nmr spectroscopy help

Hi there,

1c - a quartet is generated by an ethyl group -CH2CH3 (the two protons of the CH2 group producing the quartet). An ester with molecular formula C4H8O2 must have a CH2CH3 group in it to produce a quartet, so the contenders are methyl propanoate or ethyl ethanoate. In methyl propanoate, the ethyl group is attached to C=O and in ethyl ethanoate it is attached to -O. The chemical shifts should help you to determine which is which.

3b- the molecular formula is C12H4O2Cl4. Ignoring the Cl atoms, the molecular mass is 180. The four chlorines can be 35 or 37 so there are five possible outcomes: 4x35, 3x35 and 1x37, 2x35 and 2x37, 1x35 and 3x37, and finally 4x37.

Therefore, you will have five mass peaks with m/z ratios 320, 322, 324, 326, 328. The relative heights of each m/z peak is determined by the laws of probability. You need to consider each chlorine atom in turn and work out the probability of a molecule having that mass.

For example, if the molecule has a mass of 320 then all four Cl atoms are the 35 isotope. The probability of this is 0.75x0.75x0.75x0.75=0.316.

Now consider if the molecule has a mass of 322. The probability of the first atom being 37 and all the others being 35 is 0.25x0.75x0.75x0.75=0.1055 BUT if atom 2 had a mass of 37 and all the others were 35 this molecule would ALSO have a mass of 322. There are four ways to achieve a mass of 322, each with probability of 0.1055, so the probabilities are added together four times: 0.1055x4=0.42.

The molecule with a mass of 324 would need two 35Cl and two 37Cl. The probability is 0.75x0.75x0.25x0.25=0.035, but there are SIX ways to achieve this. Therefore the probability of the molecule having this mass is 0.035x6=0.21.

Then you have 326: 0.75x0.25x0.25x0.25=0.012. There are four ways for this to happen so the total probability is 0.012x4=0.047.

Finally you have 328: 0.25x0.25x0.25x0.25=0.0039.

Probability total= 0.316+0.42+0.21+0.047+0.0039=1 (allowing for rounding errors). This shows that we haven't missed any possibilities.

The highest peak is m/z 322.

Good luck! Reply if unsure.

Hi, thanks a lot for the reply. In the aqa formula book, for certain esters at a given chemical shift, it gives their structure as bonded to an r group. Can this r be an oxygen (followed by the rest of the chain)?

Otherwise how do we use this formula book to know the chemical shift values for esters?

Furthermore, when you have a molecule with a carbon bonded to two other carbons, which adjacent carbon should you look at?

For example, in 3,3 dimethylbutan-1-ol, for the hydrogen bonded to the carbon that is bonded to the C(CH3)3 and the CH2(OH), which adjacent carbon do you look at? If you use one of them, then you get a singlet, but if you use the other, you get a triplet? How d you know which to use.

Also, in regards to integration values, if the values are 6:1, does this mean that the hydrogens in the hydrogen environments are in the ratio 6:1 i.e. the actual number of hydrogens in each environment?

No, "R" means a hydrocarbon (alkyl) chain, CnH(2n+1)



Adjacent means "next to". If there are two carbons in different environments, then both.



(CH₃)₃C-CH₂-CH₂OH

There are two hydrogen atoms on an adjacent carbon (to the right), so the coloured hydrogens are split into a triplet



Yes, to the last question

Hi thanks, for the first question, i'm still confused as to how you identify esters. E.g. if an ester had a chemical shift of 2.3, using the formula book, how de we then identify this? It doesn't show an ester structure. You mention that if there are two carbons in different environments then use both, but what do you mean by this? I know which will split into a triplet, i'm asking which one to use. The mark scheme, when asked for the splitting pattern at this point, only said singlet.