[Mark S 2014]

Hi,

The Question is:

'Calculate the proportions of [Mn(¹²CO)₆] + and [Mn(¹²CO)₅( ¹³CO)]⁺ present in a sample of [Mn(CO)₆]⁺ containing naturally abundant carbon. Sketch the appearance of the solution-phase ⁵⁵Mn NMR spectrum of [Mn(CO)₆]⁺ , taking into account these two calculated proportions. Ensure that in your answer you fully account for the number and relative intensities of the peaks in the spectrum.'

Abundance of 12C = 0.989
Abundance of 13C = 0.011

To calculate the proportion of [Mn(¹²CO)₆]⁺ :

(0.989)^6 = 0.9358

To calculate the proportion of [Mn(¹²CO)₅(¹³CO)]⁺ :

(0.989)⁵ * (0.011) = 0.01041



If the answers for the proportions are correct, I would get the ⁵⁵Mn NMR spectrum of [Mn(CO)₆]⁺ to consist of:

​A large central singlet with relative intensity of 0.9358. Corresponding to [Mn(¹²CO)₆]⁺
A doublet of satellite peaks, a peak either side of the central singlet. Each peak would have an intensity of (0.01041/2). Corresponding to [Mn(¹²CO)₅(¹³CO)]⁺

Can anyone tell me if my way of calculating the probabilities of this is correct, and if it is, is my description of the manganese NMR spectrum correct ? I'mm pretty confident it's right. 

Appreaciate any feedback,

Mark

[mjc123]

The proportion of monosubstituted complex is wrong. (You should be able to tell this; as the abundance of ¹³C is so low, the proportion of polysubstituted complexes will be very low, so unsubstituted + monosubstituted ≈ 100%, which is not the case.)
Because there are 6 COs, there are 6 possibilities for substitution, so you multiply your calculated probability by 6. Generally, the probability of substituting r out of N ligands, each substitution having a probability p, is given by the binomial formula
P(r,N) = N!/r!(N-r)!*p^r(1-p)^N-r
Have you come across this?
If you apply this you will get P(1,6) ≈ 0.06, which is much more sensible.