetyl, one of them 13CO, are equally likely to be lost.

  • If the 13CO is the one lost (1 chance in 4), the product carries no label: 25%.
  • Otherwise the vacancy is at one of the other three cis sites. One of them lies opposite the 13CO in the equatorial plane, so methyl there ends up trans to 13CO: 25%.
  • The other two are adjacent to the 13CO, giving methyl cis to 13CO: 50%.

(A) and (B) keep the methyl in the acetyl position, as if the acyl CO itself were simply lost, and (D) has the right products in the wrong proportions.

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