Decarbonylation reaction of $\left[\text{cis}-\left(\mathrm{CH}_{3} \mathrm{CO}\right) \mathrm{Mn}\left({ }^{13} \mathrm{CO}\right)(\mathrm{CO})_{4}\right]$ yields X, Y and Z, where $\mathrm{X}=\left[(\mathrm{CH}_{3}) \mathrm{Mn}(\mathrm{CO})_{5}\right]$; $\mathrm{Y}=\left[\text{cis}-(\mathrm{CH}_{3}) \mathrm{Mn}\left({ }^{13} \mathrm{CO}\right) \mathrm{CO}_{4}\right]$; $\mathrm{Z}=\left[\text{trans}-(\mathrm{CH}_{3}) \mathrm{Mn}\left({ }^{13} \mathrm{CO}\right)(\mathrm{CO})_{4}\right]$. The molar ratio of the products $(\mathrm{X}: \mathrm{Y}: \mathrm{Z})$ in this reaction is
(a)$1: 1: 1$
(b)$1: 2: 1$
(c)$1: 1: 2$
(d)$2: 1: 1$
Answer
Answer: B ✓ checked by 4AB · confidence high
Explanation
Decarbonylation is the reverse of migratory insertion. A CO cis to the acyl is lost, and CH₃ migrates into that vacant cis site; the acyl carbonyl stays on Mn as an ordinary CO. The four cis positions hold one ¹³CO, one ¹²CO trans to ¹³CO, and two ¹²CO cis to ¹³CO, so each is lost with probability ¼. Losing ¹³CO (¼) gives unlabelled CH₃Mn(CO)₅ (X). Losing the CO trans to ¹³CO (¼) puts CH₃ trans to ¹³CO (Z). Losing either CO cis to ¹³CO (2/4) gives cis-CH₃Mn(¹³CO)(CO)₄ (Y). Therefore X : Y : Z = 1 : 2 : 1, option (b).